Definite Integrals: Study Guide
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 378
Chapter
7
Applications of
Definite Integrals
T
he art of pottery developed independently in many ancient civilizations and still exists in modern times. The desired shape of the side of a pottery vase can be described by: y ϭ 5.0 ϩ 2 sin (x/4) (0 Յ x Յ 8p) where x is the height and y is the radius at height x (in inches).
A base for the vase is preformed and placed on a potter’s wheel. How much clay should be added to the base to form this vase if the inside radius is always 1 inch less than the outside radius? Section 7.3 contains the needed mathematics.
378
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 379
Section 7.1 Integral as Net Change
379
Chapter …show more content…
7 Overview
By this point it should be apparent that finding the limits of Riemann sums is not just an intellectual exercise; it is a natural way to calculate mathematical or physical quantities that appear to be irregular when viewed as a whole, but which can be fragmented into regular pieces. We calculate values for the regular pieces using known formulas, then sum them to find a value for the irregular whole. This approach to problem solving was around for thousands of years before calculus came along, but it was tedious work and the more accurate you wanted to be the more tedious it became.
With calculus it became possible to get exact answers for these problems with almost no effort, because in the limit these sums became definite integrals and definite integrals could be evaluated with antiderivatives. With calculus, the challenge became one of fitting an integrable function to the situation at hand (the “modeling” step) and then finding an antiderivative for it.
Today we can finesse the antidifferentiation step (occasionally an insurmountable hurdle for our predecessors) with programs like NINT, but the modeling step is no less crucial. Ironically, it is the modeling step that is thousands of years old. Before either calculus or technology can be of assistance, we must still break down the irregular whole into regular parts and set up a function to be integrated. We have already seen how the process works with area, volume, and average value, for example. Now we will focus more closely on the underlying modeling step: how to set up the function to be integrated.
7.1
What you’ll learn about
• Linear Motion Revisited
• General Strategy
Integral As Net Change
Linear Motion Revisited
In many applications, the integral is viewed as net change over time. The classic example of this kind is distance traveled, a problem we discussed in Chapter 5.
• Consumption Over Time
• Net Change from Data
• Work
. . . and why
The integral is a tool that can be used to calculate net change and total accumulation.
EXAMPLE 1 Interpreting a Velocity Function
Figure 7.1 shows the velocity ds 8
ᎏᎏ ϭ v(t) ϭ t 2 Ϫ ᎏᎏ dt ͑t ϩ 1͒ 2
cm
ᎏᎏ
sec
of a particle moving along a horizontal s-axis for 0 Յ t Յ 5. Describe the motion.
SOLUTION
Solve Graphically The graph of v (Figure 7.1) starts with v͑0͒ ϭ Ϫ8, which we in-
terpret as saying that the particle has an initial velocity of 8 cm րsec to the left. It slows to a halt at about t ϭ 1.25 sec, after which it moves to the right ͑v Ͼ 0͒ with increasing speed, reaching a velocity of v͑5͒ Ϸ 24.8 cm րsec at the end.
Now try Exercise 1(a).
EXAMPLE 2 Finding Position from Displacement
Suppose the initial position of the particle in Example 1 is s͑0͒ ϭ 9. What is the particle’s position at (a) t ϭ 1 sec? (b) t ϭ 5 sec?
SOLUTION
Solve Analytically
[0, 5] by [–10, 30]
Figure 7.1 The velocity function in
Example 1.
(a) The position at t ϭ 1 is the initial position s͑0͒ plus the displacement (the amount, Δs, that the position changed from t ϭ 0 to t ϭ 1). When velocity is continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 380
380
Chapter 7
Applications of Definite Integrals
Reminder from Section 3.4
A change in position is a displacement.
If s (t) is a body’s position at time t, the displacement over the time interval from t to t ϩ Δt is s(t ϩ Δt) Ϫ s (t). The displacement may be positive, negative, or zero, depending on the motion.
constant during a motion, we can find the displacement (change in position) with the formula Displacement ϭ rate of change ϫ time.
But in our case the velocity varies, so we resort instead to partitioning the time interval
͓0, 1͔ into subintervals of length Δt so short that the velocity is effectively constant on each subinterval. If tk is any time in the kth subinterval, the particle’s velocity throughout that interval will be close to v͑tk ͒. The change in the particle’s position during the brief time this constant velocity applies is v͑tk ͒ Δt. rate of change ϫ time
If v͑tk ͒ is negative, the displacement is negative and the particle will move left. If v͑tk ͒ is positive, the particle will move right. The sum
͚ v͑t ͒ Δt k of all these small position changes approximates the displacement for the time interval
͓0, 1͔.
The sum ͚ v͑tk ͒ Δt is a Riemann sum for the continuous function v͑t͒ over ͓0, 1͔. As the norms of the partitions go to zero, the approximations improve and the sums converge to the integral of v over ͓0, 1͔, giving
͵
͵(
1
Displacement ϭ
v͑t͒ dt
0
1
ϭ
0
)
8 t 2 Ϫ ᎏᎏ dt
͑t ϩ 1͒ 2
[
t3
8
ϭ ᎏᎏ ϩ ᎏᎏ
3
tϩ1
]
1
0
1
8
11 ϭ ᎏᎏ ϩ ᎏᎏ Ϫ 8 ϭ Ϫᎏᎏ.
3
2
3
During the first second of motion, the particle moves 11 ր 3 cm to the left. It starts at s͑0͒ ϭ 9, so its position at t ϭ 1 is
11
16
New position ϭ initial position ϩ displacement ϭ 9 Ϫ ᎏᎏ ϭ ᎏᎏ.
3
3
(b) If we model the displacement from t ϭ 0 to t ϭ 5 in the same way, we arrive at
Displacement ϭ
͵
5
0
[
t3
8
v͑t͒ dt ϭ ᎏᎏ ϩ ᎏᎏ
3
tϩ1
]
5
ϭ 35.
0
The motion has the net effect of displacing the particle 35 cm to the right of its starting point. The particle’s final position is
Final position ϭ initial position ϩ displacement ϭ s͑0͒ ϩ 35 ϭ 9 ϩ 35 ϭ 44.
Support Graphically The position of the particle at any time t is given by
͵[ t s͑t͒ ϭ
0
]
8 u 2 Ϫ ᎏ 2 du ϩ 9,
ᎏ
͑u ϩ 1͒
because sЈ͑t͒ ϭ v͑t͒ and s͑0͒ ϭ 9. Figure 7.2 shows the graph of s͑t͒ given by the parametrization x͑t͒ ϭ NINT ͑v͑u͒, u, 0, t͒ ϩ 9, y͑t͒ ϭ t,
0 Յ t Յ 5. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 381
Section 7.1 Integral as Net Change
381
(a) Figure 7.2a supports that the position of the particle at t ϭ 1 is 16 ր 3.
(b) Figure 7.2b shows the position of the particle is 44 at t ϭ 5. Therefore, the displacement is 44 Ϫ 9 ϭ 35.
Now try Exercise 1(b).
T=1
X = 5.3333333 Y = 1
The reason for our method in Example 2 was to illustrate the modeling step that will be used throughout this chapter. We can also solve Example 2 using the techniques of
Chapter 6 as shown in Exploration 1.
[–10, 50] by [–2, 6]
(a)
EXPLORATION 1
Revisiting Example 2
The velocity of a particle moving along a horizontal s-axis for 0 Յ t Յ 5 is ds 8
ᎏᎏ ϭ t 2 Ϫ ᎏᎏ. dt ͑t ϩ 1͒ 2
T=5
X = 44
Y=5
1. Use the indefinite integral of dsրdt to find the solution of the initial value problem ds
8
ᎏᎏ ϭ t 2 Ϫ ᎏᎏ, dt ͑t ϩ 1͒ 2
[–10, 50] by [–2, 6]
(b)
Figure 7.2 Using TRACE and the parametrization in Example 2 you can
“see” the left and right motion of the particle. s͑0͒ ϭ 9.
2. Determine the position of the particle at t ϭ 1. Compare your answer with the answer to Example 2a.
3. Determine the position of the particle at t ϭ 5. Compare your answer with the answer to Example 2b.
We know now that the particle in Example 1 was at s͑0͒ ϭ 9 at the beginning of the motion and at s͑5͒ ϭ 44 at the end. But it did not travel from 9 to 44 directly—it began its trip by moving to the left (Figure 7.2). How much distance did the particle actually travel?
We find out in Example 3.
EXAMPLE 3 Calculating Total Distance Traveled
Find the total distance traveled by the particle in Example 1.
SOLUTION
Solve Analytically We partition the time interval as in Example 2 but record every position shift as positive by taking absolute values. The Riemann sum approximating total distance traveled is
͚ Ηv͑t ͒Η Δt, k and we are led to the integral
Total distance traveled ϭ
͵
5
͵Η
5
Ηv͑t͒Η
0
dt ϭ
0
Η
8 t 2 Ϫ ᎏᎏ dt.
͑t ϩ 1͒ 2
Evaluate Numerically We have
NINT
(Η
Η
)
8 t 2 Ϫ ᎏᎏ , t, 0, 5 Ϸ 42.59.
͑t ϩ 1͒ 2
Now try Exercise 1(c).
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 382
382
Chapter 7
Applications of Definite Integrals
What we learn from Examples 2 and 3 is this: Integrating velocity gives displacement
(net area between the velocity curve and the time axis). Integrating the absolute value of velocity gives total distance traveled (total area between the velocity curve and the time axis). General Strategy
The idea of fragmenting net effects into finite sums of easily estimated small changes is not new. We used it in Section 5.1 to estimate cardiac output, volume, and air pollution.
What is new is that we can now identify many of these sums as Riemann sums and express their limits as integrals. The advantages of doing so are twofold. First, we can evaluate one of these integrals to get an accurate result in less time than it takes to crank out even the crudest estimate from a finite sum. Second, the integral itself becomes a formula that enables us to solve similar problems without having to repeat the modeling step.
The strategy that we began in Section 5.1 and have continued here is the following:
Strategy for Modeling with Integrals
1. Approximate what you want to find as a Riemann sum of values of a continuous function multiplied by interval lengths. If f ͑x͒ is the function and ͓a, b͔ the interval, and you partition the interval into subintervals of length Δx, the approximating sums will have the form ͚ f ͑ck ͒ Δx with ck a point in the kth subinterval.
2. Write a definite integral, here ͐ab f ͑x͒ dx, to express the limit of these sums as the norms of the partitions go to zero.
3. Evaluate the integral numerically or with an antiderivative.
EXAMPLE 4 Modeling the Effects of Acceleration
A car moving with initial velocity of 5 mph accelerates at the rate of a͑t͒ ϭ 2.4t mph per second for 8 seconds.
(a) How fast is the car going when the 8 seconds are up?
(b) How far did the car travel during those 8 seconds?
SOLUTION
(a) We first model the effect of the acceleration on the car’s velocity.
Step 1: Approximate the net change in velocity as a Riemann sum. When acceleration is constant, rate of change ϫ time velocity change ϭ acceleration ϫ time applied.
To apply this formula, we partition ͓0, 8͔ into short subintervals of length Δt. On each subinterval the acceleration is nearly constant, so if tk is any point in the k th subinterval, the change in velocity imparted by the acceleration in the subinterval is approximately mph a͑tk ͒ Δt mph.
ᎏᎏ ϫ sec sec The net change in velocity for 0 Յ t Յ 8 is approximately
͚ a͑t ͒ Δt mph. k Step 2: Write a definite integral. The limit of these sums as the norms of the partitions
go to zero is
͵
8
0
a͑t͒ dt. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 383
Section 7.1 Integral as Net Change
383
Step 3: Evaluate the integral. Using an antiderivative, we have
Net velocity change ϭ
͵
8
2.4t dt ϭ 1.2t 2
0
]
8
ϭ 76.8 mph.
0
So, how fast is the car going when the 8 seconds are up? Its initial velocity is 5 mph and the acceleration adds another 76.8 mph for a total of 81.8 mph.
(b) There is nothing special about the upper limit 8 in the preceding calculation. Applying the acceleration for any length of time t adds
͵
t
2.4u du mph
u is just a dummy variable here.
0
(b) to the car’s velocity, giving v͑t͒ ϭ 5 ϩ
͵
t
2.4u du ϭ 5 ϩ 1.2t 2 mph.
0
The distance traveled from t ϭ 0 to t ϭ 8 sec is
͵
8
0
Ηv͑t͒Η
dt ϭ
͵
8
͑5 ϩ 1.2t 2 ͒ dt
Extension of Example 3
0
[
ϭ 5t ϩ 0.4t 3
]
8
0
ϭ 244.8 mph ϫ seconds.
Miles-per-hour second is not a distance unit that we normally work with! To convert to miles we multiply by hours րsecond ϭ 1 ր 3600, obtaining
1
244.8 ϫ ᎏᎏ ϭ 0.068 mile.
3600
mi h ᎏᎏ ϫ sec ϫ ᎏᎏ ϭ mi h sec
The car traveled 0.068 mi during the 8 seconds of acceleration.
Now try Exercise 9.
Consumption Over Time
The integral is a natural tool to calculate net change and total accumulation of more quantities than just distance and velocity. Integrals can be used to calculate growth, decay, and, as in the next example, consumption. Whenever you want to find the cumulative effect of a varying rate of change, integrate it.
EXAMPLE 5 Potato Consumption
From 1970 to 1980, the rate of potato consumption in a particular country was C͑t͒ ϭ
2.2 ϩ 1.1t millions of bushels per year, with t being years since the beginning of 1970.
How many bushels were consumed from the beginning of 1972 to the end of 1973?
SOLUTION
We seek the cumulative effect of the consumption rate for 2 Յ t Յ 4.
Step 1: Riemann sum. We partition ͓2, 4͔ into subintervals of length Δt and let tk be a time in the kth subinterval. The amount consumed during this interval is approximately
C͑tk ͒ Δt million bushels.
The consumption for 2 Յ t Յ 4 is approximately
͚ C͑t ͒ Δt million bushels. k continued
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 384
384
Chapter 7
Applications of Definite Integrals
Step 2: Definite integral. The amount consumed from t ϭ 2 to t ϭ 4 is the limit of
these sums as the norms of the partitions go to zero.
͵
4
C͑t͒ dt ϭ
2
͵
4
͑2.2 ϩ 1.1t ͒ dt million bushels
2
Step 3: Evaluate. Evaluating numerically, we obtain
NINT ͑2.2 ϩ 1.1t , t, 2, 4͒ Ϸ 7.066 million bushels.
Now try Exercise 21.
Net Change from Data
Table 7.1
Pumping Rates
Time (min)
Rate (gal ր min)
0
58
5
60
10
65
15
64
20
58
25
57
30
55
35
55
40
59
45
60
50
Many real applications begin with data, not a fully modeled function. In the next example, we are given data on the rate at which a pump operates in consecutive 5-minute intervals and asked to find the total amount pumped.
60
55
60
EXAMPLE 6 Finding Gallons Pumped from Rate Data
A pump connected to a generator operates at a varying rate, depending on how much power is being drawn from the generator to operate other machinery. The rate (gallons per minute) at which the pump operates is recorded at 5-minute intervals for one hour as shown in Table 7.1. How many gallons were pumped during that hour?
SOLUTION
Let R͑t͒, 0 Յ t Յ 60, be the pumping rate as a continuous function of time for the hour.
We can partition the hour into short subintervals of length Δt on which the rate is nearly constant and form the sum ͚ R͑tk ͒ Δt as an approximation to the amount pumped during the hour. This reveals the integral formula for the number of gallons pumped to be
Gallons pumped ϭ
63
63
͵
60
R͑t͒ dt.
0
We have no formula for R in this instance, but the 13 equally spaced values in Table 7.1 enable us to estimate the integral with the Trapezoidal Rule:
͵
60
0
1 joule ϭ (1 newton)(1 meter).
In symbols, 1 J ϭ 1 N • m.
It takes a force of about 1 N to lift an apple from a table. If you lift it 1 m you have done about 1 J of work on the apple. If you eat the apple, you will have consumed about 80 food calories, the heat equivalent of nearly 335,000 joules. If this energy were directly useful for mechanical work (it’s not), it would enable you to lift 335,000 more apples up 1 m.
]
ϭ 3582.5.
Joules
The joule, abbreviated J and pronounced “jewel,” is named after the
English physicist James Prescott Joule
(1818–1889). The defining equation is
[
60
R͑t͒ dt Ϸ ᎏᎏ 58 ϩ 2͑60͒ ϩ 2͑65͒ ϩ … ϩ 2͑63͒ ϩ 63
2 • 12
The total amount pumped during the hour is about 3580 gal.
Now try Exercise 27.
Work
In everyday life, work means an activity that requires muscular or mental effort. In science, the term refers specifically to a force acting on a body and the body’s subsequent displacement. When a body moves a distance d along a straight line as a result of the action of a force of constant magnitude F in the direction of motion, the work done by the force is
W ϭ Fd.
The equation W ϭ Fd is the constant-force formula for work.
The units of work are force ϫ distance. In the metric system, the unit is the newtonmeter, which, for historical reasons, is called a joule (see margin note). In the U.S. customary system, the most common unit of work is the foot-pound.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 385
Section 7.1 Integral as Net Change
385
Hooke’s Law for springs says that the force it takes to stretch or compress a spring x units from its natural (unstressed) length is a constant times x. In symbols,
F ϭ kx, where k, measured in force units per unit length, is a characteristic of the spring called the force constant.
EXAMPLE 7 A Bit of Work
It takes a force of 10 N to stretch a spring 2 m beyond its natural length. How much work is done in stretching the spring 4 m from its natural length?
SOLUTION
We let F͑x͒ represent the force in newtons required to stretch the spring x meters from its natural length. By Hooke’s Law, F͑x͒ ϭ kx for some constant k. We are told that
The force required to stretch the spring 2 m is 10 newtons.
F͑2͒ ϭ 10 ϭ k • 2,
so k ϭ 5 N րm and F͑x͒ ϭ 5x for this particular spring.
We construct an integral for the work done in applying F over the interval from x ϭ 0 to x ϭ 4.
Step 1: Riemann sum. We partition the interval into subintervals on each of which F is so nearly constant that we can apply the constant-force formula for work. If x k is any point in the kth subinterval, the value of F throughout the interval is approximately F͑x k ͒ ϭ 5x k . The work done by F across the interval is approximately 5x k Δx, where Δx is the length of the interval. The sum
͚ F͑x ͒ Δx ϭ ͚ 5x k k
Δx
approximates the work done by F from x ϭ 0 to x ϭ 4.
Steps 2 and 3: Integrate. The limit of these sums as the norms of the partitions go to
zero is
͵
Numerically, work is the area under the force graph.
4
F͑x͒ dx ϭ
0
͵
4
0
x2
5x dx ϭ 5ᎏᎏ
2
]
4
ϭ 40 N • m.
0
Now try Exercise 29.
We will revisit work in Section 7.5.
Quick Review 7.1
(For help, go to Section 1.2.)
In Exercises 1–10, find all values of x (if any) at which the function changes sign on the given interval. Sketch a number line graph of the interval, and indicate the sign of the function on each subinterval.
Example: f ͑x͒ ϭ x 2 Ϫ 1
on
͓Ϫ2, 3͔ f (x)
+
–2
–
–1
+
1
x
3
Changes sign at x ϭ Ϯ1.
1. sin 2x on
2. x 2 Ϫ 3x ϩ 2
p p Changes sign at Ϫᎏᎏ, 0, ᎏᎏ
2
2
͓Ϫ2, 4͔ Changes sign at 1, 2
͓Ϫ3, 2͔ on 3. x 2 Ϫ 2x ϩ 3
on
͓Ϫ4, 2͔ Always positive
1
Changes sign at Ϫᎏᎏ
2
p 3p 5p
5. x cos 2x on ͓0, 4͔ Changes sign at ᎏᎏ, ᎏᎏ, ᎏᎏ
4 4 4
6. xeϪx on ͓0, ∞͒ Always positive
4.
2x 3
Ϫ
3x 2
ϩ1
on
͓Ϫ2, 2͔
x
7. ᎏᎏ on ͓Ϫ5, 30͔ Changes sign at 0 x2 ϩ 1 x2 Ϫ 2
8. ᎏᎏ on ͓Ϫ3, 3͔ Changes sign at Ϫ2, Ϫ͙2, ͙2, 2
ෆ
ෆ x2 Ϫ 4
9. sec ͑1 ϩ ͙1ෆෆinෆx ͒ on ͑Ϫ∞, ∞͒
ෆ Ϫ sෆ2 ෆ
10. sin ͑1 ր x͒ on
1 1
3p 2p
͓0.1, 0.2͔ Changes sign at ᎏᎏ, ᎏᎏ
9. Changes sign at 0.9633 ϩ kp, 2.1783 ϩ kp, where k is an integer
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 386
386
Chapter 7
Applications of Definite Integrals
Section 7.1 Exercises
In Exercises 1–8, the function v(t) is the velocity in m րsec of a particle moving along the x-axis. Use analytic methods to do each of the following:
(a) Determine when the particle is moving to the right, to the left, and stopped.
(b) Find the particle’s displacement for the given time interval. If s(0) ϭ 3, what is the particle’s final position?
(c) Find the total distance traveled by the particle.
1. v͑t͒ ϭ 5 cos t,
0 Յ t Յ 2p See page 389.
0 Յ t Յ p ր2
2. v͑t͒ ϭ 6 sin 3t,
3. v͑t͒ ϭ 49 Ϫ 9.8t,
4. v͑t͒ ϭ
6t 2
Ϫ23 cm
13. What is the total distance traveled by the particle in the same time period? 33 cm a: 11
Ϫ 18t ϩ 12,
15. Approximately where does the particle achieve its greatest positive acceleration on the interval ͓0, b͔? t ϭ a
16. Approximately where does the particle achieve its greatest positive acceleration on the interval ͓0, c͔? t ϭ c
(a) Find where the particle is at the end of the trip.
0 Յ t Յ 2 See page 389.
0 Յ t Յ 2p See page 389.
(b) Find the total distance traveled by the particle.
17.
0 Յ t Յ 4 See page 389.
0
18.
1
(b) 4 meters
3
4
t (sec)
(a) 2 (b) 4 meters
344.52 feet
1
10. A particle travels with velocity v͑t͒ ϭ ͑t Ϫ 2͒ sin t m րsec
0
1
for 0 Յ t Յ 4 sec.
(b) What is the total distance traveled? Ϸ1.91411 meters
19.
11. Projectile Recall that the acceleration due to Earth’s gravity is
32 ft րsec2. From ground level, a projectile is fired straight upward with velocity 90 feet per second.
(a) What is its velocity after 3 seconds?
4
t (sec)
(a) 5
v (m/sec)
1 2 3 4 5 6 7
0
Ϫ6 ft/sec
(b) 7 meters
t (sec)
–2
(c) When it hits the ground, what is the net distance it has traveled? 0
(d) When it hits the ground, what is the total distance it has traveled? 253.125 feet
In Exercises 12–16, a particle moves along the x-axis (units in cm).
Its initial position at t ϭ 0 sec is x͑0͒ ϭ 15. The figure shows the graph of the particle’s velocity v͑t͒. The numbers are the areas of the enclosed regions.
5
c
24
3
2
5.625 sec
b
2
–1
(a) What is the particle’s displacement? ϷϪ1.44952 meters
a
2
v (m/sec)
63 mph
(b) How far does it travel in those 9 seconds?
4
(a) 6
v (m/sec)
2
7. v͑t͒ ϭ e sin t cos t, 0 Յ t Յ 2p See page 389. t 8. v͑t͒ ϭ ᎏᎏ , 0 Յ t Յ 3 See page 389.
1 ϩ t2
9. An automobile accelerates from rest at 1 ϩ 3͙t mph րsec for
ෆ
9 seconds.
(b) When does it hit the ground?
c: Ϫ8
In Exercises 17–20, the graph of the velocity of a particle moving on the x-axis is given. The particle starts at x ϭ 2 when t ϭ 0.
See page 389.
(a) What is its velocity after 9 seconds?
b: 16
14. Give the positions of the particle at times a, b, and c.
0 Յ t Յ 10 See page 389.
5. v͑t͒ ϭ 5 sin2 t cos t,
6. v͑t͒ ϭ ͙4ෆෆ,
ෆϪt
12. What is the particle’s displacement between t ϭ 0 and t ϭ c?
20.
(a) Ϫ2.5
v (m/sec)
(b) 19.5 meters
3
0
1 2 3 4 5 6 7 8 9 10
t (sec)
–3
21. U.S. Oil Consumption The rate of consumption of oil in the
United States during the 1980s (in billions of barrels per year) is modeled by the function C ϭ 27.08 • e t ր 25, where t is the number of years after January 1, 1980. Find the total consumption of oil in the United States from January 1, 1980 to January 1,
1990. Ϸ332.965 billion barrels
22. Home Electricity Use The rate at which your home consumes electricity is measured in kilowatts. If your home consumes electricity at the rate of 1 kilowatt for 1 hour, you will be charged
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 387
26. (b) The answer in (a) corresponds to the area of midpoint rectangles. Part of each rectangle is above the curve and part is below.
for 1 “kilowatt-hour” of electricity. Suppose that the average consumption rate for a certain home is modeled by the function
C͑t͒ ϭ 3.9 Ϫ 2.4 sin ͑pt ր 12͒, where C͑t͒ is measured in kilowatts and t is the number of hours past midnight. Find the average daily consumption for this home, measured in kilowatthours. 93.6 kilowatt-hours
23. Population Density Population density measures the number of people per square mile inhabiting a given living area.
Washerton’s population density, which decreases as you move away from the city center, can be approximated by the function
10,000͑2 Ϫ r͒ at a distance r miles from the city center.
(a) If the population density approaches zero at the edge of the city, what is the city’s radius? 2 miles
(b) A thin ring around the center of the city has thickness Δr and radius r. If you straighten it out, it suggests a rectangular strip.
Approximately what is its area? 2pr⌬r
(c) Writing to Learn Explain why the population of the ring in part (b) is approximately
10,000͑2 Ϫ r͒͑2p r͒ Δr.
Section 7.1 Integral as Net Change
387
(a) What was the total number of bagels sold over the 11-year period? (This is not a calculus question!) 797.5 thousand
(b) Use quadratic regression to model the annual bagel sales (in thousands) as a function B͑x͒, where x is the number of years after 1995. B(x) ϭ 1.6x2 ϩ 2.3x ϩ 5.0
(c) Integrate B͑x͒ over the interval ͓0, 11͔ to find total bagel sales for the 11-year period. Ϸ904.02
(d) Explain graphically why the answer in part (a) is smaller than the answer in part (c). See page 389.
26. Group Activity (Continuation of Exercise 25)
(a) Integrate B͑x͒ over the interval ͓Ϫ0.5, 10.5͔ to find total bagel sales for the 11-year period. Ϸ798.97 thousand
(b) Explain graphically why the answer in part (a) is better than the answer in 25(c).
27. Filling Milk Cartons A machine fills milk cartons with milk at an approximately constant rate, but backups along the assembly line cause some variation. The rates (in cases per hour) are recorded at hourly intervals during a 10-hour period, from
8:00 A.M. to 6:00 P.M.
Population ϭ Population density ϫ Area
(d) Estimate the total population of Washerton by setting up and evaluating a definite integral. Ϸ83,776
24. Oil Flow Oil flows through a cylindrical pipe of radius
3 inches, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance r inches from the center is 8͑10 Ϫ r 2 ͒ inches per second.
(a) In a plane cross section of the pipe, a thin ring with thickness
Δr at a distance r inches from the center approximates a rectangular strip when you straighten it out. What is the area of the strip (and hence the approximate area of the ring)? 2pr⌬r
(b) Explain why we know that oil passes through this ring at approximately 8͑10 Ϫ r 2 ͒͑2pr͒ Δr cubic inches per second.
(c) Set up and evaluate a definite integral that will give the rate
(in cubic inches per second) at which oil is flowing through the pipe. 396p in3/sec or Ϸ 1244.07 in3/sec
25. Group Activity Bagel Sales From 1995 to 2005, the
Konigsberg Bakery noticed a consistent increase in annual sales of its bagels. The annual sales (in thousands of bagels) are shown below.
Year
Sales
(thousands)
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
5
8.9
16
26.3
39.8
56.5
76.4
99.5
125.8
155.3
188
24. (b) 8(10 Ϫ r2) in/sec и (2pr)⌬r in2 ϭ flow in in3/sec
Time
Rate
(cases ր h)
8
9
10
11
12
1
2
3
4
5
6
120
110
115
115
119
120
120
115
112
110
121
Use the Trapezoidal Rule with n ϭ 10 to determine approximately how many cases of milk were filled by the machine over the 10-hour period.
1156.5
28. Writing to Learn As a school project, Anna accompanies her mother on a trip to the grocery store and keeps a log of the car’s speed at 10-second intervals. Explain how she can use the data to estimate the distance from her home to the store. What is the connection between this process and the definite integral?
See page 389.
29. Hooke’s Law A certain spring requires a force of 6 N to stretch it 3 cm beyond its natural length.
(a) What force would be required to stretch the string 9 cm beyond its natural length? 18 N
(b) What would be the work done in stretching the string 9 cm beyond its natural length? 81 N и cm
30. Hooke’s Law Hooke’s Law also applies to compressing springs; that is, it requires a force of kx to compress a spring a distance x from its natural length. Suppose a 10,000-lb force compressed a spring from its natural length of 12 inches to a length of 11 inches.
How much work was done in compressing the spring
(a) the first half-inch?
(b) the second
half-inch?
1250 inch-pounds
3750 inch-pounds
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 388
388
Chapter 7
Applications of Definite Integrals
Standardized Test Questions
You may use a graphing calculator to solve the following problems. 31. True or False The figure below shows the velocity for a particle moving along the x-axis. The displacement for this particle is negative. Justify your answer. False. The displacement is the
36. Multiple Choice Pollution is being removed from a lake at a rate modeled by the function y ϭ 20eϪ0.5t tons/yr, where t is the number of years since 1995. Estimate the amount of pollution removed from the lake between 1995 and 2005. Round your answer to the nearest ton. A
(A) 40
(B) 47
(C) 56
(D) 61
(E) 71
integral of the velocity from t ϭ 0 to t ϭ 5 and is positive.
Extending the Ideas
v (m/sec)
37. Inflation Although the economy is continuously changing, we analyze it with discrete measurements. The following table records the annual inflation rate as measured each month for
13 consecutive months. Use the Trapezoidal Rule with n ϭ 12 to find the overall inflation rate for the year. 0.04875
2
1
0
–1
1
2
3
4
5
6
t (sec)
–2
Month
32. True or False If the velocity of a particle moving along the x-axis is always positive, then the displacement is equal to the total distance traveled. Justify your answer.
33. Multiple Choice The graph below shows the rate at which water is pumped from a storage tank. Approximate the total gallons of water pumped from the tank in 24 hours. C
(A) 600
(B) 2400
(C) 3600
(D) 4200
(E) 4800
r (gal/hr)
250
200
150
100
50
0
6
12
18
t (hr)
24
34. Multiple Choice The data for the acceleration a(t) of a car from 0 to 15 seconds are given in the table below. If the velocity at t ϭ 0 is 5 ft/sec, which of the following gives the approximate velocity at t ϭ 15 using the Trapezoidal Rule? D
(A) 47 ft/sec
(B) 52 ft/sec
(D) 125 ft/sec
(C) 120 ft/sec
(E) 141 ft/sec
t (sec)
0
3
6
9
12
15
a(t) (ftրsec2)
4
8
6
9
10
10
35. Multiple Choice The rate at which customers arrive at a counter to be served is modeled by the function F defined by t F(t) ϭ 12 ϩ 6 cos ᎏᎏ for 0 Յ t Յ 60, where F(t) is measured in p customers per minute and t is measured in minutes. To the nearest whole number, how many customers arrive at the counter over the 60-minute period? B
(A) 720
(B) 725
(C) 732
(D) 744
(E) 756
32. True. Since the velocity is positive, the integral of the velocity is equal to the integral of its absolute value, which is the total distance traveled.
Annual Rate
January
February
March
April
May
June
July
August
September
October
November
December
January
0.04
0.04
0.05
0.06
0.05
0.04
0.04
0.05
0.04
0.06
0.06
0.05
0.05
38. Inflation Rate The table below shows the monthly inflation rate (in thousandths) for energy prices for thirteen consecutive months. Use the Trapezoidal Rule with n ϭ 12 to approximate the annual inflation rate for the 12-month period running from the middle of the first month to the middle of the last month. 40 thousandths or 0.040
Month
Monthly Rate
(in thousandths)
January
February
March
April
May
June
July
August
September
October
November
December
January
3.6
4.0
3.1
2.8
2.8
3.2
3.3
3.1
3.2
3.4
3.4
3.9
4.0
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 389
Section 7.1 Integral as Net Change
39. Center of Mass Suppose we have a finite collection of masses in the coordinate plane, the mass mk located at the point ͑xk , yk ͒ as shown in the figure. y yk
xk
mk
(xk , yk)
O
x
Each mass mk has moment mk yk with respect to the x-axis and moment mk xk about the y-axis. The moments of the entire system with respect to the two axes are
͚my,
Moment about y-axis: M ϭ ͚ m x .
Moment about x-axis: Mx ϭ y ͚
k
and
Mx
͚ mk yk y J ϭ ᎏᎏ ϭ ᎏ . m M k where dm ϭ d dA, d ϭ density, and A ϭ area of the region.
41. the region bounded by the lines y ϭ x, y ϭ Ϫx, x ϭ 2 with constant density d x ϭ 4/3, y ϭ 0
ෆ
ෆ
25. (d) The answer in (a) corresponds to the area of left hand rectangles.
These rectangles lie under the curve B(x). The answer in (c) corresponds to the area under the curve. This area is greater than the area of the rectangles.
1. (a) Right: 0 Յ t Ͻ p/2, 3p/2 Ͻ t Յ 2p
Left: p/2 Ͻ t Ͻ 3p/2
Stopped: t ϭ p/2, 3p/2
(b) 0; 3
(b) Imagine the region cut into thin strips parallel to the x-axis. Show that
͐y dm y J ϭ ᎏᎏ ,
͐dm
40. the region bounded by the parabola y ϭ x 2 and the line y ϭ 4 with constant density d x ϭ 0, ෆ ϭ 12/5 y ෆ
k k
͚
(a) Imagine the region cut into thin strips parallel to the y-axis. Show that
͐x dm x J ϭ ᎏᎏ ,
͐dm
In Exercises 40 and 41, use Exercise 39 to find the center of mass of the region with given density.
k k
The center of mass is ͑ J, J ͒ where x y
My
͚ m k xk x J ϭ ᎏᎏ ϭ ᎏ m M
Suppose we have a thin, flat plate occupying a region in the plane.
where dm ϭ d dA, d ϭ density (mass per unit area), and
A ϭ area of the region.
yk xk 389
(c) 20
2. (a) Right: 0 Ͻ t Ͻ p/3
Left: p/3 Ͻ t Յ p/2
Stopped: t ϭ 0, p/3
(b) 2; 5
3. (a) Right: 0 Յ t Ͻ 5
Left: 5 Ͻ t Յ 10
Stopped: t ϭ 5
(b) 0; 3
4. (a) Right: 0 Յ t Ͻ 1
Left: 1 Ͻ t Ͻ 2
Stopped: t ϭ 1, 2
(b) 4; 7
28. One possible answer:
Plot the speeds vs. time. Connect the points and find the area under the line graph. The definite integral also gives the area under the curve.
(c) 6
(c) 245
(c) 6
5. (a) Right: 0 Ͻ t Ͻ p/2, 3p/2 Ͻ t Ͻ 2p
Left: p/2 Ͻ t Ͻ p, p Ͻ t Ͻ 3p/2
Stopped: t ϭ 0, p/2, p, 3p/2, 2p
(b) 0 ; 3
(c) 20/3
6. (a) Right: 0 Յ t Ͻ 4
Left: never
Stopped: t ϭ 4
(b) 16/3; 25/3
(c) 16/3
7. (a) Right: 0 Յ t Ͻ p/2, 3p/2 Ͻ t Յ 2p
Left: p/2 Ͻ t Ͻ 3p/2
Stopped: t ϭ p/2, 3p/2
(b) 0; 3
(c) 2e Ϫ (2/e) Ϸ 4.7
8. (a) Right: 0 Ͻ t Յ 3
Left: never
Stopped: t ϭ 0
(b) (ln 10)/2 Ϸ 1.15; 4.15
(c) (ln 10)/2 Ϸ 1.15
39. (a, b) Take dm ϭ d dA as mk and letting dA → 0, k → ∞ in the center of mass equations.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 390
390
Chapter 7
Applications of Definite Integrals
Areas in the Plane
7.2
What you’ll learn about
Area Between Curves
• Area Between Curves
We know how to find the area of a region between a curve and the x-axis but many times we want to know the area of a region that is bounded above by one curve, y ϭ f ͑x͒, and below by another, y ϭ g͑x͒ (Figure 7.3).
We find the area as an integral by applying the first two steps of the modeling strategy developed in Section 7.1.
• Area Enclosed by Intersecting
Curves
• Boundaries with Changing
Functions
1. We partition the region into vertical strips of equal width Δx and approximate each strip with a rectangle with base parallel to ͓a, b͔ (Figure 7.4). Each rectangle has area
• Integrating with Respect to y
• Saving Time with Geometric Formulas
͓ f ͑ck ͒ Ϫ g͑ck ͔͒ Δx
. . . and why
for some ck in its respective subinterval (Figure 7.5). This expression will be nonnegative even if the region lies below the x-axis. We approximate the area of the region with the Riemann sum
The techniques of this section allow us to compute areas of complex regions of the plane.
͚ ͓ f ͑c ͒ Ϫ g͑c ͔͒ Δx. k k
y
y
y
Upper curve y ϭ f(x)
(ck, f (ck )) y ϭ f(x) f (ck) Ϫ g(ck )
a
a b Lower curve y ϭ g(x)
Figure 7.3 The region between y ϭ f (x) and y ϭ g(x) and the lines x ϭ a and x ϭ b.
a
x
b
x
ck b y ϭ g(x)
⌬x
Figure 7.4 We approximate the region with rectangles perpendicular to the x-axis.
x
(ck, g(ck ))
Figure 7.5 The area of a typical rectangle is ͓ f (ck ) Ϫ g(ck )͔ Δx.
2. The limit of these sums as Δx→0 is
͵
b
͓ f ͑x͒ Ϫ g͑x͔͒ dx.
a
This approach to finding area captures the properties of area, so it can serve as a definition. DEFINITION Area Between Curves
If f and g are continuous with f ͑x͒ Ն g͑x͒ throughout ͓a, b͔, then the area between the curves y ϭ f ( x ) and y ϭ g ( x ) from a to b is the integral of ͓ f Ϫ g͔ from a to b,
Aϭ
͵
b
a
͓ f ͑x͒ Ϫ g͑x͔͒ dx.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 391
Section 7.2 Areas in the Plane
391
EXAMPLE 1 Applying the Definition
y
Find the area of the region between y ϭ sec2 x and y ϭ sin x from x ϭ 0 to x ϭ p ր4.
y ϭ sec 2 x
2
SOLUTION
We graph the curves (Figure 7.6) to find their relative positions in the plane, and see that y ϭ sec2 x lies above y ϭ sin x on ͓0, p ր4͔. The area is therefore
Aϭ
y ϭ sin x
1
͵
p ր4
͓sec2 x Ϫ sin x͔ dx
0
[
ϭ tan x ϩ cos x
]
pր4
0
͙ෆ
2
ϭ ᎏᎏ units squared.
2
–
4
0
x
Now try Exercise 1.
Figure 7.6 The region in Example 1.
EXPLORATION 1
y1 = 2k – k sin kx y2 = k sin kx
A Family of Butterflies
For each positive integer k, let Ak denote the area of the butterfly-shaped region enclosed between the graphs of y ϭ k sin kx and y ϭ 2k Ϫ k sin kx on the interval
͓0, p րk͔. The regions for k ϭ 1 and k ϭ 2 are shown in Figure 7.7.
k=2 k=1 [0, ] by [0, 4]
Figure 7.7 Two members of the family of butterfly-shaped regions described in
Exploration 1.
1. Find the areas of the two regions in Figure 7.7.
2. Make a conjecture about the areas Ak for k Ն 3.
3. Set up a definite integral that gives the area Ak . Can you make a simple u-substitution that will transform this integral into the definite integral that gives the area A1?
4. What is lim k→∞ Ak?
5. If Pk denotes the perimeter of the kth butterfly-shaped region, what is lim k→∞ Pk ? (You can answer this question without an explicit formula for Pk .)
Area Enclosed by Intersecting Curves
When a region is enclosed by intersecting curves, the intersection points give the limits of integration. y1 = 2 – x2 y2 = – x
EXAMPLE 2 Area of an Enclosed Region
Find the area of the region enclosed by the parabola y ϭ 2 Ϫ x 2 and the line y ϭ Ϫx.
SOLUTION
We graph the curves to view the region (Figure 7.8).
The limits of integration are found by solving the equation
2 Ϫ x 2 ϭ Ϫx
[–6, 6] by [–4, 4]
Figure 7.8 The region in Example 2.
either algebraically or by calculator. The solutions are x ϭ Ϫ1 and x ϭ 2. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 392
392
Chapter 7
Applications of Definite Integrals
Since the parabola lies above the line on ͓Ϫ1, 2͔, the area integrand is 2 Ϫ x 2 Ϫ ͑Ϫx͒.
Aϭ
͵
2
͓2 Ϫ x 2 Ϫ͑Ϫx͔͒ dx
Ϫ1
[
x3 x2 ϭ 2x Ϫ ᎏᎏ ϩ ᎏᎏ
3
2
]
2
Ϫ1
9 ϭ ᎏᎏ units squared
2
Now try Exercise 5.
EXAMPLE 3 Using a Calculator
Find the area of the region enclosed by the graphs of y ϭ 2 cos x and y ϭ x 2 Ϫ 1.
y1 = 2 cos x y2 = x2 – 1
SOLUTION
The region is shown in Figure 7.9.
Using a calculator, we solve the equation
2 cos x ϭ x 2 Ϫ 1 to find the x-coordinates of the points where the curves intersect. These are the limits of integration. The solutions are x ϭ Ϯ1.265423706. We store the negative value as A and the positive value as B. The area is
NINT ͑2 cos x Ϫ ͑x 2 Ϫ 1͒, x, A, B͒ Ϸ 4.994907788.
[–3, 3] by [–2, 3]
This is the final calculation, so we are now free to round. The area is about 4.99.
Now try Exercise 7.
Figure 7.9 The region in Example 3.
Boundaries with Changing Functions
Finding Intersections by
Calculator
The coordinates of the points of intersection of two curves are sometimes needed for other calculations. To take advantage of the accuracy provided by calculators, use them to solve for the values and store the ones you want.
If a boundary of a region is defined by more than one function, we can partition the region into subregions that correspond to the function changes and proceed as usual.
EXAMPLE 4 Finding Area Using Subregions
Find the area of the region R in the first quadrant that is bounded above by y ϭ ͙x and
ෆ
below by the x-axis and the line y ϭ x Ϫ 2.
SOLUTION
The region is shown in Figure 7.10. y 2
4
⌠
Area ϭ ⎮ ⎡⎯ Ϫ x ϩ 2⎡ dx
⎯x
⎣√
⎣
yϭ⎯
√
⎯x
⌡2
⌠
(4, 2)
2
Area ϭ ⎮ √ dx
⎯
⎯x
⌡0
B
1
y ϭxϪ2
A
0
yϭ0
2
Figure 7.10 Region R split into subregions A and B. (Example 4)
4
x
continued
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 393
Section 7.2 Areas in the Plane
393
While it appears that no single integral can give the area of R (the bottom boundary is defined by two different curves), we can split the region at x ϭ 2 into two regions A and
B. The area of R can be found as the sum of the areas of A and B.
Area of R ϭ
͵
2
͙x dx ϩ
ෆ
0
͵
4
͓͙x Ϫ ͑x Ϫ 2͔͒ dx
ෆ
2
area of A
2 ϭ ᎏᎏx 3ր2
3
area of B
] [
2
2 x2 ϩ ᎏᎏx 3ր2 Ϫ ᎏᎏ ϩ 2x
3
2
0
]
4
2
10 ϭ ᎏᎏ units squared
3
Now try Exercise 9.
Integrating with Respect to y
Sometimes the boundaries of a region are more easily described by functions of y than by functions of x. We can use approximating rectangles that are horizontal rather than vertical and the resulting basic formula has y in place of x.
For regions like these y y
d
y
d
d
x ϭ f (y)
x ϭ f(y)
x ϭ g(y)
x ϭ g( y) x ϭ g(y)
c
c
0
0 c x
x ϭ f (y) x x
0
use this formula
A=
∫c
d
[f (y) – g(y)] dy.
EXAMPLE 5 Integrating with Respect to y
Find the area of the region in Example 4 by integrating with respect to y. y SOLUTION
2
(4, 2)
x ϭ y2
(g(y), y)
xϭyϩ2
1
⌬y
( f (y), y)
f (y) Ϫ g(y)
0
yϭ0
2
4
Figure 7.11 It takes two integrations to find the area of this region if we integrate with respect to x. It takes only one if we integrate with respect to y. (Example 5)
x
We remarked in solving Example 4 that “it appears that no single integral can give the area of R,” but notice how appearances change when we think of our rectangles being summed over y. The interval of integration is ͓0, 2͔, and the rectangles run between the same two curves on the entire interval. There is no need to split the region
(Figure 7.11).
We need to solve for x in terms of y in both equations: yϭxϪ2 y ϭ ͙x
ෆ
becomes becomes x ϭ y ϩ 2, x ϭ y 2,
y Ն 0. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 394
394
Chapter 7
Applications of Definite Integrals
We must still be careful to subtract the lower number from the higher number when forming the integrand. In this case, the higher numbers are the higher x-values, which are on the line x ϭ y ϩ 2 because the line lies to the right of the parabola. So, y1 = x3, y2 = √x + 2, y3 = – √x + 2
Area of R ϭ
͵
[
2
0
y2 y3 ͑y ϩ 2 Ϫ y 2 ͒ dy ϭ ᎏᎏ ϩ 2y Ϫ ᎏᎏ
2
3
]
2
10 ϭ ᎏᎏ units squared.
3
0
Now try Exercise 11.
(c, d)
(a, b)
EXAMPLE 6 Making the Choice
Find the area of the region enclosed by the graphs of y ϭ x 3 and x ϭ y 2 Ϫ 2.
SOLUTION
[–3, 3] by [–3, 3]
We can produce a graph of the region on a calculator by graphing the three curves y ϭ x 3, y ϭ ͙x ϩෆ, and y ϭ Ϫ͙x ϩෆ (Figure 7.12).
ෆෆ 2
ෆෆ 2
Figure 7.12 The region in Example 6.
This conveniently gives us all of our bounding curves as functions of x. If we integrate in terms of x, however, we need to split the region at x ϭ a (Figure 7.13).
On the other hand, we can integrate from y ϭ b to y ϭ d and handle the entire region at once. We solve the cubic for x in terms of y:
(c, d)
y ϭ x3
(a, b)
becomes
x ϭ y 1ր 3.
To find the limits of integration, we solve y 1ր 3 ϭ y 2 Ϫ 2. It is easy to see that the lower limit is b ϭ Ϫ1, but a calculator is needed to find that the upper limit d ϭ 1.793003715. We store this value as D.
The cubic lies to the right of the parabola, so
[–3, 3] by [–3, 3]
Area ϭ NINT ͑y 1ր 3 Ϫ ͑y 2 Ϫ 2͒, y, Ϫ1, D͒ ϭ 4.214939673.
Figure 7.13 If we integrate with respect to x in Example 6, we must split the region at x ϭ a.
The area is about 4.21.
Now try Exercise 27.
Saving Time with Geometry Formulas
Here is yet another way to handle Example 4. y EXAMPLE 7 Using Geometry
(4, 2)
2 y ϭ √x
⎯
yϭxϪ2
1
SOLUTION
2
Area ϭ 2
2
0
yϭ0
2
Find the area of the region in Example 4 by subtracting the area of the triangular region from the area under the square root curve.
4
x
Figure 7.14 The area of the blue region is the area under the parabola y ϭ ͙x
ෆ
minus the area of the triangle. (Example 7)
Figure 7.14 illustrates the strategy, which enables us to integrate with respect to x without splitting the region.
Area ϭ
͵
4
0
1
2
2
3
͙x dx Ϫ ᎏᎏ ͑2͒͑2͒ ϭ ᎏᎏ x 3ր2
ෆ
]
4
10
Ϫ 2 ϭ ᎏᎏ units squared
3
0
Now try Exercise 35.
The moral behind Examples 4, 5, and 7 is that you often have options for finding the area of a region, some of which may be easier than others. You can integrate with respect to x or with respect to y, you can partition the region into subregions, and sometimes you can even use traditional geometry formulas. Sketch the region first and take a moment to determine the best way to proceed.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 395
Section 7.2 Areas in the Plane
Quick Review 7.2
(For help, go to Sections 1.2 and 5.1.)
In Exercises 1–5, find the area between the x-axis and the graph of the given function over the given interval.
1. y ϭ sin x over ͓0, p͔
2. y ϭ e 2x over ͓0, 1͔
2
1
ᎏᎏ(e2
2
Ϫ 1) Ϸ 3.195
3. y ϭ sec 2 x over ͓Ϫpր4, pր4͔
4. y ϭ 4x Ϫ x 3 over ͓0, 2͔
5. y ϭ ͙9ෆෆෆ
ෆ Ϫ x2
2
6. y ϭ x 2 Ϫ 4x and
7. y ϭ e x and
9p/2
2x
9. y ϭ ᎏᎏ x2 ϩ 1
y ϭ x ϩ 6 (6, 12); (Ϫ1, 5)
y ϭ x ϩ 1 (0, 1)
8. y ϭ x 2 Ϫ px and
4
over ͓Ϫ3, 3͔
In Exercises 6–10, find the x- and y-coordinates of all points where the graphs of the given functions intersect. If the curves never intersect, write “NI.”
and
10. y ϭ sin x and
y ϭ sin x (0, 0); (p, 0) y ϭ x3
(Ϫ1, Ϫ1); (0, 0); (1, 1)
y ϭ x 3 (–0.9286, –0.8008); (0, 0); (0.9286, 0.8008)
Section 7.2 Exercises
In Exercises 1–6, find the area of the shaded region analytically. y 1.
y
4.
4/3 x ϭ 12y 2 Ϫ 12y 3
1
p/2 yϭ1 1
x ϭ 2y 2 Ϫ 2y
y ϭ cos 2 x
–
2
0
0 y 5.
4p/3
(–2, 8)
– y ϭ 1 sec 2 t
2
128/15
(2, 8)
8
y ϭ 2x 2
2
1
–
–
3
x
1
x
y
2.
y ϭ x 4 Ϫ 2x 2
0
t
–
3
–2
y ϭ Ϫ4 sin2 t
–1
–1
1
x
2
NOT TO SCALE
22/15
1
y
1
y
6.
–4
3.
1/12
x ϭ y3
y ϭ x2
–1
(1, 1)
1
0
x ϭ y2
0
1
395
x
–2
y ϭ – 2x 4
x
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 396
396
Chapter 7
Applications of Definite Integrals
In Exercises 7 and 8, use a calculator to find the area of the region enclosed by the graphs of the two functions.
7. y ϭ sin x, y ϭ 1 Ϫ
Ϸ1.670 8. y ϭ cos(2x), y ϭ
x2
x2
Ϫ 2 Ϸ4.332
In Exercises 9 and 10, find the area of the shaded region analytically.
9.
yϭx
yϭ —
4
0
and
y ϭ ͑x 2ր 2͒ ϩ 4 21ᎏ3ᎏ
1
xϭyϩ2 and 1
4ᎏ2ᎏ
4x Ϫ y ϭ 16 30ᎏ8ᎏ
3
25. x ϩ y 2 ϭ 0
5/6
y
2
ᎏᎏa3
3
and x ϩ 2y 2 ϭ 3 4 and x ϩ 3y 2 ϭ 2 8/3
26. 4x 2 ϩ y ϭ 4
10.
yϭ0
and
24. x Ϫ y 2 ϭ 0 x 2
1
a Ͼ 0,
8
2
23. y 2 Ϫ 4x ϭ 4
x2
y ϭ x2
and
ΗxΗ
20. y ϭ ͙ෆෆ and 5y ϭ x ϩ 6 1ᎏ3ᎏ (3 points of intersection)
(How many intersection points are there?)
22. x ϭ y 2
yϭ1
1
19. y ϭ x
͙a ෆϪෆෆ,
ෆ2 ෆ x 2
21. y ϭ Η x 2 Ϫ 4 Η and
5/6
y
18. y ϭ x 4 Ϫ 4x 2 ϩ 4
27. x ϩ y 2 ϭ 3
and and x 4 Ϫ y ϭ 1 6ᎏᎏ
15
14
4x ϩ y 2 ϭ 0 8
28. y ϭ 2 sin x and
1
1
Ϫp ր 3 Յ x Յ p ր 3
31.
x
2
32.
In Exercises 11 and 12, find the area enclosed by the graphs of the two curves by integrating with respect to y.
11. y2 ϭ x ϩ 1, y2 ϭ 3 Ϫ x
Ϸ7.542
12. y2 ϭ x ϩ 3, y ϭ 2x
In Exercises 13 and 14, find the total shaded area.
13.
y ϭ Ϫx 2 ϩ 3x
(2, 2)
2
– 2 –1
1
8ᎏ6ᎏ
y
y ϭ4Ϫx2
x
and y ϭ 2
40. Find the area of the region between the curve y ϭ 3 Ϫ x 2 and the line y ϭ Ϫ1 by integrating with respect to (a) x, (b) y. 32/3
͵
͵
0
c
(c) Find c by integrating with respect to x. (This puts c into the integrand as well.)
2
10ᎏ3ᎏ
16. y ϭ 2x Ϫ x 2 and
y ϭ Ϫ3
17. y ϭ 7 Ϫ 2x 2
y ϭ x2 ϩ 4 4
and
39. Find the area of the “triangular” region in the first quadrant bounded on the left by the y-axis and on the right by the curves y ϭ sin x and y ϭ cos x. ͙2 Ϫ 1 Ϸ 0.414
ෆ
(b) Find c by integrating with respect to y. (This puts c in the c 4 limits of integration.)
͙y dy ϭ ͙y dy ⇒ c ϭ 24ր3
ෆ
ෆ
In Exercises 15–34, find the area of the regions enclosed by the lines and curves.
15. y ϭ x 2 Ϫ 2
2
(a) Sketch the region and draw a line y ϭ c across it that looks about right. In terms of c, what are the coordinates of the points where the line and parabola intersect? Add them to your figure. (–͙c, c); (͙c, c)
ෆ
ෆ
y ϭ Ϫx ϩ 2
(3, –5)
–5
0 Յ y Յ pր2
x ϭ 0,
41. The region bounded below by the parabola y ϭ x 2 and above by the line y ϭ 4 is to be partitioned into two subsections of equal area by cutting across it with the horizontal line y ϭ c.
2
3
ෆ ෆෆ
34. x ϭ 3 sin y ͙cos y and
In Exercises 35 and 36, find the area of the region by subtracting the area of a triangular region from the area of a larger region.
38. Find the area of the region in the first quadrant bounded by the line y ϭ x, the line x ϭ 2, the curve y ϭ 1 ր x 2, and the x-axis. 1
–10
1 2
x2
37. Find the area of the propeller-shaped region enclosed by the curve x Ϫ y 3 ϭ 0 and the line x Ϫ y ϭ 0. 1/2
(– 2, –10)
–2 –1
x,
36. The region on or above the x-axis bounded by the curves y ϭ 4 Ϫ x2 and y ϭ 3x. 15ր2
y ϭ 2x 3 Ϫ x 2 Ϫ 5x
4
33.
4
6͙3
ෆ
4
4
Ϫ ᎏᎏ Ϸ 0.0601 y ϭ cos ͑px ր 2͒ and y ϭ 1 Ϫ
ᎏᎏ
3 p 4Ϫp y ϭ sin ͑px ր 2͒ and y ϭ x ᎏᎏ Ϸ 0.273 p p y ϭ sec 2 x, y ϭ tan2 x, x ϭ Ϫp ր4, x ϭ p ր4 ᎏᎏ
2
x ϭ tan2 y and x ϭ Ϫtan2 y, Ϫp ր4 Յ y Յ p ր4 4 Ϫ p Ϸ 0.858
sec 2
35. The region on or above the x-axis bounded by the curves y2 ϭ x ϩ 3 and y ϭ 2x. Ϸ4.333
x
2
1
Ϸ7.146
16
y
(–2, 4)
yϭ
30.
0
14.
0ՅxՅp
29. y ϭ 8 cos x and
xϩyϭ2
y ϭ x2
y ϭ sin 2x,
2
10ᎏ3ᎏ
42. Find the area of the region in the first quadrant bounded on the left by the y-axis, below by the line y ϭ x ր4, above left by the curve y ϭ 1 ϩ ͙x , and above right by the curve y ϭ 2 ր ͙x . 11/3
ෆ
ෆ
͵
͙c
ෆ
41. (c)
0
(c Ϫ x 2) dx ϭ (4 Ϫ c)͙c ϩ
ෆ
͵
2
͙c
ෆ
(4 Ϫ x 2) dx ⇒ c ϭ 24ր3
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 397
Section 7.2 Areas in the Plane
43. The figure here shows triangle AOC inscribed in the region cut from the parabola y ϭ x 2 by the line y ϭ a 2. Find the limit of the ratio of the area of the triangle to the area of the parabolic region as a approaches zero. 3/4 y ϭ x2
(–a, a )
You should solve the following problems without using a graphing calculator.
True. 36 is the value of the appropriate integral.
2
C yϭa
2
(a, a )
A
Standardized Test Questions
50. True or False The area of the region enclosed by the graph of y ϭ x2 ϩ 1 and the line y ϭ 10 is 36. Justify your answer.
y
2
397
51. True or False The area of the region in the first quadrant enclosed by the graphs of y ϭ cos x, y ϭ x, and the y-axis is
given by the definite integral ͐0 (x Ϫ cos x) dx. Justify your answer. False. It is 0.739(cos x Ϫ x) dx.
0.739
͵
0
O
–a
x
a
52. Multiple Choice Let R be the region in the first quadrant bounded by the x-axis, the graph of x ϭ y2 ϩ 2, and the line x ϭ 4. Which of the following integrals gives the area of R? A
͵
͵
͵
͙2
ෆ
44. Suppose the area of the region between the graph of a positive continuous function f and the x-axis from x ϭ a to x ϭ b is
4 square units. Find the area between the curves y ϭ f ͑x͒ and y ϭ 2 f ͑x͒ from x ϭ a to x ϭ b. 4
45. Writing to Learn Which of the following integrals, if either, calculates the area of the shaded region shown here? Give reasons for your answer. Neither; both are zero
i.
͵
͵
1
͑x Ϫ ͑Ϫx͒͒ dx ϭ
Ϫ1
ii.
͵
͵
1
͑Ϫx Ϫ ͑x͒͒ dx ϭ
Ϫ1
(B)
0
[(y2 ϩ 2) Ϫ 4] dy
0
͙2
ෆ
(C)
͵
͵
͙2
ෆ
[4 Ϫ (y2 ϩ 2)]dy
͙2
ෆ
[4 Ϫ (y2 ϩ 2)]dy
Ϫ͙2
ෆ
(D)
[(y2 ϩ 2) Ϫ 4] dy
Ϫ͙2
ෆ
4
(E)
[4 Ϫ (y2 ϩ 2)]dy
2
2x dx
Ϫ1
1
(A)
53. Multiple Choice Which of the following gives the area of the region between the graphs of y ϭ x2 and y ϭ Ϫx from x ϭ 0 to x ϭ 3? E
1
Ϫ2x dx
Ϫ1
(A) 2
(B) 9/2
(C) 13/2
(D) 13
(E) 27/2
y y ϭ –x
1
yϭx
–1
1
54. Multiple Choice Let R be the shaded region enclosed by the
2
graphs of y ϭ eϪx , y ϭ Ϫsin(3x), and the y-axis as shown in the figure below. Which of the following gives the approximate area of the region R? B x (A) 1.139
(B) 1.445
(C) 1.869
(D) 2.114 (E) 2.340
y
–1
2
46. Writing to Learn Is the following statement true, sometimes true, or never true? The area of the region between the graphs of the continuous functions y ϭ f ͑x͒ and y ϭ g͑x͒ and the vertical lines x ϭ a and x ϭ b (a Ͻ b) is
͵
0
͓ f ͑x͒ Ϫ g͑x͔͒ dx.
Give reasons for your answer.
Sometimes; If f (x) Ն g(x) on (a, b), then true.
47. Find the area of the propeller-shaped region enclosed between the graphs of ln 4 Ϫ (1/2) Ϸ 0.886
2x
y ϭ ᎏᎏ and y ϭ x 3. x2 ϩ 1
48. Find the area of the propeller-shaped region enclosed between the graphs of y ϭ sin x and y ϭ x 3. Ϸ 0.4303
49. Find the positive value of k such that the area of the region enclosed between the graph of y ϭ k cos x and the graph of y ϭ kx 2 is 2. k Ϸ 1.8269
x
–2
b
a
2
55. Multiple Choice Let f and g be the functions given by f (x) ϭ ex and g(x) ϭ 1/x. Which of the following gives the area of the region enclosed by the graphs of f and g between x ϭ 1 and x ϭ 2? A
(A) e2 Ϫ e Ϫ ln2
(B) ln 2 Ϫ e2 ϩ e
1
(C) e2 Ϫ ᎏᎏ
2
1
(D) e2 Ϫ e Ϫ ᎏᎏ
2
1
(E) ᎏᎏ Ϫ ln2 e 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 398
398
Chapter 7
Applications of Definite Integrals
Exploration
Extending the Ideas
56. Group Activity Area of Ellipse
57. Cavalieri’s Theorem Bonaventura Cavalieri (1598–1647) discovered that if two plane regions can be arranged to lie over the same interval of the x-axis in such a way that they have identical vertical cross sections at every point (see figure), then the regions have the same area. Show that this theorem is true.
An ellipse with major axis of length 2a and minor axis of length 2b can be coordinatized with its center at the origin and its major axis horizontal, in which case it is defined by the equation x2 y2 ᎏᎏ ϩ ᎏᎏ ϭ 1.
2
a b2 (a) Find the equations that define the upper and lower semiellipses as functions of x.
Cross sections have the same length at every point in [a, b].
(b) Write an integral expression that gives the area of the ellipse.
(c) With your group, use NINT to find the areas of ellipses for various lengths of a and b.
(d) There is a simple formula for the area of an ellipse with major axis of length 2a and minor axis of length 2b. Can you tell what it is from the areas you and your group have found?
(e) Work with your group to write a proof of this area formula by showing that it is the exact value of the integral expression in part (b).
a
x
b
58. Find the area of the region enclosed by the curves x y ϭ ᎏᎏ and y ϭ mx, 0 Ͻ m Ͻ 1. x2 ϩ 1
m Ϫ ln (m) Ϫ 1
Ί x (b) 2͵ bΊ1 Ϫ ᎏᎏ dx
a
x2
56. (a) y ϭ Ϯb 1 Ϫ ᎏᎏ a2 a
2
Ϫa
2
(c) Answers may vary.
(d, e) abp
57. Since f (x) Ϫ g(x) is the same for each region where f (x) and g(x) represent b the upper and lower edges, area ϭ ͐a [ f (x) Ϫ g(x)] dx will be the same for each. 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 399
Section 7.3 Volumes
399
Volumes
7.3
What you’ll learn about
Volume As an Integral
• Volume As an Integral
• Square Cross Sections
• Circular Cross Sections
• Cylindrical Shells
• Other Cross Sections
. . . and why
The techniques of this section allow us to compute volumes of certain solids in three dimensions.
In Section 5.1, Example 3, we estimated the volume of a sphere by partitioning it into thin slices that were nearly cylindrical and summing the cylinders’ volumes using MRAM.
MRAM sums are Riemann sums, and had we known how at the time, we could have continued on to express the volume of the sphere as a definite integral.
Starting the same way, we can now find the volumes of a great many solids by integration. Suppose we want to find the volume of a solid like the one in Figure 7.15. The cross section of the solid at each point x in the interval ͓a, b͔ is a region R͑x͒ of area A͑x͒. If A is a continuous function of x, we can use it to define and calculate the volume of the solid as an integral in the following way.
We partition ͓a, b͔ into subintervals of length Δx and slice the solid, as we would a loaf of bread, by planes perpendicular to the x-axis at the partition points. The kth slice, the one between the planes at x kϪ1 and x k , has approximately the same volume as the cylinder between the two planes based on the region R͑x k ͒ (Figure 7.16).
y
y
Px
Cross-section R(x) with area A(x)
Approximating cylinder based on R(xk ) has height
Δ xk ϭ xk Ϫ xkϪ1
Plane at xkϪ1
S
0
a
0
x
xkϪ1
b
Plane at xk xk x
The cylinder’s base is the region R(xk ) with area A(xk )
Figure 7.15 The cross section of an arbitrary solid at point x.
x
NOT TO SCALE
Figure 7.16 Enlarged view of the slice of the solid between the planes at xkϪ1 and xk.
The volume of the cylinder is
Vk ϭ base area ϫ height ϭ A͑x k ͒ ϫ Δx.
The sum
͚ V ϭ ͚ A͑x ͒ ϫ Δx k k
approximates the volume of the solid.
This is a Riemann sum for A͑x͒ on ͓a, b͔. We expect the approximations to improve as the norms of the partitions go to zero, so we define their limiting integral to be the volume of the solid.
DEFINITION Volume of a Solid
The volume of a solid of known integrable cross section area A͑x͒ from x ϭ a to x ϭ b is the integral of A from a to b,
Vϭ
͵
b
a
A͑x͒ dx.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 400
400
Chapter 7
Applications of Definite Integrals
To apply the formula in the previous definition, we proceed as follows.
How to Find Volume by the Method of Slicing
1. Sketch the solid and a typical cross section.
2. Find a formula for A͑x͒.
3. Find the limits of integration.
y
4. Integrate A͑x͒ to find the volume.
Typical cross-section
Square Cross Sections
x
0
3 x Let us apply the volume formula to a solid with square cross sections.
x
3
x (m)
3
EXAMPLE 1 A Square-Based Pyramid
A pyramid 3 m high has congruent triangular sides and a square base that is 3 m on each side. Each cross section of the pyramid parallel to the base is a square. Find the volume of the pyramid.
SOLUTION
Figure 7.17 A cross section of the pyramid in Example 1.
We follow the steps for the method of slicing.
1. Sketch. We draw the pyramid with its vertex at the origin and its altitude along the interval 0 Յ x Յ 3. We sketch a typical cross section at a point x between 0 and 3
(Figure 7.17).
2. Find a formula for A͑ x͒. The cross section at x is a square x meters on a side, so
A͑x͒ ϭ x 2.
3. Find the limits of integration. The squares go from x ϭ 0 to x ϭ 3.
4. Integrate to find the volume.
Vϭ
͵
3
A͑x͒ dx ϭ
0
͵
3
x2
0
x3 ϭ ᎏᎏ
3
ϭ 9 m3
0
Now try Exercise 3.
[–3, 3] by [–4, 4]
Figure 7.18 The region in Example 2.
]
3
Circular Cross Sections
y
The only thing that changes when the cross sections of a solid are circular is the formula for A͑x͒. Many such solids are solids of revolution, as in the next example. f(x) EXAMPLE 2 A Solid of Revolution x The region between the graph of f ͑x͒ ϭ 2 ϩ x cos x and the x-axis over the interval
͓Ϫ2, 2͔ is revolved about the x-axis to generate a solid. Find the volume of the solid.
SOLUTION
Figure 7.19 The region in Figure 7.18 is revolved about the x-axis to generate a solid. A typical cross section is circular, with radius f (x) ϭ 2 ϩ x cos x.
(Example 2)
Revolving the region (Figure 7.18) about the x-axis generates the vase-shaped solid in
Figure 7.19. The cross section at a typical point x is circular, with radius equal to f ͑x͒.
Its area is
A͑x͒ ϭ p͑ f ͑x͒͒ 2. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 401
Section 7.3 Volumes
401
The volume of the solid is
Vϭ
͵
2
A͑x͒ dx
Ϫ2
Ϸ NINT ͑p͑2 ϩ x cos x͒2, x, Ϫ2, 2͒ Ϸ 52.43 units cubed.
Now try Exercise 7.
EXAMPLE 3 Washer Cross Sections
The region in the first quadrant enclosed by the y-axis and the graphs of y ϭ cos x and y ϭ sin x is revolved about the x-axis to form a solid. Find its volume.
SOLUTION
The region is shown in Figure 7.20.
We revolve it about the x-axis to generate a solid with a cone-shaped cavity in its center
(Figure 7.21).
[–/4, /2] by [–1.5, 1.5]
Figure 7.20 The region in Example 3.
R r Figure 7.22 The area of a washer is pR2 Ϫ pr2. (Example 3)
Figure 7.21 The solid generated by revolving the region in Figure 7.20 about the x-axis. A typical cross section is a washer: a circular region with a circular region cut out of its center. (Example 3)
This time each cross section perpendicular to the axis of revolution is a washer, a circular region with a circular region cut from its center. The area of a washer can be found by subtracting the inner area from the outer area (Figure 7.22).
In our region the cosine curve defines the outer radius, and the curves intersect at x ϭ p ր4. The volume is
͵
͵
pր4
Vϭ
p͑cos2 x Ϫ sin2 x͒ dx
0
pր4
ϭp
CAUTION!
cos 2x dx
identity: cos2 x Ϫ sin2 x ϭ cos 2x
0
The area of a washer is pR 2 Ϫ pr 2, which you can simplify to p(R 2 Ϫ r 2), but not to p(R Ϫ r)2. No matter how tempting it is to make the latter simplification, it’s wrong. Don’t do it.
[ ]
sin 2x ϭ p ᎏᎏ
2
pր4
0
p ϭ ᎏᎏ units cubed.
2
Now try Exercise 17.
We could have done the integration in Example 3 with NINT, but we wanted to demonstrate how a trigonometric identity can be useful under unexpected circumstances in calculus. The double-angle identity turned a difficult integrand into an easy one and enabled us to get an exact answer by antidifferentiation.
Cylindrical Shells
There is another way to find volumes of solids of rotation that can be useful when the axis of revolution is perpendicular to the axis containing the natural interval of integration. Instead of summing volumes of thin slices, we sum volumes of thin cylindrical shells that grow outward from the axis of revolution like tree rings.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 402
402
Chapter 7
Applications of Definite Integrals
EXPLORATION 1
The region enclosed by the x-axis and the parabola y ϭ f ͑ x͒ ϭ 3x Ϫ x 2 is revolved about the line x ϭ Ϫ1 to generate the shape of a cake (Figures 7.23, 7.24). (Such a cake is often called a bundt cake.) What is the volume of the cake?
Integrating with respect to y would be awkward here, as it is not easy to get the original parabola in terms of y. (Try finding the volume by washers and you will soon see what we mean.) To integrate with respect to x, you can do the problem by cylindrical shells, which requires that you cut the cake in a rather unusual way.
[–6, 4] by [–3, 3]
Figure 7.23 The graph of the region in
Exploration 1, before revolution. y 0
3
x
Axis of revolution x ϭ –1
Figure 7.24 The region in Figure 7.23 is revolved about the line x ϭ Ϫ1 to form a solid cake. The natural interval of integration is along the x-axis, perpendicular to the axis of revolution. (Exploration 1)
1. Instead of cutting the usual wedge shape, cut a cylindrical slice by cutting straight down all the way around close to the inside hole. Then cut another cylindrical slice around the enlarged hole, then another, and so on. The radii of the cylinders gradually increase, and the heights of the cylinders follow the contour of the parabola: smaller to larger, then back to smaller ( Figure 7.25).
Each slice is sitting over a subinterval of the x-axis of length Δx. Its radius is approximately ͑1 ϩ x k ͒. What is its height?
2. If you unroll the cylinder at x k and flatten it out, it becomes (essentially) a rectangular slab with thickness Δx. Show that the volume of the slab is approximately
2p͑x k ϩ 1͒͑3x k Ϫ x k 2 ͒Δx.
3. ͚ 2p͑x k ϩ 1͒͑3x k Ϫ x k 2 ͒Δx is a Riemann sum. What is the limit of these Riemann sums as Δx→0?
4. Evaluate the integral you found in step 3 to find the volume of the cake!
EXAMPLE 4 Finding Volumes Using Cylindrical Shells
y
The region bounded by the curve y ϭ ͙ෆ, the x-axis, and the line x ϭ 4 is revolved x about the x-axis to generate a solid. Find the volume of the solid.
SOLUTION
xk
3
Figure 7.25 Cutting the cake into thin cylindrical slices, working from the inside out. Each slice occurs at some xk between 0 and 3 and has thickness Δx.
(Exploration 1)
x
1. Sketch the region and draw a line segment across it parallel to the axis of revolution
(Figure 7.26). Label the segment’s length (shell height) and distance from the axis of revolution (shell radius). The width of the segment is the shell thickness dy. (We drew the shell in Figure 7.27, but you need not do that.) y y
Shell height
4 Ϫ y2
Shell height
2
Interval of integration yk
0
Volume by Cylindrical Shells
⎧
⎪
⎨y
⎪
⎩
0
x ϭ y2
2
(4, 2) y 4 Ϫ y2 y ϭ ͙x
͙
(4, 2)
Shell thickness ϭ dy
y Shell radius
4
x
0 y Figure 7.26 The region, shell dimensions, and interval of integration in Example 4.
x
S
Shell radius Figure 7.27 The shell swept out by the line segment in Figure 7.26.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 403
Section 7.3 Volumes
403
2. Identify the limits of integration: y runs from 0 to 2.
3. Integrate to find the volume.
͵ ( )( )
͵
2
Vϭ
2p
0
shell shell dy radius height
2
ϭ
2p͑y͒͑4 Ϫ y 2 ͒ dy ϭ 8p
0
Now try Exercise 33(a). y EXAMPLE 5 Finding Volumes Using Cylindrical Shells
The region bounded by the curves y ϭ 4 Ϫ x2, y ϭ x, and x ϭ 0 is revolved about the y-axis to form a solid. Use cylindrical shells to find the volume of the solid.
4
SOLUTION
2
0
x
2
Figure 7.28 The region and the height of a typical shell in Example 5.
1. Sketch the region and draw a line segment across it parallel to the y-axis
(Figure 7.28). The segment’s length (shell height) is 4 Ϫ x2 Ϫ x. The distance of the segment from the axis of revolution (shell radius) is x.
2. Identify the limits of integration: The x-coordinate of the point of intersection of the curves y ϭ 4 Ϫ x2 and y ϭ x in the first quadrant is about 1.562. So x runs from 0 to
1.562.
3. Integrate to find the volume.
͵
͵
1.562
Vϭ
2p
0
( )( )
shell shell dx radius height
1.562
ϭ
2p(x)(4 Ϫ x2 Ϫ x) dx
0
Ϸ 13.327
Now try Exercise 35.
Other Cross Sections o π
xk
[–1, 3.5] by [–0.8, 2.2]
Figure 7.29 The base of the paperweight in Example 6. The segment perpendicular to the x-axis at xk is the diameter of a semicircle that is perpendicular to the base. y
2
y = 2 sin x
The method of cross-section slicing can be used to find volumes of a wide variety of unusually shaped solids, so long as the cross sections have areas that we can describe with some formula. Admittedly, it does take a special artistic talent to draw some of these solids, but a crude picture is usually enough to suggest how to set up the integral.
EXAMPLE 6 A Mathematician’s Paperweight
A mathematician has a paperweight made so that its base is the shape of the region between the x-axis and one arch of the curve y ϭ 2 sin x (linear units in inches). Each cross section cut perpendicular to the x-axis (and hence to the xy-plane) is a semicircle whose diameter runs from the x-axis to the curve. (Think of the cross section as a semicircular fin sticking up out of the plane.) Find the volume of the paperweight.
SOLUTION
0
x
Figure 7.30 Cross sections perpendicular to the region in Figure 7.29 are semicircular.
(Example 6)
The paperweight is not easily drawn, but we know what it looks like. Its base is the region in Figure 7.29, and the cross sections perpendicular to the base are semicircular fins like those in Figure 7.30.
The semicircle at each point x has
2 sin x
1
radius ϭ ᎏᎏ ϭ sin x and area A͑x͒ ϭ ᎏᎏp͑sin x͒ 2 .
2
2 continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 404
404
Chapter 7
Applications of Definite Integrals
The volume of the paperweight is
Vϭ
͵
p
A͑x͒ dx
0
p ϭ ᎏᎏ
2
͵
p
͑sin x͒ 2 dx
0
p
Ϸ ᎏᎏ NINT ͑͑sin x͒ 2, x, 0, p͒
2
p
Ϸ ᎏᎏ͑1.570796327͒.
2
Bonaventura Cavalieri
The number in parentheses looks like half of p, an observation that can be confirmed analytically, and which we support numerically by dividing by p to get 0.5. The volume of the paperweight is p p p2 ᎏᎏ • ᎏᎏ ϭ ᎏᎏ Ϸ 2.47 in3.
2 2
4
Now try Exercise 39(a).
(1598—1647)
Cavalieri, a student of
Galileo, discovered that if two plane regions can be arranged to lie over the same interval of the x-axis in such a way that they have identical vertical cross sections at every point, then the regions have the same area. This theorem and a letter of recommendation from Galileo were enough to win Cavalieri a chair at the University of Bologna in 1629. The solid geometry version in Example 7, which Cavalieri never proved, was named after him by later geometers.
EXAMPLE 7 Cavalieri’s Volume Theorem
Cavalieri’s volume theorem says that solids with equal altitudes and identical cross section areas at each height have the same volume (Figure 7.31). This follows immediately from the definition of volume, because the cross section area function A͑x͒ and the interval
͓a, b͔ are the same for both solids.
b
Same volume
a
Cross sections have the same length at every point in [a, b].
Same cross-section area at every level
Figure 7.31 Cavalieri’s volume theorem: These solids have the same volume. You can illustrate this yourself with stacks of coins. (Example 7) a x
b
Now try Exercise 43.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 405
Section 7.3 Volumes
EXPLORATION 2
405
Surface Area
We know how to find the volume of a solid of revolution, but how would we find the surface area? As before, we partition the solid into thin slices, but now we wish to form a Riemann sum of approximations to surface areas of slices (rather than of volumes of slices). y y = f(x)
b
a
x
A typical slice has a surface area that can be approximated by 2p • f ͑x͒ • Δs, where Δs is the tiny slant height of the slice. We will see in Section 7.4, when we study arc length, that Δs ϭ ͙Δx2 ϩෆ, and that this can be written as
ෆ Δy2
2 Δx.
Δs ϭ ͙1ෆෆf Јෆෆ͒
ෆ ϩ ͑ ෆ͑xk ͒ෆ
Thus, the surface area is approximated by the Riemann sum n ෆ ϩ͑ ෆ
͚ 2p f ͑x ͒ ͙1ෆෆfෆЈ͑xෆ͒͒ෆ Δx. k k
2
kϭ1
1. Write the limit of the Riemann sums as a definite integral from a to b. When will the limit exist?
2. Use the formula from part 1 to find the surface area of the solid generated by revolving a single arch of the curve y ϭ sin x about the x-axis.
3. The region enclosed by the graphs of y 2 ϭ x and x ϭ 4 is revolved about the x-axis to form a solid. Find the surface area of the solid.
Quick Review 7.3
(For help, go to Section 1.2.)
In Exercises 1–10, give a formula for the area of the plane region in terms of the single variable x.
1. a square with sides of length x x 2
2. a square with diagonals of length x x2/2
3. a semicircle of radius x px2/2
4. a semicircle of diameter x px2/8
5. an equilateral triangle with sides of length x (͙3/4)x 2
ෆ
6. an isosceles right triangle with legs of length x x2/2
7. an isosceles right triangle with hypotenuse x x2/4
8. an isosceles triangle with two sides of length 2x and one side of length x (͙15 2
ෆ/4)x
9. a triangle with sides 3x, 4x, and 5x 6x 2
10. a regular hexagon with sides of length x (3͙3/2)x 2
ෆ
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 406
406
Chapter 7
Applications of Definite Integrals
Section 7.3 Exercises
In Exercises 1 and 2, find a formula for the area A(x) of the cross sections of the solid that are perpendicular to the x-axis.
1. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis between these planes run from the semicircle y ϭ Ϫ͙ෆෆෆෆ to the semicircle y ϭ ͙ෆෆෆෆ.
1 Ϫ x2
1 Ϫ x2
2. The solid lies between planes perpendicular to the x-axis at x ϭ 0 and x ϭ 4. The cross sections perpendicular to the x-axis between these planes run from y ϭ Ϫ͙x to y ϭ ͙x .
ෆ
ෆ
(a) The cross sections are circular disks with diameters in the xy-plane. px y (a) The cross sections are circular disks with diameters in the xy-plane. p(1 Ϫ x 2) x2 ϩ y2 ϭ 1 y x ϭ y2
4
x
0
–1
1
(b) The cross sections are squares with bases in the xy-plane. 4x
x
(b) The cross sections are squares with bases in the xy-plane. 4(1 Ϫ x 2) y y
x ϭ y2
4
x
0
–1
x
1 x2 ϩ
y2
(c) The cross sections are squares with diagonals in the xy-plane. 2x
ϭ1
(c) The cross sections are squares with diagonals in the xy-plane. (The length of a square’s diagonal is ͙2 times
ෆ
the length of its sides.) 2(1 – x 2) x2 ϩ y2 ϭ 1
y
In Exercises 3–6, find the volume of the solid analytically.
3. The solid lies between planes perpendicular to the x-axis at x ϭ 0 and x ϭ 4. The cross sections perpendicular to the axis on the interval 0 Յ x Յ 4 are squares whose diagonals run from y ϭ Ϫ͙x to y ϭ ͙x . 16
ෆ
ෆ
4. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis are circular disks whose diameters run from the parabola y ϭ x 2 to the parabola y ϭ 2 Ϫ x 2. 16p/15
0
1
–1
(d) The cross sections are equilateral triangles with bases in the xy-plane. ͙3x
ෆ
x
(d) The cross sections are equilateral triangles with bases in the xy-plane. ͙3(1 Ϫ x 2)
ෆ
y
2
y ϭ x2
y
0 y ϭ 2 Ϫ x2
0
–1
x2 ϩ y2 ϭ 1
1
x
x
5. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis between these planes are squares whose bases run from the semicircle y ϭ Ϫ͙ෆෆෆ2 to the semicircle y ϭ ͙ෆෆෆ2 .
1 Ϫ xෆ
1 Ϫ xෆ
16/3
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 407
Section 7.3 Volumes
6. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis between these planes are squares whose diagonals run from the semicircle y ϭ Ϫ͙ෆෆෆෆ to the semicircle
1 Ϫ x2
2 . 8/3 y ϭ ͙ෆෆෆෆ
1Ϫx
In Exercises 7–10, find the volume of the solid generated by revolving the shaded region about the given axis.
y
x ϭ 3y/2
x ϩ 2y ϭ 2
2
9. about the y-axis 4 Ϫ p
x
1
x
3
0
y
31. Find the volume of the solid generated by revolving the region bounded by the parabola y ϭ x 2 and the line y ϭ 1 about
(a) the line y ϭ 1. 16p/15 (b) the line y ϭ 2.
y ϭ sin x cos x
(c) the line y ϭ Ϫ1.
x
0
13. y ϭ
15. y ϭ x, y ϭ 1,
17. y ϭ x 2 ϩ 1,
19. y ϭ sec x,
20. y ϭ Ϫ͙x ,
ෆ
32p/5
xϭ2
12. y ϭ x 3,
y ϭ 0,
y ϭ 0 36p 14. y ϭ x Ϫ x 2, x ϭ 0 2p/3 16. y ϭ 2x,
yϭxϩ3
117p/5
In Exercises 33 and 34, use the cylindrical shell method to find the volume of the solid generated by revolving the shaded region about the indicated axis.
(b) the line y ϭ 1
33. (a) the x-axis 6p/5
x ϭ 2 128p/7
yϭ0
(p/3)b2h
(b) the y-axis.
(c) the line y ϭ 8 ր5
4p/5
2p (d) the line y ϭ Ϫ2 ր5
2p
y
p/30
y ϭ x, x ϭ 1 p
18. y ϭ 4 Ϫ x 2,
ෆ y ϭ ͙2 , Ϫp ր4 Յ x Յ p ր4 y ϭ Ϫ2,
56p/15
64p/15
(a) the x-axis. (p/3)bh2
In Exercises 11–20, find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis. ͙9ෆෆෆ,
ෆ Ϫ x2
8p/3
32. By integration, find the volume of the solid generated by revolving the triangular region with vertices ͑0, 0͒, ͑b, 0͒,
͑0, h͒ about x 11. y ϭ x 2, y ϭ 0,
32p/5
(d) the line x ϭ 4. 224p/15
(a) the line x ϭ 1. 2p/3 (b) the line x ϭ 2.
– x ϭ tan ⎛ y⎛
⎝4 ⎝
0
(b) the y-axis.
8p/3
30. Find the volume of the solid generated by revolving the triangular region bounded by the lines y ϭ 2x, y ϭ 0, and x ϭ 1 about
10. about the x-axis p2/16
y
Group Activity In Exercises 29–32, find the volume of the solid described. (c) the line y ϭ 2.
2
0
8p
ෆ
28. the region bounded above by the curve y ϭ ͙x and below by the line y ϭ x 2p/15
(a) the x-axis. 8p
y
1
27. the region in the first quadrant bounded above by the parabola y ϭ x 2, below by the x-axis, and on the right by the line x ϭ 2
29. Find the volume of the solid generated by revolving the region bounded by y ϭ ͙x and the lines y ϭ 2 and x ϭ 0 about
ෆ
8. about the y-axis 6p
7. about the x-axis 2p/3
407
p2
y ϭ 2 Ϫ x 108p/5
1
x ϭ 12 (y2 Ϫ y 3)
0
1
Ϫ 2p
x ϭ 0 8p
In Exercises 21 and 22, find the volume of the solid generated by revolving the region about the given line.
21. the region in the first quadrant bounded above by the line y ϭ ͙2 , below by the curve y ϭ sec x tan x, and on the left by
ෆ
the y-axis, about the line y ϭ ͙2 2.301
ෆ
22. the region in the first quadrant bounded above by the line y ϭ 2, below by the curve y ϭ 2 sin x, 0 Յ x Յ p ր 2, and on the left by the y-axis, about the line y ϭ 2 p(3p Ϫ 8)
In Exercises 23–28, find the volume of the solid generated by revolving the region about the y-axis.
ෆ
23. the region enclosed by x ϭ ͙5 y 2, x ϭ 0, y ϭ Ϫ1, y ϭ 1
(c) the line y ϭ 5 8p y (d) the line y ϭ Ϫ5 ր 8 y4 y2 xϭ— Ϫ —
4
2
(2, 2)
2p
24. the region enclosed by x ϭ y 3ր 2, x ϭ 0, y ϭ 2 4p
y2 xϭ— 2
25. the region enclosed by the triangle with vertices ͑1, 0͒, ͑2, 1͒, and ͑1, 1͒ 4p/3
26. the region enclosed by the triangle with vertices ͑0, 1͒, ͑1, 0͒, and ͑1, 1͒ 2p/3
(b) the line y ϭ 2 8p/5
34. (a) the x-axis 8p/3
2
x
0
1
2
x
4p
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 408
408
Chapter 7
Applications of Definite Integrals
In Exercises 35–38, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis.
35. y ϭ x,
y ϭ Ϫx ր 2, x ϭ 2
36. y ϭ x 2,
37. y ϭ ͙x ,
ෆ
y ϭ 2 Ϫ x, y ϭ 0,
38. y ϭ 2x Ϫ 1,
43. Writing to Learn A solid lies between planes perpendicular to the x-axis at x ϭ 0 and x ϭ 12. The cross sections by planes perpendicular to the x-axis are circular disks whose diameters run from the line y ϭ x ր 2 to the line y ϭ x as shown in the figure. Explain why the solid has the same volume as a right circular cone with base radius 3 and height 12.
8p
x ϭ 0,
for x Ն 0
5p/6
x ϭ 4 128p/5
y ϭ ͙x ,
ෆ
xϭ0
The volumes are equal by Cavalieri’s
Theorem.
y
7p/15
yϭx
In Exercises 39–42, find the volume of the solid analytically.
39. The base of a solid is the region between the curve
ෆෆ x y ϭ 2 ͙sinෆ and the interval ͓0, p͔ on the x-axis. The cross sections perpendicular to the x-axis are
yϭ x
2
(a) equilateral triangles with bases running from the x-axis to the curve as shown in the figure. 2͙3
ෆ
0 x 12
y
44. A Twisted Solid A square of side length s lies in a plane perpendicular to a line L. One vertex of the square lies on L.
As this square moves a distance h along L, the square turns one revolution about L to generate a corkscrew-like column with square cross sections.
2
0
y ϭ 2 √⎯⎯⎯ x
⎯ sin
x
(b) squares with bases running from the x-axis to the curve.
40. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫp ր 3 and x ϭ p ր 3. The cross sections perpendicular to the x-axis are
(a) circular disks with diameters running from the curve y ϭ tan x to the curve y ϭ sec x. p͙3 Ϫ (p2/6)
ෆ
(b) squares whose bases run from the curve y ϭ tan x to the curve y ϭ sec x. 4͙3 Ϫ (2p/3)
ෆ
41. The solid lies between planes perpendicular to the y-axis at y ϭ 0 and y ϭ 2. The cross sections perpendicular to the y-axis are circular disks with diameters running from the y-axis to the parabola x ϭ ͙ෆy 2. 8p
5
42. The base of the solid is the disk x 2 ϩ y 2 Յ 1. The cross sections by planes perpendicular to the y-axis between y ϭ Ϫ1 and y ϭ 1 are isosceles right triangles with one leg in the disk. 8/3
(a) Find the volume of the column.
8
s2h
(b) Writing to Learn What will the volume be if the square turns twice instead of once? Give reasons for your answer. s2h
45. Find the volume of the solid generated by revolving the region in the first quadrant bounded by y ϭ x 3 and y ϭ 4x about
(a) the x-axis, 512p/21
(b) the line y ϭ 8. 832p/21
46. Find the volume of the solid generated by revolving the region bounded by y ϭ 2x Ϫ x 2 and y ϭ x about
(a) the y-axis, p/6
(b) the line x ϭ 1. p/6
47. The region in the first quadrant that is bounded above by the curve y ϭ 1 ր ͙x , on the left by the line x ϭ 1 ր4, and below
ෆ
by the line y ϭ 1 is revolved about the y-axis to generate a solid. Find the volume of the solid by (a) the washer method and
(b) the cylindrical shell method. (a) 11p/48 (b) 11p/48
48. Let f ͑x͒ ϭ
͑sin
{ 1, x͒ րx,
0ϽxՅp x ϭ 0.
(a) Show that x f ͑x͒ ϭ sin x,
0 Յ x Յ p.
(b) Find the volume of the solid generated by revolving the shaded region about the y-axis. 4p y y
⎧ sin x , 0 Ͻ x
——
yϭ⎨ x xϭ0 ⎩ 1,
≤
1
0
1 x 2 ϩ y2 ϭ 1
x
0
x
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 409
Section 7.3 Volumes
49. Designing a Plumb Bob Having been asked to design a brass plumb bob that will weigh in the neighborhood of 190 g, you decide to shape it like the solid of revolution shown here.
59. y ϭ x 2,
0 Յ x Յ 2;
60. y ϭ 3x Ϫ x 2,
y ϭ x ͙36 Ϫ x 2
12
62. y ϭ ͙x ϩෆ,
ෆෆ 1
0
cm3
(b) If you specify a brass that weighs 8.5 g րcm3, how much will the plumb bob weigh to the nearest gram? 192.3 g
50. Volume of a Bowl A bowl has a shape that can be generated by revolving the graph of y ϭ x 2ր 2 between y ϭ 0 and y ϭ 5 about the y-axis.
(a) Find the volume of the bowl.
25p
(b) If we fill the bowl with water at a constant rate of
3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep? 3/(8p)
51. The Classical Bead Problem A round hole is drilled through the center of a spherical solid of radius r. The resulting cylindrical hole has height 4 cm.
(a) What is the volume of the solid that remains?
32p/3
(b) What is unusual about the answer? The answer is independent of r.
52. Writing to Learn Explain how you could estimate the volume of a solid of revolution by measuring the shadow cast on a table parallel to its axis of revolution by a light shining directly above it. See page 410.
53. Same Volume about Each Axis The region in the first quadrant enclosed between the graph of y ϭ ax Ϫ x 2 and the x-axis generates the same volume whether it is revolved about the x-axis or the y-axis. Find the value of a. 5
54. (Continuation of Exploration 2) Let x ϭ g͑y͒ Ͼ 0 have a continuous first derivative on ͓c, d ͔. Show that the area of the surface generated by revolving the curve x ϭ g͑y͒ about the y-axis is See page 410.
Sϭ
͵
͵
0
65. Multiple Choice The base of a solid S is the region enclosed by the graph of y ϭ ln x, the line x ϭ e, and the x-axis. If the cross sections of S perpendicular to the x-axis are squares, which of the following gives the best approximation of the volume of S? A
(A) 0.718
In Exercises 55–62, find the area of the surface generated by revolving the curve about the indicated axis.
0 Յ y Յ 2; y-axis Ϸ13.614
(B) 1.718
(A) 60.3
(B) 115.2
57. x ϭ y 1ր 2 Ϫ ͑1 ր 3͒ 3 ր 2,
1 Յ y Յ 3;
58. x ϭ ͙2y Ϫෆ, ͑5ր 8͒ Յ y Յ 1;
ෆෆෆ 1
y-axis Ϸ16.110 y-axis Ϸ2.999
(D) 3.171
(E) 7.388
(C) 225.4
(D) 319.7
(E) 361.9
67. Multiple Choice Let R be the region enclosed by the graph of y ϭ x2, the line x ϭ 4, and the x-axis. Which of the following gives the best approximation of the volume of the solid generated when R is revolved about the y-axis? B
(A) 64p
(B) 128p
(C) 256p
(D) 360
(E) 512
68. Multiple Choice Let R be the region enclosed by the graphs of y ϭ eϪx, y ϭ ex, and x ϭ 1. Which of the following gives the volume of the solid generated when R is revolved about the x-axis? D
͵
͵
͵
͵
͵
1
(A)
(ex Ϫ eϪx) dx
0
1
(e2x Ϫ eϪ2x) dx
0
1
(C)
(ex Ϫ eϪx)2 dx
0
1
y-axis Ϸ0.638
(C) 2.718
66. Multiple Choice Let R be the region in the first quadrant bounded by the graph of y ϭ 8 Ϫ x3/2, the x-axis, and the y-axis.
Which of the following gives the best approximation of the volume of the solid generated when R is revolved about the x-axis? E
(B)
0 Յ y Յ 1;
x-axis Ϸ51.313
64. True or False If the region enclosed by the y-axis, the line y ϭ 2, and the curve y ϭ ͙x is revolved about the y-axis, the
ෆ
2 volume of the solid is given by the definite integral ͐0 py2 dy.
2
Justify your answer. False. The volume is given by py 4 dy.
2p g͑y͒ ͙1ෆෆg Ј͑ y͒ෆ dy.
ෆ ϩ ͑ ෆෆෆ͒ 2
c
56. x ϭ y 3ր 3,
1 Յ x Յ 5;
x-axis Ϸ6.283
63. True or False The volume of a solid of a known integrable b cross section area A(x) from x ϭ a to x ϭ b is ͐a A(x) dx. Justify your answer. True, by definition.
d
55. x ϭ ͙y ,
ෆ
0.5 Յ x Յ 1.5;
You may use a graphing calculator to solve the following problems. x (cm
36p/5
x-axis Ϸ44.877
Standardized Test Questions
6
(a) Find the plumb bob’s volume.
x-axis Ϸ53.226
0 Յ x Յ 3;
61. y ϭ ͙2 x Ϫෆ2 ,
ෆෆෆ x ෆ y (cm)
409
(e2x Ϫ eϪ2x) dx
(D) p
0
1
(E) p
0
(e x Ϫ eϪx)2 dx
5128_Ch07_pp378-433.qxd 2/3/06 4:38 PM Page 410
410
Chapter 7
Applications of Definite Integrals
Explorations
Extending the Ideas
69. Max-Min The arch y ϭ sin x, 0 Յ x Յ p, is revolved about the line y ϭ c, 0 Յ c Յ 1, to generate the solid in the figure.
71. Volume of a Hemisphere Derive the formula V ϭ ͑2 ր 3͒ pR 3 for the volume of a hemisphere of radius R by comparing its cross sections with the cross sections of a solid right circular cylinder of radius R and height R from which a solid right circular cone of base radius R and height R has been removed as suggested by the figure.
(a) Find the value of c that minimizes the volume of the solid.
2 p2
What is the minimum volume? ᎏᎏ, ᎏϪ 8
ᎏ
p
2
(b) What value of c in ͓0, 1͔ maximizes the volume of the solid? 0
R2 – h
√2
(c) Writing to Learn Graph the solid’s volume as a function of c, first for 0 Յ c Յ 1 and then on a larger domain. What happens to the volume of the solid as c moves away from ͓0, 1͔?
Does this make sense physically? Give reasons for your answers. y h
h
R
R
p(2c2p Ϫ 8c ϩ p)
V ϭ ᎏᎏ
2
Volume → ∞
h
72. Volume of a Torus The disk x 2 ϩ y 2 Յ a 2 is revolved about the line x ϭ b ͑b Ͼ a͒ to generate a solid shaped like a doughnut, a called a torus. Find its volume. (Hint: ͐ ͙ෆෆෆෆෆ dy ϭ pa 2ր 2, a2 Ϫ y2
Ϫa
since it is the area of a semicircle of radius a.) 2a2bp2
y ϭ sin x c 73. Filling a Bowl
(a) Volume A hemispherical bowl of radius a contains water to a depth h. Find the volume of water in the bowl. ph2(3a Ϫ h)/3
0 yϭc
x
70. A Vase We wish to estimate the volume of a flower vase using only a calculator, a string, and a ruler. We measure the height of the vase to be 6 inches. We then use the string and the ruler to find circumferences of the vase (in inches) at half-inch intervals. (We list them from the top down to correspond with the picture of the vase.)
6
Circumferences
0
5.4
4.5
4.4
5.1
6.3
7.8
9.4
10.8
11.6
11.6
10.8
9.0
6.3
(a) Find the areas of the cross sections that correspond to the given circumferences. 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7,
10.7, 9.3, 6.4, 3.2
(b) Express the volume of the vase as an integral with respect to
1 6 y over the interval ͓0, 6͔.
ᎏᎏ C(y)2 dy
4p
͵
52. Partition the appropriate interval on the axis of revolution and measure the radius r(x) of the shadow region at these points. Then use an approximab tion such as the trapezoidal rule to estimate the integral ͐a pr2(x) dx.
54. For a tiny horizontal slice,
(⌬x)2 ϩ (⌬y)2
1 ϩ (gЈ(y))2 slant height ϭ ⌬s ϭ ͙ෆෆ ϭ ͙ෆෆ ⌬y. So the surface area is approximated by the Riemann sum
͚ 2p
g(yk)͙ෆෆ
1 ϩ (gЈ(y))2
⌬y.
kϭ1
The limit of that is the integral.
74. Consistency of Volume Definitions The volume formulas in calculus are consistent with the standard formulas from geometry in the sense that they agree on objects to which both apply. (a) As a case in point, show that if you revolve the region enclosed by the semicircle y ϭ ͙a ෆϪෆෆ and the x-axis about
ෆ2 ෆ x 2 the x-axis to generate a solid sphere, the calculus formula for volume at the beginning of the section will give ͑4 ր 3͒pa 3 for the volume just as it should.
(b) Use calculus to find the volume of a right circular cone of height h and base radius r.
71. Hemisphere cross sectional area: p(͙ෆ2)2 ϭ A1
R2 Ϫ hෆ
Right circular cylinder with cone removed cross sectional area: pR2 Ϫ ph2 ϭ A2
Since A1 ϭ A2, the two volumes are equal by Cavalieri’s theorem.
Thus,
volume of hemisphere ϭ volume of cylinder Ϫ volume of cone
1
2
ϭ pR3 Ϫ ᎏ3ᎏpR3 ϭ ᎏ3ᎏpR3.
0
(c) Approximate the integral using the Trapezoidal Rule with n ϭ 12. Ϸ34.7 in3
n
(b) Related Rates Water runs into a sunken concrete hemispherical bowl of radius 5 m at a rate of 0.2 m3րsec. How fast is the water level in the bowl rising when the water is 4 m deep? 1/(120p) m/sec
74. (a) A cross section has radius r ϭ ͙ෆ2 and a2 Ϫ xෆ
2 ϭ p(a 2 Ϫ x 2). area A(x) ϭ pr a 4
Vϭ
p(a2 Ϫ x2) ϭ ᎏᎏpa3
3
Ϫa y (b) A cross section has radius x ϭ r 1 Ϫ ᎏᎏ and h ͵
2y y2 area A(y) ϭ px 2 ϭ pr 2 1 Ϫ ᎏᎏ ϩ ᎏᎏ . h h2
͵ h
2y y2 1
V ϭ pr 2 1 Ϫ ᎏᎏ ϩ ᎏᎏ dy ϭ ᎏᎏpr2h h h
3
0
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 411
Section 7.3 Volumes
411
Quick Quiz for AP* Preparation: Sections 7.1–7.3
You may use a graphing calculator to solve the following problems. 1. Multiple Choice The base of a solid is the region in the first quadrant bounded by the x-axis, the graph of y ϭ sinϪ1 x, and the vertical line x ϭ 1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume? C
(A) 0.117
(B) 0.285
(C) 0.467
(D) 0.571
(E) 1.571
2. Multiple Choice Let R be the region in the first quadrant bounded by the graph of y ϭ 3x Ϫ x2 and the x-axis. A solid is generated when R is revolved about the vertical line x ϭ Ϫ1.
Set up, but do not evaluate, the definite integral that gives the volume of this solid. A
͵
͵
͵
͵
͵
3
(A)
2p(x ϩ 1)(3x Ϫ x2) dx
0
3
(B)
2p(x ϩ 1)(3x Ϫ x2) dx
Ϫ1
3
(C)
(B) r(10) Ϫ r(0)
͵
͵
10
(C)
rЈ(t) dt
0
10
(D)
r(t) dt
0
(E) 10 и r(10)
4. Free Response Let R be the region bounded by the graphs of
ෆ
y ϭ ͙x, y ϭ eϪx, and the y-axis.
(a) Find the area of R.
(b) Find the volume of the solid generated when R is revolved about the horizontal line y ϭ Ϫ1.
2p(3x Ϫ x2)2 dx
(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a semicircle whose diameෆ ter runs from the graph of y ϭ ͙x to the graph of y ϭ eϪx. Find the volume of this solid.
3
0
3
(E)
(A) r(10)
2p(x)(3x Ϫ x2) dx
0
(D)
3. Multiple Choice A developing country consumes oil at a rate given by r(t) ϭ 20e0.2t million barrels per year, where t is time measured in years, for 0 Յ t Յ 10. Which of the following expressions gives the amount of oil consumed by the country during the time interval 0 Յ t Յ 10? D
(3x Ϫ x2) dx
0
4. (a) The two graphs intersect where ͙x ϭ e–x, which a calculator shows to
ෆ
be x ϭ 0.42630275. Store this value as A.
͵ (e
A
The area of R is
–x
0
Ϫ ͙x) dx ϭ 0.162.
ෆ
͵ p((e ϩ 1) Ϫ (͙x ϩ 1) ) dx ϭ 1.631.
ෆ
ෆ
1 e Ϫ ͙x
(c) Volume ϭ ͵ ᎏᎏpᎏᎏ dx ϭ 0.035.
2
2
(b) Volume ϭ
A
–x
2
2
0
A
0
Ϫx
2
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 412
412
Chapter 7
Applications of Definite Integrals
7.4
What you’ll learn about
Lengths of Curves
A Sine Wave
• A Sine Wave
• Length of a Smooth Curve
• Vertical Tangents, Corners, and
Cusps
. . . and why
How long is a sine wave (Figure 7.32)?
The usual meaning of wavelength refers to the fundamental period, which for y ϭ sin x is 2p. But how long is the curve itself? If you straightened it out like a piece of string along the positive x-axis with one end at 0, where would the other end be?
EXAMPLE 1 The Length of a Sine Wave
The length of a smooth curve can be found using a definite integral.
What is the length of the curve y ϭ sin x from x ϭ 0 to x ϭ 2p?
SOLUTION
We answer this question with integration, following our usual plan of breaking the whole into measurable parts. We partition ͓0, 2p͔ into intervals so short that the pieces of curve
(call them “arcs”) lying directly above the intervals are nearly straight. That way, each arc is nearly the same as the line segment joining its two ends and we can take the length of the segment as an approximation to the length of the arc.
Figure 7.33 shows the segment approximating the arc above the subinterval ͓x kϪ1, x k ͔.
The length of the segment is ͙Δxk2 ϩෆ. The sum
ෆ Δyk2
ෆ Δy
͚ ͙Δx ϩෆ
2
k
[0, 2] by [–2, 2]
Figure 7.32 One wave of a sine curve has to be longer than 2p. y √ ⌬xk2 + ⌬yk2
Q
y = sin x
over the entire partition approximates the length of the curve. All we need now is to find the limit of this sum as the norms of the partitions go to zero. That’s the usual plan, but this time there is a problem. Do you see it?
The problem is that the sums as written are not Riemann sums. They do not have the form ͚ f ͑ck ͒ Δx. We can rewrite them as Riemann sums if we multiply and divide each square root by Δxk .
⌬yk
P
͚
⌬xk
O
xk –1
xk
͙⌬ ෆ2 ϩෆෆෆ ϭ
ෆxkෆෆ ⌬yk2 ϭ x
Figure 7.33 The line segment approximating the arc PQ of the sine curve above the subinterval ͓x kϪ1, x k ͔. (Example 1)
2 k ͚
͙ෆxk ͒ 2ෆෆ⌬yk ͒ 2
͑⌬ ෆෆ ϩ ͑ෆෆෆ
ᎏᎏ ⌬xk
⌬ xk
( )
͚ Ί ⌬x
⌬yk
1 ϩ ᎏᎏ
⌬ xk
2
k
This is better, but we still need to write the last square root as a function evaluated at some ck in the kth subinterval. For this, we call on the Mean Value Theorem for differentiable functions (Section 4.2), which says that since sin x is continuous on ͓x kϪ1, x k ͔ and is differentiable on ͑x kϪ1, x k ͒ there is a point ck in ͓xkϪ1, xk ͔ at which Δyk րΔxk ϭ sinЈ ck (Figure 7.34). That gives us
ෆ ϩ ෆෆЈ c ͒
͚ ͙1ෆෆ͑sinෆෆෆෆ Δx , k Group Exploration
Later in this section we will use an integral to find the length of the sine wave with great precision. But there are ways to get good approximations without integrating. Take five minutes to come up with a written estimate of the curve’s length. No fair looking ahead.
2
k
which is a Riemann sum.
Now we take the limit as the norms of the subdivisions go to zero and find that the length of one wave of the sine function is
͵
2p
͙1ෆෆsinෆx͒ ෆ
ෆ ϩ ͑ෆෆЈ ෆ2
dx ϭ
0
How close was your estimate?
͵
2p
͙1ෆෆos 2ෆ dx Ϸ 7.64.
ෆ ϩ cෆෆ x
Using NINT
0
Now try Exercise 9.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 413
Section 7.4 Lengths of Curves
Length of a Smooth Curve
Slope sin' (ck)
Q
⌬yk
P
⌬xk
xk –1
ck
xk
Figure 7.34 The portion of the sine curve above ͓x kϪ1, x k ͔. At some ck in the interval, sinЈ ͑ck ͒ ϭ ⌬yk ր⌬x k , the slope of segment PQ. (Example 1)
We are almost ready to define the length of a curve as a definite integral, using the procedure of Example 1. We first call attention to two properties of the sine function that came into play along the way.
We obviously used differentiability when we invoked the Mean Value Theorem to replace Δyk րΔxk by sinЈ͑ck ͒ for some ck in the interval ͓xkϪ1, xk ͔. Less obviously, we used the continuity of the derivative of sine in passing from ͚ ͙1ෆෆsinЈ ͑ck ͒͒2 Δx k to the
ෆ ϩ ͑ ෆෆ ෆ ෆ ෆ ෆ
Riemann integral. The requirement for finding the length of a curve by this method, then, is that the function have a continuous first derivative. We call this property smoothness. A function with a continuous first derivative is smooth and its graph is a smooth curve.
Let us review the process, this time with a general smooth function f ͑x͒. Suppose the graph of f begins at the point ͑a, c͒ and ends at ͑b, d͒, as shown in Figure 7.35. We partition the interval a Յ x Յ b into subintervals so short that the arcs of the curve above them are nearly straight. The length of the segment approximating the arc above the subinterval
͓xkϪ1, xk ͔ is ͙Δxk2 ϩෆ. The sum ͚ ͙Δxk2 ϩෆ approximates the length of the curve. We
ෆ Δyk2
ෆ Δyk2 apply the Mean Value Theorem to f on each subinterval to rewrite the sum as a Riemann sum,
͚
͙⌬ ෆෆϩෆෆෆ ϭ
ෆxk2 ෆ ⌬yk2 ϭ y
(b, d)
d
⎯ (⌬xk ) ϩ (⌬yk
√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯)
2
2
Q
y ϭ f(x) c 0
Lϭ
⌬xk
xk – 1
xk
b
x
Figure 7.35 The graph of f, approximated by line segments.
2
k
ෆϩ͑ ෆ
͚ ͙1ෆෆfෆЈ͑cෆ͒͒ෆ ⌬x .
2
k
͵
b
k
͵ Ί
( ) b ͙1ෆෆf Ј͑x͒ෆ dx ϭ
ෆ ϩ ͑ ෆෆ ͒ 2
ෆ
a
(a, c)
a
͚ Ί ⌬x
( )
⌬yk
1 ϩ ᎏᎏ
⌬ xk
For some point ck in (xkϪ1, xk)
Passing to the limit as the norms of the subdivisions go to zero gives the length of the curve as
⌬yk
P
413
a
dy
1 ϩ ᎏᎏ dx 2
dx.
We could as easily have transformed ͚ ͙Δxk2 ϩෆ into a Riemann sum by dividing
ෆ Δyk2 and multiplying by Δyk , giving a formula that involves x as a function of y ͑say, x ϭ g͑y͒͒ on the interval ͓c, d͔:
Lഠ
͙ෆxk ͒ 2ෆෆ⌬yk ͒ 2
͑⌬ ෆෆ ϩ ͑ෆෆෆ
ᎏᎏ ⌬yk ϭ
⌬ yk
͚
ϭ
͚ Ί ⌬y
( )
⌬ xk
1 ϩ ᎏᎏ
⌬ yk
2
k
ෆ ϩ ͑ ෆෆෆෆ
͚ ͙1ෆෆgЈ͑c ͒͒ ෆ ⌬y . k 2
k
For some ck in (ykϪ1, yk)
The limit of these sums, as the norms of the subdivisions go to zero, gives another reasonable way to calculate the curve’s length,
Lϭ
͵
d
͵ Ί
( ) d ͙1ෆෆgЈ͑ y͒ෆ dy ϭ
ෆ ϩ ͑ ෆෆෆ͒ 2
c
c
dx
1 ϩ ᎏᎏ dy 2
dy.
Putting these two formulas together, we have the following definition for the length of a smooth curve.
DEFINITION Arc Length: Length of a Smooth Curve
If a smooth curve begins at ͑a, c͒ and ends at ͑b, d ͒, a Ͻ b, c Ͻ d, then the length
(arc length) of the curve is
͵ Ί
( )
͵ Ί
( ) b Lϭ
a
d
Lϭ
c
dy
1 ϩ ᎏᎏ dx dx
1 ϩ ᎏᎏ dy 2
dx
if y is a smooth function of x on ͓a, b͔;
dy
if x is a smooth function of y on ͓c, d͔.
2
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 414
414
Chapter 7
Applications of Definite Integrals
EXAMPLE 2 Applying the Definition
Find the exact length of the curve
4͙2
ෆ y ϭ ᎏᎏx 3ր2 Ϫ 1
3
SOLUTION
0 Յ x Յ 1.
for
dy
4͙2 3
ෆ
ᎏᎏ ϭ ᎏᎏ • ᎏᎏx 1ր 2 ϭ 2͙2 x 1ր 2,
ෆ
dx
3
2
which is continuous on ͓0, 1͔. Therefore,
͵ Ί
( )
͵ Ί
( )
͵
1
Lϭ
dy
1 ϩ ᎏᎏ dx 0
2
dx
1
ϭ
1ϩ
2͙2 x 1ր 2
ෆ
2
dx
0
1
ϭ
͙1ෆෆx dx
ෆ ϩ 8ෆ
0
]
2 1 ϭ ᎏᎏ • ᎏᎏ ͑1 ϩ 8x͒ 3ր2
3 8
1
0
13 ϭ ᎏᎏ .
6
Now try Exercise 11.
We asked for an exact length in Example 2 to take advantage of the rare opportunity it afforded of taking the antiderivative of an arc length integrand. When you add 1 to the square of the derivative of an arbitrary smooth function and then take the square root of that sum, the result is rarely antidifferentiable by reasonable methods. We know a few more functions that give “nice” integrands, but we are saving those for the exercises.
y
(8, 2) x Vertical Tangents, Corners, and Cusps
(–8, –2)
Figure 7.36 The graph of y ϭ x 1ր 3 has a vertical tangent line at the origin where dyրdx does not exist. (Example 3) x Sometimes a curve has a vertical tangent, corner, or cusp where the derivative we need to work with is undefined. We can sometimes get around such a difficulty in ways illustrated by the following examples.
EXAMPLE 3 A Vertical Tangent
Find the length of the curve y ϭ x1ր3 between ͑Ϫ8, Ϫ2͒ and ͑8, 2͒.
SOLUTION
(2, 8)
The derivative dy 1
1
ᎏᎏ ϭ ᎏᎏxϪ2 ր3 ϭ ᎏᎏ dx 3
3x 2 ր3 y (–2, –8)
is not defined at x ϭ 0. Graphically, there is a vertical tangent at x ϭ 0 where the derivative becomes infinite (Figure 7.36). If we change to x as a function of y, the tangent at the origin will be horizontal (Figure 7.37) and the derivative will be zero instead of undefined. Solving y ϭ x1ր3 for x gives x ϭ y 3, and we have
͵ Ί ͵
( )
2
Figure 7.37 The curve in Figure 7.36 plotted with x as a function of y. The tangent at the origin is now horizontal.
(Example 3)
Lϭ
Ϫ2
dx
1 ϩ ᎏᎏ dy 2
dy ϭ
2
͙1ෆෆ3y 2 ͒ෆ dy ഠ 17.26.
ෆ ϩ ͑ෆෆෆ2
Using NINT
Ϫ2
Now try Exercise 25.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 415
Section 7.4 Lengths of Curves
415
What happens if you fail to notice that dyրdx is undefined at x ϭ 0 and ask your calculator to compute
(
(Ί)
)
2
NINT
1 ϩ ͑1 ր 3͒ xϪ2 ր3 , x, Ϫ8, 8 ?
This actually depends on your calculator. If, in the process of its calculations, it tries to evaluate the function at x ϭ 0, then some sort of domain error will result. If it tries to find convergent Riemann sums near x ϭ 0, it might get into a long, futile loop of computations that you will have to interrupt. Or it might actually produce an answer—in which case you hope it would be sufficiently bizarre for you to realize that it should not be trusted. EXAMPLE 4 Getting Around a Corner
Find the length of the curve y ϭ x 2 Ϫ 4ΗxΗ Ϫ x from x ϭ Ϫ4 to x ϭ 4.
SOLUTION
We should always be alert for abrupt slope changes when absolute value is involved. We graph the function to check (Figure 7.38).
There is clearly a corner at x ϭ 0 where neither dyրdx nor dx րdy can exist. To find the length, we split the curve at x ϭ 0 to write the function without absolute values:
[–5, 5] by [–7, 5]
x 2 Ϫ 4ΗxΗ Ϫ x ϭ
Figure 7.38 The graph of y ϭ x 2 Ϫ 4 Η x Η Ϫ x, Ϫ4 Յ x Յ 4,
ϩ 3x
{ xx Ϫ 5x
2
2
if if x Ͻ 0, x Ն 0.
Then,
has a corner at x ϭ 0 where neither dy/dx nor dx/dy exists. We find the lengths of the two smooth pieces and add them together.
(Example 4)
Lϭ
͵
0
͙1ෆෆ2x ϩෆෆ dx ϩ
ෆ ϩ ͑ෆෆෆ 3͒2
Ϫ4
͵
4
͙1ෆෆ2x Ϫෆෆ dx
ෆ ϩ ͑ෆෆෆ 5͒2
0
ഠ 19.56.
By NINT
Now try Exercise 27.
Finally, cusps are handled the same way corners are: split the curve into smooth pieces and add the lengths of those pieces.
Quick Review 7.4
(For help, go to Sections 1.3 and 3.2.)
In Exercises 1–5, simplify the function.
1.
͙1ෆෆx ϩෆෆ
ෆ ϩ 2 ෆෆ x 2
2.
Ί1ᎏ
Ϫ x ϩ 4ᎏ x2 3. ͙1ෆෆtanෆෆ
ෆ ϩ ͑ෆෆ x͒ 2
on
In Exercises 6–10, identify all values of x for which the function fails to be differentiable.
͓1, 5͔ x ϩ 1
6. f ͑x͒ ϭ Η x Ϫ 4 Η
2Ϫx
ᎏᎏ
2
on
͓Ϫ3, Ϫ1͔
on
͓0, p ր 3͔ sec x
4. ͙1ෆෆx ր4ෆෆր xෆ
ෆ ϩ ͑ ෆෆ Ϫ 1 ෆ͒ 2
5. ͙1ෆෆos 2 x on
ෆ ϩ cෆෆෆෆ
on
x2
7. f ͑x͒ ϭ 5x 2 ր 3
4
0
8. f ͑x͒ ϭ ͙x ϩෆ Ϫ3
ෆෆ 3
5
ϩ4
4x
͓4, 12͔ ᎏᎏ
͓0, p ր 2͔ ͙2 cos x
ෆ
9. f ͑x͒ ϭ ͙x ෆϪෆx ϩෆ 2
ෆ2 ෆ 4 ෆෆ 4
10. f ͑x͒ ϭ 1 ϩ ͙siෆෆ
ෆn x
3
kp, k any integer
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 416
416
Chapter 7
Applications of Definite Integrals
Section 7.4 Exercises
21. Find the length of the curve
In Exercises 1–10,
(a) set up an integral for the length of the curve;
yϭ
(b) graph the curve to see what it looks like;
0ՅyՅp
ෆ Ϫ y2
4. x ϭ ͙1ෆෆෆ, Ϫ1 ր 2 Յ y Յ 1 ր 2
5. y 2 ϩ 2y ϭ 2x ϩ 1,
from ͑Ϫ1, Ϫ1͒ to ͑7, 3͒
6. y ϭ sin x Ϫ x cos x,
7. y ϭ ͐0 tan t dt, x 10. y ϭ
͑e x
ϩ
1
0 Յ x Յ p ր6
Ϫp ր 3 Յ y Յ p ր4
y
9. y ϭ sec x,
y
0ՅxՅp
8. x ϭ ͐0 ͙sec 2ෆෆෆ dt,
ෆෆ t Ϫ 1
1
22. The Length of an Astroid The graph of the equation x 2 ր3 ϩ y 2 ր3 ϭ 1 is one of the family of curves called astroids
(not “asteroids”) because of their starlike appearance (see figure).
Find the length of this particular astroid by finding the length of half the first quadrant portion, y ϭ ͑1 Ϫ x 2ր3 ͒ 3ր2, ͙2 ր4 Յ x Յ 1,
ෆ
and multiplying by 8. 6
Ϫp ր 3 Յ x Յ 0
3. x ϭ sin y,
͙coෆ2t dt
ෆs ෆෆ
from x ϭ 0 to x ϭ p ր4.
Ϫ1 Յ x Յ 2
2. y ϭ tan x,
x
0
(c) use NINT to find the length of the curve.
1. y ϭ x 2,
͵
x 2/3 ϩ y 2/3 ϭ 1
Ϫp ր 3 Յ x Յ p ր 3 eϪx ͒
ր 2, Ϫ3 Յ x Յ 3
–1
0
1
x
In Exercises 11–18, find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative.
11. y ϭ ͑1 ր 3͒͑x 2 ϩ 2͒ 3ր2
–1
from x ϭ 0 to x ϭ 3 12
12. y ϭ x 3ր2 from x ϭ 0 to x ϭ 4 (80͙10 Ϫ 8)/27
ෆ
13. x ϭ ͑y 3ր 3͒ ϩ 1 ր ͑4y͒ from y ϭ 1 to y ϭ 3
[Hint: 1 ϩ ͑dx րdy͒ 2 is a perfect square.] 53/6
14. x ϭ ͑y 4ր4͒ ϩ 1 ր ͑8y 2 ͒ from y ϭ 1 to y ϭ 2
[Hint: 1 ϩ ͑dx րdy͒ 2 is a perfect square.] 123/32
15. x ϭ ͑y 3ր6͒ ϩ 1 ր ͑2y͒ from y ϭ 1 to y ϭ 2
[Hint: 1 ϩ ͑dx րdy͒ 2 is a perfect square.] 17/12
16. y ϭ ͑x 3ր 3͒ ϩ x 2 ϩ x ϩ 1 ր ͑4x ϩ 4͒,
17. x ϭ ͐0 ͙sec 4ෆෆෆ dt,
ෆෆ t Ϫ 1
0 Յ x Յ 2 53/6
Ϫp ր4 Յ y Յ p ր4
y
2
18. y ϭ ͐ ͙3ෆෆෆ dt, Ϫ2 Յ x Յ Ϫ1 7͙3/3
ෆt 4 Ϫ 1
ෆ
23. Fabricating Metal Sheets Your metal fabrication company is bidding for a contract to make sheets of corrugated steel roofing like the one shown here. The cross sections of the corrugated sheets are to conform to the curve
( )
3p y ϭ sin ᎏ ᎏx ,
20
0 Յ x Յ 20 in.
If the roofing is to be stamped from flat sheets by a process that does not stretch the material, how wide should the original material be? Give your answer to two decimal places. Ϸ21.07 inches
Original sheet
y
x
Ϫ2
Corrugated sheet
19. (a) Group Activity Find a curve through the point ͑1, 1͒ whose length integral is y ϭ ͙x
ෆ
͵ Ί
4
Lϭ
1
1
1 ϩ ᎏᎏ dx.
4x
(b) Writing to Learn How many such curves are there? Give reasons for your answer.
20. (a) Group Activity Find a curve through the point ͑0, 1͒ whose length integral is y ϭ 1/(1 Ϫ x)
͵ Ί
2
Lϭ
1
1
1 ϩ ᎏᎏ dy. y4 (b) Writing to Learn How many such curves are there? Give reasons for your answer.
19. (b) Only one. We know the derivative of the function and the value of the function at one value of x.
20 in.
O
3 20 y ϭ sin — x
–
20
x (in.)
24. Tunnel Construction Your engineering firm is bidding for the contract to construct the tunnel shown on the next page. The tunnel is 300 ft long and 50 ft wide at the base. The cross section is shaped like one arch of the curve y ϭ 25 cos ͑p x ր50͒.
Upon completion, the tunnel’s inside surface (excluding the roadway) will be treated with a waterproof sealer that costs
$1.75 per square foot to apply. How much will it cost to apply the sealer? $38,422
20. (b) Only one. We know the derivative of the function and the value of the function at one value of x.
5128_Ch07_pp378-433.qxd 2/3/06 5:59 PM Page 417
30. Because the limit of the sum ͚⌬xk as the norm of the partition goes to zero will always be the length (b Ϫ a) of the interval (a, b). y y ϭ 25 cos ( x/50)
Section 7.4 Lengths of Curves
34. Multiple Choice Which of the following gives the best approximation of the length of the arc of y ϭ cos(2x) from x ϭ 0 to x ϭ p/4? D
(A) 0.785
(B) 0.955
(C) 1.0
(D) 1.318 (E) 1.977
35. Multiple Choice Which of the following expressions gives the length of the graph of x ϭ y3 from y ϭ Ϫ2 to y ϭ 2? C
– 25
0
417
300 ft
͵
͵
͵
2
25 x (ft)
(A)
(B)
Ϫ2
NOT TO SCALE
͙1 ϩ y6 dy
ෆ
Ϫ2
2
(C)
͵
͵
2
(1 ϩ y6) dy
2
͙1 ϩ 9y ෆ dy
ෆ4
(D)
Ϫ2
͙1 ϩ x2 dx
ෆ
Ϫ2
2
(E)
͙1 ϩ x 4 dx
ෆ
Ϫ2
36. Multiple Choice Find the length of the curve described by
2
y ϭ ᎏᎏ x 3/2 from x ϭ 0 to x ϭ 8. B
3
26
52
512͙2
ෆ
(A) ᎏᎏ
(B) ᎏᎏ
(C) ᎏᎏ
3
3
15
512͙2
ෆ
(D) ᎏᎏ ϩ 8
(E) 96
15
In Exercises 25 and 26, find the length of the curve.
25. f ͑x͒ ϭ x 1 ր 3 ϩ x 2 ր 3, 0 Յ x Յ 2 Ϸ3.6142 xϪ1 1
26. f ͑x͒ ϭ ᎏᎏ, Ϫ ᎏᎏ Յ x Յ 1 Ϸ2.1089
4x 2 ϩ 1
2
In Exercises 27–29, find the length of the nonsmooth curve.
37. Multiple Choice Which of the following expressions should be used to find the length of the curve y ϭ x2/3 from x ϭ Ϫ1 to x ϭ 1? A
1
0
(C)
29. y ϭ ͙x from x ϭ 0 to x ϭ 16 Ϸ16.647
ෆ
30. Writing to Learn Explain geometrically why it does not work to use short horizontal line segments to approximate the lengths of small arcs when we search for a Riemann sum that leads to the formula for arc length.
31. Writing to Learn A curve is totally contained inside the square with vertices ͑0, 0͒, ͑1, 0͒, ͑1, 1͒, and ͑0, 1͒. Is there any limit to the possible length of the curve? Explain.
Standardized Test Questions
͙1 ϩ y3 dy
ෆ
Ϫ1
9
1 ϩ ᎏᎏy dy
4
1
(D)
͙1 ϩ y6 dy
ෆ
0
͙1 ϩ y9/4 dy
ෆෆ
0
Exploration
38. Modeling Running Tracks Two lanes of a running track are modeled by the semiellipses as shown. The equation for lane 1 is y ϭ ͙100 Ϫෆෆx 2 , and the equation for lane 2
ෆෆෆ 0.2ෆෆ is y ϭ ͙15ෆϪෆ.2 ෆ2 The starting point for lane 1 is at the
ෆ0 ෆ 0ෆx ෆ. negative x-intercept ͑Ϫ͙50ෆ, 0͒. The finish points for both lanes
ෆ0
are the positive x-intercepts. Where should the starting point be placed on lane 2 so that the two lane lengths will be equal
(running clockwise)? Ϸ(–19.909, 8.410)
You should solve the following problems without using a graphing calculator.
32. True or False If a function y ϭ f (x) is continuous on an interval [a, b], then the length of its curve is given by
͵ Ί
͵
1
(B)
0
1
(E)
4
9
1 ϩ ᎏᎏy dy
4
1
27. y ϭ x 3 ϩ 5 Η x Η from x ϭ Ϫ2 to x ϭ 1 Ϸ13.132
28. ͙x ϩ ͙y ϭ 1 Ϸ1.623
ෆ
ෆ
͵ Ί
͵
͵
(A) 2
y
Start lane 2
͵ Ί
b a
dy 2
1 ϩ ᎏᎏ dx. Justify your answer. dx False. The function must be differentiable.
33. True or False If a function y ϭ f (x) is differentiable on an interval [a, b], then the length of its curve is given by
͵ Ί
b a
dy 2
1 ϩ ᎏᎏ dx. Justify your answer. dx True, by definition.
1
1
31. No. Consider the curve y ϭ ᎏᎏ sin ᎏᎏ ϩ 0.5 for 0 Ͻ x Ͻ 1.
3
x
1
10
Start lane 1
x
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 418
418
Chapter 7
Applications of Definite Integrals
Extending the Ideas
39. Using Tangent Fins to Find Arc Length Assume f is smooth on ͓a, b͔ and partition the interval ͓a, b͔ in the usual way. In each subinterval ͓x kϪ1, x k ͔ construct the tangent fin at the point ͑x kϪ1, f ͑x kϪ1 ͒͒ as shown in the figure.
(a) Show that the length of the kth tangent fin over the interval
͓x kϪ1, x k ͔ equals
͙͑⌬ xk ෆෆෆf Ј͑xෆෆ⌬ xk ෆ.
ෆෆ͒ 2 ϩ ͑ ෆෆkϪ1 ͒ ෆෆ͒ 2
(b) Show that n lim
n→ ∞
Tangent fin with slope f'(xk – 1)
(xk – 1, f (xk – 1))
⌬xk
xk
x
39. (a) The fin is the hypotenuse of a right triangle with leg lengths ⌬xk and df ⌬ x ϭ fЈ(xk–1) ⌬xk.
ᎏᎏ
dx xϭxkϪ1 k
Έ
n
ෆ ϩ ( fЈෆෆ
͚ ͙(⌬ xk)2ෆ(xkϪ1)⌬xk)2 n→∞ kϭ1
(b) lim
n
ϭ lim
ෆЈ(xkϪ1)ෆ
͚ ⌬xk ͙1 ϩ ( f ෆ)2
n→∞ kϭ1
ϭ
͵
b
a
͙1 ϩ ( f ෆ dx
ෆЈ(x))2
͵
b
͙1ෆෆf Ј͑x͒͒ 2 dx,
ෆ ϩ ͑ ෆෆෆෆ
a
kϭ1
which is the length L of the curve y ϭ f ͑x͒ from x ϭ a to x ϭ b.
y ϭ f (x)
xk – 1
͚
(length of kth tangent fin) ϭ
40. Is there a smooth curve y ϭ f ͑x͒ whose length over
ෆ
the interval 0 Յ x Յ a is always a͙2 ? Give reasons for your answer. Yes. Any curve of the form y ϭ Ϯx ϩ c, c a constant. 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 419
Section 7.5 Applications from Science and Statistics
7.5
What you’ll learn about
• Work Revisited
419
Applications from Science and Statistics
Our goal in this section is to hint at the diversity of ways in which the definite integral can be used. The contexts may be new to you, but we will explain what you need to know as we go along.
• Fluid Force and Fluid Pressure
• Normal Probabilities
Work Revisited
. . . and why
It is important to see applications of integrals as various accumulation functions.
4.4 newtons Ϸ 1 lb
Recall from Section 7.1 that work is defined as force (in the direction of motion) times displacement. A familiar example is to move against the force of gravity to lift an object. The object has to move, incidentally, before “work” is done, no matter how tired you get trying.
If the force F(x) is not constant, then the work done in moving an object from x ϭ a to b x ϭ b is the definite integral W ϭ ͐a F(x)dx.
EXAMPLE 1 Finding the Work Done by a Force
(1 newton)(1 meter) ϭ 1 N • m ϭ 1 Joule
Find the work done by the force F(x) ϭ cos(px) newtons along the x-axis from x ϭ 0 meters to x ϭ 1ր2 meter.
SOLUTION
͵
1ր2
Wϭ
cos(px) dx
0
Έ
1/2
1
ϭ ᎏᎏ sin(px) p 0
1
p ϭ ᎏᎏ sin ᎏᎏ Ϫ sin(0) p 2
1
ϭ ᎏᎏ Ϸ 0.318 p
Now try Exercise 1.
EXAMPLE 2 Work Done Lifting
22
(N)
A leaky bucket weighs 22 newtons (N) empty. It is lifted from the ground at a constant rate to a point 20 m above the ground by a rope weighing 0.4 N/m. The bucket starts with 70 N (approximately 7.1 liters) of water, but it leaks at a constant rate and just finishes draining as the bucket reaches the top. Find the amount of work done
(a) lifting the bucket alone;
(b) lifting the water alone;
(c) lifting the rope alone;
(d) lifting the bucket, water, and rope together.
SOLUTION
Work
440
N m
(a) The bucket alone. This is easy because the bucket’s weight is constant. To lift it, you must exert a force of 22 N through the entire 20-meter interval.
.
(m)
20
Figure 7.39 The work done by a constant 22-N force lifting a bucket 20 m is 440 N • m. (Example 2)
Work ϭ ͑22 N͒ ϫ ͑20 m͒ ϭ 440 N • m ϭ 440 J
Figure 7.39 shows the graph of force vs. distance applied. The work corresponds to the area under the force graph. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 420
420
Chapter 7
Applications of Definite Integrals
(b) The water alone. The force needed to lift the water is equal to the water’s weight, which decreases steadily from 70 N to 0 N over the 20-m lift. When the bucket is x m off the ground, the water weighs
y
70
(N)
(
(m)
20
) (
x
Figure 7.40 The force required to lift the water varies with distance but the work still corresponds to the area under the force graph. (Example 2)
original weight of water
proportion left at elevation x
The work done is (Figure 7.40)
͵
͵
Wϭ
b
F͑x͒ dx
a
ϭ
[
20
͑70 Ϫ 3.5x͒ dx ϭ 70x Ϫ 1.75x 2
0
]
20
ϭ 1400 Ϫ 700 ϭ 700 J.
0
(c) The rope alone. The force needed to lift the rope is also variable, starting at
͑0.4͒͑20͒ ϭ 8 N when the bucket is on the ground and ending at 0 N when the bucket and rope are all at the top. As with the leaky bucket, the rate of decrease is constant.
At elevation x meters, the ͑20 Ϫ x͒ meters of rope still there to lift weigh F͑x͒ ϭ ͑0.4͒
͑20 Ϫ x͒ N. Figure 7.41 shows the graph of F. The work done lifting the rope is
8
(N)
͵
Work
20
0
(m)
20
F͑x͒ dx ϭ
͵
20
͑0.4͒͑20 Ϫ x͒ dx
0
[
ϭ 8x Ϫ
Figure 7.41 The work done lifting the rope to the top corresponds to the area of another triangle. (Example 2)
y
10 Ϫ y
)
x
20 Ϫ x
F͑x͒ ϭ 70 ᎏᎏ ϭ 70 1 Ϫ ᎏᎏ ϭ 70 Ϫ 3.5x N.
20
20
F(x) = 70 – 3.5x
y ϭ 2x or x ϭ 1 y
2
10
8
(5, 10)
1y
2
y
Δy
0
5
x
0.2x 2
]
20
ϭ 160 Ϫ 80 ϭ 80 N • m ϭ 80 J.
0
(d) The bucket, water, and rope together. The total work is
440 ϩ 700 ϩ 80 ϭ 1220 J.
Now try Exercise 5.
EXAMPLE 3 Work Done Pumping
The conical tank in Figure 7.42 is filled to within 2 ft of the top with olive oil weighing
57 lb ր ft 3. How much work does it take to pump the oil to the rim of the tank?
SOLUTION
We imagine the oil partitioned into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval ͓0, 8͔. ( The 8 represents the top of the oil, not the top of the tank.)
The typical slab between the planes at y and y ϩ Δy has a volume of about
2
Figure 7.42 The conical tank in
Example 3.
( )
1 p ΔV ϭ p(radius) 2 (thickness) ϭ p ᎏᎏy Δy ϭ ᎏᎏ y 2 Δy ft 3.
2
4
The force F͑y͒ required to lift this slab is equal to its weight,
57p
weight per
Weight ϭ
F͑y͒ ϭ 57 ΔV ϭ ᎏᎏ y 2 Δy lb. unit volume
4
(
) ϫ volume
The distance through which F͑y͒ must act to lift this slab to the level of the rim of the cone is about ͑10 Ϫ y͒ ft, so the work done lifting the slab is about
57p
ΔW ϭ ᎏᎏ ͑10 Ϫ y͒y 2 Δy ft • lb.
4
The work done lifting all the slabs from y ϭ 0 to y ϭ 8 to the rim is approximately
Wഠ
͚ ᎏ4ᎏ ͑10 Ϫ y͒ y
57p
2 Δy
ft • lb. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 421
Section 7.5 Applications from Science and Statistics
421
This is a Riemann sum for the function ͑57p ր4͒͑10 Ϫ y͒y 2 on the interval from y ϭ 0 to y ϭ 8. The work of pumping the oil to the rim is the limit of these sums as the norms of the partitions go to zero.
Wϭ
͵
8
0
57p
57p
ᎏᎏ ͑10 Ϫ y͒y 2 dy ϭ ᎏᎏ
4
4
[
57p 10 y 3 y4 ϭ ᎏᎏ ᎏᎏ Ϫ ᎏᎏ
4
3
4
]
͵
8
͑10y 2 Ϫ y 3 ͒ dy
0
8
ഠ 30,561 ft • lb
0
Now try Exercise 17.
Fluid Force and Fluid Pressure
We make dams thicker at the bottom than at the top (Figure 7.43) because the pressure against them increases with depth. It is a remarkable fact that the pressure at any point on a dam depends only on how far below the surface the point lies and not on how much water the dam is holding back. In any liquid, the fluid pressure p (force per unit area) at depth h is
Figure 7.43 To withstand the increasing pressure, dams are built thicker toward the bottom. p ϭ wh,
lb lb Dimensions check: ᎏ ᎏ ϭ ᎏ ᎏ ϫ ft, for example ft2 ft3
where w is the weight-density (weight per unit volume) of the liquid.
EXAMPLE 4 The Great Molasses Flood of 1919
Typical Weight-densities (lb/ft3)
Gasoline
Mercury
Milk
Molasses
Seawater
Water
42
849
64.5
100
64
62.4
90 ft
90 ft
1 ft
SHADED BAND NOT TO SCALE
Figure 7.44 The molasses tank of
Example 4.
At 1:00 P.M. on January 15, 1919 (an unseasonably warm day), a 90-ft-high, 90-footdiameter cylindrical metal tank in which the Puritan Distilling Company stored molasses at the corner of Foster and Commercial streets in Boston’s North End exploded. Molasses flooded the streets 30 feet deep, trapping pedestrians and horses, knocking down buildings, and oozing into homes. It was eventually tracked all over town and even made its way into the suburbs via trolley cars and people’s shoes. It took weeks to clean up.
(a) Given that the tank was full of molasses weighing 100 lb ր ft 3, what was the total force exerted by the molasses on the bottom of the tank at the time it ruptured?
(b) What was the total force against the bottom foot-wide band of the tank wall
(Figure 7.44)? continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 422
422
Chapter 7
Applications of Definite Integrals
SOLUTION
(a) At the bottom of the tank, the molasses exerted a constant pressure of
(
)
lb lb p ϭ wh ϭ 100 ᎏᎏ (90 ft) ϭ 9000 ᎏᎏ.
3
ft ft 2
yk
90
Since the area of the base was p͑45͒ 2, the total force on the base was
(
45
Figure 7.45 The 1-ft band at the bottom of the tank wall can be partitioned into thin strips on which the pressure is approximately constant. (Example 4)
)
lb
9000 ᎏᎏ (2025 p ft 2 ) Ϸ 57,225,526 lb. ft 2
(b) We partition the band from depth 89 ft to depth 90 ft into narrower bands of width
Δy and choose a depth yk in each one. The pressure at this depth yk is p ϭ wh ϭ 100 yk lb ր ft 2 (Figure 7.45). The force against each narrow band is approximately pressure ϫ area ϭ ͑100yk ͒͑90p Δy͒ ϭ 9000p yk Δy lb.
Adding the forces against all the bands in the partition and passing to the limit as the norms go to zero, we arrive at
Fϭ
͵
90
9000py dy ϭ 9000p
89
͵
90
y dy Ϸ 2,530,553 lb
89
for the force against the bottom foot of tank wall. y Now try Exercise 25.
Normal Probabilities
1
12
Area
= 1
4
2
5
12
Figure 7.46 The probability that the clock has stopped between 2:00 and 5:00 can be represented as an area of 1 ր4.
The rectangle over the entire interval has area 1.
x
Suppose you find an old clock in the attic. What is the probability that it has stopped somewhere between 2:00 and 5:00?
If you imagine time being measured continuously over a 12-hour interval, it is easy to conclude that the answer is 1 ր4 (since the interval from 2:00 to 5:00 contains one-fourth of the time), and that is correct. Mathematically, however, the situation is not quite that clear because both the 12-hour interval and the 3-hour interval contain an infinite number of times. In what sense does the ratio of one infinity to another infinity equal 1 ր4?
The easiest way to resolve that question is to look at area. We represent the total probability of the 12-hour interval as a rectangle of area 1 sitting above the interval (Figure 7.46).
Not only does it make perfect sense to say that the rectangle over the time interval ͓2, 5͔ has an area that is one-fourth the area of the total rectangle, the area actually equals 1 ր4, since the total rectangle has area 1. That is why mathematicians represent probabilities as areas, and that is where definite integrals enter the picture.
DEFINITION Probability Density Function (pdf)
Improper Integrals
More information about improper ϱ integrals like ͐ f͑x͒ dx can be found in
Ϫϱ
Section 8.3. ( You will not need that information here.)
A probability density function is a function f ͑x͒ with domain all reals such that f ͑x͒ Ն 0 for all x
and
͵
ϱ
f ͑x͒ dx ϭ 1.
Ϫϱ
Then the probability associated with an interval ͓a, b͔ is
͵
b
f ͑x͒ dx.
a
Probabilities of events, such as the clock stopping between 2:00 and 5:00, are integrals of an appropriate pdf.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 423
Section 7.5 Applications from Science and Statistics
423
EXAMPLE 5 Probability of the Clock Stopping
Find the probability that the clock stopped between 2:00 and 5:00.
SOLUTION
The pdf of the clock is f ͑t͒ ϭ
{10,ր12,
0 Յ t Յ 12 otherwise. The probability that the clock stopped at some time t with 2 Յ t Յ 5 is
͵
5
2
a
1 f ͑t͒ dt ϭ ᎏᎏ .
4
Now try Exercise 27.
By far the most useful kind of pdf is the normal kind. (“Normal” here is a technical term, referring to a curve with the shape in Figure 7.47.) The normal curve, often called the “bell curve,” is one of the most significant curves in applied mathematics because it enables us to describe entire populations based on the statistical measurements taken from a reasonably-sized sample. The measurements needed are the mean ͑m͒ and the standard deviation ͑s͒, which your calculators will approximate for you from the data. The symbols on the calculator will probably be J and s (see your Owner’s Manual), but go ahead and use x them as m and s, respectively. Once you have the numbers, you can find the curve by using the following remarkable formula discovered by Karl Friedrich Gauss.
b
∫f(x)dx = .17287148
Figure 7.47 A normal probability density function. The probability associated with the interval ͓a, b͔ is the area under the curve, as shown.
DEFINITION Normal Probability Density Function (pdf)
The normal probability density function (Gaussian curve) for a population with mean m and standard deviation s is
1
2
2
f ͑x͒ ϭ ᎏᎏ eϪ͑xϪm͒ ր͑2s ͒. s͙2p ෆෆ
68% of area
95% of the area
The mean m represents the average value of the variable x. The standard deviation s measures the “scatter” around the mean. For a normal curve, the mean and standard deviation tell you where most of the probability lies. The rule of thumb, illustrated in Figure 7.48, is this:
99.7% of the area
–3 –2 –1
0
1
2
3
Figure 7.48 The 68-95-99.7 rule for normal distributions.
The 68-95-99.7 Rule for Normal Distributions y Given a normal curve,
= 0.5
• 68% of the area will lie within s of the mean m,
• 95% of the area will lie within 2s of the mean m,
• 99.7% of the area will lie within 3s of the mean m.
=1
=2
0
Figure 7.49 Normal pdf curves with mean m ϭ 2 and s ϭ 0.5, 1, and 2.
x
Even with the 68-95-99.7 rule, the area under the curve can spread quite a bit, depending on the size of s. Figure 7.49 shows three normal pdfs with mean m ϭ 2 and standard deviations equal to 0.5, 1, and 2.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 424
424
Chapter 7
Applications of Definite Integrals
EXAMPLE 6 A Telephone Help Line
Suppose a telephone help line takes a mean of 2 minutes to answer calls. If the standard deviation is s ϭ 0.5, then 68% of the calls are answered in the range of 1.5 to 2.5 minutes and 99.7% of the calls are answered in the range of 0.5 to 3.5 minutes.
Now try Exercise 29.
EXAMPLE 7 Weights of Spinach Boxes
Suppose that frozen spinach boxes marked as “10 ounces” of spinach have a mean weight of 10.3 ounces and a standard deviation of 0.2 ounce.
(a) What percentage of all such spinach boxes can be expected to weigh between 10 and
11 ounces?
(b) What percentage would we expect to weigh less than 10 ounces?
(c) What is the probability that a box weighs exactly 10 ounces?
SOLUTION
Assuming that some person or machine is trying to pack 10 ounces of spinach into these boxes, we expect that most of the weights will be around 10, with probabilities tailing off for boxes being heavier or lighter. We expect, in other words, that a normal pdf will model these probabilities. First, we define f ͑x͒ using the formula:
10.3
1
2
f ͑x͒ ϭ ᎏᎏ eϪ͑xϪ10.3͒ ր͑0.08͒.
0.2͙2p
ෆෆ
The graph (Figure 7.50) has the look we are expecting.
(a) For an arbitrary box of this spinach, the probability that it weighs between 10 and 11 ounces is the area under the curve from 10 to 11, which is
11
[9, 11.5] by [–1, 2.5]
Figure 7.50 The normal pdf for the spinach weights in Example 7. The mean is at the center.
NINT ͑ f ͑x͒, x, 10, 11͒ ഠ 0.933.
So without doing any more measuring, we can predict that about 93.3% of all such spinach boxes will weigh between 10 and 11 ounces.
(b) For the probability that a box weighs less than 10 ounces, we use the entire area under the curve to the left of x ϭ 10. The curve actually approaches the x-axis as an asymptote, but you can see from the graph (Figure 7.50) that f ͑x͒ approaches zero quite quickly. Indeed, f ͑9͒ is only slightly larger than a billionth. So getting the area from 9 to
10 should do it:
NINT ͑ f ͑x͒, x, 9, 10͒ ഠ 0.067.
We would expect only about 6.7% of the boxes to weigh less than 10 ounces.
(c) This would be the integral from 10 to 10, which is zero. This zero probability might seem strange at first, but remember that we are assuming a continuous, unbroken interval of possible spinach weights, and 10 is but one of an infinite number of them.
Now try Exercise 31.
Quick Review 7.5
(For help, go to Section 5.2.)
In Exercises 1–5, find the definite integral by (a) antiderivatives and
(b) using NINT.
1.
͵
1
0
eϪx dx
a. 1 Ϫ (1/e) b. Ϸ0.632
2.
͵
3.
0
2
5.
1
a. ͙2/2
ෆ
sin x dx
pր4
1
e x dx a. e Ϫ 1 b. Ϸ1.718
͵
͵
pր 2
b. Ϸ0.707
4.
͵
3
͑x 2 ϩ 2͒ dx 15
0
x2
ᎏᎏ dx a. (1/3) ln (9/2) b. Ϸ0.501
3ϩ1
x
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 425
Section 7.5 Applications from Science and Statistics
In Exercises 6–10 find, but do not evaluate, the definite integral that is the limit as the norms of the partitions go to zero of the Riemann sums on the closed interval ͓0, 7͔.
͚ 2p͑x ϩ 2͒͑sin x ͒ Δ x ͵ 2p(x ϩ 2) sin x dx
7. ͚ ͑1 Ϫ x ͒͑2p x ͒Δx ͵ (1 Ϫ x )(2px) dx
7
6.
k
k
8.
9.
0
7
2 k 2
k
0
10.
425
͚ p͑cos x ͒ Δx ͵ p cos x dx
2
k
͚
7
2
0
2
( )
͵ p(y/2)
yk p ᎏᎏ ͑10 Ϫ yk ͒ Δy
2
͙3
ෆ
͚ ᎏ ͑sin
4
2
xk ͒ Δ x
7
2
0
͵ (͙3/4) sin
ෆ
(10 Ϫ y) dy
7
2
x dx
0
Section 7.5 Exercises
In Exercises 1–4, find the work done by the force of F(x) newtons along the x-axis from x ϭ a meters to x ϭ b meters.
1. F͑x͒ ϭ xeϪx ր 3,
a ϭ 0,
b ϭ 5 Ϸ4.4670 J
2. F͑x͒ ϭ x sin ͑px ր4͒, a ϭ 0,
3. F͑x͒ ϭ x ͙9ෆෆෆ,
ෆ Ϫ x2
4. F͑x͒ ϭ
e sin x
a ϭ 0,
cos x ϩ 2,
b ϭ 3 Ϸ3.8473 J bϭ3 9J
a ϭ 0,
b ϭ 10 Ϸ19.5804 J
5. Leaky Bucket The workers in Example 2 changed to a larger bucket that held 50 L (490 N) of water, but the new bucket had an even larger leak so that it too was empty by the time it reached the top. Assuming the water leaked out at a steady rate, how much work was done lifting the water to a point 20 meters above the ground? (Do not include the rope and bucket.) 4900 J
6. Leaky Bucket The bucket in Exercise 5 is hauled up more quickly so that there is still 10 L (98 N) of water left when the bucket reaches the top. How much work is done lifting the water this time? (Do not include the rope and bucket.) 5880 J
(b) How much work does it take to compress the assembly the first half inch? the second half inch? Answer to the nearest inchpound. Ϸ905 in.-lb and Ϸ2714 in.-lb
(Source: Data courtesy of Bombardier, Inc., Mass Transit
Division, for spring assemblies in subway cars delivered to the
New York City Transit Authority from 1985 to 1987.)
10. Bathroom Scale A bathroom scale is compressed 1 ր 16 in. when a 150-lb person stands on it. Assuming the scale behaves like a spring that obeys Hooke’s Law,
(a) how much does someone who compresses the scale
1 ր 8 in. weigh? 300 lb
18.75 in.-lb
(b) how much work is done in compressing the scale 1 ր 8 in.?
11. Hauling a Rope A mountain climber is about to haul up a
50-m length of hanging rope. How much work will it take if the rope weighs 0.624 N ր m? 780 J
12. Compressing Gas Suppose that gas in a circular cylinder of cross section area A is being compressed by a piston
(see figure). y 7. Leaky Sand Bag A bag of sand originally weighing 144 lb was lifted at a constant rate. As it rose, sand leaked out at a constant rate. The sand was half gone by the time the bag had been lifted 18 ft. How much work was done lifting the sand this far? (Neglect the weights of the bag and lifting equipment.) 1944 ft-lb
8. Stretching a Spring A spring has a natural length of 10 in.
An 800-lb force stretches the spring to 14 in.
(a) Find the force constant.
200 lb/in.
(b) How much work is done in stretching the spring from 10 in. to 12 in.? 400 in.-lb
(c) How far beyond its natural length will a 1600-lb force stretch the spring? 8 in.
9. Subway Car Springs It takes a force of 21,714 lb to compress a coil spring assembly on a New York City Transit
Authority subway car from its free height of 8 in. to its fully compressed height of 5 in.
(a) What is the assembly’s force constant?
7238 lb/in.
x
(a) If p is the pressure of the gas in pounds per square inch and
V is the volume in cubic inches, show that the work done in compressing the gas from state ͑ p1, V1 ͒ to state ͑ p 2, V2 ͒ is given by the equation
Work ϭ
͵
͑ p 2, V2 ͒
p dV in. • lb,
͑ p1, V1͒
where the force against the piston is pA.
(b) Find the work done in compressing the gas from
V1 ϭ 243 in3 to V2 ϭ 32 in3 if p1 ϭ 50 lb ր in3 and p and V obey the gas law pV 1.4 ϭ constant (for adiabatic processes). –37,968.75 in.-lb
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 426
426
Chapter 7
Applications of Definite Integrals
Group Activity In Exercises 13–16, the vertical end of a tank containing water (blue shading) weighing 62.4 lb րft 3 has the given shape. (a) Writing to Learn Explain how to approximate the force against the end of the tank by a Riemann sum.
(b) Find the force as an integral and evaluate it.
13. semicircle
(b) 1123.2 lb
14. semiellipse (b) 7987.2 lb
3 ft
(d) The Weight of Water Because of differences in the strength of Earth’s gravitational field, the weight of a cubic foot of water at sea level can vary from as little as 62.26 lb at the equator to as much as 62.59 lb near the poles, a variation of about 0.5%. A cubic foot of water that weighs 62.4 lb in
Melbourne or New York City will weigh 62.5 lb in Juneau or
Stockholm. What are the answers to parts (a) and (b) in a location where water weighs 62.26 lb ր ft 3 ? 62.5 lb ր ft 3 ?
19. Writing to Learn The cylindrical tank shown here is to be filled by pumping water from a lake 15 ft below the bottom of the tank. There are two ways to go about this. One is to pump the water through a hose attached to a valve in the bottom of the tank. The other is to attach the hose to the rim of the tank and let the water pour in. Which way will require less work? Give reasons for your answer.
8 ft
16. parabola (b) Ϸ1506.1 lb
6 ft
3
4 ft
4.5 ft
8 ft
Open top
Through valve:
Ϸ84,687.3 ft-lb
Over the rim:
Ϸ98,801.8 ft-lb
Through a hose attached to a valve in the bottom is faster, because it takes more time to do more work. 2 ft
6 ft
Valve at base
6 ft
17. Pumping Water The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.4 lb ր ft 3.
Grou
nd level 1,500,000 ft-lb, 100 min
18. Emptying a Tank A vertical right cylindrical tank measures
30 ft high and 20 ft in diameter. It is full of kerosene weighing
51.2 lb րft 3. How much work does it take to pump the kerosene to the level of the top of the tank? Ϸ7,238,229 ft-lb
6 ft
15. triangle (b) 3705 lb
17. (d) 1,494,240 ft-lb, Ϸ100 min;
10 ft
12 ft
0
20. Drinking a Milkshake The truncated conical container shown here is full of strawberry milkshake that weighs ͑4 ր 9͒ oz ր in 3.
As you can see, the container is 7 in. deep, 2.5 in. across at the base, and 3.5 in. across at the top (a standard size at Brigham’s in Boston). The straw sticks up an inch above the top. About how much work does it take to drink the milkshake through the straw (neglecting friction)? Answer in inch-ounces.
Ϸ91.3244 in.-oz y y
⌬y
8
8Ϫy
7
20 y y
(a) How much work does it take to empty the tank by pumping the water back to ground level once the tank is full? 1,497,600 ft-lb
(b) If the water is pumped to ground level with a
͑5 ր 11͒-horsepower motor (work output 250 ft • lb րsec), how long will it take to empty the full tank (to the nearest minute)? Ϸ100 min
(c) Show that the pump in part (b) will lower the water level
10 ft (halfway) during the first 25 min of pumping.
y ϭ 14x Ϫ 17.5
(1.75, 7) y ϩ 17.5
14
Δy
0
x
1.25
Dimensions in inches
21. Revisiting Example 3 How much work will it take to pump the oil in Example 3 to a level 3 ft above the cone’s rim?
Ϸ53,482.5 ft-lb
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 427
Section 7.5 Applications from Science and Statistics
427
26. Milk Carton A rectangular milk carton measures 3.75 in.
22. Pumping Milk Suppose the conical tank in Example 3 by 3.75 in. at the base and is 7.75 in. tall. Find the force contains milk weighing 64.5 lb ր ft 3 instead of olive oil. How of the milk ͑weighing 64.5 lb րft 3 ͒ on one side when the much work will it take to pump the contents to the rim? Ϸ34,582.65 ft-lb carton is full. Ϸ4.2 lb
23. Writing to Learn You are in charge of the evacuation and repair of the storage tank shown here. The tank is a hemisphere of radius 10 ft and is full of benzene weighing 56 lb ր ft 3.
y
Outlet pipe
x 2 ϩ y 2 ϭ 100
10
30. Test Scores The mean score on a national aptitude test is 498 with a standard deviation of 100 points.
2 ft
(a) What percentage of the population has scores between 400 and 500? Ϸ0.34 (34%)
0
10
27. Find the probability that a clock stopped between 1:00 and 5:00.
1/3
28. Find the probability that a clock stopped between 3:00 and 6:00.
1/4
29. Suppose a telephone help line takes a mean of 2 minutes to answer calls. If the standard deviation is s ϭ 2, what percentage of the calls are answered in the range of 0 to 4 minutes? 68%
x
(b) If we sample 300 test-takers at random, about how many should have scores above 700? 6.5
z
A firm you contacted says it can empty the tank for 1 ր 2 cent per foot-pound of work. Find the work required to empty the tank by pumping the benzene to an outlet 2 ft above the tank.
If you have budgeted $5000 for the job, can you afford to hire the firm? Ϸ967,611 ft-lb, yes
24. Water Tower Your town has decided to drill a well to increase its water supply. As the town engineer, you have determined that a water tower will be necessary to provide the pressure needed for distribution, and you have designed the system shown here.
The water is to be pumped from a 300-ft well through a vertical
4-in. pipe into the base of a cylindrical tank 20 ft in diameter and 25 ft high. The base of the tank will be 60 ft above ground.
The pump is a 3-hp pump, rated at 1650 ft • lb րsec. To the nearest hour, how long will it take to fill the tank the first time?
(Include the time it takes to fill the pipe.) Assume water weighs
62.4 lb ր ft 3. Ϸ31 hr
31. Heights of Females The mean height of an adult female in
New York City is estimated to be 63.4 inches with a standard deviation of 3.2 inches. What proportion of the adult females in
New York City are
(a) less than 63.4 inches tall?
0.5 (50%)
Ϸ0.24 (24%)
(b) between 63 and 65 inches tall?
(c) taller than 6 feet? Ϸ0.0036 (0.36%)
(d) exactly 5 feet tall?
0 if we assume a continuous distribution;
Ϸ0.071; 7.1% between 59.5 in. and 60.5 in.
32. Writing to Learn Exercises 30 and 31 are subtly different, in that the heights in Exercise 31 are measured continuously and the scores in Exercise 30 are measured discretely. The discrete probabilities determine rectangles above the individual test scores, so that there actually is a nonzero probability of scoring, say, 560. The rectangles would look like the figure below, and would have total area 1.
10 ft
25 ft
Ground
60 ft
Explain why integration gives a good estimate for the probability, even in the discrete case. Integration is a good
4 in.
approximation to the area.
33. Writing to Learn Suppose that f ͑t͒ is the probability density function for the lifetime of a certain type of lightbulb where t is in hours. What is the meaning of the integral
300 ft
Water surface
͵
Submersible pump
800
100
NOT TO SCALE
Standardized Test Questions
25. Fish Tank A rectangular freshwater fish tank with base
2 ϫ 4 ft and height 2 ft (interior dimensions) is filled to within
2 in. of the top.
(a) Find the fluid force against each end of the tank.
f ͑t͒ dt?
The proportion of lightbulbs that last between 100 and 800 hours. Ϸ209.73 lb
(b) Suppose the tank is sealed and stood on end (without spilling) so that one of the square ends is the base. What does that do to the fluid forces on the rectangular sides?
Ϸ838.93 lb; the fluid force doubles
You may use a graphing calculator to solve the following problems. 34. True or False A force is applied to compress a spring several inches. Assume the spring obeys Hooke’s Law. Twice as much work is required to compress the spring the second inch than is required to compress the spring the first inch. Justify your answer. False. Three times as much work is required.
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 428
428
Chapter 7
Applications of Definite Integrals
35. True. The force against each vertical side is 842.4 lb
35. True or False An aquarium contains water weighing 62.4 lb/ft3.
The aquarium is in the shape of a cube where the length of each edge is 3 ft. Each side of the aquarium is engineered to withstand 1000 pounds of force. This should be sufficient to withstand the force from water pressure. Justify your answer.
36. Multiple Choice A force of F(x) ϭ 350x newtons moves a particle along a line from x ϭ 0 m to x ϭ 5 m. Which of the following gives the best approximation of the work done by the force? E
(A) 1750 J
(B) 2187.5 J
(D) 3281.25 J
(C) 2916.67 J
(E) 4375 J
37. Multiple Choice A leaky bag of sand weighs 50 n. It is lifted from the ground at a constant rate, to a height of 20 m above the ground. The sand leaks at a constant rate and just finishes draining as the bag reaches the top. Which of the following gives the work done to lift the sand to the top? (Neglect the bag.) D
(A) 50 J
(B) 100 J (C) 250 J
(D) 500 J
(E) 1000 J
38. Multiple Choice A spring has a natural length of 0.10 m.
A 200-n force stretches the spring to a length of 0.15 m. Which of the following gives the work done in stretching the spring from 0.10 m to 0.15 m? B
(A) 0.05 J
(B) 5 J
(C) 10 J
(D) 200 J
(E) 4000 J
39. Multiple Choice A vertical right cylindrical tank measures
12 ft high and 16 ft in diameter. It is full of water weighing
62.4 lb/ft3. How much work does it take to pump the water to the level of the top of the tank? Round your answer to the nearest ft-lb. E
41. Forcing Electrons Together Two electrons r meters apart repel each other with a force of
23 ϫ 10Ϫ29
F ϭ ᎏᎏ newton. r2 (a) Suppose one electron is held fixed at the point ͑1, 0͒ on the x-axis (units in meters). How much work does it take to move a second electron along the x-axis from the point
͑Ϫ1, 0͒ to the origin? 1.15 ϫ 10Ϫ28 J
(b) Suppose an electron is held fixed at each of the points
͑Ϫ1, 0͒ and ͑1, 0͒. How much work does it take to move a third electron along the x-axis from ͑5, 0͒ to ͑3, 0͒? Ϸ7.6667 ϫ 10Ϫ29 J
42. Kinetic Energy If a variable force of magnitude F͑x͒ moves a body of mass m along the x-axis from x1 to x 2 , the body’s velocity v can be written as dx րdt (where t represents time). Use Newton’s second law of motion, F ϭ m͑dvրdt͒, and the Chain Rule dv dv d x dv See page 429.
ᎏᎏ ϭ ᎏᎏ ᎏᎏ ϭ v ᎏᎏ dt d x dt dx to show that the net work done by the force in moving the body from x1 to x 2 is
Wϭ
͵
x2
x1
1
1
F͑x͒ dx ϭ ᎏᎏ mv22 Ϫ ᎏᎏmv12,
2
2
(1)
where v1 and v2 are the body’s velocities at x1 and x 2 . In physics the expression ͑1 ր 2͒mv 2 is the kinetic energy of the body moving with velocity v. Therefore, the work done by the force equals the change in the body’s kinetic energy, and we can find the work by calculating this change.
(A) 149,490 ft-lb
Weight vs. Mass
(B) 285,696 ft-lb
Weight is the force that results from gravity pulling on a mass. The two are related by the equation in
Newton’s second law,
(C) 360,240 ft-lb
(D) 448,776 ft-lb
weight ϭ mass ϫ acceleration.
(E) 903,331 ft-lb
Thus,
Extending the Ideas
40. Putting a Satellite into Orbit The strength of Earth’s gravitational field varies with the distance r from Earth’s center, and the magnitude of the gravitational force experienced by a satellite of mass m during and after launch is m MG
F͑r͒ ϭ ᎏᎏ . r2 24 kg is Earth’s mass,
Here, M ϭ 5.975 ϫ 10
G ϭ 6.6726 ϫ 10Ϫ11 N • m2 kgϪ2 is the universal gravitational constant, and r is measured in meters. The work it takes to lift a
1000-kg satellite from Earth’s surface to a circular orbit 35,780 km above Earth’s center is therefore given by the integral
Work ϭ
͵
35,780,000
6,370,000
1000 MG
ᎏᎏ dr joules. r2 The lower limit of integration is Earth’s radius in meters at the launch site. Evaluate the integral. (This calculation does not take into account energy spent lifting the launch vehicle or energy spent bringing the satellite to orbit velocity.) 5.1446 ϫ 1010 J
newtons ϭ kilograms ϫ m րsec 2, pounds ϭ slugs ϫ ft րsec 2.
To convert mass to weight, multiply by the acceleration of gravity. To convert weight to mass, divide by the acceleration of gravity.
In Exercises 43–49, use Equation 1 from Exercise 42.
43. Tennis A 2-oz tennis ball was served at 160 ft րsec (about
109 mph). How much work was done on the ball to make it go this fast? 50 ft-lb
44. Baseball How many foot-pounds of work does it take to throw a baseball 90 mph? A baseball weighs 5 oz ϭ 0.3125 lb.
Ϸ85.1 ft-lb
45. Golf A 1.6-oz golf ball is driven off the tee at a speed of
280 ft ր sec (about 191 mph). How many foot-pounds of work are done getting the ball into the air? 122.5 ft-lb
46. Tennis During the match in which Pete Sampras won the 1990
U.S. Open men’s tennis championship, Sampras hit a serve that was clocked at a phenomenal 124 mph. How much work did
Sampras have to do on the 2-oz ball to get it to that speed?
Ϸ64.6 ft-lb
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 429
Section 7.5 Applications from Science and Statistics
47. Football A quarterback threw a 14.5-oz football 88 ft ր sec
(60 mph). How many foot-pounds of work were done on the ball to get it to that speed? Ϸ109.7 ft-lb
48. Softball How much work has to be performed on a
6.5-oz softball to pitch it at 132 ft ր sec (90 mph)? Ϸ110.6 ft-lb x ͵ ϭ͵ dv dv 42. F ϭ mᎏᎏ ϭ mvᎏᎏ, so W ϭ dt dx
2
x1
49. A Ball Bearing A 2-oz steel ball bearing is placed on a vertical spring whose force constant is k ϭ 18 lb րft. The spring is compressed 3 in. and released. About how high does the ball bearing go? (Hint: The kinetic (compression) energy, mgh, of a
1
spring is ᎏ2ᎏks2, where s is the distance the spring is compressed, m is the mass, g is the acceleration of gravity, and h is the height.)
4.5 ft
F(x) dx
x1 x2 dv mvᎏᎏ dx ϭ dx ͵
429
v2
v1
1
1
mv dv ϭ ᎏᎏmv22 Ϫ ᎏᎏmv12
2
2
Quick Quiz for AP* Preparation: Sections 7.4 and 7.5
You should solve the following problems without using a graphing calculator.
1. Multiple Choice The length of a curve from x ϭ 0 to x ϭ 1 is
ෆx6
given by ͐0 ͙1 ϩ 16ෆ dx. If the curve contains the point (1, 4), which of the following could be an equation for this curve? A
1
(A) y ϭ x4 ϩ 3
4. Free Response The front of a fish tank is rectangular in shape and measures 2 ft wide by 1.5 ft tall. The water in the tank exerts pressure on the front of the tank. The pressure at any point on the front of the tank depends only on how far below the surface the point lies and is given by the equation p ϭ 62.4h, where h is depth below the surface measured in feet and p is pressure measured in pounds/ft2.
(B) y ϭ x4 ϩ 1
(C) y ϭ 1 ϩ 16x6
1.5 ft
(D) y ϭ ͙1 ϩ 16ෆ
ෆx6
x7
(E) y ϭ x ϩ ᎏᎏ
7
2. Multiple Choice Which of the following gives the length of
1
the path described by the parametric equations x ϭ ᎏᎏt 4 and
4
y ϭ t3, where 0 Յ t Յ 2? D
͵
͵
͵
͵
͵
h
⌬h
2 ft
The front of the tank can be partitioned into narrow horizontal bands of height Δh. The force exerted by the water on a band at depth hi is approximately
2
(A)
t 6 ϩ 9t 4 dt
0
pressure ؒ area = 62.4hi ؒ 2Δh.
2
(B)
͙t 6 ϩ 1 dt
ෆ
(a) Write the Riemann sum that approximates the force exerted on the entire front of the tank.
0
2
(C)
͙1 ϩ 9t ෆ dt
ෆ4
(b) Use the Riemann sum from part (a) to write and evaluate a definite integral that gives the force exerted on the front of the tank. Include correct units.
0
2
(D)
͙t 6 ϩ 9ෆ dt
ෆt 4
(c) Find the total force exerted on the front of the tank if the front
(and back) are semicircles with diameter 2 ft. Include correct units.
0
2
(E)
͙t3 ϩ 3t2 dt
ෆෆ
2 ft
0
3. Multiple Choice The base of a solid is a circle of radius
2 inches. Each cross section perpendicular to a certain diameter is a square with one side lying in the circle. The volume of the solid in cubic inches is C
128
128p
(A) 16 (B) 16p (C) ᎏᎏ (D) ᎏᎏ (E) 32p
3
3
n
(a) ͚ 62.4hi ؒ 2 ⌬h iϭ1 ͵
1.5
(b)
0
62.4h ؒ 2 dh ϭ 140.4 lbs
͵
1.5
(c)
0
62.4h ؒ 2͙1Ϫh2 dh ϭ 41.6 lbs
ෆ
5128_Ch07_pp378-433.qxd 2/3/06 6:25 PM Page 430
430
Chapter 7
Applications of Definite Integrals
Calculus at Work
I am working toward becoming an archeaoastronomer and ethnoastronomer of Africa. I have a Bachelor’s degree in
Physics, a Master’s degree in Astronomy, and a Ph.D. in Astronomy and Astrophysics. From 1988 to 1990 I was a member of the Peace Corps, and I taught mathematics to high school students in the Fiji Islands. Calculus is a required course in high schools there.
For my Ph.D. dissertation, I investigated the possibility of the birthrate of stars being related to the composition of star formation clouds. I collected data on the absorption of electromagnetic emissions emanating from these regions. The intensity of emissions graphed versus wave-
length produces a flat curve with downward spikes at the characteristic wavelengths of the elements present. An estimate of the area between a spike and the flat curve results in a concentration in molecules/cm3 of an element. This area is the difference in the integrals of the flat and spike curves. In particular, I was looking for a large concentration of water-ice, which increases the probability of planets forming in a region.
Currently, I am applying for two research grants. One will allow me to use the NASA infrared telescope on Mauna Kea to search for C3S2 in comets. The other will help me study the history of astronomy in
Tunisia.
Jarita Holbrook
Los Angeles, CA
Chapter 7 Key Terms arc length (p. 413) area between curves (p. 390)
Cavalieri’s theorems (p. 404) center of mass (p. 389) constant-force formula (p. 384) cylindrical shells (p. 402) displacement (p. 380) fluid force (p. 421) fluid pressure (p. 421) foot-pound (p. 384) force constant (p. 385)
Gaussian curve (p. 423)
Hooke’s Law (p. 385) inflation rate (p. 388) joule (p. 384) length of a curve (p. 413) mean (p. 423) moment (p. 389) net change (p. 379) newton (p. 384) normal curve (p. 423) normal pdf (p. 423) probability density function (pdf) (p. 422)
68-95-99.7 rule (p. 423)
smooth curve (p. 413) smooth function (p. 413) solid of revolution (p. 400) standard deviation (p. 423) surface area (p. 405) total distance traveled (p. 381) universal gravitational constant (p. 428) volume by cylindrical shells (p. 402) volume by slicing (p. 400) volume of a solid (p. 399) weight-density (p. 421) work (p. 384)
Chapter 7 Review Exercises
The collection of exercises marked in red could be used as a chapter test. In Exercises 1–5, the application involves the accumulation of small changes over an interval to give the net change over that entire interval. Set up an integral to model the accumulation and evaluate it to answer the question.
1. A toy car slides down a ramp and coasts to a stop after
5 sec. Its velocity from t ϭ 0 to t ϭ 5 is modeled by v͑t͒ ϭ t 2 Ϫ 0.2t 3 ft ր sec. How far does it travel? Ϸ10.417 ft
2. The fuel consumption of a diesel motor between weekly maintenance periods is modeled by the function c͑t͒ ϭ
4 ϩ 0.001t 4 gal րday, 0 Յ t Յ 7. How many gallons does it consume in a week? Ϸ31.361 gal
3. The number of billboards per mile along a 100-mile stretch of an interstate highway approaching a certain city is modeled by the function B͑x͒ ϭ 21 Ϫ e 0.03x, where x is the distance from the city in miles. About how many billboards are along that stretch of highway? Ϸ1464
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 431
Chapter 7 Review Exercises
431
4. A 2-meter rod has a variable density modeled by the function r͑x͒ ϭ 11 Ϫ 4x g ր m, where x is the distance in meters from the base of the rod. What is the total mass of the rod? 14 g
5. The electrical power consumption (measured in kilowatts) at a factory t hours after midnight during a typical day is modeled by
E͑t͒ ϭ 300͑2 Ϫ cos ͑pt ր12͒͒. How many kilowatt-hours of electrical energy does the company consume in a typical day? 14,400
In Exercises 6–19, find the area of the region enclosed by the lines and curves.
6. y ϭ x, y ϭ 1 ր x 2,
7. y ϭ x ϩ 1,
xϭ2
18. The Bell the region enclosed by the graphs of x2 Ϫ 3
2
y ϭ 31Ϫx and y ϭ ᎏᎏ
10
9
2
y ϭ 3 Ϫ x 2 ᎏᎏ
8. ͙x ϩ ͙y ϭ 1,
ෆ
ෆ
[–2, 2] by [–1.5, 1.5]
1
x ϭ 0, y ϭ 0 1/6
Ϸ5.7312
y
1
⎯
√x ϩ ⎯ y ϭ 1
√
0
1
x
[–4, 4] by [–2, 3.5]
9. x ϭ
x ϭ 0,
2y 2,
10. 4x ϭ
y2
Ϫ 4,
11. y ϭ sin x,
yϭ3
14. y ϭ
sec 2
19. The Kissing Fish the region enclosed between the graphs of y ϭ x sin x and y ϭ Ϫx sin x over the interval ͓Ϫp, p͔ 4p
4x ϭ y ϩ 16 30.375
y ϭ x,
x ϭ p ր4
12. y ϭ 2 sin x, y ϭ sin 2x,
13. y ϭ cos x,
18
yϭ4Ϫ
x2
Ϸ0.0155
0ՅxՅp
4
Ϸ8.9023
x, y ϭ 3 Ϫ Η x Η Ϸ2.1043
15. The Necklace one of the smaller bead-shaped regions enclosed by the graphs of y ϭ 1 ϩ cos x and y ϭ 2 Ϫ cos x 2͙3 Ϫ 2p/3 Ϸ 1.370
ෆ
[–5, 5] by [–3, 3]
20. Find the volume of the solid generated by revolving the region bounded by the x-axis, the curve y ϭ 3x 4, and the lines x ϭ Ϫ1 and x ϭ 1 about the x-axis. 2p
21. Find the volume of the solid generated by revolving the region enclosed by the parabola y 2 ϭ 4x and the line y ϭ x about
(a) the x-axis. 32p/3
[–4, 4] by [–4, 8]
(b) the y-axis.
128p/15
(c) the line x ϭ 4. 64p/5 (d) the line y ϭ 4.
32p/3
22. The section of the parabola y ϭ ր 2 from y ϭ 0 to y ϭ 2 is revolved about the y-axis to form a bowl. x2 16. one of the larger bead-shaped regions enclosed by the curves in
Exercise 15 2͙3 ϩ 4p/3 Ϸ 7.653
ෆ
17. The Bow Tie the region enclosed by the graphs of x y ϭ x 3 Ϫ x and y ϭ ᎏᎏ x2 ϩ 1
(shown in the next column).
Ϸ1.2956
(a) Find the volume of the bowl.
4p
(b) Find how much the bowl is holding when it is filled to a depth of k units ͑0 Ͻ k Ͻ 2͒. pk2
(c) If the bowl is filled at a rate of 2 cubic units per second, how fast is the depth k increasing when k ϭ 1? 1/p
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 432
432
Chapter 7
Applications of Definite Integrals
23. The profile of a football resembles the ellipse shown here (all dimensions in inches). Find the volume of the football to the nearest cubic inch. 88p Ϸ 276 in3 y 4x 2 y2 —– ϩ — ϭ 1
121
12
– 11
—
2
0
11
—
2
x
UNITS IN INCHES
24. The base of a solid is the region enclosed between the graphs of y ϭ sin x and y ϭ Ϫsin x from x ϭ 0 to x ϭ p. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume of the solid. p2/4
35. No, the work going uphill is positive, but the work going downhill is negative. 34. Stretching a Spring If a force of 80 N is required to hold a spring 0.3 m beyond its unstressed length, how much work does it take to stretch the spring this far? How much work does it take to stretch the spring an additional meter? 12 J, Ϸ213.3 J
35. Writing to Learn It takes a lot more effort to roll a stone up a hill than to roll the stone down the hill, but the weight of the stone and the distance it covers are the same. Does this mean that the same amount of work is done? Explain.
36. Emptying a Bowl A hemispherical bowl with radius 8 inches is filled with punch (weighing 0.04 pound per cubic inch) to within 2 inches of the top. How much work is done emptying the bowl if the contents are pumped just high enough to get over the rim? Ϸ113.097 in.-lb
37. Fluid Force The vertical triangular plate shown below is the end plate of a feeding trough full of hog slop, weighing
80 pounds per cubic foot. What is the force against the plate?
25. The region enclosed by the graphs of y ϭ e x ր 2, y ϭ 1, and x ϭ ln 3 is revolved about the x-axis. Find the volume of the solid generated. p(2 Ϫ ln 3)
2
28. Find the perimeter of the bow-tie-shaped region enclosed between the graphs of y ϭ x 3 Ϫ x and y ϭ x Ϫ x 3. Ϸ5.2454
29. A particle travels at 2 units per second along the curve y ϭ x 3 Ϫ 3x 2 ϩ 2. How long does it take to travel from the local maximum to the local minimum? 2.296 sec
–4
x yϭ –
2
0
26. A round hole of radius ͙3 feet is bored through the center of a 3
ෆ
28p/3 ft Ϸ 29.3215 ft3 sphere of radius 2 feet. Find the volume of the piece cut out.
27. Find the length of the arch of the parabola y ϭ 9 Ϫ x 2 that lies above the x-axis. Ϸ19.4942
Ϸ426.67 lbs
y
4
UNITS IN FEET
38. Fluid Force A standard olive oil can measures 5.75 in. by
3.5 in. by 10 in. Find the fluid force against the base and each side of the can when it is full. (Olive oil has a weight-density base Ϸ 6.6385 lb, of 57 pounds per cubic foot.) front and back:
5.7726 lb, sides Ϸ 9.4835 lb
30. Group Activity One of the following statements is true for all k Ͼ 0 and one is false. Which is which? Explain. (a) is true
(a) The graphs of y ϭ k sin x and y ϭ sin kx have the same length on the interval ͓0, 2p͔.
(b) The graph of y ϭ k sin x is k times as long as the graph of y ϭ sin x on the interval ͓0, 2p͔.
x
OLIVE
OIL
t4 Ϫ 1
31. Let F͑x͒ ϭ ͐ ͙ෆෆෆ dt. Find the exact length of the graph of
1
F from x ϭ 2 to x ϭ 5 without using a calculator. 39 x 32. Rock Climbing A rock climber is about to haul up 100 N
(about 22.5 lb) of equipment that has been hanging beneath her on 40 m of rope weighing 0.8 N ր m. How much work will it take to lift
(a) the equipment? 4000 J
(b) the rope? 640 J
(c) the rope and equipment together?
39. Volume A solid lies between planes perpendicular to the x-axis at x ϭ 0 and at x ϭ 6. The cross sections between the planes are squares whose bases run from the x-axis up to the curve ͙x ϩ ͙y ϭ ͙6 . Find the volume of the solid. Ϸ14.4
ෆ
ෆ
ෆ
y
4640 J
33. Hauling Water You drove an 800-gallon tank truck from the base of Mt. Washington to the summit and discovered on arrival that the tank was only half full. You had started out with a full tank of water, had climbed at a steady rate, and had taken 50 minutes to accomplish the 4750-ft elevation change. Assuming that the water leaked out at a steady rate, how much work was spent in carrying the water to the summit? Water weighs 8 lb րgal.
( Do not count the work done getting you and the truck to the top.) 22,800,000 ft-lb
6
x1/2 ϩ y1/2 ϭ √⎯6
⎯
6
x
5128_Ch07_pp378-433.qxd 01/16/06 12:19 PM Page 433
Chapter 7 Review Exercises
40. Yellow Perch A researcher measures the lengths of 3-year-old yellow perch in a fish hatchery and finds that they have a mean length of 17.2 cm with a standard deviation of 3.4 cm. What proportion of 3-year-old yellow perch raised under similar conditions can be expected to reach a length of 20 cm or more?
Ϸ0.2051 (20.5%)
41. Group Activity Using as large a sample of classmates as possible, measure the span of each person’s fully stretched hand, from the tip of the pinky finger to the tip of the thumb. Based on the mean and standard deviation of your sample, what percentage of students your age would have a finger span of more than 10 inches? Answers will vary.
42. The 68-95-99.7 Rule (a) Verify that for every normal pdf, the proportion of the population lying within one standard deviation of the mean is close to 68%. ͑Hint: Since it is the same for every pdf, you can simplify the function by assuming that m ϭ 0 and s ϭ 1. Then integrate from Ϫ1 to 1.͒
(a) Ϸ0.6827 (68.27%)
(b) Verify the two remaining parts of the rule.
(b) Ϸ0.9545 (95.45%)
43. Writing to Learn Explain why the area under the graph of a probability density function has to equal 1. The probability that the variable has some value in the range of all possible values is 1.
In Exercises 44–48, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis.
44. y ϭ 2x,
y ϭ x ր 2,
xϭ1
45. y ϭ 1 ր x,
y ϭ 0,
x ϭ 1 ր 2,
46. y ϭ sin x,
y ϭ 0,
47. y ϭ x Ϫ 3,
p
xϭ2
0ՅxՅp
3p
2p2
y ϭ x 2 Ϫ 3x 16p/3
48. the bell-shaped region in Exercise 18
Ϸ9.7717
49. Bundt Cake A bundt cake (see Exploration 1, Section 7.3) has a hole of radius 2 inches and an outer radius of 6 inches at the base. It is 5 inches high, and each cross-sectional slice is parabolic. (a) Model a typical slice by finding the equation of the parabola
5
with y-intercept 5 and x-intercepts Ϯ2. y ϭ 5 Ϫ ᎏ4ᎏx2
(b) Revolve the parabolic region about an appropriate line to generate the bundt cake and find its volume. Ϸ335.1032 in3
50. Finding a Function Find a function f that has a continuous derivative on ͑0, ∞͒ and that has both of the following properties. i. The graph of f goes through the point ͑1, 1͒. ii. The length L of the curve from ͑1, 1͒ to any point
͑x, f ͑x͒͒ is given by the formula L ϭ ln x ϩ f ͑x͒ Ϫ 1.
In Exercises 51 and 52, find the area of the surface generated by revolving the curve about the indicated axis.
51. y ϭ tan x,
52. xy ϭ 1,
0 Յ x Յ p ր4;
1 Յ y Յ 2;
x 2 Ϫ 2 ln x ϩ 3
50. f (x) ϭ ᎏᎏ
4
x-axis Ϸ3.84
y-axis Ϸ5.02
433
AP* Examination Preparation
You may use a graphing calculator to solve the following problems. 53. Let R be the region in the first quadrant enclosed by the y-axis and the graphs of y ϭ 2 ϩ sin x and y ϭ sec x.
(a) Find the area of R.
(b) Find the volume of the solid generated when R is revolved about the x-axis.
(c) Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares.
54. The temperature outside a house during a 24-hour period is given by pt F(t) ϭ 80 Ϫ 10 cos ᎏᎏ , 0 Յ t Յ 24,
12
where F(t) is measured in degrees Fahrenheit and t is measured in hours.
(a) Find the average temperature, to the nearest degree Fahrenheit, between t ϭ 6 and t ϭ 14.
(b) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of t was the air conditioner cooling the house?
(c) The cost of cooling the house accumulates at the rate of
$0.05 per hour for each degree the outside temperature exceeds
78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 24-hour period?
55. The rate at which people enter an amusement park on a given day is modeled by the function E defined by
15600
E(t) ϭ ᎏᎏ. t2 Ϫ 24t ϩ 160
The rate at which people leave the same amusement park on the same day is modeled by the function L defined by
9890
L(t) ϭ ᎏᎏ. t2 Ϫ 38t ϩ 370
Both E(t) and L(t) are measured in people per hour, and time t is measured in hours after midnight. These functions are valid for
9 Յ t Յ 23, which are the hours that the park is open. At time t ϭ 9, there are no people in the park.
(a) How many people have entered the park by 5:00 P.M.
(t ϭ 17)? Round your answer to the nearest whole number.
(b) The price of admission to the park is $15 until 5:00 P.M.
(t ϭ 17). After 5:00 P.M., the price of admission to the park is $11.
How many dollars are collected from admissions to the park on the given day? Round your answer to the nearest whole number.
(c) Let H(t) ϭ ͐9(E(x) Ϫ L(x))dx for 9 Յ t Յ 23. The value of
H(17) to the nearest whole number is 3725. Find the value of
HЈ(17) and explain the meaning of H(17) and HЈ(17) in the context of the park. t (d) At what time t, for 9 Յ t Յ 23, does the model predict that the number of people in the park is a maximum?
Chapter
7
Applications of
Definite Integrals
T
he art of pottery developed independently in many ancient civilizations and still exists in modern times. The desired shape of the side of a pottery vase can be described by: y ϭ 5.0 ϩ 2 sin (x/4) (0 Յ x Յ 8p) where x is the height and y is the radius at height x (in inches).
A base for the vase is preformed and placed on a potter’s wheel. How much clay should be added to the base to form this vase if the inside radius is always 1 inch less than the outside radius? Section 7.3 contains the needed mathematics.
378
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 379
Section 7.1 Integral as Net Change
379
Chapter …show more content…
7 Overview
By this point it should be apparent that finding the limits of Riemann sums is not just an intellectual exercise; it is a natural way to calculate mathematical or physical quantities that appear to be irregular when viewed as a whole, but which can be fragmented into regular pieces. We calculate values for the regular pieces using known formulas, then sum them to find a value for the irregular whole. This approach to problem solving was around for thousands of years before calculus came along, but it was tedious work and the more accurate you wanted to be the more tedious it became.
With calculus it became possible to get exact answers for these problems with almost no effort, because in the limit these sums became definite integrals and definite integrals could be evaluated with antiderivatives. With calculus, the challenge became one of fitting an integrable function to the situation at hand (the “modeling” step) and then finding an antiderivative for it.
Today we can finesse the antidifferentiation step (occasionally an insurmountable hurdle for our predecessors) with programs like NINT, but the modeling step is no less crucial. Ironically, it is the modeling step that is thousands of years old. Before either calculus or technology can be of assistance, we must still break down the irregular whole into regular parts and set up a function to be integrated. We have already seen how the process works with area, volume, and average value, for example. Now we will focus more closely on the underlying modeling step: how to set up the function to be integrated.
7.1
What you’ll learn about
• Linear Motion Revisited
• General Strategy
Integral As Net Change
Linear Motion Revisited
In many applications, the integral is viewed as net change over time. The classic example of this kind is distance traveled, a problem we discussed in Chapter 5.
• Consumption Over Time
• Net Change from Data
• Work
. . . and why
The integral is a tool that can be used to calculate net change and total accumulation.
EXAMPLE 1 Interpreting a Velocity Function
Figure 7.1 shows the velocity ds 8
ᎏᎏ ϭ v(t) ϭ t 2 Ϫ ᎏᎏ dt ͑t ϩ 1͒ 2
cm
ᎏᎏ
sec
of a particle moving along a horizontal s-axis for 0 Յ t Յ 5. Describe the motion.
SOLUTION
Solve Graphically The graph of v (Figure 7.1) starts with v͑0͒ ϭ Ϫ8, which we in-
terpret as saying that the particle has an initial velocity of 8 cm րsec to the left. It slows to a halt at about t ϭ 1.25 sec, after which it moves to the right ͑v Ͼ 0͒ with increasing speed, reaching a velocity of v͑5͒ Ϸ 24.8 cm րsec at the end.
Now try Exercise 1(a).
EXAMPLE 2 Finding Position from Displacement
Suppose the initial position of the particle in Example 1 is s͑0͒ ϭ 9. What is the particle’s position at (a) t ϭ 1 sec? (b) t ϭ 5 sec?
SOLUTION
Solve Analytically
[0, 5] by [–10, 30]
Figure 7.1 The velocity function in
Example 1.
(a) The position at t ϭ 1 is the initial position s͑0͒ plus the displacement (the amount, Δs, that the position changed from t ϭ 0 to t ϭ 1). When velocity is continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 380
380
Chapter 7
Applications of Definite Integrals
Reminder from Section 3.4
A change in position is a displacement.
If s (t) is a body’s position at time t, the displacement over the time interval from t to t ϩ Δt is s(t ϩ Δt) Ϫ s (t). The displacement may be positive, negative, or zero, depending on the motion.
constant during a motion, we can find the displacement (change in position) with the formula Displacement ϭ rate of change ϫ time.
But in our case the velocity varies, so we resort instead to partitioning the time interval
͓0, 1͔ into subintervals of length Δt so short that the velocity is effectively constant on each subinterval. If tk is any time in the kth subinterval, the particle’s velocity throughout that interval will be close to v͑tk ͒. The change in the particle’s position during the brief time this constant velocity applies is v͑tk ͒ Δt. rate of change ϫ time
If v͑tk ͒ is negative, the displacement is negative and the particle will move left. If v͑tk ͒ is positive, the particle will move right. The sum
͚ v͑t ͒ Δt k of all these small position changes approximates the displacement for the time interval
͓0, 1͔.
The sum ͚ v͑tk ͒ Δt is a Riemann sum for the continuous function v͑t͒ over ͓0, 1͔. As the norms of the partitions go to zero, the approximations improve and the sums converge to the integral of v over ͓0, 1͔, giving
͵
͵(
1
Displacement ϭ
v͑t͒ dt
0
1
ϭ
0
)
8 t 2 Ϫ ᎏᎏ dt
͑t ϩ 1͒ 2
[
t3
8
ϭ ᎏᎏ ϩ ᎏᎏ
3
tϩ1
]
1
0
1
8
11 ϭ ᎏᎏ ϩ ᎏᎏ Ϫ 8 ϭ Ϫᎏᎏ.
3
2
3
During the first second of motion, the particle moves 11 ր 3 cm to the left. It starts at s͑0͒ ϭ 9, so its position at t ϭ 1 is
11
16
New position ϭ initial position ϩ displacement ϭ 9 Ϫ ᎏᎏ ϭ ᎏᎏ.
3
3
(b) If we model the displacement from t ϭ 0 to t ϭ 5 in the same way, we arrive at
Displacement ϭ
͵
5
0
[
t3
8
v͑t͒ dt ϭ ᎏᎏ ϩ ᎏᎏ
3
tϩ1
]
5
ϭ 35.
0
The motion has the net effect of displacing the particle 35 cm to the right of its starting point. The particle’s final position is
Final position ϭ initial position ϩ displacement ϭ s͑0͒ ϩ 35 ϭ 9 ϩ 35 ϭ 44.
Support Graphically The position of the particle at any time t is given by
͵[ t s͑t͒ ϭ
0
]
8 u 2 Ϫ ᎏ 2 du ϩ 9,
ᎏ
͑u ϩ 1͒
because sЈ͑t͒ ϭ v͑t͒ and s͑0͒ ϭ 9. Figure 7.2 shows the graph of s͑t͒ given by the parametrization x͑t͒ ϭ NINT ͑v͑u͒, u, 0, t͒ ϩ 9, y͑t͒ ϭ t,
0 Յ t Յ 5. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 381
Section 7.1 Integral as Net Change
381
(a) Figure 7.2a supports that the position of the particle at t ϭ 1 is 16 ր 3.
(b) Figure 7.2b shows the position of the particle is 44 at t ϭ 5. Therefore, the displacement is 44 Ϫ 9 ϭ 35.
Now try Exercise 1(b).
T=1
X = 5.3333333 Y = 1
The reason for our method in Example 2 was to illustrate the modeling step that will be used throughout this chapter. We can also solve Example 2 using the techniques of
Chapter 6 as shown in Exploration 1.
[–10, 50] by [–2, 6]
(a)
EXPLORATION 1
Revisiting Example 2
The velocity of a particle moving along a horizontal s-axis for 0 Յ t Յ 5 is ds 8
ᎏᎏ ϭ t 2 Ϫ ᎏᎏ. dt ͑t ϩ 1͒ 2
T=5
X = 44
Y=5
1. Use the indefinite integral of dsրdt to find the solution of the initial value problem ds
8
ᎏᎏ ϭ t 2 Ϫ ᎏᎏ, dt ͑t ϩ 1͒ 2
[–10, 50] by [–2, 6]
(b)
Figure 7.2 Using TRACE and the parametrization in Example 2 you can
“see” the left and right motion of the particle. s͑0͒ ϭ 9.
2. Determine the position of the particle at t ϭ 1. Compare your answer with the answer to Example 2a.
3. Determine the position of the particle at t ϭ 5. Compare your answer with the answer to Example 2b.
We know now that the particle in Example 1 was at s͑0͒ ϭ 9 at the beginning of the motion and at s͑5͒ ϭ 44 at the end. But it did not travel from 9 to 44 directly—it began its trip by moving to the left (Figure 7.2). How much distance did the particle actually travel?
We find out in Example 3.
EXAMPLE 3 Calculating Total Distance Traveled
Find the total distance traveled by the particle in Example 1.
SOLUTION
Solve Analytically We partition the time interval as in Example 2 but record every position shift as positive by taking absolute values. The Riemann sum approximating total distance traveled is
͚ Ηv͑t ͒Η Δt, k and we are led to the integral
Total distance traveled ϭ
͵
5
͵Η
5
Ηv͑t͒Η
0
dt ϭ
0
Η
8 t 2 Ϫ ᎏᎏ dt.
͑t ϩ 1͒ 2
Evaluate Numerically We have
NINT
(Η
Η
)
8 t 2 Ϫ ᎏᎏ , t, 0, 5 Ϸ 42.59.
͑t ϩ 1͒ 2
Now try Exercise 1(c).
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 382
382
Chapter 7
Applications of Definite Integrals
What we learn from Examples 2 and 3 is this: Integrating velocity gives displacement
(net area between the velocity curve and the time axis). Integrating the absolute value of velocity gives total distance traveled (total area between the velocity curve and the time axis). General Strategy
The idea of fragmenting net effects into finite sums of easily estimated small changes is not new. We used it in Section 5.1 to estimate cardiac output, volume, and air pollution.
What is new is that we can now identify many of these sums as Riemann sums and express their limits as integrals. The advantages of doing so are twofold. First, we can evaluate one of these integrals to get an accurate result in less time than it takes to crank out even the crudest estimate from a finite sum. Second, the integral itself becomes a formula that enables us to solve similar problems without having to repeat the modeling step.
The strategy that we began in Section 5.1 and have continued here is the following:
Strategy for Modeling with Integrals
1. Approximate what you want to find as a Riemann sum of values of a continuous function multiplied by interval lengths. If f ͑x͒ is the function and ͓a, b͔ the interval, and you partition the interval into subintervals of length Δx, the approximating sums will have the form ͚ f ͑ck ͒ Δx with ck a point in the kth subinterval.
2. Write a definite integral, here ͐ab f ͑x͒ dx, to express the limit of these sums as the norms of the partitions go to zero.
3. Evaluate the integral numerically or with an antiderivative.
EXAMPLE 4 Modeling the Effects of Acceleration
A car moving with initial velocity of 5 mph accelerates at the rate of a͑t͒ ϭ 2.4t mph per second for 8 seconds.
(a) How fast is the car going when the 8 seconds are up?
(b) How far did the car travel during those 8 seconds?
SOLUTION
(a) We first model the effect of the acceleration on the car’s velocity.
Step 1: Approximate the net change in velocity as a Riemann sum. When acceleration is constant, rate of change ϫ time velocity change ϭ acceleration ϫ time applied.
To apply this formula, we partition ͓0, 8͔ into short subintervals of length Δt. On each subinterval the acceleration is nearly constant, so if tk is any point in the k th subinterval, the change in velocity imparted by the acceleration in the subinterval is approximately mph a͑tk ͒ Δt mph.
ᎏᎏ ϫ sec sec The net change in velocity for 0 Յ t Յ 8 is approximately
͚ a͑t ͒ Δt mph. k Step 2: Write a definite integral. The limit of these sums as the norms of the partitions
go to zero is
͵
8
0
a͑t͒ dt. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 383
Section 7.1 Integral as Net Change
383
Step 3: Evaluate the integral. Using an antiderivative, we have
Net velocity change ϭ
͵
8
2.4t dt ϭ 1.2t 2
0
]
8
ϭ 76.8 mph.
0
So, how fast is the car going when the 8 seconds are up? Its initial velocity is 5 mph and the acceleration adds another 76.8 mph for a total of 81.8 mph.
(b) There is nothing special about the upper limit 8 in the preceding calculation. Applying the acceleration for any length of time t adds
͵
t
2.4u du mph
u is just a dummy variable here.
0
(b) to the car’s velocity, giving v͑t͒ ϭ 5 ϩ
͵
t
2.4u du ϭ 5 ϩ 1.2t 2 mph.
0
The distance traveled from t ϭ 0 to t ϭ 8 sec is
͵
8
0
Ηv͑t͒Η
dt ϭ
͵
8
͑5 ϩ 1.2t 2 ͒ dt
Extension of Example 3
0
[
ϭ 5t ϩ 0.4t 3
]
8
0
ϭ 244.8 mph ϫ seconds.
Miles-per-hour second is not a distance unit that we normally work with! To convert to miles we multiply by hours րsecond ϭ 1 ր 3600, obtaining
1
244.8 ϫ ᎏᎏ ϭ 0.068 mile.
3600
mi h ᎏᎏ ϫ sec ϫ ᎏᎏ ϭ mi h sec
The car traveled 0.068 mi during the 8 seconds of acceleration.
Now try Exercise 9.
Consumption Over Time
The integral is a natural tool to calculate net change and total accumulation of more quantities than just distance and velocity. Integrals can be used to calculate growth, decay, and, as in the next example, consumption. Whenever you want to find the cumulative effect of a varying rate of change, integrate it.
EXAMPLE 5 Potato Consumption
From 1970 to 1980, the rate of potato consumption in a particular country was C͑t͒ ϭ
2.2 ϩ 1.1t millions of bushels per year, with t being years since the beginning of 1970.
How many bushels were consumed from the beginning of 1972 to the end of 1973?
SOLUTION
We seek the cumulative effect of the consumption rate for 2 Յ t Յ 4.
Step 1: Riemann sum. We partition ͓2, 4͔ into subintervals of length Δt and let tk be a time in the kth subinterval. The amount consumed during this interval is approximately
C͑tk ͒ Δt million bushels.
The consumption for 2 Յ t Յ 4 is approximately
͚ C͑t ͒ Δt million bushels. k continued
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 384
384
Chapter 7
Applications of Definite Integrals
Step 2: Definite integral. The amount consumed from t ϭ 2 to t ϭ 4 is the limit of
these sums as the norms of the partitions go to zero.
͵
4
C͑t͒ dt ϭ
2
͵
4
͑2.2 ϩ 1.1t ͒ dt million bushels
2
Step 3: Evaluate. Evaluating numerically, we obtain
NINT ͑2.2 ϩ 1.1t , t, 2, 4͒ Ϸ 7.066 million bushels.
Now try Exercise 21.
Net Change from Data
Table 7.1
Pumping Rates
Time (min)
Rate (gal ր min)
0
58
5
60
10
65
15
64
20
58
25
57
30
55
35
55
40
59
45
60
50
Many real applications begin with data, not a fully modeled function. In the next example, we are given data on the rate at which a pump operates in consecutive 5-minute intervals and asked to find the total amount pumped.
60
55
60
EXAMPLE 6 Finding Gallons Pumped from Rate Data
A pump connected to a generator operates at a varying rate, depending on how much power is being drawn from the generator to operate other machinery. The rate (gallons per minute) at which the pump operates is recorded at 5-minute intervals for one hour as shown in Table 7.1. How many gallons were pumped during that hour?
SOLUTION
Let R͑t͒, 0 Յ t Յ 60, be the pumping rate as a continuous function of time for the hour.
We can partition the hour into short subintervals of length Δt on which the rate is nearly constant and form the sum ͚ R͑tk ͒ Δt as an approximation to the amount pumped during the hour. This reveals the integral formula for the number of gallons pumped to be
Gallons pumped ϭ
63
63
͵
60
R͑t͒ dt.
0
We have no formula for R in this instance, but the 13 equally spaced values in Table 7.1 enable us to estimate the integral with the Trapezoidal Rule:
͵
60
0
1 joule ϭ (1 newton)(1 meter).
In symbols, 1 J ϭ 1 N • m.
It takes a force of about 1 N to lift an apple from a table. If you lift it 1 m you have done about 1 J of work on the apple. If you eat the apple, you will have consumed about 80 food calories, the heat equivalent of nearly 335,000 joules. If this energy were directly useful for mechanical work (it’s not), it would enable you to lift 335,000 more apples up 1 m.
]
ϭ 3582.5.
Joules
The joule, abbreviated J and pronounced “jewel,” is named after the
English physicist James Prescott Joule
(1818–1889). The defining equation is
[
60
R͑t͒ dt Ϸ ᎏᎏ 58 ϩ 2͑60͒ ϩ 2͑65͒ ϩ … ϩ 2͑63͒ ϩ 63
2 • 12
The total amount pumped during the hour is about 3580 gal.
Now try Exercise 27.
Work
In everyday life, work means an activity that requires muscular or mental effort. In science, the term refers specifically to a force acting on a body and the body’s subsequent displacement. When a body moves a distance d along a straight line as a result of the action of a force of constant magnitude F in the direction of motion, the work done by the force is
W ϭ Fd.
The equation W ϭ Fd is the constant-force formula for work.
The units of work are force ϫ distance. In the metric system, the unit is the newtonmeter, which, for historical reasons, is called a joule (see margin note). In the U.S. customary system, the most common unit of work is the foot-pound.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 385
Section 7.1 Integral as Net Change
385
Hooke’s Law for springs says that the force it takes to stretch or compress a spring x units from its natural (unstressed) length is a constant times x. In symbols,
F ϭ kx, where k, measured in force units per unit length, is a characteristic of the spring called the force constant.
EXAMPLE 7 A Bit of Work
It takes a force of 10 N to stretch a spring 2 m beyond its natural length. How much work is done in stretching the spring 4 m from its natural length?
SOLUTION
We let F͑x͒ represent the force in newtons required to stretch the spring x meters from its natural length. By Hooke’s Law, F͑x͒ ϭ kx for some constant k. We are told that
The force required to stretch the spring 2 m is 10 newtons.
F͑2͒ ϭ 10 ϭ k • 2,
so k ϭ 5 N րm and F͑x͒ ϭ 5x for this particular spring.
We construct an integral for the work done in applying F over the interval from x ϭ 0 to x ϭ 4.
Step 1: Riemann sum. We partition the interval into subintervals on each of which F is so nearly constant that we can apply the constant-force formula for work. If x k is any point in the kth subinterval, the value of F throughout the interval is approximately F͑x k ͒ ϭ 5x k . The work done by F across the interval is approximately 5x k Δx, where Δx is the length of the interval. The sum
͚ F͑x ͒ Δx ϭ ͚ 5x k k
Δx
approximates the work done by F from x ϭ 0 to x ϭ 4.
Steps 2 and 3: Integrate. The limit of these sums as the norms of the partitions go to
zero is
͵
Numerically, work is the area under the force graph.
4
F͑x͒ dx ϭ
0
͵
4
0
x2
5x dx ϭ 5ᎏᎏ
2
]
4
ϭ 40 N • m.
0
Now try Exercise 29.
We will revisit work in Section 7.5.
Quick Review 7.1
(For help, go to Section 1.2.)
In Exercises 1–10, find all values of x (if any) at which the function changes sign on the given interval. Sketch a number line graph of the interval, and indicate the sign of the function on each subinterval.
Example: f ͑x͒ ϭ x 2 Ϫ 1
on
͓Ϫ2, 3͔ f (x)
+
–2
–
–1
+
1
x
3
Changes sign at x ϭ Ϯ1.
1. sin 2x on
2. x 2 Ϫ 3x ϩ 2
p p Changes sign at Ϫᎏᎏ, 0, ᎏᎏ
2
2
͓Ϫ2, 4͔ Changes sign at 1, 2
͓Ϫ3, 2͔ on 3. x 2 Ϫ 2x ϩ 3
on
͓Ϫ4, 2͔ Always positive
1
Changes sign at Ϫᎏᎏ
2
p 3p 5p
5. x cos 2x on ͓0, 4͔ Changes sign at ᎏᎏ, ᎏᎏ, ᎏᎏ
4 4 4
6. xeϪx on ͓0, ∞͒ Always positive
4.
2x 3
Ϫ
3x 2
ϩ1
on
͓Ϫ2, 2͔
x
7. ᎏᎏ on ͓Ϫ5, 30͔ Changes sign at 0 x2 ϩ 1 x2 Ϫ 2
8. ᎏᎏ on ͓Ϫ3, 3͔ Changes sign at Ϫ2, Ϫ͙2, ͙2, 2
ෆ
ෆ x2 Ϫ 4
9. sec ͑1 ϩ ͙1ෆෆinෆx ͒ on ͑Ϫ∞, ∞͒
ෆ Ϫ sෆ2 ෆ
10. sin ͑1 ր x͒ on
1 1
3p 2p
͓0.1, 0.2͔ Changes sign at ᎏᎏ, ᎏᎏ
9. Changes sign at 0.9633 ϩ kp, 2.1783 ϩ kp, where k is an integer
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 386
386
Chapter 7
Applications of Definite Integrals
Section 7.1 Exercises
In Exercises 1–8, the function v(t) is the velocity in m րsec of a particle moving along the x-axis. Use analytic methods to do each of the following:
(a) Determine when the particle is moving to the right, to the left, and stopped.
(b) Find the particle’s displacement for the given time interval. If s(0) ϭ 3, what is the particle’s final position?
(c) Find the total distance traveled by the particle.
1. v͑t͒ ϭ 5 cos t,
0 Յ t Յ 2p See page 389.
0 Յ t Յ p ր2
2. v͑t͒ ϭ 6 sin 3t,
3. v͑t͒ ϭ 49 Ϫ 9.8t,
4. v͑t͒ ϭ
6t 2
Ϫ23 cm
13. What is the total distance traveled by the particle in the same time period? 33 cm a: 11
Ϫ 18t ϩ 12,
15. Approximately where does the particle achieve its greatest positive acceleration on the interval ͓0, b͔? t ϭ a
16. Approximately where does the particle achieve its greatest positive acceleration on the interval ͓0, c͔? t ϭ c
(a) Find where the particle is at the end of the trip.
0 Յ t Յ 2 See page 389.
0 Յ t Յ 2p See page 389.
(b) Find the total distance traveled by the particle.
17.
0 Յ t Յ 4 See page 389.
0
18.
1
(b) 4 meters
3
4
t (sec)
(a) 2 (b) 4 meters
344.52 feet
1
10. A particle travels with velocity v͑t͒ ϭ ͑t Ϫ 2͒ sin t m րsec
0
1
for 0 Յ t Յ 4 sec.
(b) What is the total distance traveled? Ϸ1.91411 meters
19.
11. Projectile Recall that the acceleration due to Earth’s gravity is
32 ft րsec2. From ground level, a projectile is fired straight upward with velocity 90 feet per second.
(a) What is its velocity after 3 seconds?
4
t (sec)
(a) 5
v (m/sec)
1 2 3 4 5 6 7
0
Ϫ6 ft/sec
(b) 7 meters
t (sec)
–2
(c) When it hits the ground, what is the net distance it has traveled? 0
(d) When it hits the ground, what is the total distance it has traveled? 253.125 feet
In Exercises 12–16, a particle moves along the x-axis (units in cm).
Its initial position at t ϭ 0 sec is x͑0͒ ϭ 15. The figure shows the graph of the particle’s velocity v͑t͒. The numbers are the areas of the enclosed regions.
5
c
24
3
2
5.625 sec
b
2
–1
(a) What is the particle’s displacement? ϷϪ1.44952 meters
a
2
v (m/sec)
63 mph
(b) How far does it travel in those 9 seconds?
4
(a) 6
v (m/sec)
2
7. v͑t͒ ϭ e sin t cos t, 0 Յ t Յ 2p See page 389. t 8. v͑t͒ ϭ ᎏᎏ , 0 Յ t Յ 3 See page 389.
1 ϩ t2
9. An automobile accelerates from rest at 1 ϩ 3͙t mph րsec for
ෆ
9 seconds.
(b) When does it hit the ground?
c: Ϫ8
In Exercises 17–20, the graph of the velocity of a particle moving on the x-axis is given. The particle starts at x ϭ 2 when t ϭ 0.
See page 389.
(a) What is its velocity after 9 seconds?
b: 16
14. Give the positions of the particle at times a, b, and c.
0 Յ t Յ 10 See page 389.
5. v͑t͒ ϭ 5 sin2 t cos t,
6. v͑t͒ ϭ ͙4ෆෆ,
ෆϪt
12. What is the particle’s displacement between t ϭ 0 and t ϭ c?
20.
(a) Ϫ2.5
v (m/sec)
(b) 19.5 meters
3
0
1 2 3 4 5 6 7 8 9 10
t (sec)
–3
21. U.S. Oil Consumption The rate of consumption of oil in the
United States during the 1980s (in billions of barrels per year) is modeled by the function C ϭ 27.08 • e t ր 25, where t is the number of years after January 1, 1980. Find the total consumption of oil in the United States from January 1, 1980 to January 1,
1990. Ϸ332.965 billion barrels
22. Home Electricity Use The rate at which your home consumes electricity is measured in kilowatts. If your home consumes electricity at the rate of 1 kilowatt for 1 hour, you will be charged
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 387
26. (b) The answer in (a) corresponds to the area of midpoint rectangles. Part of each rectangle is above the curve and part is below.
for 1 “kilowatt-hour” of electricity. Suppose that the average consumption rate for a certain home is modeled by the function
C͑t͒ ϭ 3.9 Ϫ 2.4 sin ͑pt ր 12͒, where C͑t͒ is measured in kilowatts and t is the number of hours past midnight. Find the average daily consumption for this home, measured in kilowatthours. 93.6 kilowatt-hours
23. Population Density Population density measures the number of people per square mile inhabiting a given living area.
Washerton’s population density, which decreases as you move away from the city center, can be approximated by the function
10,000͑2 Ϫ r͒ at a distance r miles from the city center.
(a) If the population density approaches zero at the edge of the city, what is the city’s radius? 2 miles
(b) A thin ring around the center of the city has thickness Δr and radius r. If you straighten it out, it suggests a rectangular strip.
Approximately what is its area? 2pr⌬r
(c) Writing to Learn Explain why the population of the ring in part (b) is approximately
10,000͑2 Ϫ r͒͑2p r͒ Δr.
Section 7.1 Integral as Net Change
387
(a) What was the total number of bagels sold over the 11-year period? (This is not a calculus question!) 797.5 thousand
(b) Use quadratic regression to model the annual bagel sales (in thousands) as a function B͑x͒, where x is the number of years after 1995. B(x) ϭ 1.6x2 ϩ 2.3x ϩ 5.0
(c) Integrate B͑x͒ over the interval ͓0, 11͔ to find total bagel sales for the 11-year period. Ϸ904.02
(d) Explain graphically why the answer in part (a) is smaller than the answer in part (c). See page 389.
26. Group Activity (Continuation of Exercise 25)
(a) Integrate B͑x͒ over the interval ͓Ϫ0.5, 10.5͔ to find total bagel sales for the 11-year period. Ϸ798.97 thousand
(b) Explain graphically why the answer in part (a) is better than the answer in 25(c).
27. Filling Milk Cartons A machine fills milk cartons with milk at an approximately constant rate, but backups along the assembly line cause some variation. The rates (in cases per hour) are recorded at hourly intervals during a 10-hour period, from
8:00 A.M. to 6:00 P.M.
Population ϭ Population density ϫ Area
(d) Estimate the total population of Washerton by setting up and evaluating a definite integral. Ϸ83,776
24. Oil Flow Oil flows through a cylindrical pipe of radius
3 inches, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance r inches from the center is 8͑10 Ϫ r 2 ͒ inches per second.
(a) In a plane cross section of the pipe, a thin ring with thickness
Δr at a distance r inches from the center approximates a rectangular strip when you straighten it out. What is the area of the strip (and hence the approximate area of the ring)? 2pr⌬r
(b) Explain why we know that oil passes through this ring at approximately 8͑10 Ϫ r 2 ͒͑2pr͒ Δr cubic inches per second.
(c) Set up and evaluate a definite integral that will give the rate
(in cubic inches per second) at which oil is flowing through the pipe. 396p in3/sec or Ϸ 1244.07 in3/sec
25. Group Activity Bagel Sales From 1995 to 2005, the
Konigsberg Bakery noticed a consistent increase in annual sales of its bagels. The annual sales (in thousands of bagels) are shown below.
Year
Sales
(thousands)
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
5
8.9
16
26.3
39.8
56.5
76.4
99.5
125.8
155.3
188
24. (b) 8(10 Ϫ r2) in/sec и (2pr)⌬r in2 ϭ flow in in3/sec
Time
Rate
(cases ր h)
8
9
10
11
12
1
2
3
4
5
6
120
110
115
115
119
120
120
115
112
110
121
Use the Trapezoidal Rule with n ϭ 10 to determine approximately how many cases of milk were filled by the machine over the 10-hour period.
1156.5
28. Writing to Learn As a school project, Anna accompanies her mother on a trip to the grocery store and keeps a log of the car’s speed at 10-second intervals. Explain how she can use the data to estimate the distance from her home to the store. What is the connection between this process and the definite integral?
See page 389.
29. Hooke’s Law A certain spring requires a force of 6 N to stretch it 3 cm beyond its natural length.
(a) What force would be required to stretch the string 9 cm beyond its natural length? 18 N
(b) What would be the work done in stretching the string 9 cm beyond its natural length? 81 N и cm
30. Hooke’s Law Hooke’s Law also applies to compressing springs; that is, it requires a force of kx to compress a spring a distance x from its natural length. Suppose a 10,000-lb force compressed a spring from its natural length of 12 inches to a length of 11 inches.
How much work was done in compressing the spring
(a) the first half-inch?
(b) the second
half-inch?
1250 inch-pounds
3750 inch-pounds
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 388
388
Chapter 7
Applications of Definite Integrals
Standardized Test Questions
You may use a graphing calculator to solve the following problems. 31. True or False The figure below shows the velocity for a particle moving along the x-axis. The displacement for this particle is negative. Justify your answer. False. The displacement is the
36. Multiple Choice Pollution is being removed from a lake at a rate modeled by the function y ϭ 20eϪ0.5t tons/yr, where t is the number of years since 1995. Estimate the amount of pollution removed from the lake between 1995 and 2005. Round your answer to the nearest ton. A
(A) 40
(B) 47
(C) 56
(D) 61
(E) 71
integral of the velocity from t ϭ 0 to t ϭ 5 and is positive.
Extending the Ideas
v (m/sec)
37. Inflation Although the economy is continuously changing, we analyze it with discrete measurements. The following table records the annual inflation rate as measured each month for
13 consecutive months. Use the Trapezoidal Rule with n ϭ 12 to find the overall inflation rate for the year. 0.04875
2
1
0
–1
1
2
3
4
5
6
t (sec)
–2
Month
32. True or False If the velocity of a particle moving along the x-axis is always positive, then the displacement is equal to the total distance traveled. Justify your answer.
33. Multiple Choice The graph below shows the rate at which water is pumped from a storage tank. Approximate the total gallons of water pumped from the tank in 24 hours. C
(A) 600
(B) 2400
(C) 3600
(D) 4200
(E) 4800
r (gal/hr)
250
200
150
100
50
0
6
12
18
t (hr)
24
34. Multiple Choice The data for the acceleration a(t) of a car from 0 to 15 seconds are given in the table below. If the velocity at t ϭ 0 is 5 ft/sec, which of the following gives the approximate velocity at t ϭ 15 using the Trapezoidal Rule? D
(A) 47 ft/sec
(B) 52 ft/sec
(D) 125 ft/sec
(C) 120 ft/sec
(E) 141 ft/sec
t (sec)
0
3
6
9
12
15
a(t) (ftրsec2)
4
8
6
9
10
10
35. Multiple Choice The rate at which customers arrive at a counter to be served is modeled by the function F defined by t F(t) ϭ 12 ϩ 6 cos ᎏᎏ for 0 Յ t Յ 60, where F(t) is measured in p customers per minute and t is measured in minutes. To the nearest whole number, how many customers arrive at the counter over the 60-minute period? B
(A) 720
(B) 725
(C) 732
(D) 744
(E) 756
32. True. Since the velocity is positive, the integral of the velocity is equal to the integral of its absolute value, which is the total distance traveled.
Annual Rate
January
February
March
April
May
June
July
August
September
October
November
December
January
0.04
0.04
0.05
0.06
0.05
0.04
0.04
0.05
0.04
0.06
0.06
0.05
0.05
38. Inflation Rate The table below shows the monthly inflation rate (in thousandths) for energy prices for thirteen consecutive months. Use the Trapezoidal Rule with n ϭ 12 to approximate the annual inflation rate for the 12-month period running from the middle of the first month to the middle of the last month. 40 thousandths or 0.040
Month
Monthly Rate
(in thousandths)
January
February
March
April
May
June
July
August
September
October
November
December
January
3.6
4.0
3.1
2.8
2.8
3.2
3.3
3.1
3.2
3.4
3.4
3.9
4.0
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 389
Section 7.1 Integral as Net Change
39. Center of Mass Suppose we have a finite collection of masses in the coordinate plane, the mass mk located at the point ͑xk , yk ͒ as shown in the figure. y yk
xk
mk
(xk , yk)
O
x
Each mass mk has moment mk yk with respect to the x-axis and moment mk xk about the y-axis. The moments of the entire system with respect to the two axes are
͚my,
Moment about y-axis: M ϭ ͚ m x .
Moment about x-axis: Mx ϭ y ͚
k
and
Mx
͚ mk yk y J ϭ ᎏᎏ ϭ ᎏ . m M k where dm ϭ d dA, d ϭ density, and A ϭ area of the region.
41. the region bounded by the lines y ϭ x, y ϭ Ϫx, x ϭ 2 with constant density d x ϭ 4/3, y ϭ 0
ෆ
ෆ
25. (d) The answer in (a) corresponds to the area of left hand rectangles.
These rectangles lie under the curve B(x). The answer in (c) corresponds to the area under the curve. This area is greater than the area of the rectangles.
1. (a) Right: 0 Յ t Ͻ p/2, 3p/2 Ͻ t Յ 2p
Left: p/2 Ͻ t Ͻ 3p/2
Stopped: t ϭ p/2, 3p/2
(b) 0; 3
(b) Imagine the region cut into thin strips parallel to the x-axis. Show that
͐y dm y J ϭ ᎏᎏ ,
͐dm
40. the region bounded by the parabola y ϭ x 2 and the line y ϭ 4 with constant density d x ϭ 0, ෆ ϭ 12/5 y ෆ
k k
͚
(a) Imagine the region cut into thin strips parallel to the y-axis. Show that
͐x dm x J ϭ ᎏᎏ ,
͐dm
In Exercises 40 and 41, use Exercise 39 to find the center of mass of the region with given density.
k k
The center of mass is ͑ J, J ͒ where x y
My
͚ m k xk x J ϭ ᎏᎏ ϭ ᎏ m M
Suppose we have a thin, flat plate occupying a region in the plane.
where dm ϭ d dA, d ϭ density (mass per unit area), and
A ϭ area of the region.
yk xk 389
(c) 20
2. (a) Right: 0 Ͻ t Ͻ p/3
Left: p/3 Ͻ t Յ p/2
Stopped: t ϭ 0, p/3
(b) 2; 5
3. (a) Right: 0 Յ t Ͻ 5
Left: 5 Ͻ t Յ 10
Stopped: t ϭ 5
(b) 0; 3
4. (a) Right: 0 Յ t Ͻ 1
Left: 1 Ͻ t Ͻ 2
Stopped: t ϭ 1, 2
(b) 4; 7
28. One possible answer:
Plot the speeds vs. time. Connect the points and find the area under the line graph. The definite integral also gives the area under the curve.
(c) 6
(c) 245
(c) 6
5. (a) Right: 0 Ͻ t Ͻ p/2, 3p/2 Ͻ t Ͻ 2p
Left: p/2 Ͻ t Ͻ p, p Ͻ t Ͻ 3p/2
Stopped: t ϭ 0, p/2, p, 3p/2, 2p
(b) 0 ; 3
(c) 20/3
6. (a) Right: 0 Յ t Ͻ 4
Left: never
Stopped: t ϭ 4
(b) 16/3; 25/3
(c) 16/3
7. (a) Right: 0 Յ t Ͻ p/2, 3p/2 Ͻ t Յ 2p
Left: p/2 Ͻ t Ͻ 3p/2
Stopped: t ϭ p/2, 3p/2
(b) 0; 3
(c) 2e Ϫ (2/e) Ϸ 4.7
8. (a) Right: 0 Ͻ t Յ 3
Left: never
Stopped: t ϭ 0
(b) (ln 10)/2 Ϸ 1.15; 4.15
(c) (ln 10)/2 Ϸ 1.15
39. (a, b) Take dm ϭ d dA as mk and letting dA → 0, k → ∞ in the center of mass equations.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 390
390
Chapter 7
Applications of Definite Integrals
Areas in the Plane
7.2
What you’ll learn about
Area Between Curves
• Area Between Curves
We know how to find the area of a region between a curve and the x-axis but many times we want to know the area of a region that is bounded above by one curve, y ϭ f ͑x͒, and below by another, y ϭ g͑x͒ (Figure 7.3).
We find the area as an integral by applying the first two steps of the modeling strategy developed in Section 7.1.
• Area Enclosed by Intersecting
Curves
• Boundaries with Changing
Functions
1. We partition the region into vertical strips of equal width Δx and approximate each strip with a rectangle with base parallel to ͓a, b͔ (Figure 7.4). Each rectangle has area
• Integrating with Respect to y
• Saving Time with Geometric Formulas
͓ f ͑ck ͒ Ϫ g͑ck ͔͒ Δx
. . . and why
for some ck in its respective subinterval (Figure 7.5). This expression will be nonnegative even if the region lies below the x-axis. We approximate the area of the region with the Riemann sum
The techniques of this section allow us to compute areas of complex regions of the plane.
͚ ͓ f ͑c ͒ Ϫ g͑c ͔͒ Δx. k k
y
y
y
Upper curve y ϭ f(x)
(ck, f (ck )) y ϭ f(x) f (ck) Ϫ g(ck )
a
a b Lower curve y ϭ g(x)
Figure 7.3 The region between y ϭ f (x) and y ϭ g(x) and the lines x ϭ a and x ϭ b.
a
x
b
x
ck b y ϭ g(x)
⌬x
Figure 7.4 We approximate the region with rectangles perpendicular to the x-axis.
x
(ck, g(ck ))
Figure 7.5 The area of a typical rectangle is ͓ f (ck ) Ϫ g(ck )͔ Δx.
2. The limit of these sums as Δx→0 is
͵
b
͓ f ͑x͒ Ϫ g͑x͔͒ dx.
a
This approach to finding area captures the properties of area, so it can serve as a definition. DEFINITION Area Between Curves
If f and g are continuous with f ͑x͒ Ն g͑x͒ throughout ͓a, b͔, then the area between the curves y ϭ f ( x ) and y ϭ g ( x ) from a to b is the integral of ͓ f Ϫ g͔ from a to b,
Aϭ
͵
b
a
͓ f ͑x͒ Ϫ g͑x͔͒ dx.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 391
Section 7.2 Areas in the Plane
391
EXAMPLE 1 Applying the Definition
y
Find the area of the region between y ϭ sec2 x and y ϭ sin x from x ϭ 0 to x ϭ p ր4.
y ϭ sec 2 x
2
SOLUTION
We graph the curves (Figure 7.6) to find their relative positions in the plane, and see that y ϭ sec2 x lies above y ϭ sin x on ͓0, p ր4͔. The area is therefore
Aϭ
y ϭ sin x
1
͵
p ր4
͓sec2 x Ϫ sin x͔ dx
0
[
ϭ tan x ϩ cos x
]
pր4
0
͙ෆ
2
ϭ ᎏᎏ units squared.
2
–
4
0
x
Now try Exercise 1.
Figure 7.6 The region in Example 1.
EXPLORATION 1
y1 = 2k – k sin kx y2 = k sin kx
A Family of Butterflies
For each positive integer k, let Ak denote the area of the butterfly-shaped region enclosed between the graphs of y ϭ k sin kx and y ϭ 2k Ϫ k sin kx on the interval
͓0, p րk͔. The regions for k ϭ 1 and k ϭ 2 are shown in Figure 7.7.
k=2 k=1 [0, ] by [0, 4]
Figure 7.7 Two members of the family of butterfly-shaped regions described in
Exploration 1.
1. Find the areas of the two regions in Figure 7.7.
2. Make a conjecture about the areas Ak for k Ն 3.
3. Set up a definite integral that gives the area Ak . Can you make a simple u-substitution that will transform this integral into the definite integral that gives the area A1?
4. What is lim k→∞ Ak?
5. If Pk denotes the perimeter of the kth butterfly-shaped region, what is lim k→∞ Pk ? (You can answer this question without an explicit formula for Pk .)
Area Enclosed by Intersecting Curves
When a region is enclosed by intersecting curves, the intersection points give the limits of integration. y1 = 2 – x2 y2 = – x
EXAMPLE 2 Area of an Enclosed Region
Find the area of the region enclosed by the parabola y ϭ 2 Ϫ x 2 and the line y ϭ Ϫx.
SOLUTION
We graph the curves to view the region (Figure 7.8).
The limits of integration are found by solving the equation
2 Ϫ x 2 ϭ Ϫx
[–6, 6] by [–4, 4]
Figure 7.8 The region in Example 2.
either algebraically or by calculator. The solutions are x ϭ Ϫ1 and x ϭ 2. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 392
392
Chapter 7
Applications of Definite Integrals
Since the parabola lies above the line on ͓Ϫ1, 2͔, the area integrand is 2 Ϫ x 2 Ϫ ͑Ϫx͒.
Aϭ
͵
2
͓2 Ϫ x 2 Ϫ͑Ϫx͔͒ dx
Ϫ1
[
x3 x2 ϭ 2x Ϫ ᎏᎏ ϩ ᎏᎏ
3
2
]
2
Ϫ1
9 ϭ ᎏᎏ units squared
2
Now try Exercise 5.
EXAMPLE 3 Using a Calculator
Find the area of the region enclosed by the graphs of y ϭ 2 cos x and y ϭ x 2 Ϫ 1.
y1 = 2 cos x y2 = x2 – 1
SOLUTION
The region is shown in Figure 7.9.
Using a calculator, we solve the equation
2 cos x ϭ x 2 Ϫ 1 to find the x-coordinates of the points where the curves intersect. These are the limits of integration. The solutions are x ϭ Ϯ1.265423706. We store the negative value as A and the positive value as B. The area is
NINT ͑2 cos x Ϫ ͑x 2 Ϫ 1͒, x, A, B͒ Ϸ 4.994907788.
[–3, 3] by [–2, 3]
This is the final calculation, so we are now free to round. The area is about 4.99.
Now try Exercise 7.
Figure 7.9 The region in Example 3.
Boundaries with Changing Functions
Finding Intersections by
Calculator
The coordinates of the points of intersection of two curves are sometimes needed for other calculations. To take advantage of the accuracy provided by calculators, use them to solve for the values and store the ones you want.
If a boundary of a region is defined by more than one function, we can partition the region into subregions that correspond to the function changes and proceed as usual.
EXAMPLE 4 Finding Area Using Subregions
Find the area of the region R in the first quadrant that is bounded above by y ϭ ͙x and
ෆ
below by the x-axis and the line y ϭ x Ϫ 2.
SOLUTION
The region is shown in Figure 7.10. y 2
4
⌠
Area ϭ ⎮ ⎡⎯ Ϫ x ϩ 2⎡ dx
⎯x
⎣√
⎣
yϭ⎯
√
⎯x
⌡2
⌠
(4, 2)
2
Area ϭ ⎮ √ dx
⎯
⎯x
⌡0
B
1
y ϭxϪ2
A
0
yϭ0
2
Figure 7.10 Region R split into subregions A and B. (Example 4)
4
x
continued
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 393
Section 7.2 Areas in the Plane
393
While it appears that no single integral can give the area of R (the bottom boundary is defined by two different curves), we can split the region at x ϭ 2 into two regions A and
B. The area of R can be found as the sum of the areas of A and B.
Area of R ϭ
͵
2
͙x dx ϩ
ෆ
0
͵
4
͓͙x Ϫ ͑x Ϫ 2͔͒ dx
ෆ
2
area of A
2 ϭ ᎏᎏx 3ր2
3
area of B
] [
2
2 x2 ϩ ᎏᎏx 3ր2 Ϫ ᎏᎏ ϩ 2x
3
2
0
]
4
2
10 ϭ ᎏᎏ units squared
3
Now try Exercise 9.
Integrating with Respect to y
Sometimes the boundaries of a region are more easily described by functions of y than by functions of x. We can use approximating rectangles that are horizontal rather than vertical and the resulting basic formula has y in place of x.
For regions like these y y
d
y
d
d
x ϭ f (y)
x ϭ f(y)
x ϭ g(y)
x ϭ g( y) x ϭ g(y)
c
c
0
0 c x
x ϭ f (y) x x
0
use this formula
A=
∫c
d
[f (y) – g(y)] dy.
EXAMPLE 5 Integrating with Respect to y
Find the area of the region in Example 4 by integrating with respect to y. y SOLUTION
2
(4, 2)
x ϭ y2
(g(y), y)
xϭyϩ2
1
⌬y
( f (y), y)
f (y) Ϫ g(y)
0
yϭ0
2
4
Figure 7.11 It takes two integrations to find the area of this region if we integrate with respect to x. It takes only one if we integrate with respect to y. (Example 5)
x
We remarked in solving Example 4 that “it appears that no single integral can give the area of R,” but notice how appearances change when we think of our rectangles being summed over y. The interval of integration is ͓0, 2͔, and the rectangles run between the same two curves on the entire interval. There is no need to split the region
(Figure 7.11).
We need to solve for x in terms of y in both equations: yϭxϪ2 y ϭ ͙x
ෆ
becomes becomes x ϭ y ϩ 2, x ϭ y 2,
y Ն 0. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 394
394
Chapter 7
Applications of Definite Integrals
We must still be careful to subtract the lower number from the higher number when forming the integrand. In this case, the higher numbers are the higher x-values, which are on the line x ϭ y ϩ 2 because the line lies to the right of the parabola. So, y1 = x3, y2 = √x + 2, y3 = – √x + 2
Area of R ϭ
͵
[
2
0
y2 y3 ͑y ϩ 2 Ϫ y 2 ͒ dy ϭ ᎏᎏ ϩ 2y Ϫ ᎏᎏ
2
3
]
2
10 ϭ ᎏᎏ units squared.
3
0
Now try Exercise 11.
(c, d)
(a, b)
EXAMPLE 6 Making the Choice
Find the area of the region enclosed by the graphs of y ϭ x 3 and x ϭ y 2 Ϫ 2.
SOLUTION
[–3, 3] by [–3, 3]
We can produce a graph of the region on a calculator by graphing the three curves y ϭ x 3, y ϭ ͙x ϩෆ, and y ϭ Ϫ͙x ϩෆ (Figure 7.12).
ෆෆ 2
ෆෆ 2
Figure 7.12 The region in Example 6.
This conveniently gives us all of our bounding curves as functions of x. If we integrate in terms of x, however, we need to split the region at x ϭ a (Figure 7.13).
On the other hand, we can integrate from y ϭ b to y ϭ d and handle the entire region at once. We solve the cubic for x in terms of y:
(c, d)
y ϭ x3
(a, b)
becomes
x ϭ y 1ր 3.
To find the limits of integration, we solve y 1ր 3 ϭ y 2 Ϫ 2. It is easy to see that the lower limit is b ϭ Ϫ1, but a calculator is needed to find that the upper limit d ϭ 1.793003715. We store this value as D.
The cubic lies to the right of the parabola, so
[–3, 3] by [–3, 3]
Area ϭ NINT ͑y 1ր 3 Ϫ ͑y 2 Ϫ 2͒, y, Ϫ1, D͒ ϭ 4.214939673.
Figure 7.13 If we integrate with respect to x in Example 6, we must split the region at x ϭ a.
The area is about 4.21.
Now try Exercise 27.
Saving Time with Geometry Formulas
Here is yet another way to handle Example 4. y EXAMPLE 7 Using Geometry
(4, 2)
2 y ϭ √x
⎯
yϭxϪ2
1
SOLUTION
2
Area ϭ 2
2
0
yϭ0
2
Find the area of the region in Example 4 by subtracting the area of the triangular region from the area under the square root curve.
4
x
Figure 7.14 The area of the blue region is the area under the parabola y ϭ ͙x
ෆ
minus the area of the triangle. (Example 7)
Figure 7.14 illustrates the strategy, which enables us to integrate with respect to x without splitting the region.
Area ϭ
͵
4
0
1
2
2
3
͙x dx Ϫ ᎏᎏ ͑2͒͑2͒ ϭ ᎏᎏ x 3ր2
ෆ
]
4
10
Ϫ 2 ϭ ᎏᎏ units squared
3
0
Now try Exercise 35.
The moral behind Examples 4, 5, and 7 is that you often have options for finding the area of a region, some of which may be easier than others. You can integrate with respect to x or with respect to y, you can partition the region into subregions, and sometimes you can even use traditional geometry formulas. Sketch the region first and take a moment to determine the best way to proceed.
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 395
Section 7.2 Areas in the Plane
Quick Review 7.2
(For help, go to Sections 1.2 and 5.1.)
In Exercises 1–5, find the area between the x-axis and the graph of the given function over the given interval.
1. y ϭ sin x over ͓0, p͔
2. y ϭ e 2x over ͓0, 1͔
2
1
ᎏᎏ(e2
2
Ϫ 1) Ϸ 3.195
3. y ϭ sec 2 x over ͓Ϫpր4, pր4͔
4. y ϭ 4x Ϫ x 3 over ͓0, 2͔
5. y ϭ ͙9ෆෆෆ
ෆ Ϫ x2
2
6. y ϭ x 2 Ϫ 4x and
7. y ϭ e x and
9p/2
2x
9. y ϭ ᎏᎏ x2 ϩ 1
y ϭ x ϩ 6 (6, 12); (Ϫ1, 5)
y ϭ x ϩ 1 (0, 1)
8. y ϭ x 2 Ϫ px and
4
over ͓Ϫ3, 3͔
In Exercises 6–10, find the x- and y-coordinates of all points where the graphs of the given functions intersect. If the curves never intersect, write “NI.”
and
10. y ϭ sin x and
y ϭ sin x (0, 0); (p, 0) y ϭ x3
(Ϫ1, Ϫ1); (0, 0); (1, 1)
y ϭ x 3 (–0.9286, –0.8008); (0, 0); (0.9286, 0.8008)
Section 7.2 Exercises
In Exercises 1–6, find the area of the shaded region analytically. y 1.
y
4.
4/3 x ϭ 12y 2 Ϫ 12y 3
1
p/2 yϭ1 1
x ϭ 2y 2 Ϫ 2y
y ϭ cos 2 x
–
2
0
0 y 5.
4p/3
(–2, 8)
– y ϭ 1 sec 2 t
2
128/15
(2, 8)
8
y ϭ 2x 2
2
1
–
–
3
x
1
x
y
2.
y ϭ x 4 Ϫ 2x 2
0
t
–
3
–2
y ϭ Ϫ4 sin2 t
–1
–1
1
x
2
NOT TO SCALE
22/15
1
y
1
y
6.
–4
3.
1/12
x ϭ y3
y ϭ x2
–1
(1, 1)
1
0
x ϭ y2
0
1
395
x
–2
y ϭ – 2x 4
x
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 396
396
Chapter 7
Applications of Definite Integrals
In Exercises 7 and 8, use a calculator to find the area of the region enclosed by the graphs of the two functions.
7. y ϭ sin x, y ϭ 1 Ϫ
Ϸ1.670 8. y ϭ cos(2x), y ϭ
x2
x2
Ϫ 2 Ϸ4.332
In Exercises 9 and 10, find the area of the shaded region analytically.
9.
yϭx
yϭ —
4
0
and
y ϭ ͑x 2ր 2͒ ϩ 4 21ᎏ3ᎏ
1
xϭyϩ2 and 1
4ᎏ2ᎏ
4x Ϫ y ϭ 16 30ᎏ8ᎏ
3
25. x ϩ y 2 ϭ 0
5/6
y
2
ᎏᎏa3
3
and x ϩ 2y 2 ϭ 3 4 and x ϩ 3y 2 ϭ 2 8/3
26. 4x 2 ϩ y ϭ 4
10.
yϭ0
and
24. x Ϫ y 2 ϭ 0 x 2
1
a Ͼ 0,
8
2
23. y 2 Ϫ 4x ϭ 4
x2
y ϭ x2
and
ΗxΗ
20. y ϭ ͙ෆෆ and 5y ϭ x ϩ 6 1ᎏ3ᎏ (3 points of intersection)
(How many intersection points are there?)
22. x ϭ y 2
yϭ1
1
19. y ϭ x
͙a ෆϪෆෆ,
ෆ2 ෆ x 2
21. y ϭ Η x 2 Ϫ 4 Η and
5/6
y
18. y ϭ x 4 Ϫ 4x 2 ϩ 4
27. x ϩ y 2 ϭ 3
and and x 4 Ϫ y ϭ 1 6ᎏᎏ
15
14
4x ϩ y 2 ϭ 0 8
28. y ϭ 2 sin x and
1
1
Ϫp ր 3 Յ x Յ p ր 3
31.
x
2
32.
In Exercises 11 and 12, find the area enclosed by the graphs of the two curves by integrating with respect to y.
11. y2 ϭ x ϩ 1, y2 ϭ 3 Ϫ x
Ϸ7.542
12. y2 ϭ x ϩ 3, y ϭ 2x
In Exercises 13 and 14, find the total shaded area.
13.
y ϭ Ϫx 2 ϩ 3x
(2, 2)
2
– 2 –1
1
8ᎏ6ᎏ
y
y ϭ4Ϫx2
x
and y ϭ 2
40. Find the area of the region between the curve y ϭ 3 Ϫ x 2 and the line y ϭ Ϫ1 by integrating with respect to (a) x, (b) y. 32/3
͵
͵
0
c
(c) Find c by integrating with respect to x. (This puts c into the integrand as well.)
2
10ᎏ3ᎏ
16. y ϭ 2x Ϫ x 2 and
y ϭ Ϫ3
17. y ϭ 7 Ϫ 2x 2
y ϭ x2 ϩ 4 4
and
39. Find the area of the “triangular” region in the first quadrant bounded on the left by the y-axis and on the right by the curves y ϭ sin x and y ϭ cos x. ͙2 Ϫ 1 Ϸ 0.414
ෆ
(b) Find c by integrating with respect to y. (This puts c in the c 4 limits of integration.)
͙y dy ϭ ͙y dy ⇒ c ϭ 24ր3
ෆ
ෆ
In Exercises 15–34, find the area of the regions enclosed by the lines and curves.
15. y ϭ x 2 Ϫ 2
2
(a) Sketch the region and draw a line y ϭ c across it that looks about right. In terms of c, what are the coordinates of the points where the line and parabola intersect? Add them to your figure. (–͙c, c); (͙c, c)
ෆ
ෆ
y ϭ Ϫx ϩ 2
(3, –5)
–5
0 Յ y Յ pր2
x ϭ 0,
41. The region bounded below by the parabola y ϭ x 2 and above by the line y ϭ 4 is to be partitioned into two subsections of equal area by cutting across it with the horizontal line y ϭ c.
2
3
ෆ ෆෆ
34. x ϭ 3 sin y ͙cos y and
In Exercises 35 and 36, find the area of the region by subtracting the area of a triangular region from the area of a larger region.
38. Find the area of the region in the first quadrant bounded by the line y ϭ x, the line x ϭ 2, the curve y ϭ 1 ր x 2, and the x-axis. 1
–10
1 2
x2
37. Find the area of the propeller-shaped region enclosed by the curve x Ϫ y 3 ϭ 0 and the line x Ϫ y ϭ 0. 1/2
(– 2, –10)
–2 –1
x,
36. The region on or above the x-axis bounded by the curves y ϭ 4 Ϫ x2 and y ϭ 3x. 15ր2
y ϭ 2x 3 Ϫ x 2 Ϫ 5x
4
33.
4
6͙3
ෆ
4
4
Ϫ ᎏᎏ Ϸ 0.0601 y ϭ cos ͑px ր 2͒ and y ϭ 1 Ϫ
ᎏᎏ
3 p 4Ϫp y ϭ sin ͑px ր 2͒ and y ϭ x ᎏᎏ Ϸ 0.273 p p y ϭ sec 2 x, y ϭ tan2 x, x ϭ Ϫp ր4, x ϭ p ր4 ᎏᎏ
2
x ϭ tan2 y and x ϭ Ϫtan2 y, Ϫp ր4 Յ y Յ p ր4 4 Ϫ p Ϸ 0.858
sec 2
35. The region on or above the x-axis bounded by the curves y2 ϭ x ϩ 3 and y ϭ 2x. Ϸ4.333
x
2
1
Ϸ7.146
16
y
(–2, 4)
yϭ
30.
0
14.
0ՅxՅp
29. y ϭ 8 cos x and
xϩyϭ2
y ϭ x2
y ϭ sin 2x,
2
10ᎏ3ᎏ
42. Find the area of the region in the first quadrant bounded on the left by the y-axis, below by the line y ϭ x ր4, above left by the curve y ϭ 1 ϩ ͙x , and above right by the curve y ϭ 2 ր ͙x . 11/3
ෆ
ෆ
͵
͙c
ෆ
41. (c)
0
(c Ϫ x 2) dx ϭ (4 Ϫ c)͙c ϩ
ෆ
͵
2
͙c
ෆ
(4 Ϫ x 2) dx ⇒ c ϭ 24ր3
5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 397
Section 7.2 Areas in the Plane
43. The figure here shows triangle AOC inscribed in the region cut from the parabola y ϭ x 2 by the line y ϭ a 2. Find the limit of the ratio of the area of the triangle to the area of the parabolic region as a approaches zero. 3/4 y ϭ x2
(–a, a )
You should solve the following problems without using a graphing calculator.
True. 36 is the value of the appropriate integral.
2
C yϭa
2
(a, a )
A
Standardized Test Questions
50. True or False The area of the region enclosed by the graph of y ϭ x2 ϩ 1 and the line y ϭ 10 is 36. Justify your answer.
y
2
397
51. True or False The area of the region in the first quadrant enclosed by the graphs of y ϭ cos x, y ϭ x, and the y-axis is
given by the definite integral ͐0 (x Ϫ cos x) dx. Justify your answer. False. It is 0.739(cos x Ϫ x) dx.
0.739
͵
0
O
–a
x
a
52. Multiple Choice Let R be the region in the first quadrant bounded by the x-axis, the graph of x ϭ y2 ϩ 2, and the line x ϭ 4. Which of the following integrals gives the area of R? A
͵
͵
͵
͙2
ෆ
44. Suppose the area of the region between the graph of a positive continuous function f and the x-axis from x ϭ a to x ϭ b is
4 square units. Find the area between the curves y ϭ f ͑x͒ and y ϭ 2 f ͑x͒ from x ϭ a to x ϭ b. 4
45. Writing to Learn Which of the following integrals, if either, calculates the area of the shaded region shown here? Give reasons for your answer. Neither; both are zero
i.
͵
͵
1
͑x Ϫ ͑Ϫx͒͒ dx ϭ
Ϫ1
ii.
͵
͵
1
͑Ϫx Ϫ ͑x͒͒ dx ϭ
Ϫ1
(B)
0
[(y2 ϩ 2) Ϫ 4] dy
0
͙2
ෆ
(C)
͵
͵
͙2
ෆ
[4 Ϫ (y2 ϩ 2)]dy
͙2
ෆ
[4 Ϫ (y2 ϩ 2)]dy
Ϫ͙2
ෆ
(D)
[(y2 ϩ 2) Ϫ 4] dy
Ϫ͙2
ෆ
4
(E)
[4 Ϫ (y2 ϩ 2)]dy
2
2x dx
Ϫ1
1
(A)
53. Multiple Choice Which of the following gives the area of the region between the graphs of y ϭ x2 and y ϭ Ϫx from x ϭ 0 to x ϭ 3? E
1
Ϫ2x dx
Ϫ1
(A) 2
(B) 9/2
(C) 13/2
(D) 13
(E) 27/2
y y ϭ –x
1
yϭx
–1
1
54. Multiple Choice Let R be the shaded region enclosed by the
2
graphs of y ϭ eϪx , y ϭ Ϫsin(3x), and the y-axis as shown in the figure below. Which of the following gives the approximate area of the region R? B x (A) 1.139
(B) 1.445
(C) 1.869
(D) 2.114 (E) 2.340
y
–1
2
46. Writing to Learn Is the following statement true, sometimes true, or never true? The area of the region between the graphs of the continuous functions y ϭ f ͑x͒ and y ϭ g͑x͒ and the vertical lines x ϭ a and x ϭ b (a Ͻ b) is
͵
0
͓ f ͑x͒ Ϫ g͑x͔͒ dx.
Give reasons for your answer.
Sometimes; If f (x) Ն g(x) on (a, b), then true.
47. Find the area of the propeller-shaped region enclosed between the graphs of ln 4 Ϫ (1/2) Ϸ 0.886
2x
y ϭ ᎏᎏ and y ϭ x 3. x2 ϩ 1
48. Find the area of the propeller-shaped region enclosed between the graphs of y ϭ sin x and y ϭ x 3. Ϸ 0.4303
49. Find the positive value of k such that the area of the region enclosed between the graph of y ϭ k cos x and the graph of y ϭ kx 2 is 2. k Ϸ 1.8269
x
–2
b
a
2
55. Multiple Choice Let f and g be the functions given by f (x) ϭ ex and g(x) ϭ 1/x. Which of the following gives the area of the region enclosed by the graphs of f and g between x ϭ 1 and x ϭ 2? A
(A) e2 Ϫ e Ϫ ln2
(B) ln 2 Ϫ e2 ϩ e
1
(C) e2 Ϫ ᎏᎏ
2
1
(D) e2 Ϫ e Ϫ ᎏᎏ
2
1
(E) ᎏᎏ Ϫ ln2 e 5128_Ch07_pp378-433.qxd 1/13/06 1:13 PM Page 398
398
Chapter 7
Applications of Definite Integrals
Exploration
Extending the Ideas
56. Group Activity Area of Ellipse
57. Cavalieri’s Theorem Bonaventura Cavalieri (1598–1647) discovered that if two plane regions can be arranged to lie over the same interval of the x-axis in such a way that they have identical vertical cross sections at every point (see figure), then the regions have the same area. Show that this theorem is true.
An ellipse with major axis of length 2a and minor axis of length 2b can be coordinatized with its center at the origin and its major axis horizontal, in which case it is defined by the equation x2 y2 ᎏᎏ ϩ ᎏᎏ ϭ 1.
2
a b2 (a) Find the equations that define the upper and lower semiellipses as functions of x.
Cross sections have the same length at every point in [a, b].
(b) Write an integral expression that gives the area of the ellipse.
(c) With your group, use NINT to find the areas of ellipses for various lengths of a and b.
(d) There is a simple formula for the area of an ellipse with major axis of length 2a and minor axis of length 2b. Can you tell what it is from the areas you and your group have found?
(e) Work with your group to write a proof of this area formula by showing that it is the exact value of the integral expression in part (b).
a
x
b
58. Find the area of the region enclosed by the curves x y ϭ ᎏᎏ and y ϭ mx, 0 Ͻ m Ͻ 1. x2 ϩ 1
m Ϫ ln (m) Ϫ 1
Ί x (b) 2͵ bΊ1 Ϫ ᎏᎏ dx
a
x2
56. (a) y ϭ Ϯb 1 Ϫ ᎏᎏ a2 a
2
Ϫa
2
(c) Answers may vary.
(d, e) abp
57. Since f (x) Ϫ g(x) is the same for each region where f (x) and g(x) represent b the upper and lower edges, area ϭ ͐a [ f (x) Ϫ g(x)] dx will be the same for each. 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 399
Section 7.3 Volumes
399
Volumes
7.3
What you’ll learn about
Volume As an Integral
• Volume As an Integral
• Square Cross Sections
• Circular Cross Sections
• Cylindrical Shells
• Other Cross Sections
. . . and why
The techniques of this section allow us to compute volumes of certain solids in three dimensions.
In Section 5.1, Example 3, we estimated the volume of a sphere by partitioning it into thin slices that were nearly cylindrical and summing the cylinders’ volumes using MRAM.
MRAM sums are Riemann sums, and had we known how at the time, we could have continued on to express the volume of the sphere as a definite integral.
Starting the same way, we can now find the volumes of a great many solids by integration. Suppose we want to find the volume of a solid like the one in Figure 7.15. The cross section of the solid at each point x in the interval ͓a, b͔ is a region R͑x͒ of area A͑x͒. If A is a continuous function of x, we can use it to define and calculate the volume of the solid as an integral in the following way.
We partition ͓a, b͔ into subintervals of length Δx and slice the solid, as we would a loaf of bread, by planes perpendicular to the x-axis at the partition points. The kth slice, the one between the planes at x kϪ1 and x k , has approximately the same volume as the cylinder between the two planes based on the region R͑x k ͒ (Figure 7.16).
y
y
Px
Cross-section R(x) with area A(x)
Approximating cylinder based on R(xk ) has height
Δ xk ϭ xk Ϫ xkϪ1
Plane at xkϪ1
S
0
a
0
x
xkϪ1
b
Plane at xk xk x
The cylinder’s base is the region R(xk ) with area A(xk )
Figure 7.15 The cross section of an arbitrary solid at point x.
x
NOT TO SCALE
Figure 7.16 Enlarged view of the slice of the solid between the planes at xkϪ1 and xk.
The volume of the cylinder is
Vk ϭ base area ϫ height ϭ A͑x k ͒ ϫ Δx.
The sum
͚ V ϭ ͚ A͑x ͒ ϫ Δx k k
approximates the volume of the solid.
This is a Riemann sum for A͑x͒ on ͓a, b͔. We expect the approximations to improve as the norms of the partitions go to zero, so we define their limiting integral to be the volume of the solid.
DEFINITION Volume of a Solid
The volume of a solid of known integrable cross section area A͑x͒ from x ϭ a to x ϭ b is the integral of A from a to b,
Vϭ
͵
b
a
A͑x͒ dx.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 400
400
Chapter 7
Applications of Definite Integrals
To apply the formula in the previous definition, we proceed as follows.
How to Find Volume by the Method of Slicing
1. Sketch the solid and a typical cross section.
2. Find a formula for A͑x͒.
3. Find the limits of integration.
y
4. Integrate A͑x͒ to find the volume.
Typical cross-section
Square Cross Sections
x
0
3 x Let us apply the volume formula to a solid with square cross sections.
x
3
x (m)
3
EXAMPLE 1 A Square-Based Pyramid
A pyramid 3 m high has congruent triangular sides and a square base that is 3 m on each side. Each cross section of the pyramid parallel to the base is a square. Find the volume of the pyramid.
SOLUTION
Figure 7.17 A cross section of the pyramid in Example 1.
We follow the steps for the method of slicing.
1. Sketch. We draw the pyramid with its vertex at the origin and its altitude along the interval 0 Յ x Յ 3. We sketch a typical cross section at a point x between 0 and 3
(Figure 7.17).
2. Find a formula for A͑ x͒. The cross section at x is a square x meters on a side, so
A͑x͒ ϭ x 2.
3. Find the limits of integration. The squares go from x ϭ 0 to x ϭ 3.
4. Integrate to find the volume.
Vϭ
͵
3
A͑x͒ dx ϭ
0
͵
3
x2
0
x3 ϭ ᎏᎏ
3
ϭ 9 m3
0
Now try Exercise 3.
[–3, 3] by [–4, 4]
Figure 7.18 The region in Example 2.
]
3
Circular Cross Sections
y
The only thing that changes when the cross sections of a solid are circular is the formula for A͑x͒. Many such solids are solids of revolution, as in the next example. f(x) EXAMPLE 2 A Solid of Revolution x The region between the graph of f ͑x͒ ϭ 2 ϩ x cos x and the x-axis over the interval
͓Ϫ2, 2͔ is revolved about the x-axis to generate a solid. Find the volume of the solid.
SOLUTION
Figure 7.19 The region in Figure 7.18 is revolved about the x-axis to generate a solid. A typical cross section is circular, with radius f (x) ϭ 2 ϩ x cos x.
(Example 2)
Revolving the region (Figure 7.18) about the x-axis generates the vase-shaped solid in
Figure 7.19. The cross section at a typical point x is circular, with radius equal to f ͑x͒.
Its area is
A͑x͒ ϭ p͑ f ͑x͒͒ 2. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 401
Section 7.3 Volumes
401
The volume of the solid is
Vϭ
͵
2
A͑x͒ dx
Ϫ2
Ϸ NINT ͑p͑2 ϩ x cos x͒2, x, Ϫ2, 2͒ Ϸ 52.43 units cubed.
Now try Exercise 7.
EXAMPLE 3 Washer Cross Sections
The region in the first quadrant enclosed by the y-axis and the graphs of y ϭ cos x and y ϭ sin x is revolved about the x-axis to form a solid. Find its volume.
SOLUTION
The region is shown in Figure 7.20.
We revolve it about the x-axis to generate a solid with a cone-shaped cavity in its center
(Figure 7.21).
[–/4, /2] by [–1.5, 1.5]
Figure 7.20 The region in Example 3.
R r Figure 7.22 The area of a washer is pR2 Ϫ pr2. (Example 3)
Figure 7.21 The solid generated by revolving the region in Figure 7.20 about the x-axis. A typical cross section is a washer: a circular region with a circular region cut out of its center. (Example 3)
This time each cross section perpendicular to the axis of revolution is a washer, a circular region with a circular region cut from its center. The area of a washer can be found by subtracting the inner area from the outer area (Figure 7.22).
In our region the cosine curve defines the outer radius, and the curves intersect at x ϭ p ր4. The volume is
͵
͵
pր4
Vϭ
p͑cos2 x Ϫ sin2 x͒ dx
0
pր4
ϭp
CAUTION!
cos 2x dx
identity: cos2 x Ϫ sin2 x ϭ cos 2x
0
The area of a washer is pR 2 Ϫ pr 2, which you can simplify to p(R 2 Ϫ r 2), but not to p(R Ϫ r)2. No matter how tempting it is to make the latter simplification, it’s wrong. Don’t do it.
[ ]
sin 2x ϭ p ᎏᎏ
2
pր4
0
p ϭ ᎏᎏ units cubed.
2
Now try Exercise 17.
We could have done the integration in Example 3 with NINT, but we wanted to demonstrate how a trigonometric identity can be useful under unexpected circumstances in calculus. The double-angle identity turned a difficult integrand into an easy one and enabled us to get an exact answer by antidifferentiation.
Cylindrical Shells
There is another way to find volumes of solids of rotation that can be useful when the axis of revolution is perpendicular to the axis containing the natural interval of integration. Instead of summing volumes of thin slices, we sum volumes of thin cylindrical shells that grow outward from the axis of revolution like tree rings.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 402
402
Chapter 7
Applications of Definite Integrals
EXPLORATION 1
The region enclosed by the x-axis and the parabola y ϭ f ͑ x͒ ϭ 3x Ϫ x 2 is revolved about the line x ϭ Ϫ1 to generate the shape of a cake (Figures 7.23, 7.24). (Such a cake is often called a bundt cake.) What is the volume of the cake?
Integrating with respect to y would be awkward here, as it is not easy to get the original parabola in terms of y. (Try finding the volume by washers and you will soon see what we mean.) To integrate with respect to x, you can do the problem by cylindrical shells, which requires that you cut the cake in a rather unusual way.
[–6, 4] by [–3, 3]
Figure 7.23 The graph of the region in
Exploration 1, before revolution. y 0
3
x
Axis of revolution x ϭ –1
Figure 7.24 The region in Figure 7.23 is revolved about the line x ϭ Ϫ1 to form a solid cake. The natural interval of integration is along the x-axis, perpendicular to the axis of revolution. (Exploration 1)
1. Instead of cutting the usual wedge shape, cut a cylindrical slice by cutting straight down all the way around close to the inside hole. Then cut another cylindrical slice around the enlarged hole, then another, and so on. The radii of the cylinders gradually increase, and the heights of the cylinders follow the contour of the parabola: smaller to larger, then back to smaller ( Figure 7.25).
Each slice is sitting over a subinterval of the x-axis of length Δx. Its radius is approximately ͑1 ϩ x k ͒. What is its height?
2. If you unroll the cylinder at x k and flatten it out, it becomes (essentially) a rectangular slab with thickness Δx. Show that the volume of the slab is approximately
2p͑x k ϩ 1͒͑3x k Ϫ x k 2 ͒Δx.
3. ͚ 2p͑x k ϩ 1͒͑3x k Ϫ x k 2 ͒Δx is a Riemann sum. What is the limit of these Riemann sums as Δx→0?
4. Evaluate the integral you found in step 3 to find the volume of the cake!
EXAMPLE 4 Finding Volumes Using Cylindrical Shells
y
The region bounded by the curve y ϭ ͙ෆ, the x-axis, and the line x ϭ 4 is revolved x about the x-axis to generate a solid. Find the volume of the solid.
SOLUTION
xk
3
Figure 7.25 Cutting the cake into thin cylindrical slices, working from the inside out. Each slice occurs at some xk between 0 and 3 and has thickness Δx.
(Exploration 1)
x
1. Sketch the region and draw a line segment across it parallel to the axis of revolution
(Figure 7.26). Label the segment’s length (shell height) and distance from the axis of revolution (shell radius). The width of the segment is the shell thickness dy. (We drew the shell in Figure 7.27, but you need not do that.) y y
Shell height
4 Ϫ y2
Shell height
2
Interval of integration yk
0
Volume by Cylindrical Shells
⎧
⎪
⎨y
⎪
⎩
0
x ϭ y2
2
(4, 2) y 4 Ϫ y2 y ϭ ͙x
͙
(4, 2)
Shell thickness ϭ dy
y Shell radius
4
x
0 y Figure 7.26 The region, shell dimensions, and interval of integration in Example 4.
x
S
Shell radius Figure 7.27 The shell swept out by the line segment in Figure 7.26.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 403
Section 7.3 Volumes
403
2. Identify the limits of integration: y runs from 0 to 2.
3. Integrate to find the volume.
͵ ( )( )
͵
2
Vϭ
2p
0
shell shell dy radius height
2
ϭ
2p͑y͒͑4 Ϫ y 2 ͒ dy ϭ 8p
0
Now try Exercise 33(a). y EXAMPLE 5 Finding Volumes Using Cylindrical Shells
The region bounded by the curves y ϭ 4 Ϫ x2, y ϭ x, and x ϭ 0 is revolved about the y-axis to form a solid. Use cylindrical shells to find the volume of the solid.
4
SOLUTION
2
0
x
2
Figure 7.28 The region and the height of a typical shell in Example 5.
1. Sketch the region and draw a line segment across it parallel to the y-axis
(Figure 7.28). The segment’s length (shell height) is 4 Ϫ x2 Ϫ x. The distance of the segment from the axis of revolution (shell radius) is x.
2. Identify the limits of integration: The x-coordinate of the point of intersection of the curves y ϭ 4 Ϫ x2 and y ϭ x in the first quadrant is about 1.562. So x runs from 0 to
1.562.
3. Integrate to find the volume.
͵
͵
1.562
Vϭ
2p
0
( )( )
shell shell dx radius height
1.562
ϭ
2p(x)(4 Ϫ x2 Ϫ x) dx
0
Ϸ 13.327
Now try Exercise 35.
Other Cross Sections o π
xk
[–1, 3.5] by [–0.8, 2.2]
Figure 7.29 The base of the paperweight in Example 6. The segment perpendicular to the x-axis at xk is the diameter of a semicircle that is perpendicular to the base. y
2
y = 2 sin x
The method of cross-section slicing can be used to find volumes of a wide variety of unusually shaped solids, so long as the cross sections have areas that we can describe with some formula. Admittedly, it does take a special artistic talent to draw some of these solids, but a crude picture is usually enough to suggest how to set up the integral.
EXAMPLE 6 A Mathematician’s Paperweight
A mathematician has a paperweight made so that its base is the shape of the region between the x-axis and one arch of the curve y ϭ 2 sin x (linear units in inches). Each cross section cut perpendicular to the x-axis (and hence to the xy-plane) is a semicircle whose diameter runs from the x-axis to the curve. (Think of the cross section as a semicircular fin sticking up out of the plane.) Find the volume of the paperweight.
SOLUTION
0
x
Figure 7.30 Cross sections perpendicular to the region in Figure 7.29 are semicircular.
(Example 6)
The paperweight is not easily drawn, but we know what it looks like. Its base is the region in Figure 7.29, and the cross sections perpendicular to the base are semicircular fins like those in Figure 7.30.
The semicircle at each point x has
2 sin x
1
radius ϭ ᎏᎏ ϭ sin x and area A͑x͒ ϭ ᎏᎏp͑sin x͒ 2 .
2
2 continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 404
404
Chapter 7
Applications of Definite Integrals
The volume of the paperweight is
Vϭ
͵
p
A͑x͒ dx
0
p ϭ ᎏᎏ
2
͵
p
͑sin x͒ 2 dx
0
p
Ϸ ᎏᎏ NINT ͑͑sin x͒ 2, x, 0, p͒
2
p
Ϸ ᎏᎏ͑1.570796327͒.
2
Bonaventura Cavalieri
The number in parentheses looks like half of p, an observation that can be confirmed analytically, and which we support numerically by dividing by p to get 0.5. The volume of the paperweight is p p p2 ᎏᎏ • ᎏᎏ ϭ ᎏᎏ Ϸ 2.47 in3.
2 2
4
Now try Exercise 39(a).
(1598—1647)
Cavalieri, a student of
Galileo, discovered that if two plane regions can be arranged to lie over the same interval of the x-axis in such a way that they have identical vertical cross sections at every point, then the regions have the same area. This theorem and a letter of recommendation from Galileo were enough to win Cavalieri a chair at the University of Bologna in 1629. The solid geometry version in Example 7, which Cavalieri never proved, was named after him by later geometers.
EXAMPLE 7 Cavalieri’s Volume Theorem
Cavalieri’s volume theorem says that solids with equal altitudes and identical cross section areas at each height have the same volume (Figure 7.31). This follows immediately from the definition of volume, because the cross section area function A͑x͒ and the interval
͓a, b͔ are the same for both solids.
b
Same volume
a
Cross sections have the same length at every point in [a, b].
Same cross-section area at every level
Figure 7.31 Cavalieri’s volume theorem: These solids have the same volume. You can illustrate this yourself with stacks of coins. (Example 7) a x
b
Now try Exercise 43.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 405
Section 7.3 Volumes
EXPLORATION 2
405
Surface Area
We know how to find the volume of a solid of revolution, but how would we find the surface area? As before, we partition the solid into thin slices, but now we wish to form a Riemann sum of approximations to surface areas of slices (rather than of volumes of slices). y y = f(x)
b
a
x
A typical slice has a surface area that can be approximated by 2p • f ͑x͒ • Δs, where Δs is the tiny slant height of the slice. We will see in Section 7.4, when we study arc length, that Δs ϭ ͙Δx2 ϩෆ, and that this can be written as
ෆ Δy2
2 Δx.
Δs ϭ ͙1ෆෆf Јෆෆ͒
ෆ ϩ ͑ ෆ͑xk ͒ෆ
Thus, the surface area is approximated by the Riemann sum n ෆ ϩ͑ ෆ
͚ 2p f ͑x ͒ ͙1ෆෆfෆЈ͑xෆ͒͒ෆ Δx. k k
2
kϭ1
1. Write the limit of the Riemann sums as a definite integral from a to b. When will the limit exist?
2. Use the formula from part 1 to find the surface area of the solid generated by revolving a single arch of the curve y ϭ sin x about the x-axis.
3. The region enclosed by the graphs of y 2 ϭ x and x ϭ 4 is revolved about the x-axis to form a solid. Find the surface area of the solid.
Quick Review 7.3
(For help, go to Section 1.2.)
In Exercises 1–10, give a formula for the area of the plane region in terms of the single variable x.
1. a square with sides of length x x 2
2. a square with diagonals of length x x2/2
3. a semicircle of radius x px2/2
4. a semicircle of diameter x px2/8
5. an equilateral triangle with sides of length x (͙3/4)x 2
ෆ
6. an isosceles right triangle with legs of length x x2/2
7. an isosceles right triangle with hypotenuse x x2/4
8. an isosceles triangle with two sides of length 2x and one side of length x (͙15 2
ෆ/4)x
9. a triangle with sides 3x, 4x, and 5x 6x 2
10. a regular hexagon with sides of length x (3͙3/2)x 2
ෆ
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 406
406
Chapter 7
Applications of Definite Integrals
Section 7.3 Exercises
In Exercises 1 and 2, find a formula for the area A(x) of the cross sections of the solid that are perpendicular to the x-axis.
1. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis between these planes run from the semicircle y ϭ Ϫ͙ෆෆෆෆ to the semicircle y ϭ ͙ෆෆෆෆ.
1 Ϫ x2
1 Ϫ x2
2. The solid lies between planes perpendicular to the x-axis at x ϭ 0 and x ϭ 4. The cross sections perpendicular to the x-axis between these planes run from y ϭ Ϫ͙x to y ϭ ͙x .
ෆ
ෆ
(a) The cross sections are circular disks with diameters in the xy-plane. px y (a) The cross sections are circular disks with diameters in the xy-plane. p(1 Ϫ x 2) x2 ϩ y2 ϭ 1 y x ϭ y2
4
x
0
–1
1
(b) The cross sections are squares with bases in the xy-plane. 4x
x
(b) The cross sections are squares with bases in the xy-plane. 4(1 Ϫ x 2) y y
x ϭ y2
4
x
0
–1
x
1 x2 ϩ
y2
(c) The cross sections are squares with diagonals in the xy-plane. 2x
ϭ1
(c) The cross sections are squares with diagonals in the xy-plane. (The length of a square’s diagonal is ͙2 times
ෆ
the length of its sides.) 2(1 – x 2) x2 ϩ y2 ϭ 1
y
In Exercises 3–6, find the volume of the solid analytically.
3. The solid lies between planes perpendicular to the x-axis at x ϭ 0 and x ϭ 4. The cross sections perpendicular to the axis on the interval 0 Յ x Յ 4 are squares whose diagonals run from y ϭ Ϫ͙x to y ϭ ͙x . 16
ෆ
ෆ
4. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis are circular disks whose diameters run from the parabola y ϭ x 2 to the parabola y ϭ 2 Ϫ x 2. 16p/15
0
1
–1
(d) The cross sections are equilateral triangles with bases in the xy-plane. ͙3x
ෆ
x
(d) The cross sections are equilateral triangles with bases in the xy-plane. ͙3(1 Ϫ x 2)
ෆ
y
2
y ϭ x2
y
0 y ϭ 2 Ϫ x2
0
–1
x2 ϩ y2 ϭ 1
1
x
x
5. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis between these planes are squares whose bases run from the semicircle y ϭ Ϫ͙ෆෆෆ2 to the semicircle y ϭ ͙ෆෆෆ2 .
1 Ϫ xෆ
1 Ϫ xෆ
16/3
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 407
Section 7.3 Volumes
6. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫ1 and x ϭ 1. The cross sections perpendicular to the x-axis between these planes are squares whose diagonals run from the semicircle y ϭ Ϫ͙ෆෆෆෆ to the semicircle
1 Ϫ x2
2 . 8/3 y ϭ ͙ෆෆෆෆ
1Ϫx
In Exercises 7–10, find the volume of the solid generated by revolving the shaded region about the given axis.
y
x ϭ 3y/2
x ϩ 2y ϭ 2
2
9. about the y-axis 4 Ϫ p
x
1
x
3
0
y
31. Find the volume of the solid generated by revolving the region bounded by the parabola y ϭ x 2 and the line y ϭ 1 about
(a) the line y ϭ 1. 16p/15 (b) the line y ϭ 2.
y ϭ sin x cos x
(c) the line y ϭ Ϫ1.
x
0
13. y ϭ
15. y ϭ x, y ϭ 1,
17. y ϭ x 2 ϩ 1,
19. y ϭ sec x,
20. y ϭ Ϫ͙x ,
ෆ
32p/5
xϭ2
12. y ϭ x 3,
y ϭ 0,
y ϭ 0 36p 14. y ϭ x Ϫ x 2, x ϭ 0 2p/3 16. y ϭ 2x,
yϭxϩ3
117p/5
In Exercises 33 and 34, use the cylindrical shell method to find the volume of the solid generated by revolving the shaded region about the indicated axis.
(b) the line y ϭ 1
33. (a) the x-axis 6p/5
x ϭ 2 128p/7
yϭ0
(p/3)b2h
(b) the y-axis.
(c) the line y ϭ 8 ր5
4p/5
2p (d) the line y ϭ Ϫ2 ր5
2p
y
p/30
y ϭ x, x ϭ 1 p
18. y ϭ 4 Ϫ x 2,
ෆ y ϭ ͙2 , Ϫp ր4 Յ x Յ p ր4 y ϭ Ϫ2,
56p/15
64p/15
(a) the x-axis. (p/3)bh2
In Exercises 11–20, find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis. ͙9ෆෆෆ,
ෆ Ϫ x2
8p/3
32. By integration, find the volume of the solid generated by revolving the triangular region with vertices ͑0, 0͒, ͑b, 0͒,
͑0, h͒ about x 11. y ϭ x 2, y ϭ 0,
32p/5
(d) the line x ϭ 4. 224p/15
(a) the line x ϭ 1. 2p/3 (b) the line x ϭ 2.
– x ϭ tan ⎛ y⎛
⎝4 ⎝
0
(b) the y-axis.
8p/3
30. Find the volume of the solid generated by revolving the triangular region bounded by the lines y ϭ 2x, y ϭ 0, and x ϭ 1 about
10. about the x-axis p2/16
y
Group Activity In Exercises 29–32, find the volume of the solid described. (c) the line y ϭ 2.
2
0
8p
ෆ
28. the region bounded above by the curve y ϭ ͙x and below by the line y ϭ x 2p/15
(a) the x-axis. 8p
y
1
27. the region in the first quadrant bounded above by the parabola y ϭ x 2, below by the x-axis, and on the right by the line x ϭ 2
29. Find the volume of the solid generated by revolving the region bounded by y ϭ ͙x and the lines y ϭ 2 and x ϭ 0 about
ෆ
8. about the y-axis 6p
7. about the x-axis 2p/3
407
p2
y ϭ 2 Ϫ x 108p/5
1
x ϭ 12 (y2 Ϫ y 3)
0
1
Ϫ 2p
x ϭ 0 8p
In Exercises 21 and 22, find the volume of the solid generated by revolving the region about the given line.
21. the region in the first quadrant bounded above by the line y ϭ ͙2 , below by the curve y ϭ sec x tan x, and on the left by
ෆ
the y-axis, about the line y ϭ ͙2 2.301
ෆ
22. the region in the first quadrant bounded above by the line y ϭ 2, below by the curve y ϭ 2 sin x, 0 Յ x Յ p ր 2, and on the left by the y-axis, about the line y ϭ 2 p(3p Ϫ 8)
In Exercises 23–28, find the volume of the solid generated by revolving the region about the y-axis.
ෆ
23. the region enclosed by x ϭ ͙5 y 2, x ϭ 0, y ϭ Ϫ1, y ϭ 1
(c) the line y ϭ 5 8p y (d) the line y ϭ Ϫ5 ր 8 y4 y2 xϭ— Ϫ —
4
2
(2, 2)
2p
24. the region enclosed by x ϭ y 3ր 2, x ϭ 0, y ϭ 2 4p
y2 xϭ— 2
25. the region enclosed by the triangle with vertices ͑1, 0͒, ͑2, 1͒, and ͑1, 1͒ 4p/3
26. the region enclosed by the triangle with vertices ͑0, 1͒, ͑1, 0͒, and ͑1, 1͒ 2p/3
(b) the line y ϭ 2 8p/5
34. (a) the x-axis 8p/3
2
x
0
1
2
x
4p
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 408
408
Chapter 7
Applications of Definite Integrals
In Exercises 35–38, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis.
35. y ϭ x,
y ϭ Ϫx ր 2, x ϭ 2
36. y ϭ x 2,
37. y ϭ ͙x ,
ෆ
y ϭ 2 Ϫ x, y ϭ 0,
38. y ϭ 2x Ϫ 1,
43. Writing to Learn A solid lies between planes perpendicular to the x-axis at x ϭ 0 and x ϭ 12. The cross sections by planes perpendicular to the x-axis are circular disks whose diameters run from the line y ϭ x ր 2 to the line y ϭ x as shown in the figure. Explain why the solid has the same volume as a right circular cone with base radius 3 and height 12.
8p
x ϭ 0,
for x Ն 0
5p/6
x ϭ 4 128p/5
y ϭ ͙x ,
ෆ
xϭ0
The volumes are equal by Cavalieri’s
Theorem.
y
7p/15
yϭx
In Exercises 39–42, find the volume of the solid analytically.
39. The base of a solid is the region between the curve
ෆෆ x y ϭ 2 ͙sinෆ and the interval ͓0, p͔ on the x-axis. The cross sections perpendicular to the x-axis are
yϭ x
2
(a) equilateral triangles with bases running from the x-axis to the curve as shown in the figure. 2͙3
ෆ
0 x 12
y
44. A Twisted Solid A square of side length s lies in a plane perpendicular to a line L. One vertex of the square lies on L.
As this square moves a distance h along L, the square turns one revolution about L to generate a corkscrew-like column with square cross sections.
2
0
y ϭ 2 √⎯⎯⎯ x
⎯ sin
x
(b) squares with bases running from the x-axis to the curve.
40. The solid lies between planes perpendicular to the x-axis at x ϭ Ϫp ր 3 and x ϭ p ր 3. The cross sections perpendicular to the x-axis are
(a) circular disks with diameters running from the curve y ϭ tan x to the curve y ϭ sec x. p͙3 Ϫ (p2/6)
ෆ
(b) squares whose bases run from the curve y ϭ tan x to the curve y ϭ sec x. 4͙3 Ϫ (2p/3)
ෆ
41. The solid lies between planes perpendicular to the y-axis at y ϭ 0 and y ϭ 2. The cross sections perpendicular to the y-axis are circular disks with diameters running from the y-axis to the parabola x ϭ ͙ෆy 2. 8p
5
42. The base of the solid is the disk x 2 ϩ y 2 Յ 1. The cross sections by planes perpendicular to the y-axis between y ϭ Ϫ1 and y ϭ 1 are isosceles right triangles with one leg in the disk. 8/3
(a) Find the volume of the column.
8
s2h
(b) Writing to Learn What will the volume be if the square turns twice instead of once? Give reasons for your answer. s2h
45. Find the volume of the solid generated by revolving the region in the first quadrant bounded by y ϭ x 3 and y ϭ 4x about
(a) the x-axis, 512p/21
(b) the line y ϭ 8. 832p/21
46. Find the volume of the solid generated by revolving the region bounded by y ϭ 2x Ϫ x 2 and y ϭ x about
(a) the y-axis, p/6
(b) the line x ϭ 1. p/6
47. The region in the first quadrant that is bounded above by the curve y ϭ 1 ր ͙x , on the left by the line x ϭ 1 ր4, and below
ෆ
by the line y ϭ 1 is revolved about the y-axis to generate a solid. Find the volume of the solid by (a) the washer method and
(b) the cylindrical shell method. (a) 11p/48 (b) 11p/48
48. Let f ͑x͒ ϭ
͑sin
{ 1, x͒ րx,
0ϽxՅp x ϭ 0.
(a) Show that x f ͑x͒ ϭ sin x,
0 Յ x Յ p.
(b) Find the volume of the solid generated by revolving the shaded region about the y-axis. 4p y y
⎧ sin x , 0 Ͻ x
——
yϭ⎨ x xϭ0 ⎩ 1,
≤
1
0
1 x 2 ϩ y2 ϭ 1
x
0
x
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 409
Section 7.3 Volumes
49. Designing a Plumb Bob Having been asked to design a brass plumb bob that will weigh in the neighborhood of 190 g, you decide to shape it like the solid of revolution shown here.
59. y ϭ x 2,
0 Յ x Յ 2;
60. y ϭ 3x Ϫ x 2,
y ϭ x ͙36 Ϫ x 2
12
62. y ϭ ͙x ϩෆ,
ෆෆ 1
0
cm3
(b) If you specify a brass that weighs 8.5 g րcm3, how much will the plumb bob weigh to the nearest gram? 192.3 g
50. Volume of a Bowl A bowl has a shape that can be generated by revolving the graph of y ϭ x 2ր 2 between y ϭ 0 and y ϭ 5 about the y-axis.
(a) Find the volume of the bowl.
25p
(b) If we fill the bowl with water at a constant rate of
3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep? 3/(8p)
51. The Classical Bead Problem A round hole is drilled through the center of a spherical solid of radius r. The resulting cylindrical hole has height 4 cm.
(a) What is the volume of the solid that remains?
32p/3
(b) What is unusual about the answer? The answer is independent of r.
52. Writing to Learn Explain how you could estimate the volume of a solid of revolution by measuring the shadow cast on a table parallel to its axis of revolution by a light shining directly above it. See page 410.
53. Same Volume about Each Axis The region in the first quadrant enclosed between the graph of y ϭ ax Ϫ x 2 and the x-axis generates the same volume whether it is revolved about the x-axis or the y-axis. Find the value of a. 5
54. (Continuation of Exploration 2) Let x ϭ g͑y͒ Ͼ 0 have a continuous first derivative on ͓c, d ͔. Show that the area of the surface generated by revolving the curve x ϭ g͑y͒ about the y-axis is See page 410.
Sϭ
͵
͵
0
65. Multiple Choice The base of a solid S is the region enclosed by the graph of y ϭ ln x, the line x ϭ e, and the x-axis. If the cross sections of S perpendicular to the x-axis are squares, which of the following gives the best approximation of the volume of S? A
(A) 0.718
In Exercises 55–62, find the area of the surface generated by revolving the curve about the indicated axis.
0 Յ y Յ 2; y-axis Ϸ13.614
(B) 1.718
(A) 60.3
(B) 115.2
57. x ϭ y 1ր 2 Ϫ ͑1 ր 3͒ 3 ր 2,
1 Յ y Յ 3;
58. x ϭ ͙2y Ϫෆ, ͑5ր 8͒ Յ y Յ 1;
ෆෆෆ 1
y-axis Ϸ16.110 y-axis Ϸ2.999
(D) 3.171
(E) 7.388
(C) 225.4
(D) 319.7
(E) 361.9
67. Multiple Choice Let R be the region enclosed by the graph of y ϭ x2, the line x ϭ 4, and the x-axis. Which of the following gives the best approximation of the volume of the solid generated when R is revolved about the y-axis? B
(A) 64p
(B) 128p
(C) 256p
(D) 360
(E) 512
68. Multiple Choice Let R be the region enclosed by the graphs of y ϭ eϪx, y ϭ ex, and x ϭ 1. Which of the following gives the volume of the solid generated when R is revolved about the x-axis? D
͵
͵
͵
͵
͵
1
(A)
(ex Ϫ eϪx) dx
0
1
(e2x Ϫ eϪ2x) dx
0
1
(C)
(ex Ϫ eϪx)2 dx
0
1
y-axis Ϸ0.638
(C) 2.718
66. Multiple Choice Let R be the region in the first quadrant bounded by the graph of y ϭ 8 Ϫ x3/2, the x-axis, and the y-axis.
Which of the following gives the best approximation of the volume of the solid generated when R is revolved about the x-axis? E
(B)
0 Յ y Յ 1;
x-axis Ϸ51.313
64. True or False If the region enclosed by the y-axis, the line y ϭ 2, and the curve y ϭ ͙x is revolved about the y-axis, the
ෆ
2 volume of the solid is given by the definite integral ͐0 py2 dy.
2
Justify your answer. False. The volume is given by py 4 dy.
2p g͑y͒ ͙1ෆෆg Ј͑ y͒ෆ dy.
ෆ ϩ ͑ ෆෆෆ͒ 2
c
56. x ϭ y 3ր 3,
1 Յ x Յ 5;
x-axis Ϸ6.283
63. True or False The volume of a solid of a known integrable b cross section area A(x) from x ϭ a to x ϭ b is ͐a A(x) dx. Justify your answer. True, by definition.
d
55. x ϭ ͙y ,
ෆ
0.5 Յ x Յ 1.5;
You may use a graphing calculator to solve the following problems. x (cm
36p/5
x-axis Ϸ44.877
Standardized Test Questions
6
(a) Find the plumb bob’s volume.
x-axis Ϸ53.226
0 Յ x Յ 3;
61. y ϭ ͙2 x Ϫෆ2 ,
ෆෆෆ x ෆ y (cm)
409
(e2x Ϫ eϪ2x) dx
(D) p
0
1
(E) p
0
(e x Ϫ eϪx)2 dx
5128_Ch07_pp378-433.qxd 2/3/06 4:38 PM Page 410
410
Chapter 7
Applications of Definite Integrals
Explorations
Extending the Ideas
69. Max-Min The arch y ϭ sin x, 0 Յ x Յ p, is revolved about the line y ϭ c, 0 Յ c Յ 1, to generate the solid in the figure.
71. Volume of a Hemisphere Derive the formula V ϭ ͑2 ր 3͒ pR 3 for the volume of a hemisphere of radius R by comparing its cross sections with the cross sections of a solid right circular cylinder of radius R and height R from which a solid right circular cone of base radius R and height R has been removed as suggested by the figure.
(a) Find the value of c that minimizes the volume of the solid.
2 p2
What is the minimum volume? ᎏᎏ, ᎏϪ 8
ᎏ
p
2
(b) What value of c in ͓0, 1͔ maximizes the volume of the solid? 0
R2 – h
√2
(c) Writing to Learn Graph the solid’s volume as a function of c, first for 0 Յ c Յ 1 and then on a larger domain. What happens to the volume of the solid as c moves away from ͓0, 1͔?
Does this make sense physically? Give reasons for your answers. y h
h
R
R
p(2c2p Ϫ 8c ϩ p)
V ϭ ᎏᎏ
2
Volume → ∞
h
72. Volume of a Torus The disk x 2 ϩ y 2 Յ a 2 is revolved about the line x ϭ b ͑b Ͼ a͒ to generate a solid shaped like a doughnut, a called a torus. Find its volume. (Hint: ͐ ͙ෆෆෆෆෆ dy ϭ pa 2ր 2, a2 Ϫ y2
Ϫa
since it is the area of a semicircle of radius a.) 2a2bp2
y ϭ sin x c 73. Filling a Bowl
(a) Volume A hemispherical bowl of radius a contains water to a depth h. Find the volume of water in the bowl. ph2(3a Ϫ h)/3
0 yϭc
x
70. A Vase We wish to estimate the volume of a flower vase using only a calculator, a string, and a ruler. We measure the height of the vase to be 6 inches. We then use the string and the ruler to find circumferences of the vase (in inches) at half-inch intervals. (We list them from the top down to correspond with the picture of the vase.)
6
Circumferences
0
5.4
4.5
4.4
5.1
6.3
7.8
9.4
10.8
11.6
11.6
10.8
9.0
6.3
(a) Find the areas of the cross sections that correspond to the given circumferences. 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7,
10.7, 9.3, 6.4, 3.2
(b) Express the volume of the vase as an integral with respect to
1 6 y over the interval ͓0, 6͔.
ᎏᎏ C(y)2 dy
4p
͵
52. Partition the appropriate interval on the axis of revolution and measure the radius r(x) of the shadow region at these points. Then use an approximab tion such as the trapezoidal rule to estimate the integral ͐a pr2(x) dx.
54. For a tiny horizontal slice,
(⌬x)2 ϩ (⌬y)2
1 ϩ (gЈ(y))2 slant height ϭ ⌬s ϭ ͙ෆෆ ϭ ͙ෆෆ ⌬y. So the surface area is approximated by the Riemann sum
͚ 2p
g(yk)͙ෆෆ
1 ϩ (gЈ(y))2
⌬y.
kϭ1
The limit of that is the integral.
74. Consistency of Volume Definitions The volume formulas in calculus are consistent with the standard formulas from geometry in the sense that they agree on objects to which both apply. (a) As a case in point, show that if you revolve the region enclosed by the semicircle y ϭ ͙a ෆϪෆෆ and the x-axis about
ෆ2 ෆ x 2 the x-axis to generate a solid sphere, the calculus formula for volume at the beginning of the section will give ͑4 ր 3͒pa 3 for the volume just as it should.
(b) Use calculus to find the volume of a right circular cone of height h and base radius r.
71. Hemisphere cross sectional area: p(͙ෆ2)2 ϭ A1
R2 Ϫ hෆ
Right circular cylinder with cone removed cross sectional area: pR2 Ϫ ph2 ϭ A2
Since A1 ϭ A2, the two volumes are equal by Cavalieri’s theorem.
Thus,
volume of hemisphere ϭ volume of cylinder Ϫ volume of cone
1
2
ϭ pR3 Ϫ ᎏ3ᎏpR3 ϭ ᎏ3ᎏpR3.
0
(c) Approximate the integral using the Trapezoidal Rule with n ϭ 12. Ϸ34.7 in3
n
(b) Related Rates Water runs into a sunken concrete hemispherical bowl of radius 5 m at a rate of 0.2 m3րsec. How fast is the water level in the bowl rising when the water is 4 m deep? 1/(120p) m/sec
74. (a) A cross section has radius r ϭ ͙ෆ2 and a2 Ϫ xෆ
2 ϭ p(a 2 Ϫ x 2). area A(x) ϭ pr a 4
Vϭ
p(a2 Ϫ x2) ϭ ᎏᎏpa3
3
Ϫa y (b) A cross section has radius x ϭ r 1 Ϫ ᎏᎏ and h ͵
2y y2 area A(y) ϭ px 2 ϭ pr 2 1 Ϫ ᎏᎏ ϩ ᎏᎏ . h h2
͵ h
2y y2 1
V ϭ pr 2 1 Ϫ ᎏᎏ ϩ ᎏᎏ dy ϭ ᎏᎏpr2h h h
3
0
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 411
Section 7.3 Volumes
411
Quick Quiz for AP* Preparation: Sections 7.1–7.3
You may use a graphing calculator to solve the following problems. 1. Multiple Choice The base of a solid is the region in the first quadrant bounded by the x-axis, the graph of y ϭ sinϪ1 x, and the vertical line x ϭ 1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume? C
(A) 0.117
(B) 0.285
(C) 0.467
(D) 0.571
(E) 1.571
2. Multiple Choice Let R be the region in the first quadrant bounded by the graph of y ϭ 3x Ϫ x2 and the x-axis. A solid is generated when R is revolved about the vertical line x ϭ Ϫ1.
Set up, but do not evaluate, the definite integral that gives the volume of this solid. A
͵
͵
͵
͵
͵
3
(A)
2p(x ϩ 1)(3x Ϫ x2) dx
0
3
(B)
2p(x ϩ 1)(3x Ϫ x2) dx
Ϫ1
3
(C)
(B) r(10) Ϫ r(0)
͵
͵
10
(C)
rЈ(t) dt
0
10
(D)
r(t) dt
0
(E) 10 и r(10)
4. Free Response Let R be the region bounded by the graphs of
ෆ
y ϭ ͙x, y ϭ eϪx, and the y-axis.
(a) Find the area of R.
(b) Find the volume of the solid generated when R is revolved about the horizontal line y ϭ Ϫ1.
2p(3x Ϫ x2)2 dx
(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a semicircle whose diameෆ ter runs from the graph of y ϭ ͙x to the graph of y ϭ eϪx. Find the volume of this solid.
3
0
3
(E)
(A) r(10)
2p(x)(3x Ϫ x2) dx
0
(D)
3. Multiple Choice A developing country consumes oil at a rate given by r(t) ϭ 20e0.2t million barrels per year, where t is time measured in years, for 0 Յ t Յ 10. Which of the following expressions gives the amount of oil consumed by the country during the time interval 0 Յ t Յ 10? D
(3x Ϫ x2) dx
0
4. (a) The two graphs intersect where ͙x ϭ e–x, which a calculator shows to
ෆ
be x ϭ 0.42630275. Store this value as A.
͵ (e
A
The area of R is
–x
0
Ϫ ͙x) dx ϭ 0.162.
ෆ
͵ p((e ϩ 1) Ϫ (͙x ϩ 1) ) dx ϭ 1.631.
ෆ
ෆ
1 e Ϫ ͙x
(c) Volume ϭ ͵ ᎏᎏpᎏᎏ dx ϭ 0.035.
2
2
(b) Volume ϭ
A
–x
2
2
0
A
0
Ϫx
2
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 412
412
Chapter 7
Applications of Definite Integrals
7.4
What you’ll learn about
Lengths of Curves
A Sine Wave
• A Sine Wave
• Length of a Smooth Curve
• Vertical Tangents, Corners, and
Cusps
. . . and why
How long is a sine wave (Figure 7.32)?
The usual meaning of wavelength refers to the fundamental period, which for y ϭ sin x is 2p. But how long is the curve itself? If you straightened it out like a piece of string along the positive x-axis with one end at 0, where would the other end be?
EXAMPLE 1 The Length of a Sine Wave
The length of a smooth curve can be found using a definite integral.
What is the length of the curve y ϭ sin x from x ϭ 0 to x ϭ 2p?
SOLUTION
We answer this question with integration, following our usual plan of breaking the whole into measurable parts. We partition ͓0, 2p͔ into intervals so short that the pieces of curve
(call them “arcs”) lying directly above the intervals are nearly straight. That way, each arc is nearly the same as the line segment joining its two ends and we can take the length of the segment as an approximation to the length of the arc.
Figure 7.33 shows the segment approximating the arc above the subinterval ͓x kϪ1, x k ͔.
The length of the segment is ͙Δxk2 ϩෆ. The sum
ෆ Δyk2
ෆ Δy
͚ ͙Δx ϩෆ
2
k
[0, 2] by [–2, 2]
Figure 7.32 One wave of a sine curve has to be longer than 2p. y √ ⌬xk2 + ⌬yk2
Q
y = sin x
over the entire partition approximates the length of the curve. All we need now is to find the limit of this sum as the norms of the partitions go to zero. That’s the usual plan, but this time there is a problem. Do you see it?
The problem is that the sums as written are not Riemann sums. They do not have the form ͚ f ͑ck ͒ Δx. We can rewrite them as Riemann sums if we multiply and divide each square root by Δxk .
⌬yk
P
͚
⌬xk
O
xk –1
xk
͙⌬ ෆ2 ϩෆෆෆ ϭ
ෆxkෆෆ ⌬yk2 ϭ x
Figure 7.33 The line segment approximating the arc PQ of the sine curve above the subinterval ͓x kϪ1, x k ͔. (Example 1)
2 k ͚
͙ෆxk ͒ 2ෆෆ⌬yk ͒ 2
͑⌬ ෆෆ ϩ ͑ෆෆෆ
ᎏᎏ ⌬xk
⌬ xk
( )
͚ Ί ⌬x
⌬yk
1 ϩ ᎏᎏ
⌬ xk
2
k
This is better, but we still need to write the last square root as a function evaluated at some ck in the kth subinterval. For this, we call on the Mean Value Theorem for differentiable functions (Section 4.2), which says that since sin x is continuous on ͓x kϪ1, x k ͔ and is differentiable on ͑x kϪ1, x k ͒ there is a point ck in ͓xkϪ1, xk ͔ at which Δyk րΔxk ϭ sinЈ ck (Figure 7.34). That gives us
ෆ ϩ ෆෆЈ c ͒
͚ ͙1ෆෆ͑sinෆෆෆෆ Δx , k Group Exploration
Later in this section we will use an integral to find the length of the sine wave with great precision. But there are ways to get good approximations without integrating. Take five minutes to come up with a written estimate of the curve’s length. No fair looking ahead.
2
k
which is a Riemann sum.
Now we take the limit as the norms of the subdivisions go to zero and find that the length of one wave of the sine function is
͵
2p
͙1ෆෆsinෆx͒ ෆ
ෆ ϩ ͑ෆෆЈ ෆ2
dx ϭ
0
How close was your estimate?
͵
2p
͙1ෆෆos 2ෆ dx Ϸ 7.64.
ෆ ϩ cෆෆ x
Using NINT
0
Now try Exercise 9.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 413
Section 7.4 Lengths of Curves
Length of a Smooth Curve
Slope sin' (ck)
Q
⌬yk
P
⌬xk
xk –1
ck
xk
Figure 7.34 The portion of the sine curve above ͓x kϪ1, x k ͔. At some ck in the interval, sinЈ ͑ck ͒ ϭ ⌬yk ր⌬x k , the slope of segment PQ. (Example 1)
We are almost ready to define the length of a curve as a definite integral, using the procedure of Example 1. We first call attention to two properties of the sine function that came into play along the way.
We obviously used differentiability when we invoked the Mean Value Theorem to replace Δyk րΔxk by sinЈ͑ck ͒ for some ck in the interval ͓xkϪ1, xk ͔. Less obviously, we used the continuity of the derivative of sine in passing from ͚ ͙1ෆෆsinЈ ͑ck ͒͒2 Δx k to the
ෆ ϩ ͑ ෆෆ ෆ ෆ ෆ ෆ
Riemann integral. The requirement for finding the length of a curve by this method, then, is that the function have a continuous first derivative. We call this property smoothness. A function with a continuous first derivative is smooth and its graph is a smooth curve.
Let us review the process, this time with a general smooth function f ͑x͒. Suppose the graph of f begins at the point ͑a, c͒ and ends at ͑b, d͒, as shown in Figure 7.35. We partition the interval a Յ x Յ b into subintervals so short that the arcs of the curve above them are nearly straight. The length of the segment approximating the arc above the subinterval
͓xkϪ1, xk ͔ is ͙Δxk2 ϩෆ. The sum ͚ ͙Δxk2 ϩෆ approximates the length of the curve. We
ෆ Δyk2
ෆ Δyk2 apply the Mean Value Theorem to f on each subinterval to rewrite the sum as a Riemann sum,
͚
͙⌬ ෆෆϩෆෆෆ ϭ
ෆxk2 ෆ ⌬yk2 ϭ y
(b, d)
d
⎯ (⌬xk ) ϩ (⌬yk
√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯)
2
2
Q
y ϭ f(x) c 0
Lϭ
⌬xk
xk – 1
xk
b
x
Figure 7.35 The graph of f, approximated by line segments.
2
k
ෆϩ͑ ෆ
͚ ͙1ෆෆfෆЈ͑cෆ͒͒ෆ ⌬x .
2
k
͵
b
k
͵ Ί
( ) b ͙1ෆෆf Ј͑x͒ෆ dx ϭ
ෆ ϩ ͑ ෆෆ ͒ 2
ෆ
a
(a, c)
a
͚ Ί ⌬x
( )
⌬yk
1 ϩ ᎏᎏ
⌬ xk
For some point ck in (xkϪ1, xk)
Passing to the limit as the norms of the subdivisions go to zero gives the length of the curve as
⌬yk
P
413
a
dy
1 ϩ ᎏᎏ dx 2
dx.
We could as easily have transformed ͚ ͙Δxk2 ϩෆ into a Riemann sum by dividing
ෆ Δyk2 and multiplying by Δyk , giving a formula that involves x as a function of y ͑say, x ϭ g͑y͒͒ on the interval ͓c, d͔:
Lഠ
͙ෆxk ͒ 2ෆෆ⌬yk ͒ 2
͑⌬ ෆෆ ϩ ͑ෆෆෆ
ᎏᎏ ⌬yk ϭ
⌬ yk
͚
ϭ
͚ Ί ⌬y
( )
⌬ xk
1 ϩ ᎏᎏ
⌬ yk
2
k
ෆ ϩ ͑ ෆෆෆෆ
͚ ͙1ෆෆgЈ͑c ͒͒ ෆ ⌬y . k 2
k
For some ck in (ykϪ1, yk)
The limit of these sums, as the norms of the subdivisions go to zero, gives another reasonable way to calculate the curve’s length,
Lϭ
͵
d
͵ Ί
( ) d ͙1ෆෆgЈ͑ y͒ෆ dy ϭ
ෆ ϩ ͑ ෆෆෆ͒ 2
c
c
dx
1 ϩ ᎏᎏ dy 2
dy.
Putting these two formulas together, we have the following definition for the length of a smooth curve.
DEFINITION Arc Length: Length of a Smooth Curve
If a smooth curve begins at ͑a, c͒ and ends at ͑b, d ͒, a Ͻ b, c Ͻ d, then the length
(arc length) of the curve is
͵ Ί
( )
͵ Ί
( ) b Lϭ
a
d
Lϭ
c
dy
1 ϩ ᎏᎏ dx dx
1 ϩ ᎏᎏ dy 2
dx
if y is a smooth function of x on ͓a, b͔;
dy
if x is a smooth function of y on ͓c, d͔.
2
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 414
414
Chapter 7
Applications of Definite Integrals
EXAMPLE 2 Applying the Definition
Find the exact length of the curve
4͙2
ෆ y ϭ ᎏᎏx 3ր2 Ϫ 1
3
SOLUTION
0 Յ x Յ 1.
for
dy
4͙2 3
ෆ
ᎏᎏ ϭ ᎏᎏ • ᎏᎏx 1ր 2 ϭ 2͙2 x 1ր 2,
ෆ
dx
3
2
which is continuous on ͓0, 1͔. Therefore,
͵ Ί
( )
͵ Ί
( )
͵
1
Lϭ
dy
1 ϩ ᎏᎏ dx 0
2
dx
1
ϭ
1ϩ
2͙2 x 1ր 2
ෆ
2
dx
0
1
ϭ
͙1ෆෆx dx
ෆ ϩ 8ෆ
0
]
2 1 ϭ ᎏᎏ • ᎏᎏ ͑1 ϩ 8x͒ 3ր2
3 8
1
0
13 ϭ ᎏᎏ .
6
Now try Exercise 11.
We asked for an exact length in Example 2 to take advantage of the rare opportunity it afforded of taking the antiderivative of an arc length integrand. When you add 1 to the square of the derivative of an arbitrary smooth function and then take the square root of that sum, the result is rarely antidifferentiable by reasonable methods. We know a few more functions that give “nice” integrands, but we are saving those for the exercises.
y
(8, 2) x Vertical Tangents, Corners, and Cusps
(–8, –2)
Figure 7.36 The graph of y ϭ x 1ր 3 has a vertical tangent line at the origin where dyրdx does not exist. (Example 3) x Sometimes a curve has a vertical tangent, corner, or cusp where the derivative we need to work with is undefined. We can sometimes get around such a difficulty in ways illustrated by the following examples.
EXAMPLE 3 A Vertical Tangent
Find the length of the curve y ϭ x1ր3 between ͑Ϫ8, Ϫ2͒ and ͑8, 2͒.
SOLUTION
(2, 8)
The derivative dy 1
1
ᎏᎏ ϭ ᎏᎏxϪ2 ր3 ϭ ᎏᎏ dx 3
3x 2 ր3 y (–2, –8)
is not defined at x ϭ 0. Graphically, there is a vertical tangent at x ϭ 0 where the derivative becomes infinite (Figure 7.36). If we change to x as a function of y, the tangent at the origin will be horizontal (Figure 7.37) and the derivative will be zero instead of undefined. Solving y ϭ x1ր3 for x gives x ϭ y 3, and we have
͵ Ί ͵
( )
2
Figure 7.37 The curve in Figure 7.36 plotted with x as a function of y. The tangent at the origin is now horizontal.
(Example 3)
Lϭ
Ϫ2
dx
1 ϩ ᎏᎏ dy 2
dy ϭ
2
͙1ෆෆ3y 2 ͒ෆ dy ഠ 17.26.
ෆ ϩ ͑ෆෆෆ2
Using NINT
Ϫ2
Now try Exercise 25.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 415
Section 7.4 Lengths of Curves
415
What happens if you fail to notice that dyրdx is undefined at x ϭ 0 and ask your calculator to compute
(
(Ί)
)
2
NINT
1 ϩ ͑1 ր 3͒ xϪ2 ր3 , x, Ϫ8, 8 ?
This actually depends on your calculator. If, in the process of its calculations, it tries to evaluate the function at x ϭ 0, then some sort of domain error will result. If it tries to find convergent Riemann sums near x ϭ 0, it might get into a long, futile loop of computations that you will have to interrupt. Or it might actually produce an answer—in which case you hope it would be sufficiently bizarre for you to realize that it should not be trusted. EXAMPLE 4 Getting Around a Corner
Find the length of the curve y ϭ x 2 Ϫ 4ΗxΗ Ϫ x from x ϭ Ϫ4 to x ϭ 4.
SOLUTION
We should always be alert for abrupt slope changes when absolute value is involved. We graph the function to check (Figure 7.38).
There is clearly a corner at x ϭ 0 where neither dyրdx nor dx րdy can exist. To find the length, we split the curve at x ϭ 0 to write the function without absolute values:
[–5, 5] by [–7, 5]
x 2 Ϫ 4ΗxΗ Ϫ x ϭ
Figure 7.38 The graph of y ϭ x 2 Ϫ 4 Η x Η Ϫ x, Ϫ4 Յ x Յ 4,
ϩ 3x
{ xx Ϫ 5x
2
2
if if x Ͻ 0, x Ն 0.
Then,
has a corner at x ϭ 0 where neither dy/dx nor dx/dy exists. We find the lengths of the two smooth pieces and add them together.
(Example 4)
Lϭ
͵
0
͙1ෆෆ2x ϩෆෆ dx ϩ
ෆ ϩ ͑ෆෆෆ 3͒2
Ϫ4
͵
4
͙1ෆෆ2x Ϫෆෆ dx
ෆ ϩ ͑ෆෆෆ 5͒2
0
ഠ 19.56.
By NINT
Now try Exercise 27.
Finally, cusps are handled the same way corners are: split the curve into smooth pieces and add the lengths of those pieces.
Quick Review 7.4
(For help, go to Sections 1.3 and 3.2.)
In Exercises 1–5, simplify the function.
1.
͙1ෆෆx ϩෆෆ
ෆ ϩ 2 ෆෆ x 2
2.
Ί1ᎏ
Ϫ x ϩ 4ᎏ x2 3. ͙1ෆෆtanෆෆ
ෆ ϩ ͑ෆෆ x͒ 2
on
In Exercises 6–10, identify all values of x for which the function fails to be differentiable.
͓1, 5͔ x ϩ 1
6. f ͑x͒ ϭ Η x Ϫ 4 Η
2Ϫx
ᎏᎏ
2
on
͓Ϫ3, Ϫ1͔
on
͓0, p ր 3͔ sec x
4. ͙1ෆෆx ր4ෆෆր xෆ
ෆ ϩ ͑ ෆෆ Ϫ 1 ෆ͒ 2
5. ͙1ෆෆos 2 x on
ෆ ϩ cෆෆෆෆ
on
x2
7. f ͑x͒ ϭ 5x 2 ր 3
4
0
8. f ͑x͒ ϭ ͙x ϩෆ Ϫ3
ෆෆ 3
5
ϩ4
4x
͓4, 12͔ ᎏᎏ
͓0, p ր 2͔ ͙2 cos x
ෆ
9. f ͑x͒ ϭ ͙x ෆϪෆx ϩෆ 2
ෆ2 ෆ 4 ෆෆ 4
10. f ͑x͒ ϭ 1 ϩ ͙siෆෆ
ෆn x
3
kp, k any integer
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 416
416
Chapter 7
Applications of Definite Integrals
Section 7.4 Exercises
21. Find the length of the curve
In Exercises 1–10,
(a) set up an integral for the length of the curve;
yϭ
(b) graph the curve to see what it looks like;
0ՅyՅp
ෆ Ϫ y2
4. x ϭ ͙1ෆෆෆ, Ϫ1 ր 2 Յ y Յ 1 ր 2
5. y 2 ϩ 2y ϭ 2x ϩ 1,
from ͑Ϫ1, Ϫ1͒ to ͑7, 3͒
6. y ϭ sin x Ϫ x cos x,
7. y ϭ ͐0 tan t dt, x 10. y ϭ
͑e x
ϩ
1
0 Յ x Յ p ր6
Ϫp ր 3 Յ y Յ p ր4
y
9. y ϭ sec x,
y
0ՅxՅp
8. x ϭ ͐0 ͙sec 2ෆෆෆ dt,
ෆෆ t Ϫ 1
1
22. The Length of an Astroid The graph of the equation x 2 ր3 ϩ y 2 ր3 ϭ 1 is one of the family of curves called astroids
(not “asteroids”) because of their starlike appearance (see figure).
Find the length of this particular astroid by finding the length of half the first quadrant portion, y ϭ ͑1 Ϫ x 2ր3 ͒ 3ր2, ͙2 ր4 Յ x Յ 1,
ෆ
and multiplying by 8. 6
Ϫp ր 3 Յ x Յ 0
3. x ϭ sin y,
͙coෆ2t dt
ෆs ෆෆ
from x ϭ 0 to x ϭ p ր4.
Ϫ1 Յ x Յ 2
2. y ϭ tan x,
x
0
(c) use NINT to find the length of the curve.
1. y ϭ x 2,
͵
x 2/3 ϩ y 2/3 ϭ 1
Ϫp ր 3 Յ x Յ p ր 3 eϪx ͒
ր 2, Ϫ3 Յ x Յ 3
–1
0
1
x
In Exercises 11–18, find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative.
11. y ϭ ͑1 ր 3͒͑x 2 ϩ 2͒ 3ր2
–1
from x ϭ 0 to x ϭ 3 12
12. y ϭ x 3ր2 from x ϭ 0 to x ϭ 4 (80͙10 Ϫ 8)/27
ෆ
13. x ϭ ͑y 3ր 3͒ ϩ 1 ր ͑4y͒ from y ϭ 1 to y ϭ 3
[Hint: 1 ϩ ͑dx րdy͒ 2 is a perfect square.] 53/6
14. x ϭ ͑y 4ր4͒ ϩ 1 ր ͑8y 2 ͒ from y ϭ 1 to y ϭ 2
[Hint: 1 ϩ ͑dx րdy͒ 2 is a perfect square.] 123/32
15. x ϭ ͑y 3ր6͒ ϩ 1 ր ͑2y͒ from y ϭ 1 to y ϭ 2
[Hint: 1 ϩ ͑dx րdy͒ 2 is a perfect square.] 17/12
16. y ϭ ͑x 3ր 3͒ ϩ x 2 ϩ x ϩ 1 ր ͑4x ϩ 4͒,
17. x ϭ ͐0 ͙sec 4ෆෆෆ dt,
ෆෆ t Ϫ 1
0 Յ x Յ 2 53/6
Ϫp ր4 Յ y Յ p ր4
y
2
18. y ϭ ͐ ͙3ෆෆෆ dt, Ϫ2 Յ x Յ Ϫ1 7͙3/3
ෆt 4 Ϫ 1
ෆ
23. Fabricating Metal Sheets Your metal fabrication company is bidding for a contract to make sheets of corrugated steel roofing like the one shown here. The cross sections of the corrugated sheets are to conform to the curve
( )
3p y ϭ sin ᎏ ᎏx ,
20
0 Յ x Յ 20 in.
If the roofing is to be stamped from flat sheets by a process that does not stretch the material, how wide should the original material be? Give your answer to two decimal places. Ϸ21.07 inches
Original sheet
y
x
Ϫ2
Corrugated sheet
19. (a) Group Activity Find a curve through the point ͑1, 1͒ whose length integral is y ϭ ͙x
ෆ
͵ Ί
4
Lϭ
1
1
1 ϩ ᎏᎏ dx.
4x
(b) Writing to Learn How many such curves are there? Give reasons for your answer.
20. (a) Group Activity Find a curve through the point ͑0, 1͒ whose length integral is y ϭ 1/(1 Ϫ x)
͵ Ί
2
Lϭ
1
1
1 ϩ ᎏᎏ dy. y4 (b) Writing to Learn How many such curves are there? Give reasons for your answer.
19. (b) Only one. We know the derivative of the function and the value of the function at one value of x.
20 in.
O
3 20 y ϭ sin — x
–
20
x (in.)
24. Tunnel Construction Your engineering firm is bidding for the contract to construct the tunnel shown on the next page. The tunnel is 300 ft long and 50 ft wide at the base. The cross section is shaped like one arch of the curve y ϭ 25 cos ͑p x ր50͒.
Upon completion, the tunnel’s inside surface (excluding the roadway) will be treated with a waterproof sealer that costs
$1.75 per square foot to apply. How much will it cost to apply the sealer? $38,422
20. (b) Only one. We know the derivative of the function and the value of the function at one value of x.
5128_Ch07_pp378-433.qxd 2/3/06 5:59 PM Page 417
30. Because the limit of the sum ͚⌬xk as the norm of the partition goes to zero will always be the length (b Ϫ a) of the interval (a, b). y y ϭ 25 cos ( x/50)
Section 7.4 Lengths of Curves
34. Multiple Choice Which of the following gives the best approximation of the length of the arc of y ϭ cos(2x) from x ϭ 0 to x ϭ p/4? D
(A) 0.785
(B) 0.955
(C) 1.0
(D) 1.318 (E) 1.977
35. Multiple Choice Which of the following expressions gives the length of the graph of x ϭ y3 from y ϭ Ϫ2 to y ϭ 2? C
– 25
0
417
300 ft
͵
͵
͵
2
25 x (ft)
(A)
(B)
Ϫ2
NOT TO SCALE
͙1 ϩ y6 dy
ෆ
Ϫ2
2
(C)
͵
͵
2
(1 ϩ y6) dy
2
͙1 ϩ 9y ෆ dy
ෆ4
(D)
Ϫ2
͙1 ϩ x2 dx
ෆ
Ϫ2
2
(E)
͙1 ϩ x 4 dx
ෆ
Ϫ2
36. Multiple Choice Find the length of the curve described by
2
y ϭ ᎏᎏ x 3/2 from x ϭ 0 to x ϭ 8. B
3
26
52
512͙2
ෆ
(A) ᎏᎏ
(B) ᎏᎏ
(C) ᎏᎏ
3
3
15
512͙2
ෆ
(D) ᎏᎏ ϩ 8
(E) 96
15
In Exercises 25 and 26, find the length of the curve.
25. f ͑x͒ ϭ x 1 ր 3 ϩ x 2 ր 3, 0 Յ x Յ 2 Ϸ3.6142 xϪ1 1
26. f ͑x͒ ϭ ᎏᎏ, Ϫ ᎏᎏ Յ x Յ 1 Ϸ2.1089
4x 2 ϩ 1
2
In Exercises 27–29, find the length of the nonsmooth curve.
37. Multiple Choice Which of the following expressions should be used to find the length of the curve y ϭ x2/3 from x ϭ Ϫ1 to x ϭ 1? A
1
0
(C)
29. y ϭ ͙x from x ϭ 0 to x ϭ 16 Ϸ16.647
ෆ
30. Writing to Learn Explain geometrically why it does not work to use short horizontal line segments to approximate the lengths of small arcs when we search for a Riemann sum that leads to the formula for arc length.
31. Writing to Learn A curve is totally contained inside the square with vertices ͑0, 0͒, ͑1, 0͒, ͑1, 1͒, and ͑0, 1͒. Is there any limit to the possible length of the curve? Explain.
Standardized Test Questions
͙1 ϩ y3 dy
ෆ
Ϫ1
9
1 ϩ ᎏᎏy dy
4
1
(D)
͙1 ϩ y6 dy
ෆ
0
͙1 ϩ y9/4 dy
ෆෆ
0
Exploration
38. Modeling Running Tracks Two lanes of a running track are modeled by the semiellipses as shown. The equation for lane 1 is y ϭ ͙100 Ϫෆෆx 2 , and the equation for lane 2
ෆෆෆ 0.2ෆෆ is y ϭ ͙15ෆϪෆ.2 ෆ2 The starting point for lane 1 is at the
ෆ0 ෆ 0ෆx ෆ. negative x-intercept ͑Ϫ͙50ෆ, 0͒. The finish points for both lanes
ෆ0
are the positive x-intercepts. Where should the starting point be placed on lane 2 so that the two lane lengths will be equal
(running clockwise)? Ϸ(–19.909, 8.410)
You should solve the following problems without using a graphing calculator.
32. True or False If a function y ϭ f (x) is continuous on an interval [a, b], then the length of its curve is given by
͵ Ί
͵
1
(B)
0
1
(E)
4
9
1 ϩ ᎏᎏy dy
4
1
27. y ϭ x 3 ϩ 5 Η x Η from x ϭ Ϫ2 to x ϭ 1 Ϸ13.132
28. ͙x ϩ ͙y ϭ 1 Ϸ1.623
ෆ
ෆ
͵ Ί
͵
͵
(A) 2
y
Start lane 2
͵ Ί
b a
dy 2
1 ϩ ᎏᎏ dx. Justify your answer. dx False. The function must be differentiable.
33. True or False If a function y ϭ f (x) is differentiable on an interval [a, b], then the length of its curve is given by
͵ Ί
b a
dy 2
1 ϩ ᎏᎏ dx. Justify your answer. dx True, by definition.
1
1
31. No. Consider the curve y ϭ ᎏᎏ sin ᎏᎏ ϩ 0.5 for 0 Ͻ x Ͻ 1.
3
x
1
10
Start lane 1
x
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 418
418
Chapter 7
Applications of Definite Integrals
Extending the Ideas
39. Using Tangent Fins to Find Arc Length Assume f is smooth on ͓a, b͔ and partition the interval ͓a, b͔ in the usual way. In each subinterval ͓x kϪ1, x k ͔ construct the tangent fin at the point ͑x kϪ1, f ͑x kϪ1 ͒͒ as shown in the figure.
(a) Show that the length of the kth tangent fin over the interval
͓x kϪ1, x k ͔ equals
͙͑⌬ xk ෆෆෆf Ј͑xෆෆ⌬ xk ෆ.
ෆෆ͒ 2 ϩ ͑ ෆෆkϪ1 ͒ ෆෆ͒ 2
(b) Show that n lim
n→ ∞
Tangent fin with slope f'(xk – 1)
(xk – 1, f (xk – 1))
⌬xk
xk
x
39. (a) The fin is the hypotenuse of a right triangle with leg lengths ⌬xk and df ⌬ x ϭ fЈ(xk–1) ⌬xk.
ᎏᎏ
dx xϭxkϪ1 k
Έ
n
ෆ ϩ ( fЈෆෆ
͚ ͙(⌬ xk)2ෆ(xkϪ1)⌬xk)2 n→∞ kϭ1
(b) lim
n
ϭ lim
ෆЈ(xkϪ1)ෆ
͚ ⌬xk ͙1 ϩ ( f ෆ)2
n→∞ kϭ1
ϭ
͵
b
a
͙1 ϩ ( f ෆ dx
ෆЈ(x))2
͵
b
͙1ෆෆf Ј͑x͒͒ 2 dx,
ෆ ϩ ͑ ෆෆෆෆ
a
kϭ1
which is the length L of the curve y ϭ f ͑x͒ from x ϭ a to x ϭ b.
y ϭ f (x)
xk – 1
͚
(length of kth tangent fin) ϭ
40. Is there a smooth curve y ϭ f ͑x͒ whose length over
ෆ
the interval 0 Յ x Յ a is always a͙2 ? Give reasons for your answer. Yes. Any curve of the form y ϭ Ϯx ϩ c, c a constant. 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 419
Section 7.5 Applications from Science and Statistics
7.5
What you’ll learn about
• Work Revisited
419
Applications from Science and Statistics
Our goal in this section is to hint at the diversity of ways in which the definite integral can be used. The contexts may be new to you, but we will explain what you need to know as we go along.
• Fluid Force and Fluid Pressure
• Normal Probabilities
Work Revisited
. . . and why
It is important to see applications of integrals as various accumulation functions.
4.4 newtons Ϸ 1 lb
Recall from Section 7.1 that work is defined as force (in the direction of motion) times displacement. A familiar example is to move against the force of gravity to lift an object. The object has to move, incidentally, before “work” is done, no matter how tired you get trying.
If the force F(x) is not constant, then the work done in moving an object from x ϭ a to b x ϭ b is the definite integral W ϭ ͐a F(x)dx.
EXAMPLE 1 Finding the Work Done by a Force
(1 newton)(1 meter) ϭ 1 N • m ϭ 1 Joule
Find the work done by the force F(x) ϭ cos(px) newtons along the x-axis from x ϭ 0 meters to x ϭ 1ր2 meter.
SOLUTION
͵
1ր2
Wϭ
cos(px) dx
0
Έ
1/2
1
ϭ ᎏᎏ sin(px) p 0
1
p ϭ ᎏᎏ sin ᎏᎏ Ϫ sin(0) p 2
1
ϭ ᎏᎏ Ϸ 0.318 p
Now try Exercise 1.
EXAMPLE 2 Work Done Lifting
22
(N)
A leaky bucket weighs 22 newtons (N) empty. It is lifted from the ground at a constant rate to a point 20 m above the ground by a rope weighing 0.4 N/m. The bucket starts with 70 N (approximately 7.1 liters) of water, but it leaks at a constant rate and just finishes draining as the bucket reaches the top. Find the amount of work done
(a) lifting the bucket alone;
(b) lifting the water alone;
(c) lifting the rope alone;
(d) lifting the bucket, water, and rope together.
SOLUTION
Work
440
N m
(a) The bucket alone. This is easy because the bucket’s weight is constant. To lift it, you must exert a force of 22 N through the entire 20-meter interval.
.
(m)
20
Figure 7.39 The work done by a constant 22-N force lifting a bucket 20 m is 440 N • m. (Example 2)
Work ϭ ͑22 N͒ ϫ ͑20 m͒ ϭ 440 N • m ϭ 440 J
Figure 7.39 shows the graph of force vs. distance applied. The work corresponds to the area under the force graph. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 420
420
Chapter 7
Applications of Definite Integrals
(b) The water alone. The force needed to lift the water is equal to the water’s weight, which decreases steadily from 70 N to 0 N over the 20-m lift. When the bucket is x m off the ground, the water weighs
y
70
(N)
(
(m)
20
) (
x
Figure 7.40 The force required to lift the water varies with distance but the work still corresponds to the area under the force graph. (Example 2)
original weight of water
proportion left at elevation x
The work done is (Figure 7.40)
͵
͵
Wϭ
b
F͑x͒ dx
a
ϭ
[
20
͑70 Ϫ 3.5x͒ dx ϭ 70x Ϫ 1.75x 2
0
]
20
ϭ 1400 Ϫ 700 ϭ 700 J.
0
(c) The rope alone. The force needed to lift the rope is also variable, starting at
͑0.4͒͑20͒ ϭ 8 N when the bucket is on the ground and ending at 0 N when the bucket and rope are all at the top. As with the leaky bucket, the rate of decrease is constant.
At elevation x meters, the ͑20 Ϫ x͒ meters of rope still there to lift weigh F͑x͒ ϭ ͑0.4͒
͑20 Ϫ x͒ N. Figure 7.41 shows the graph of F. The work done lifting the rope is
8
(N)
͵
Work
20
0
(m)
20
F͑x͒ dx ϭ
͵
20
͑0.4͒͑20 Ϫ x͒ dx
0
[
ϭ 8x Ϫ
Figure 7.41 The work done lifting the rope to the top corresponds to the area of another triangle. (Example 2)
y
10 Ϫ y
)
x
20 Ϫ x
F͑x͒ ϭ 70 ᎏᎏ ϭ 70 1 Ϫ ᎏᎏ ϭ 70 Ϫ 3.5x N.
20
20
F(x) = 70 – 3.5x
y ϭ 2x or x ϭ 1 y
2
10
8
(5, 10)
1y
2
y
Δy
0
5
x
0.2x 2
]
20
ϭ 160 Ϫ 80 ϭ 80 N • m ϭ 80 J.
0
(d) The bucket, water, and rope together. The total work is
440 ϩ 700 ϩ 80 ϭ 1220 J.
Now try Exercise 5.
EXAMPLE 3 Work Done Pumping
The conical tank in Figure 7.42 is filled to within 2 ft of the top with olive oil weighing
57 lb ր ft 3. How much work does it take to pump the oil to the rim of the tank?
SOLUTION
We imagine the oil partitioned into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval ͓0, 8͔. ( The 8 represents the top of the oil, not the top of the tank.)
The typical slab between the planes at y and y ϩ Δy has a volume of about
2
Figure 7.42 The conical tank in
Example 3.
( )
1 p ΔV ϭ p(radius) 2 (thickness) ϭ p ᎏᎏy Δy ϭ ᎏᎏ y 2 Δy ft 3.
2
4
The force F͑y͒ required to lift this slab is equal to its weight,
57p
weight per
Weight ϭ
F͑y͒ ϭ 57 ΔV ϭ ᎏᎏ y 2 Δy lb. unit volume
4
(
) ϫ volume
The distance through which F͑y͒ must act to lift this slab to the level of the rim of the cone is about ͑10 Ϫ y͒ ft, so the work done lifting the slab is about
57p
ΔW ϭ ᎏᎏ ͑10 Ϫ y͒y 2 Δy ft • lb.
4
The work done lifting all the slabs from y ϭ 0 to y ϭ 8 to the rim is approximately
Wഠ
͚ ᎏ4ᎏ ͑10 Ϫ y͒ y
57p
2 Δy
ft • lb. continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 421
Section 7.5 Applications from Science and Statistics
421
This is a Riemann sum for the function ͑57p ր4͒͑10 Ϫ y͒y 2 on the interval from y ϭ 0 to y ϭ 8. The work of pumping the oil to the rim is the limit of these sums as the norms of the partitions go to zero.
Wϭ
͵
8
0
57p
57p
ᎏᎏ ͑10 Ϫ y͒y 2 dy ϭ ᎏᎏ
4
4
[
57p 10 y 3 y4 ϭ ᎏᎏ ᎏᎏ Ϫ ᎏᎏ
4
3
4
]
͵
8
͑10y 2 Ϫ y 3 ͒ dy
0
8
ഠ 30,561 ft • lb
0
Now try Exercise 17.
Fluid Force and Fluid Pressure
We make dams thicker at the bottom than at the top (Figure 7.43) because the pressure against them increases with depth. It is a remarkable fact that the pressure at any point on a dam depends only on how far below the surface the point lies and not on how much water the dam is holding back. In any liquid, the fluid pressure p (force per unit area) at depth h is
Figure 7.43 To withstand the increasing pressure, dams are built thicker toward the bottom. p ϭ wh,
lb lb Dimensions check: ᎏ ᎏ ϭ ᎏ ᎏ ϫ ft, for example ft2 ft3
where w is the weight-density (weight per unit volume) of the liquid.
EXAMPLE 4 The Great Molasses Flood of 1919
Typical Weight-densities (lb/ft3)
Gasoline
Mercury
Milk
Molasses
Seawater
Water
42
849
64.5
100
64
62.4
90 ft
90 ft
1 ft
SHADED BAND NOT TO SCALE
Figure 7.44 The molasses tank of
Example 4.
At 1:00 P.M. on January 15, 1919 (an unseasonably warm day), a 90-ft-high, 90-footdiameter cylindrical metal tank in which the Puritan Distilling Company stored molasses at the corner of Foster and Commercial streets in Boston’s North End exploded. Molasses flooded the streets 30 feet deep, trapping pedestrians and horses, knocking down buildings, and oozing into homes. It was eventually tracked all over town and even made its way into the suburbs via trolley cars and people’s shoes. It took weeks to clean up.
(a) Given that the tank was full of molasses weighing 100 lb ր ft 3, what was the total force exerted by the molasses on the bottom of the tank at the time it ruptured?
(b) What was the total force against the bottom foot-wide band of the tank wall
(Figure 7.44)? continued 5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 422
422
Chapter 7
Applications of Definite Integrals
SOLUTION
(a) At the bottom of the tank, the molasses exerted a constant pressure of
(
)
lb lb p ϭ wh ϭ 100 ᎏᎏ (90 ft) ϭ 9000 ᎏᎏ.
3
ft ft 2
yk
90
Since the area of the base was p͑45͒ 2, the total force on the base was
(
45
Figure 7.45 The 1-ft band at the bottom of the tank wall can be partitioned into thin strips on which the pressure is approximately constant. (Example 4)
)
lb
9000 ᎏᎏ (2025 p ft 2 ) Ϸ 57,225,526 lb. ft 2
(b) We partition the band from depth 89 ft to depth 90 ft into narrower bands of width
Δy and choose a depth yk in each one. The pressure at this depth yk is p ϭ wh ϭ 100 yk lb ր ft 2 (Figure 7.45). The force against each narrow band is approximately pressure ϫ area ϭ ͑100yk ͒͑90p Δy͒ ϭ 9000p yk Δy lb.
Adding the forces against all the bands in the partition and passing to the limit as the norms go to zero, we arrive at
Fϭ
͵
90
9000py dy ϭ 9000p
89
͵
90
y dy Ϸ 2,530,553 lb
89
for the force against the bottom foot of tank wall. y Now try Exercise 25.
Normal Probabilities
1
12
Area
= 1
4
2
5
12
Figure 7.46 The probability that the clock has stopped between 2:00 and 5:00 can be represented as an area of 1 ր4.
The rectangle over the entire interval has area 1.
x
Suppose you find an old clock in the attic. What is the probability that it has stopped somewhere between 2:00 and 5:00?
If you imagine time being measured continuously over a 12-hour interval, it is easy to conclude that the answer is 1 ր4 (since the interval from 2:00 to 5:00 contains one-fourth of the time), and that is correct. Mathematically, however, the situation is not quite that clear because both the 12-hour interval and the 3-hour interval contain an infinite number of times. In what sense does the ratio of one infinity to another infinity equal 1 ր4?
The easiest way to resolve that question is to look at area. We represent the total probability of the 12-hour interval as a rectangle of area 1 sitting above the interval (Figure 7.46).
Not only does it make perfect sense to say that the rectangle over the time interval ͓2, 5͔ has an area that is one-fourth the area of the total rectangle, the area actually equals 1 ր4, since the total rectangle has area 1. That is why mathematicians represent probabilities as areas, and that is where definite integrals enter the picture.
DEFINITION Probability Density Function (pdf)
Improper Integrals
More information about improper ϱ integrals like ͐ f͑x͒ dx can be found in
Ϫϱ
Section 8.3. ( You will not need that information here.)
A probability density function is a function f ͑x͒ with domain all reals such that f ͑x͒ Ն 0 for all x
and
͵
ϱ
f ͑x͒ dx ϭ 1.
Ϫϱ
Then the probability associated with an interval ͓a, b͔ is
͵
b
f ͑x͒ dx.
a
Probabilities of events, such as the clock stopping between 2:00 and 5:00, are integrals of an appropriate pdf.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 423
Section 7.5 Applications from Science and Statistics
423
EXAMPLE 5 Probability of the Clock Stopping
Find the probability that the clock stopped between 2:00 and 5:00.
SOLUTION
The pdf of the clock is f ͑t͒ ϭ
{10,ր12,
0 Յ t Յ 12 otherwise. The probability that the clock stopped at some time t with 2 Յ t Յ 5 is
͵
5
2
a
1 f ͑t͒ dt ϭ ᎏᎏ .
4
Now try Exercise 27.
By far the most useful kind of pdf is the normal kind. (“Normal” here is a technical term, referring to a curve with the shape in Figure 7.47.) The normal curve, often called the “bell curve,” is one of the most significant curves in applied mathematics because it enables us to describe entire populations based on the statistical measurements taken from a reasonably-sized sample. The measurements needed are the mean ͑m͒ and the standard deviation ͑s͒, which your calculators will approximate for you from the data. The symbols on the calculator will probably be J and s (see your Owner’s Manual), but go ahead and use x them as m and s, respectively. Once you have the numbers, you can find the curve by using the following remarkable formula discovered by Karl Friedrich Gauss.
b
∫f(x)dx = .17287148
Figure 7.47 A normal probability density function. The probability associated with the interval ͓a, b͔ is the area under the curve, as shown.
DEFINITION Normal Probability Density Function (pdf)
The normal probability density function (Gaussian curve) for a population with mean m and standard deviation s is
1
2
2
f ͑x͒ ϭ ᎏᎏ eϪ͑xϪm͒ ր͑2s ͒. s͙2p ෆෆ
68% of area
95% of the area
The mean m represents the average value of the variable x. The standard deviation s measures the “scatter” around the mean. For a normal curve, the mean and standard deviation tell you where most of the probability lies. The rule of thumb, illustrated in Figure 7.48, is this:
99.7% of the area
–3 –2 –1
0
1
2
3
Figure 7.48 The 68-95-99.7 rule for normal distributions.
The 68-95-99.7 Rule for Normal Distributions y Given a normal curve,
= 0.5
• 68% of the area will lie within s of the mean m,
• 95% of the area will lie within 2s of the mean m,
• 99.7% of the area will lie within 3s of the mean m.
=1
=2
0
Figure 7.49 Normal pdf curves with mean m ϭ 2 and s ϭ 0.5, 1, and 2.
x
Even with the 68-95-99.7 rule, the area under the curve can spread quite a bit, depending on the size of s. Figure 7.49 shows three normal pdfs with mean m ϭ 2 and standard deviations equal to 0.5, 1, and 2.
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 424
424
Chapter 7
Applications of Definite Integrals
EXAMPLE 6 A Telephone Help Line
Suppose a telephone help line takes a mean of 2 minutes to answer calls. If the standard deviation is s ϭ 0.5, then 68% of the calls are answered in the range of 1.5 to 2.5 minutes and 99.7% of the calls are answered in the range of 0.5 to 3.5 minutes.
Now try Exercise 29.
EXAMPLE 7 Weights of Spinach Boxes
Suppose that frozen spinach boxes marked as “10 ounces” of spinach have a mean weight of 10.3 ounces and a standard deviation of 0.2 ounce.
(a) What percentage of all such spinach boxes can be expected to weigh between 10 and
11 ounces?
(b) What percentage would we expect to weigh less than 10 ounces?
(c) What is the probability that a box weighs exactly 10 ounces?
SOLUTION
Assuming that some person or machine is trying to pack 10 ounces of spinach into these boxes, we expect that most of the weights will be around 10, with probabilities tailing off for boxes being heavier or lighter. We expect, in other words, that a normal pdf will model these probabilities. First, we define f ͑x͒ using the formula:
10.3
1
2
f ͑x͒ ϭ ᎏᎏ eϪ͑xϪ10.3͒ ր͑0.08͒.
0.2͙2p
ෆෆ
The graph (Figure 7.50) has the look we are expecting.
(a) For an arbitrary box of this spinach, the probability that it weighs between 10 and 11 ounces is the area under the curve from 10 to 11, which is
11
[9, 11.5] by [–1, 2.5]
Figure 7.50 The normal pdf for the spinach weights in Example 7. The mean is at the center.
NINT ͑ f ͑x͒, x, 10, 11͒ ഠ 0.933.
So without doing any more measuring, we can predict that about 93.3% of all such spinach boxes will weigh between 10 and 11 ounces.
(b) For the probability that a box weighs less than 10 ounces, we use the entire area under the curve to the left of x ϭ 10. The curve actually approaches the x-axis as an asymptote, but you can see from the graph (Figure 7.50) that f ͑x͒ approaches zero quite quickly. Indeed, f ͑9͒ is only slightly larger than a billionth. So getting the area from 9 to
10 should do it:
NINT ͑ f ͑x͒, x, 9, 10͒ ഠ 0.067.
We would expect only about 6.7% of the boxes to weigh less than 10 ounces.
(c) This would be the integral from 10 to 10, which is zero. This zero probability might seem strange at first, but remember that we are assuming a continuous, unbroken interval of possible spinach weights, and 10 is but one of an infinite number of them.
Now try Exercise 31.
Quick Review 7.5
(For help, go to Section 5.2.)
In Exercises 1–5, find the definite integral by (a) antiderivatives and
(b) using NINT.
1.
͵
1
0
eϪx dx
a. 1 Ϫ (1/e) b. Ϸ0.632
2.
͵
3.
0
2
5.
1
a. ͙2/2
ෆ
sin x dx
pր4
1
e x dx a. e Ϫ 1 b. Ϸ1.718
͵
͵
pր 2
b. Ϸ0.707
4.
͵
3
͑x 2 ϩ 2͒ dx 15
0
x2
ᎏᎏ dx a. (1/3) ln (9/2) b. Ϸ0.501
3ϩ1
x
5128_Ch07_pp378-433.qxd 1/13/06 1:14 PM Page 425
Section 7.5 Applications from Science and Statistics
In Exercises 6–10 find, but do not evaluate, the definite integral that is the limit as the norms of the partitions go to zero of the Riemann sums on the closed interval ͓0, 7͔.
͚ 2p͑x ϩ 2͒͑sin x ͒ Δ x ͵ 2p(x ϩ 2) sin x dx
7. ͚ ͑1 Ϫ x ͒͑2p x ͒Δx ͵ (1 Ϫ x )(2px) dx
7
6.
k
k
8.
9.
0
7
2 k 2
k
0
10.
425
͚ p͑cos x ͒ Δx ͵ p cos x dx
2
k
͚
7
2
0
2
( )
͵ p(y/2)
yk p ᎏᎏ ͑10 Ϫ yk ͒ Δy
2
͙3
ෆ
͚ ᎏ ͑sin
4
2
xk ͒ Δ x
7
2
0
͵ (͙3/4) sin
ෆ
(10 Ϫ y) dy
7
2
x dx
0
Section 7.5 Exercises
In Exercises 1–4, find the work done by the force of F(x) newtons along the x-axis from x ϭ a meters to x ϭ b meters.
1. F͑x͒ ϭ xeϪx ր 3,
a ϭ 0,
b ϭ 5 Ϸ4.4670 J
2. F͑x͒ ϭ x sin ͑px ր4͒, a ϭ 0,
3. F͑x͒ ϭ x ͙9ෆෆෆ,
ෆ Ϫ x2
4. F͑x͒ ϭ
e sin x
a ϭ 0,
cos x ϩ 2,
b ϭ 3 Ϸ3.8473 J bϭ3 9J
a ϭ 0,
b ϭ 10 Ϸ19.5804 J
5. Leaky Bucket The workers in Example 2 changed to a larger bucket that held 50 L (490 N) of water, but the new bucket had an even larger leak so that it too was empty by the time it reached the top. Assuming the water leaked out at a steady rate, how much work was done lifting the water to a point 20 meters above the ground? (Do not include the rope and bucket.) 4900 J
6. Leaky Bucket The bucket in Exercise 5 is hauled up more quickly so that there is still 10 L (98 N) of water left when the bucket reaches the top. How much work is done lifting the water this time? (Do not include the rope and bucket.) 5880 J
(b) How much work does it take to compress the assembly the first half inch? the second half inch? Answer to the nearest inchpound. Ϸ905 in.-lb and Ϸ2714 in.-lb
(Source: Data courtesy of Bombardier, Inc., Mass Transit
Division, for spring assemblies in subway cars delivered to the
New York City Transit Authority from 1985 to 1987.)
10. Bathroom Scale A bathroom scale is compressed 1 ր 16 in. when a 150-lb person stands on it. Assuming the scale behaves like a spring that obeys Hooke’s Law,
(a) how much does someone who compresses the scale
1 ր 8 in. weigh? 300 lb
18.75 in.-lb
(b) how much work is done in compressing the scale 1 ր 8 in.?
11. Hauling a Rope A mountain climber is about to haul up a
50-m length of hanging rope. How much work will it take if the rope weighs 0.624 N ր m? 780 J
12. Compressing Gas Suppose that gas in a circular cylinder of cross section area A is being compressed by a piston
(see figure). y 7. Leaky Sand Bag A bag of sand originally weighing 144 lb was lifted at a constant rate. As it rose, sand leaked out at a constant rate. The sand was half gone by the time the bag had been lifted 18 ft. How much work was done lifting the sand this far? (Neglect the weights of the bag and lifting equipment.) 1944 ft-lb
8. Stretching a Spring A spring has a natural length of 10 in.
An 800-lb force stretches the spring to 14 in.
(a) Find the force constant.
200 lb/in.
(b) How much work is done in stretching the spring from 10 in. to 12 in.? 400 in.-lb
(c) How far beyond its natural length will a 1600-lb force stretch the spring? 8 in.
9. Subway Car Springs It takes a force of 21,714 lb to compress a coil spring assembly on a New York City Transit
Authority subway car from its free height of 8 in. to its fully compressed height of 5 in.
(a) What is the assembly’s force constant?
7238 lb/in.
x
(a) If p is the pressure of the gas in pounds per square inch and
V is the volume in cubic inches, show that the work done in compressing the gas from state ͑ p1, V1 ͒ to state ͑ p 2, V2 ͒ is given by the equation
Work ϭ
͵
͑ p 2, V2 ͒
p dV in. • lb,
͑ p1, V1͒
where the force against the piston is pA.
(b) Find the work done in compressing the gas from
V1 ϭ 243 in3 to V2 ϭ 32 in3 if p1 ϭ 50 lb ր in3 and p and V obey the gas law pV 1.4 ϭ constant (for adiabatic processes). –37,968.75 in.-lb
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 426
426
Chapter 7
Applications of Definite Integrals
Group Activity In Exercises 13–16, the vertical end of a tank containing water (blue shading) weighing 62.4 lb րft 3 has the given shape. (a) Writing to Learn Explain how to approximate the force against the end of the tank by a Riemann sum.
(b) Find the force as an integral and evaluate it.
13. semicircle
(b) 1123.2 lb
14. semiellipse (b) 7987.2 lb
3 ft
(d) The Weight of Water Because of differences in the strength of Earth’s gravitational field, the weight of a cubic foot of water at sea level can vary from as little as 62.26 lb at the equator to as much as 62.59 lb near the poles, a variation of about 0.5%. A cubic foot of water that weighs 62.4 lb in
Melbourne or New York City will weigh 62.5 lb in Juneau or
Stockholm. What are the answers to parts (a) and (b) in a location where water weighs 62.26 lb ր ft 3 ? 62.5 lb ր ft 3 ?
19. Writing to Learn The cylindrical tank shown here is to be filled by pumping water from a lake 15 ft below the bottom of the tank. There are two ways to go about this. One is to pump the water through a hose attached to a valve in the bottom of the tank. The other is to attach the hose to the rim of the tank and let the water pour in. Which way will require less work? Give reasons for your answer.
8 ft
16. parabola (b) Ϸ1506.1 lb
6 ft
3
4 ft
4.5 ft
8 ft
Open top
Through valve:
Ϸ84,687.3 ft-lb
Over the rim:
Ϸ98,801.8 ft-lb
Through a hose attached to a valve in the bottom is faster, because it takes more time to do more work. 2 ft
6 ft
Valve at base
6 ft
17. Pumping Water The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.4 lb ր ft 3.
Grou
nd level 1,500,000 ft-lb, 100 min
18. Emptying a Tank A vertical right cylindrical tank measures
30 ft high and 20 ft in diameter. It is full of kerosene weighing
51.2 lb րft 3. How much work does it take to pump the kerosene to the level of the top of the tank? Ϸ7,238,229 ft-lb
6 ft
15. triangle (b) 3705 lb
17. (d) 1,494,240 ft-lb, Ϸ100 min;
10 ft
12 ft
0
20. Drinking a Milkshake The truncated conical container shown here is full of strawberry milkshake that weighs ͑4 ր 9͒ oz ր in 3.
As you can see, the container is 7 in. deep, 2.5 in. across at the base, and 3.5 in. across at the top (a standard size at Brigham’s in Boston). The straw sticks up an inch above the top. About how much work does it take to drink the milkshake through the straw (neglecting friction)? Answer in inch-ounces.
Ϸ91.3244 in.-oz y y
⌬y
8
8Ϫy
7
20 y y
(a) How much work does it take to empty the tank by pumping the water back to ground level once the tank is full? 1,497,600 ft-lb
(b) If the water is pumped to ground level with a
͑5 ր 11͒-horsepower motor (work output 250 ft • lb րsec), how long will it take to empty the full tank (to the nearest minute)? Ϸ100 min
(c) Show that the pump in part (b) will lower the water level
10 ft (halfway) during the first 25 min of pumping.
y ϭ 14x Ϫ 17.5
(1.75, 7) y ϩ 17.5
14
Δy
0
x
1.25
Dimensions in inches
21. Revisiting Example 3 How much work will it take to pump the oil in Example 3 to a level 3 ft above the cone’s rim?
Ϸ53,482.5 ft-lb
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 427
Section 7.5 Applications from Science and Statistics
427
26. Milk Carton A rectangular milk carton measures 3.75 in.
22. Pumping Milk Suppose the conical tank in Example 3 by 3.75 in. at the base and is 7.75 in. tall. Find the force contains milk weighing 64.5 lb ր ft 3 instead of olive oil. How of the milk ͑weighing 64.5 lb րft 3 ͒ on one side when the much work will it take to pump the contents to the rim? Ϸ34,582.65 ft-lb carton is full. Ϸ4.2 lb
23. Writing to Learn You are in charge of the evacuation and repair of the storage tank shown here. The tank is a hemisphere of radius 10 ft and is full of benzene weighing 56 lb ր ft 3.
y
Outlet pipe
x 2 ϩ y 2 ϭ 100
10
30. Test Scores The mean score on a national aptitude test is 498 with a standard deviation of 100 points.
2 ft
(a) What percentage of the population has scores between 400 and 500? Ϸ0.34 (34%)
0
10
27. Find the probability that a clock stopped between 1:00 and 5:00.
1/3
28. Find the probability that a clock stopped between 3:00 and 6:00.
1/4
29. Suppose a telephone help line takes a mean of 2 minutes to answer calls. If the standard deviation is s ϭ 2, what percentage of the calls are answered in the range of 0 to 4 minutes? 68%
x
(b) If we sample 300 test-takers at random, about how many should have scores above 700? 6.5
z
A firm you contacted says it can empty the tank for 1 ր 2 cent per foot-pound of work. Find the work required to empty the tank by pumping the benzene to an outlet 2 ft above the tank.
If you have budgeted $5000 for the job, can you afford to hire the firm? Ϸ967,611 ft-lb, yes
24. Water Tower Your town has decided to drill a well to increase its water supply. As the town engineer, you have determined that a water tower will be necessary to provide the pressure needed for distribution, and you have designed the system shown here.
The water is to be pumped from a 300-ft well through a vertical
4-in. pipe into the base of a cylindrical tank 20 ft in diameter and 25 ft high. The base of the tank will be 60 ft above ground.
The pump is a 3-hp pump, rated at 1650 ft • lb րsec. To the nearest hour, how long will it take to fill the tank the first time?
(Include the time it takes to fill the pipe.) Assume water weighs
62.4 lb ր ft 3. Ϸ31 hr
31. Heights of Females The mean height of an adult female in
New York City is estimated to be 63.4 inches with a standard deviation of 3.2 inches. What proportion of the adult females in
New York City are
(a) less than 63.4 inches tall?
0.5 (50%)
Ϸ0.24 (24%)
(b) between 63 and 65 inches tall?
(c) taller than 6 feet? Ϸ0.0036 (0.36%)
(d) exactly 5 feet tall?
0 if we assume a continuous distribution;
Ϸ0.071; 7.1% between 59.5 in. and 60.5 in.
32. Writing to Learn Exercises 30 and 31 are subtly different, in that the heights in Exercise 31 are measured continuously and the scores in Exercise 30 are measured discretely. The discrete probabilities determine rectangles above the individual test scores, so that there actually is a nonzero probability of scoring, say, 560. The rectangles would look like the figure below, and would have total area 1.
10 ft
25 ft
Ground
60 ft
Explain why integration gives a good estimate for the probability, even in the discrete case. Integration is a good
4 in.
approximation to the area.
33. Writing to Learn Suppose that f ͑t͒ is the probability density function for the lifetime of a certain type of lightbulb where t is in hours. What is the meaning of the integral
300 ft
Water surface
͵
Submersible pump
800
100
NOT TO SCALE
Standardized Test Questions
25. Fish Tank A rectangular freshwater fish tank with base
2 ϫ 4 ft and height 2 ft (interior dimensions) is filled to within
2 in. of the top.
(a) Find the fluid force against each end of the tank.
f ͑t͒ dt?
The proportion of lightbulbs that last between 100 and 800 hours. Ϸ209.73 lb
(b) Suppose the tank is sealed and stood on end (without spilling) so that one of the square ends is the base. What does that do to the fluid forces on the rectangular sides?
Ϸ838.93 lb; the fluid force doubles
You may use a graphing calculator to solve the following problems. 34. True or False A force is applied to compress a spring several inches. Assume the spring obeys Hooke’s Law. Twice as much work is required to compress the spring the second inch than is required to compress the spring the first inch. Justify your answer. False. Three times as much work is required.
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 428
428
Chapter 7
Applications of Definite Integrals
35. True. The force against each vertical side is 842.4 lb
35. True or False An aquarium contains water weighing 62.4 lb/ft3.
The aquarium is in the shape of a cube where the length of each edge is 3 ft. Each side of the aquarium is engineered to withstand 1000 pounds of force. This should be sufficient to withstand the force from water pressure. Justify your answer.
36. Multiple Choice A force of F(x) ϭ 350x newtons moves a particle along a line from x ϭ 0 m to x ϭ 5 m. Which of the following gives the best approximation of the work done by the force? E
(A) 1750 J
(B) 2187.5 J
(D) 3281.25 J
(C) 2916.67 J
(E) 4375 J
37. Multiple Choice A leaky bag of sand weighs 50 n. It is lifted from the ground at a constant rate, to a height of 20 m above the ground. The sand leaks at a constant rate and just finishes draining as the bag reaches the top. Which of the following gives the work done to lift the sand to the top? (Neglect the bag.) D
(A) 50 J
(B) 100 J (C) 250 J
(D) 500 J
(E) 1000 J
38. Multiple Choice A spring has a natural length of 0.10 m.
A 200-n force stretches the spring to a length of 0.15 m. Which of the following gives the work done in stretching the spring from 0.10 m to 0.15 m? B
(A) 0.05 J
(B) 5 J
(C) 10 J
(D) 200 J
(E) 4000 J
39. Multiple Choice A vertical right cylindrical tank measures
12 ft high and 16 ft in diameter. It is full of water weighing
62.4 lb/ft3. How much work does it take to pump the water to the level of the top of the tank? Round your answer to the nearest ft-lb. E
41. Forcing Electrons Together Two electrons r meters apart repel each other with a force of
23 ϫ 10Ϫ29
F ϭ ᎏᎏ newton. r2 (a) Suppose one electron is held fixed at the point ͑1, 0͒ on the x-axis (units in meters). How much work does it take to move a second electron along the x-axis from the point
͑Ϫ1, 0͒ to the origin? 1.15 ϫ 10Ϫ28 J
(b) Suppose an electron is held fixed at each of the points
͑Ϫ1, 0͒ and ͑1, 0͒. How much work does it take to move a third electron along the x-axis from ͑5, 0͒ to ͑3, 0͒? Ϸ7.6667 ϫ 10Ϫ29 J
42. Kinetic Energy If a variable force of magnitude F͑x͒ moves a body of mass m along the x-axis from x1 to x 2 , the body’s velocity v can be written as dx րdt (where t represents time). Use Newton’s second law of motion, F ϭ m͑dvրdt͒, and the Chain Rule dv dv d x dv See page 429.
ᎏᎏ ϭ ᎏᎏ ᎏᎏ ϭ v ᎏᎏ dt d x dt dx to show that the net work done by the force in moving the body from x1 to x 2 is
Wϭ
͵
x2
x1
1
1
F͑x͒ dx ϭ ᎏᎏ mv22 Ϫ ᎏᎏmv12,
2
2
(1)
where v1 and v2 are the body’s velocities at x1 and x 2 . In physics the expression ͑1 ր 2͒mv 2 is the kinetic energy of the body moving with velocity v. Therefore, the work done by the force equals the change in the body’s kinetic energy, and we can find the work by calculating this change.
(A) 149,490 ft-lb
Weight vs. Mass
(B) 285,696 ft-lb
Weight is the force that results from gravity pulling on a mass. The two are related by the equation in
Newton’s second law,
(C) 360,240 ft-lb
(D) 448,776 ft-lb
weight ϭ mass ϫ acceleration.
(E) 903,331 ft-lb
Thus,
Extending the Ideas
40. Putting a Satellite into Orbit The strength of Earth’s gravitational field varies with the distance r from Earth’s center, and the magnitude of the gravitational force experienced by a satellite of mass m during and after launch is m MG
F͑r͒ ϭ ᎏᎏ . r2 24 kg is Earth’s mass,
Here, M ϭ 5.975 ϫ 10
G ϭ 6.6726 ϫ 10Ϫ11 N • m2 kgϪ2 is the universal gravitational constant, and r is measured in meters. The work it takes to lift a
1000-kg satellite from Earth’s surface to a circular orbit 35,780 km above Earth’s center is therefore given by the integral
Work ϭ
͵
35,780,000
6,370,000
1000 MG
ᎏᎏ dr joules. r2 The lower limit of integration is Earth’s radius in meters at the launch site. Evaluate the integral. (This calculation does not take into account energy spent lifting the launch vehicle or energy spent bringing the satellite to orbit velocity.) 5.1446 ϫ 1010 J
newtons ϭ kilograms ϫ m րsec 2, pounds ϭ slugs ϫ ft րsec 2.
To convert mass to weight, multiply by the acceleration of gravity. To convert weight to mass, divide by the acceleration of gravity.
In Exercises 43–49, use Equation 1 from Exercise 42.
43. Tennis A 2-oz tennis ball was served at 160 ft րsec (about
109 mph). How much work was done on the ball to make it go this fast? 50 ft-lb
44. Baseball How many foot-pounds of work does it take to throw a baseball 90 mph? A baseball weighs 5 oz ϭ 0.3125 lb.
Ϸ85.1 ft-lb
45. Golf A 1.6-oz golf ball is driven off the tee at a speed of
280 ft ր sec (about 191 mph). How many foot-pounds of work are done getting the ball into the air? 122.5 ft-lb
46. Tennis During the match in which Pete Sampras won the 1990
U.S. Open men’s tennis championship, Sampras hit a serve that was clocked at a phenomenal 124 mph. How much work did
Sampras have to do on the 2-oz ball to get it to that speed?
Ϸ64.6 ft-lb
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 429
Section 7.5 Applications from Science and Statistics
47. Football A quarterback threw a 14.5-oz football 88 ft ր sec
(60 mph). How many foot-pounds of work were done on the ball to get it to that speed? Ϸ109.7 ft-lb
48. Softball How much work has to be performed on a
6.5-oz softball to pitch it at 132 ft ր sec (90 mph)? Ϸ110.6 ft-lb x ͵ ϭ͵ dv dv 42. F ϭ mᎏᎏ ϭ mvᎏᎏ, so W ϭ dt dx
2
x1
49. A Ball Bearing A 2-oz steel ball bearing is placed on a vertical spring whose force constant is k ϭ 18 lb րft. The spring is compressed 3 in. and released. About how high does the ball bearing go? (Hint: The kinetic (compression) energy, mgh, of a
1
spring is ᎏ2ᎏks2, where s is the distance the spring is compressed, m is the mass, g is the acceleration of gravity, and h is the height.)
4.5 ft
F(x) dx
x1 x2 dv mvᎏᎏ dx ϭ dx ͵
429
v2
v1
1
1
mv dv ϭ ᎏᎏmv22 Ϫ ᎏᎏmv12
2
2
Quick Quiz for AP* Preparation: Sections 7.4 and 7.5
You should solve the following problems without using a graphing calculator.
1. Multiple Choice The length of a curve from x ϭ 0 to x ϭ 1 is
ෆx6
given by ͐0 ͙1 ϩ 16ෆ dx. If the curve contains the point (1, 4), which of the following could be an equation for this curve? A
1
(A) y ϭ x4 ϩ 3
4. Free Response The front of a fish tank is rectangular in shape and measures 2 ft wide by 1.5 ft tall. The water in the tank exerts pressure on the front of the tank. The pressure at any point on the front of the tank depends only on how far below the surface the point lies and is given by the equation p ϭ 62.4h, where h is depth below the surface measured in feet and p is pressure measured in pounds/ft2.
(B) y ϭ x4 ϩ 1
(C) y ϭ 1 ϩ 16x6
1.5 ft
(D) y ϭ ͙1 ϩ 16ෆ
ෆx6
x7
(E) y ϭ x ϩ ᎏᎏ
7
2. Multiple Choice Which of the following gives the length of
1
the path described by the parametric equations x ϭ ᎏᎏt 4 and
4
y ϭ t3, where 0 Յ t Յ 2? D
͵
͵
͵
͵
͵
h
⌬h
2 ft
The front of the tank can be partitioned into narrow horizontal bands of height Δh. The force exerted by the water on a band at depth hi is approximately
2
(A)
t 6 ϩ 9t 4 dt
0
pressure ؒ area = 62.4hi ؒ 2Δh.
2
(B)
͙t 6 ϩ 1 dt
ෆ
(a) Write the Riemann sum that approximates the force exerted on the entire front of the tank.
0
2
(C)
͙1 ϩ 9t ෆ dt
ෆ4
(b) Use the Riemann sum from part (a) to write and evaluate a definite integral that gives the force exerted on the front of the tank. Include correct units.
0
2
(D)
͙t 6 ϩ 9ෆ dt
ෆt 4
(c) Find the total force exerted on the front of the tank if the front
(and back) are semicircles with diameter 2 ft. Include correct units.
0
2
(E)
͙t3 ϩ 3t2 dt
ෆෆ
2 ft
0
3. Multiple Choice The base of a solid is a circle of radius
2 inches. Each cross section perpendicular to a certain diameter is a square with one side lying in the circle. The volume of the solid in cubic inches is C
128
128p
(A) 16 (B) 16p (C) ᎏᎏ (D) ᎏᎏ (E) 32p
3
3
n
(a) ͚ 62.4hi ؒ 2 ⌬h iϭ1 ͵
1.5
(b)
0
62.4h ؒ 2 dh ϭ 140.4 lbs
͵
1.5
(c)
0
62.4h ؒ 2͙1Ϫh2 dh ϭ 41.6 lbs
ෆ
5128_Ch07_pp378-433.qxd 2/3/06 6:25 PM Page 430
430
Chapter 7
Applications of Definite Integrals
Calculus at Work
I am working toward becoming an archeaoastronomer and ethnoastronomer of Africa. I have a Bachelor’s degree in
Physics, a Master’s degree in Astronomy, and a Ph.D. in Astronomy and Astrophysics. From 1988 to 1990 I was a member of the Peace Corps, and I taught mathematics to high school students in the Fiji Islands. Calculus is a required course in high schools there.
For my Ph.D. dissertation, I investigated the possibility of the birthrate of stars being related to the composition of star formation clouds. I collected data on the absorption of electromagnetic emissions emanating from these regions. The intensity of emissions graphed versus wave-
length produces a flat curve with downward spikes at the characteristic wavelengths of the elements present. An estimate of the area between a spike and the flat curve results in a concentration in molecules/cm3 of an element. This area is the difference in the integrals of the flat and spike curves. In particular, I was looking for a large concentration of water-ice, which increases the probability of planets forming in a region.
Currently, I am applying for two research grants. One will allow me to use the NASA infrared telescope on Mauna Kea to search for C3S2 in comets. The other will help me study the history of astronomy in
Tunisia.
Jarita Holbrook
Los Angeles, CA
Chapter 7 Key Terms arc length (p. 413) area between curves (p. 390)
Cavalieri’s theorems (p. 404) center of mass (p. 389) constant-force formula (p. 384) cylindrical shells (p. 402) displacement (p. 380) fluid force (p. 421) fluid pressure (p. 421) foot-pound (p. 384) force constant (p. 385)
Gaussian curve (p. 423)
Hooke’s Law (p. 385) inflation rate (p. 388) joule (p. 384) length of a curve (p. 413) mean (p. 423) moment (p. 389) net change (p. 379) newton (p. 384) normal curve (p. 423) normal pdf (p. 423) probability density function (pdf) (p. 422)
68-95-99.7 rule (p. 423)
smooth curve (p. 413) smooth function (p. 413) solid of revolution (p. 400) standard deviation (p. 423) surface area (p. 405) total distance traveled (p. 381) universal gravitational constant (p. 428) volume by cylindrical shells (p. 402) volume by slicing (p. 400) volume of a solid (p. 399) weight-density (p. 421) work (p. 384)
Chapter 7 Review Exercises
The collection of exercises marked in red could be used as a chapter test. In Exercises 1–5, the application involves the accumulation of small changes over an interval to give the net change over that entire interval. Set up an integral to model the accumulation and evaluate it to answer the question.
1. A toy car slides down a ramp and coasts to a stop after
5 sec. Its velocity from t ϭ 0 to t ϭ 5 is modeled by v͑t͒ ϭ t 2 Ϫ 0.2t 3 ft ր sec. How far does it travel? Ϸ10.417 ft
2. The fuel consumption of a diesel motor between weekly maintenance periods is modeled by the function c͑t͒ ϭ
4 ϩ 0.001t 4 gal րday, 0 Յ t Յ 7. How many gallons does it consume in a week? Ϸ31.361 gal
3. The number of billboards per mile along a 100-mile stretch of an interstate highway approaching a certain city is modeled by the function B͑x͒ ϭ 21 Ϫ e 0.03x, where x is the distance from the city in miles. About how many billboards are along that stretch of highway? Ϸ1464
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 431
Chapter 7 Review Exercises
431
4. A 2-meter rod has a variable density modeled by the function r͑x͒ ϭ 11 Ϫ 4x g ր m, where x is the distance in meters from the base of the rod. What is the total mass of the rod? 14 g
5. The electrical power consumption (measured in kilowatts) at a factory t hours after midnight during a typical day is modeled by
E͑t͒ ϭ 300͑2 Ϫ cos ͑pt ր12͒͒. How many kilowatt-hours of electrical energy does the company consume in a typical day? 14,400
In Exercises 6–19, find the area of the region enclosed by the lines and curves.
6. y ϭ x, y ϭ 1 ր x 2,
7. y ϭ x ϩ 1,
xϭ2
18. The Bell the region enclosed by the graphs of x2 Ϫ 3
2
y ϭ 31Ϫx and y ϭ ᎏᎏ
10
9
2
y ϭ 3 Ϫ x 2 ᎏᎏ
8. ͙x ϩ ͙y ϭ 1,
ෆ
ෆ
[–2, 2] by [–1.5, 1.5]
1
x ϭ 0, y ϭ 0 1/6
Ϸ5.7312
y
1
⎯
√x ϩ ⎯ y ϭ 1
√
0
1
x
[–4, 4] by [–2, 3.5]
9. x ϭ
x ϭ 0,
2y 2,
10. 4x ϭ
y2
Ϫ 4,
11. y ϭ sin x,
yϭ3
14. y ϭ
sec 2
19. The Kissing Fish the region enclosed between the graphs of y ϭ x sin x and y ϭ Ϫx sin x over the interval ͓Ϫp, p͔ 4p
4x ϭ y ϩ 16 30.375
y ϭ x,
x ϭ p ր4
12. y ϭ 2 sin x, y ϭ sin 2x,
13. y ϭ cos x,
18
yϭ4Ϫ
x2
Ϸ0.0155
0ՅxՅp
4
Ϸ8.9023
x, y ϭ 3 Ϫ Η x Η Ϸ2.1043
15. The Necklace one of the smaller bead-shaped regions enclosed by the graphs of y ϭ 1 ϩ cos x and y ϭ 2 Ϫ cos x 2͙3 Ϫ 2p/3 Ϸ 1.370
ෆ
[–5, 5] by [–3, 3]
20. Find the volume of the solid generated by revolving the region bounded by the x-axis, the curve y ϭ 3x 4, and the lines x ϭ Ϫ1 and x ϭ 1 about the x-axis. 2p
21. Find the volume of the solid generated by revolving the region enclosed by the parabola y 2 ϭ 4x and the line y ϭ x about
(a) the x-axis. 32p/3
[–4, 4] by [–4, 8]
(b) the y-axis.
128p/15
(c) the line x ϭ 4. 64p/5 (d) the line y ϭ 4.
32p/3
22. The section of the parabola y ϭ ր 2 from y ϭ 0 to y ϭ 2 is revolved about the y-axis to form a bowl. x2 16. one of the larger bead-shaped regions enclosed by the curves in
Exercise 15 2͙3 ϩ 4p/3 Ϸ 7.653
ෆ
17. The Bow Tie the region enclosed by the graphs of x y ϭ x 3 Ϫ x and y ϭ ᎏᎏ x2 ϩ 1
(shown in the next column).
Ϸ1.2956
(a) Find the volume of the bowl.
4p
(b) Find how much the bowl is holding when it is filled to a depth of k units ͑0 Ͻ k Ͻ 2͒. pk2
(c) If the bowl is filled at a rate of 2 cubic units per second, how fast is the depth k increasing when k ϭ 1? 1/p
5128_Ch07_pp378-433.qxd 1/13/06 1:15 PM Page 432
432
Chapter 7
Applications of Definite Integrals
23. The profile of a football resembles the ellipse shown here (all dimensions in inches). Find the volume of the football to the nearest cubic inch. 88p Ϸ 276 in3 y 4x 2 y2 —– ϩ — ϭ 1
121
12
– 11
—
2
0
11
—
2
x
UNITS IN INCHES
24. The base of a solid is the region enclosed between the graphs of y ϭ sin x and y ϭ Ϫsin x from x ϭ 0 to x ϭ p. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume of the solid. p2/4
35. No, the work going uphill is positive, but the work going downhill is negative. 34. Stretching a Spring If a force of 80 N is required to hold a spring 0.3 m beyond its unstressed length, how much work does it take to stretch the spring this far? How much work does it take to stretch the spring an additional meter? 12 J, Ϸ213.3 J
35. Writing to Learn It takes a lot more effort to roll a stone up a hill than to roll the stone down the hill, but the weight of the stone and the distance it covers are the same. Does this mean that the same amount of work is done? Explain.
36. Emptying a Bowl A hemispherical bowl with radius 8 inches is filled with punch (weighing 0.04 pound per cubic inch) to within 2 inches of the top. How much work is done emptying the bowl if the contents are pumped just high enough to get over the rim? Ϸ113.097 in.-lb
37. Fluid Force The vertical triangular plate shown below is the end plate of a feeding trough full of hog slop, weighing
80 pounds per cubic foot. What is the force against the plate?
25. The region enclosed by the graphs of y ϭ e x ր 2, y ϭ 1, and x ϭ ln 3 is revolved about the x-axis. Find the volume of the solid generated. p(2 Ϫ ln 3)
2
28. Find the perimeter of the bow-tie-shaped region enclosed between the graphs of y ϭ x 3 Ϫ x and y ϭ x Ϫ x 3. Ϸ5.2454
29. A particle travels at 2 units per second along the curve y ϭ x 3 Ϫ 3x 2 ϩ 2. How long does it take to travel from the local maximum to the local minimum? 2.296 sec
–4
x yϭ –
2
0
26. A round hole of radius ͙3 feet is bored through the center of a 3
ෆ
28p/3 ft Ϸ 29.3215 ft3 sphere of radius 2 feet. Find the volume of the piece cut out.
27. Find the length of the arch of the parabola y ϭ 9 Ϫ x 2 that lies above the x-axis. Ϸ19.4942
Ϸ426.67 lbs
y
4
UNITS IN FEET
38. Fluid Force A standard olive oil can measures 5.75 in. by
3.5 in. by 10 in. Find the fluid force against the base and each side of the can when it is full. (Olive oil has a weight-density base Ϸ 6.6385 lb, of 57 pounds per cubic foot.) front and back:
5.7726 lb, sides Ϸ 9.4835 lb
30. Group Activity One of the following statements is true for all k Ͼ 0 and one is false. Which is which? Explain. (a) is true
(a) The graphs of y ϭ k sin x and y ϭ sin kx have the same length on the interval ͓0, 2p͔.
(b) The graph of y ϭ k sin x is k times as long as the graph of y ϭ sin x on the interval ͓0, 2p͔.
x
OLIVE
OIL
t4 Ϫ 1
31. Let F͑x͒ ϭ ͐ ͙ෆෆෆ dt. Find the exact length of the graph of
1
F from x ϭ 2 to x ϭ 5 without using a calculator. 39 x 32. Rock Climbing A rock climber is about to haul up 100 N
(about 22.5 lb) of equipment that has been hanging beneath her on 40 m of rope weighing 0.8 N ր m. How much work will it take to lift
(a) the equipment? 4000 J
(b) the rope? 640 J
(c) the rope and equipment together?
39. Volume A solid lies between planes perpendicular to the x-axis at x ϭ 0 and at x ϭ 6. The cross sections between the planes are squares whose bases run from the x-axis up to the curve ͙x ϩ ͙y ϭ ͙6 . Find the volume of the solid. Ϸ14.4
ෆ
ෆ
ෆ
y
4640 J
33. Hauling Water You drove an 800-gallon tank truck from the base of Mt. Washington to the summit and discovered on arrival that the tank was only half full. You had started out with a full tank of water, had climbed at a steady rate, and had taken 50 minutes to accomplish the 4750-ft elevation change. Assuming that the water leaked out at a steady rate, how much work was spent in carrying the water to the summit? Water weighs 8 lb րgal.
( Do not count the work done getting you and the truck to the top.) 22,800,000 ft-lb
6
x1/2 ϩ y1/2 ϭ √⎯6
⎯
6
x
5128_Ch07_pp378-433.qxd 01/16/06 12:19 PM Page 433
Chapter 7 Review Exercises
40. Yellow Perch A researcher measures the lengths of 3-year-old yellow perch in a fish hatchery and finds that they have a mean length of 17.2 cm with a standard deviation of 3.4 cm. What proportion of 3-year-old yellow perch raised under similar conditions can be expected to reach a length of 20 cm or more?
Ϸ0.2051 (20.5%)
41. Group Activity Using as large a sample of classmates as possible, measure the span of each person’s fully stretched hand, from the tip of the pinky finger to the tip of the thumb. Based on the mean and standard deviation of your sample, what percentage of students your age would have a finger span of more than 10 inches? Answers will vary.
42. The 68-95-99.7 Rule (a) Verify that for every normal pdf, the proportion of the population lying within one standard deviation of the mean is close to 68%. ͑Hint: Since it is the same for every pdf, you can simplify the function by assuming that m ϭ 0 and s ϭ 1. Then integrate from Ϫ1 to 1.͒
(a) Ϸ0.6827 (68.27%)
(b) Verify the two remaining parts of the rule.
(b) Ϸ0.9545 (95.45%)
43. Writing to Learn Explain why the area under the graph of a probability density function has to equal 1. The probability that the variable has some value in the range of all possible values is 1.
In Exercises 44–48, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis.
44. y ϭ 2x,
y ϭ x ր 2,
xϭ1
45. y ϭ 1 ր x,
y ϭ 0,
x ϭ 1 ր 2,
46. y ϭ sin x,
y ϭ 0,
47. y ϭ x Ϫ 3,
p
xϭ2
0ՅxՅp
3p
2p2
y ϭ x 2 Ϫ 3x 16p/3
48. the bell-shaped region in Exercise 18
Ϸ9.7717
49. Bundt Cake A bundt cake (see Exploration 1, Section 7.3) has a hole of radius 2 inches and an outer radius of 6 inches at the base. It is 5 inches high, and each cross-sectional slice is parabolic. (a) Model a typical slice by finding the equation of the parabola
5
with y-intercept 5 and x-intercepts Ϯ2. y ϭ 5 Ϫ ᎏ4ᎏx2
(b) Revolve the parabolic region about an appropriate line to generate the bundt cake and find its volume. Ϸ335.1032 in3
50. Finding a Function Find a function f that has a continuous derivative on ͑0, ∞͒ and that has both of the following properties. i. The graph of f goes through the point ͑1, 1͒. ii. The length L of the curve from ͑1, 1͒ to any point
͑x, f ͑x͒͒ is given by the formula L ϭ ln x ϩ f ͑x͒ Ϫ 1.
In Exercises 51 and 52, find the area of the surface generated by revolving the curve about the indicated axis.
51. y ϭ tan x,
52. xy ϭ 1,
0 Յ x Յ p ր4;
1 Յ y Յ 2;
x 2 Ϫ 2 ln x ϩ 3
50. f (x) ϭ ᎏᎏ
4
x-axis Ϸ3.84
y-axis Ϸ5.02
433
AP* Examination Preparation
You may use a graphing calculator to solve the following problems. 53. Let R be the region in the first quadrant enclosed by the y-axis and the graphs of y ϭ 2 ϩ sin x and y ϭ sec x.
(a) Find the area of R.
(b) Find the volume of the solid generated when R is revolved about the x-axis.
(c) Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares.
54. The temperature outside a house during a 24-hour period is given by pt F(t) ϭ 80 Ϫ 10 cos ᎏᎏ , 0 Յ t Յ 24,
12
where F(t) is measured in degrees Fahrenheit and t is measured in hours.
(a) Find the average temperature, to the nearest degree Fahrenheit, between t ϭ 6 and t ϭ 14.
(b) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of t was the air conditioner cooling the house?
(c) The cost of cooling the house accumulates at the rate of
$0.05 per hour for each degree the outside temperature exceeds
78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 24-hour period?
55. The rate at which people enter an amusement park on a given day is modeled by the function E defined by
15600
E(t) ϭ ᎏᎏ. t2 Ϫ 24t ϩ 160
The rate at which people leave the same amusement park on the same day is modeled by the function L defined by
9890
L(t) ϭ ᎏᎏ. t2 Ϫ 38t ϩ 370
Both E(t) and L(t) are measured in people per hour, and time t is measured in hours after midnight. These functions are valid for
9 Յ t Յ 23, which are the hours that the park is open. At time t ϭ 9, there are no people in the park.
(a) How many people have entered the park by 5:00 P.M.
(t ϭ 17)? Round your answer to the nearest whole number.
(b) The price of admission to the park is $15 until 5:00 P.M.
(t ϭ 17). After 5:00 P.M., the price of admission to the park is $11.
How many dollars are collected from admissions to the park on the given day? Round your answer to the nearest whole number.
(c) Let H(t) ϭ ͐9(E(x) Ϫ L(x))dx for 9 Յ t Յ 23. The value of
H(17) to the nearest whole number is 3725. Find the value of
HЈ(17) and explain the meaning of H(17) and HЈ(17) in the context of the park. t (d) At what time t, for 9 Յ t Յ 23, does the model predict that the number of people in the park is a maximum?