Investigating Ratios of Areas and Volumes
Investigating Ratios of Areas and Volumes
In this portfolio, I will be investigating the ratios of the areas and volumes formed from a curve in the form y = xn between two arbitrary parameters x = a and x = b, such that a < b. This will be done by using integration to find the area under the curve or volume of revolution about an axis. The two areas that will be compared will be labeled ‘A’ and ‘B’ (see figure A). In order to prove or disprove my conjectures, several different values for n will be used, including irrational, real numbers (π, √2). In addition, the values for a and b will be altered to different values to prove or disprove my conjectures. In order to aid in the calculation, a TI-84 Plus calculator will be used, and Microsoft Excel and WolframAlpha (http://www.wolframalpha.com/) will be used to create and display graphs.
Figure A
1. In the first problem, region B is the area under the curve y = x2 and is bounded by x = 0, x = 1, and the x-axis. Region A is the region bounded by the curve, y = 0, y = 1, and the y-axis. In order to find the ratio of the two areas, I first had to calculate the areas of both regions, which is seen below. For region A, I integrated in relation to y, while for region B, I integrated in relation to x. Therefore, the two formulas that I used were y = x2 and x = √y, or x = y1/2.
The ratio of region A to region B was 2:1.
Next, I calculated the ratio for other functions of the type y = xn where n ∈ ℤ+ between x = 0 and x = 1. The first value of n that I tested was 3. Because the formula is y = x3, the inverse of that is x = y1/3.
In this case, the value for n was 3, and the ratio was 3:1 or 3.
I then used 4 for the value of n. In this case, the formula was y = x4 and its inverse was x = y1/4.
For the value n = 4, the ratio was 4:1, or 4.
After I analyzed these 3 values of n and their corresponding ratios of areas, I came up with my first conjecture:
Conjecture 1: For all positive integers n, in
In this portfolio, I will be investigating the ratios of the areas and volumes formed from a curve in the form y = xn between two arbitrary parameters x = a and x = b, such that a < b. This will be done by using integration to find the area under the curve or volume of revolution about an axis. The two areas that will be compared will be labeled ‘A’ and ‘B’ (see figure A). In order to prove or disprove my conjectures, several different values for n will be used, including irrational, real numbers (π, √2). In addition, the values for a and b will be altered to different values to prove or disprove my conjectures. In order to aid in the calculation, a TI-84 Plus calculator will be used, and Microsoft Excel and WolframAlpha (http://www.wolframalpha.com/) will be used to create and display graphs.
Figure A
1. In the first problem, region B is the area under the curve y = x2 and is bounded by x = 0, x = 1, and the x-axis. Region A is the region bounded by the curve, y = 0, y = 1, and the y-axis. In order to find the ratio of the two areas, I first had to calculate the areas of both regions, which is seen below. For region A, I integrated in relation to y, while for region B, I integrated in relation to x. Therefore, the two formulas that I used were y = x2 and x = √y, or x = y1/2.
The ratio of region A to region B was 2:1.
Next, I calculated the ratio for other functions of the type y = xn where n ∈ ℤ+ between x = 0 and x = 1. The first value of n that I tested was 3. Because the formula is y = x3, the inverse of that is x = y1/3.
In this case, the value for n was 3, and the ratio was 3:1 or 3.
I then used 4 for the value of n. In this case, the formula was y = x4 and its inverse was x = y1/4.
For the value n = 4, the ratio was 4:1, or 4.
After I analyzed these 3 values of n and their corresponding ratios of areas, I came up with my first conjecture:
Conjecture 1: For all positive integers n, in