Investigating Ratios of Areas and Volumes

Topics: Real number, Number, Mathematics Pages: 5 (2092 words) Published: November 3, 2011
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 ( 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 the form y = xn, where the graph is between x = 0 and x = 1, the ratio of region A to region B is equal to n. In order to test this conjecture further, I used other numbers that were not necessarily integers as n and placed them in the function y = xn. In this case, I used n = ½. The two equations were y = x1/2 and x = y2.

For n = ½, the ratio was 1:2, or ½.
I also used π as a value of n. In this case, the two functions were y = xπ and x = y1/π.

Again, the value of n was π, and the ratio was π:1, or π. As a result, I concluded that Conjecture 1 was true for all positive real numbers n, in the form y = xn, between x = 0 and x = 1.

2. After proving that Conjecture 1 was true, I used other parameters to check if my conjecture was only true for x = 0 to x = 1, or if it could be applied to all possible parameters. First, I tested the formula y = xn for all positive real numbers n from x = 0 to x = 2. My first value for n was 2. The two formulas used were y = x2 and x = y1/2. In this case, the parameters were from x = 0 to x = 2, but the y parameters were from y = 0 to y = 4, because 02 = 0 and 22 = 4.

In this case, n was 2, and the ratio was 2:1, or 2.
I also tested a different value for n, 3, with the same x-parameters. The two formulas were y = x3 and x = y1/3. The y-parameters were y = 0 to y = 8.

Again, the n value, 3, was the same as the ratio, 3:1.
In order to test the conjecture further, I decided to use different values for the x-parameters, from x = 1 to x = 2. Using the general formula y = xn, I used 2 for the n value.

Again, the ratio was equal to the n value. After testing the conjecture multiple times with different parameters, I decided to update my conjecture to reflect my findings. The n value did not necessarily have to be an integer; using fractions such as ½ and irrational numbers such as π did not affect the outcome. Regardless of the value for n, as long as it was positive, the ratio was always equal to n. In addition,...
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