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 (? , v2). 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 = vy, 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 ? ?+ 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 = ?. The two equations were y = x1/2 and x = y2. For n = ? , the ratio was 1:2, or ?. I also used ? as a value of n. In this case, the two functions were y = x? and x = y1/?. Again, the value of n was ? , and the ratio was ? :1, or ?. 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 ? and irrational numbers such as ? 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, the parameters did not have an effect on the ratio; it remained equal to the value used for n. Conjecture 2: For all positive real numbers n, in the form y = xn, where the graph is between x = a and x = b and a < b, the ratio of region A to region B is equal to n. . In order to prove my second conjecture true, I used values from the general case in order to prove than any values a and b will work. So, instead of specific values, I made the x-parameters from x = a to x = b. By doing this, region A will be the region bounded by y = xn, y = an, y = bn, and the y-axis. Region B is the region enclosed by y = xn, x = a, x = b, and the x-axis. The formulas used were y = xn and x = y1/n. The ratio of region A to region B is n:1, or n. This proves my conjecture correct, because the value for n was equivalent to the ratio of the two regions. . The next part of the portfolio was to determine the ratio of the volumes of revolution of regions A and B when rotated around the x-axis and the y-axis. First, I determined the ratio of the volumes of revolutions when the function is rotated about the x-axis. For the first example, I will integrate from x = 0 to x = 1 with the formula y = x2. In this case, n = 2. When region B is rotated about the x-axis, it can be easily solved with the volume of rotation formula. When region A is rotated about the x-axis, the resulting volume will be bounded by y = 4 and y = x2. The value for n is 2, while the ratio is 4:1. In this case, I was able to figure out the volume of A by subtracting the volume of B from the cylinder formed when the entire section (A and B) is rotated about the x-axis. For the next example, I integrated the function y = x2 from x = 1 to x = 2. In this case, I would have to calculate region A using a different method. By finding the volume of A rotated around the x-axis, I would also find the volume of the portion shown in figure B labeled Q. This is because region A is bounded by y = 4, y = x2, and y = 1. Therefore, I would have to then subtract the volume of region Q rotated around the x-axis in order to get the volume of only region A. In this case, the value for n was 2, and the ratio was 4:1. After this, I decided to try one more example, this time with y = x3 but using the same parameters as the previous problem. So, the value for n is 3 and the parameters are from x = 1 to x = 2. In this case, n was equal to 3, and the ratio was 6:1. In the next example that I did, I chose a non-integer number for n, to determine whether the current pattern of the ratio being two times the value of n was valid. For this one, I chose n = ? with the parameters being from x = 0 to x = 1. In this case, n = ? and the ratio was 2? :1, or 2?. After this, I decided to make a conjecture based on the 4 examples that I had completed. Because I had used multiple variations for the parameters, I have established that they do not play a role in the ratio; only the value for n seems to have an effect. Conjecture 3: For all positive real numbers n, in the form y = xn, where the function is limited from x = a to x = b and a < b, the ratio of region A to region B is equal to two times the value of n. In order to prove this conjecture, I used values from the general case in order to prove than any values a and b will work. So, instead of specific values, I made the x-parameters from x = a to x = b. By doing this, region A will be the region bounded by y = xn, y = an, y = bn, and the y-axis. Region B is the region enclosed by y = xn, x = a, x = b, and the x-axis. In this example, n = n and the ratio was equal to 2n:1. This proves my conjecture that the ratio is two times the value for n. When the two regions are rotated about the x-axis, the ratio is two times the value for n. However, this does not apply to when they are rotated about the y-axis. In order to test that, I did 3 examples, one being the general equation. The first one I did was for y = x2 from x = 1 to x =2. When finding the volume of revolution in terms of the y-axis, it is important to note that the function must be changed into terms of x. Therefore, the function that I will use is x = y1/2. In addition, the y-parameters are from y = 1 to y = 4, because the x values are from 1 to 2. In this example, n = 2 and the ratio was 1:1. The next example that I did was a simpler one, but the value for n was not an integer. Instead, I chose ? , and the x-parameters were from x = 0 to x = 1. The formula used was x = y1/?. In this example, the ratio was ? :2, or ? /2. After doing this example, and using prior knowledge of the regions revolved around the x-axis, I was able to come up with a conjecture for the ratio of regions A and B revolving around the y-axis. Conjecture 4: For all positive real numbers n, in the form y = xn, where the function is limited from x = a to x = b and a < b, the ratio of region A to region B is equal to one half the value of n. In order to prove this conjecture, I used values from the general case in order to prove than any values a and b will work. This is similar to what I did to prove Conjecture 3. So, instead of specific values, I made the x-parameters from x = a to x = b. By doing this, region A will be the region bounded by y = xn, y = an, y = bn, and the y-axis. Region B is the region enclosed by y = xn, x = a, x = b, and the x-axis. The ratio that I got at the end was n:2, which is n/2. Because the value of n is n, this proves that my conjecture is correct. In conclusion, the ratio of the areas formed by region A and region B is equal to the value of n. n can be any positive real number, when it is in the form y = xn. The parameters for this function are x = a and x = b, where a < b. In terms of volumes of revolution, when both regions are revolved around the x-axis, the ratio is two times the value of n, or 2n. However, when both regions A and B are revolved around the y-axis, the ratio is one half the value of n, or n/2. In both situations, n includes the set of all positive real numbers.