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Investigating Ratios of Areas and Volumes

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Introduction

Introduction

        In this portfolio, I will investigate the ratio of areas formed when y = xn is graphed between two arbitrary parameters x = a and x = b such that a < b.

Ratio of Area

        Given the function y = x2, I will first consider the region formed by this function from

x = 0 to x = 1 and the x-axis. The region will be labeled B. The region from y = 0 to y = 1 and the y-axis will be labeled A. A diagram of this is show below:

image04.png

Finding the ratio of area A: area B:

By integrating this function from x = 0 to x = 1, we can find the area under the curve, area B.

B = image05.png = image39.png = image48.png − 0 = image48.png

Since the total area of area A and B is 1, we can simply subtract the area of B from 1 to get area A.

A = 1 − image48.png = image58.png

Thus the ratio of area A: area B is 2:1.


...read more.

Middle

1 − image58.png = image48.png

1:2 = ½

y = x5/3

image59.png = image60.png = image61.png − 0 = image61.png

1 − image61.png = image62.png

5:3 = 5/3

The graphs of these functions are shown below:

image63.png

One important subset of real numbers has been excluded from further consideration is negative numbers. This is because no corresponding y-value exists at x = 0. There is a vertical asymptote at x = 0 and a horizontal asymptote at y = 0, as they approach infinity. Thus y = x-ncannot be integrated as infinity is not a valid area.  The graph below demonstrates them phenomena for

y = x-2:

image64.png

To further test the conjecture, I will now consider the areas between other x-values. I will examine if the conjecture holds for x = 0 and x = 2, x = 1 and x = 2, and x = 2 and x = 5.

First, I will consider functions of the type y = xn, image54.png between x = 0 and x = 2. The conjecture seems to hold for the bounds x = 0 and x = 2.

Function

Area of B (under the curve)

Area of A

Ratio

y = x0

image65.png = 2

0

0:2 = 0

y = x2

image66.png = image67.png = image68.png − 0 = image68.png

8 − image68.png = image69.png

2

y = x2/3

image70.png= image71.png= 1.9048… − 0 = 1.9048…

image72.png − 1.9038… = 1.2699…

1.2699…/ 1.9038… = 2/3

image73.png

Area of A and B is now defined as the following:

Area A: y = xn, y = an, y = bn, and the y-axis

Area B: y =x, x = a, y = b, and the x-axis

The graph below is an example of the bounds for y = x2 from x=1 to x=2image74.pngimage02.pngimage03.pngimage00.pngimage01.png

Next, I will consider functions of the type y = xn, image54.png between x = 1 and x =2. The conjecture seems to hold for the bounds x = 1 and x = 2.

Function

Area of B (under the curve)

Area of A

Ratio

y = x0

image75.png = 1

0

0:1 = 0

y = x2

image76.png = image77.png = image68.png − image48.png = image78.png

7 − image78.png = image79.png

14:7 = 2

y = x2/3

image80.png=image81.png= 1.9048… − image82.png = 1.3048…

(image72.png−1)  − 1.3048…= 0.8699…

0.869…/ 1.3048… = 2/3

...read more.

Conclusion

y = xn from x = a to x = b such that a < b and for the regions defined below:

Area A: y = xn, y = an, y = bn, and the y-axis

Area B: y =x, x = a, y = b, and the x-axis

Proof:

B = image92.png = image93.png = image51.pngimage94.png

The formula for area A can be defined as: image95.png

image96.png = image97.png= image98.png=  image99.pngimage07.png= image08.png


Thus the ratio of A: B is n:

image09.png=image10.png= n

Ratios of Volume

After considering and proving that the ratio of the area A: area B is simply n, I will now consider a general formula for the ratios of the volumes of revolution generated by the regions A and B when they are rotated about the x-axis and y-axis.

Proof for x-axis conjecture:

I will use the disc method to solve for volumes of revolution: V = image11.png, y is the function

y = xn

VB = image11.png= image12.png= image13.png= image13.png=image14.png

VA = image15.png= image16.png=image17.png

Thus image18.png=image19.png= image20.png

Proof for the y-axis:

VA = image21.png=image22.png=image23.png=image24.png=image25.png =image26.png

VB = image27.png=image28.png=image29.png= image30.png

image31.png=image32.png =image33.png

Conclusion

Thus the ratio of area A: area B is image20.pngfor the x-axis and image33.pngfor the y-axis.

...read more.

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