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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: 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 = = = − 0 = 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 − = Thus the ratio of area A: area B is 2:1.

Middle

1 − = 1:2 = ½

y = x5/3 = = − 0 = 1 − = 5:3 = 5/3

The graphs of these functions are shown below: 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: 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, 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 = 2 0 0:2 = 0 y = x2 = = − 0 = 8 − = 2 y = x2/3 = = 1.9048… − 0 = 1.9048… − 1.9038… = 1.2699… 1.2699…/ 1.9038… = 2/3 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=2     Next, I will consider functions of the type y = xn, 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 = 1 0 0:1 = 0 y = x2 = = − = 7 − = 14:7 = 2 y = x2/3 = = 1.9048… − = 1.3048… ( −1)  − 1.3048…= 0.8699… 0.869…/ 1.3048… = 2/3

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 = = =  The formula for area A can be defined as:  = = =  = Thus the ratio of A: B is n: = = 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 = , y is the function

y = xn

VB = = = = = VA = = = Thus = = Proof for the y-axis:

VA = = = = = = VB = = = =  = = Conclusion

Thus the ratio of area A: area B is for the x-axis and for the y-axis.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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