View source for 009B Sample Final 2, Problem 3

<span class="exam">Find the volume of the solid obtained by rotating the region bounded by the curves &nbsp;<math style="vertical-align: -4px">y=x</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=x^2</math>&nbsp; about the line &nbsp;<math>y=2.</math>

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!Foundations: &nbsp; 
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|'''1.''' You can find the intersection points of two functions, say &nbsp; <math style="vertical-align: -5px">f(x),g(x),</math>
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&nbsp; &nbsp; &nbsp; &nbsp; by setting &nbsp;<math style="vertical-align: -5px">f(x)=g(x)</math>&nbsp; and solving for &nbsp;<math style="vertical-align: 0px">x.</math>
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|'''2.''' The volume of a solid obtained by rotating an area around the &nbsp;<math style="vertical-align: 0px">x</math>-axis using the washer method is given by  
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&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -18px">\int \pi(r_{\text{outer}}^2-r_{\text{inner}}^2)~dx,</math>&nbsp; where &nbsp;<math style="vertical-align: -4px">r_{\text{inner}}</math>&nbsp; is the inner radius of the washer and &nbsp;<math style="vertical-align: -4px">r_{\text{outer}}</math>&nbsp; is the outer radius of the washer.
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'''Solution:'''

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!Step 1: &nbsp; 
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|First, we need to find the intersection points of &nbsp;<math style="vertical-align: -5px">y=x</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">y=x^2.</math>
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|To do this, we need to solve 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: 0px">x=x^2.</math>
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|Moving all the terms on one side of the equation, we get 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{0} & = & \displaystyle{x^2-x}\\
&&\\
& = & \displaystyle{x(x-1).}
\end{array}</math>
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|Hence, these two curves intersect at &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=1.</math>
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|So, we are interested in the region between &nbsp;<math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">x=1.</math>
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!Step 2: &nbsp;
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|We use the washer method to calculate this volume. 
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|The outer radius is 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">r_{\text{outer}}=2-x^2</math>&nbsp; 
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|and the inner radius is 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -4px">r_{\text{inner}}=2-x.</math>&nbsp;
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|Therefore, the volume of the solid is 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{V} & = & \displaystyle{\int_0^1 \pi(r_{\text{outer}}^2-r_{\text{inner}}^2)~dx}\\
&&\\
& = & \displaystyle{\int_0^1 \pi((2-x^2)^2-(2-x)^2)~dx.}
\end{array}</math> 
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!Step 3: &nbsp;
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|Now, we integrate to get
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{V} & = & \displaystyle{\pi \int_0^1 ((4-4x^2+x^4)-(4-4x+x^2))~dx}\\
&&\\
& = & \displaystyle{\pi \int_0^1 (4x-5x^2+x^4)~dx}\\
&&\\
& = & \displaystyle{\pi\bigg(2x^2-\frac{5x^3}{3}+\frac{x^5}{5}\bigg)\bigg|_0^1}\\
&&\\
& = & \displaystyle{\pi\bigg(2-\frac{5}{3}+\frac{1}{5}\bigg)-0}\\
&&\\
& = & \displaystyle{\frac{8\pi}{15}.}
\end{array}</math>
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!Final Answer: &nbsp; 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{8\pi}{15}</math>
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[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]