009B Sample Final 2, Problem 3 - Revision history

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Created page with "<span class="exam">Find the volume of the solid obtained by rotating the region bounded by the curves <math style="vertical-align: -4px">y=x</math> and <math..."

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

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

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!Step 1:  
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|First, we need to find the intersection points of  <math style="vertical-align: -5px">y=x</math>  and  <math style="vertical-align: -5px">y=x^2.</math>
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|To do this, we need to solve
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|       <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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|        <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  <math style="vertical-align: 0px">x=0</math>  and  <math style="vertical-align: 0px">x=1.</math>
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|So, we are interested in the region between  <math style="vertical-align: 0px">x=0</math>  and  <math style="vertical-align: -1px">x=1.</math>
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!Step 2:  
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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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|       <math style="vertical-align: -4px">r_{\text{outer}}=2-x^2</math> 
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|and the inner radius is
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|       <math style="vertical-align: -4px">r_{\text{inner}}=2-x.</math> 
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|Therefore, the volume of the solid is
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|        <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:  
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|Now, we integrate to get
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|        <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:  
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|       <math>\frac{8\pi}{15}</math>
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[[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]