MathAdmin: Created page with " Find the volume of the solid obtained by rotating about the $x$ -axis the region bounded by $y=\sqr..."$

2017-04-10T16:53:59Z

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

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<span class="exam"> Find the volume of the solid obtained by rotating about the  <math>x</math>-axis the region bounded by  <math style="vertical-align: -4px">y=\sqrt{1-x^2}</math>  and  <math style="vertical-align: -4px">y=0.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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 a region around the  <math style="vertical-align: 0px">x</math>-axis using disk method is given by
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        <math style="vertical-align: -13px">\int \pi r^2~dx,</math>  where  <math style="vertical-align: 0px">r</math>  is the radius of the disk.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|We start by finding the intersection points of the functions  <math style="vertical-align: -4px">y=\sqrt{1-x^2}</math>  and  <math style="vertical-align: -4px">y=0.</math>
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|We need to solve
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|       <math>0=\sqrt{1-x^2}.</math>
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|If we square both sides, we get
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|       <math>0=1-x^2.</math>
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|The solutions to this equation are  <math style="vertical-align: -1px">x=-1</math>  and  <math style="vertical-align: -1px">x=1.</math>
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|Hence, we are interested in the region between  <math style="vertical-align: -1px">x=-1</math>  and  <math style="vertical-align: -1px">x=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Using the disk method, the radius of each disk is given by
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|       <math style="vertical-align: -3px">r=\sqrt{1-x^2}.</math>
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|Therefore, the volume of the solid is
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|        <math>\begin{array}{rcl}
\displaystyle{V} & = & \displaystyle{\int_{-1}^1 \pi (\sqrt{1-x^2})^2~dx}\\
&&\\
& = & \displaystyle{\int_{-1}^1 \pi (1-x^2)~dx}\\
&&\\
& = & \displaystyle{\pi\bigg(x-\frac{x^3}{3}\bigg)\bigg|_{-1}^1}\\
&&\\
& = & \displaystyle{\pi\bigg(1-\frac{1}{3}\bigg)-\pi\bigg(-1+\frac{1}{3}\bigg)}\\
&&\\
& = & \displaystyle{\frac{4\pi}{3}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|       <math>\frac{4\pi}{3}</math>
|}
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

009B Sample Final 3, Problem 4 - Revision history

MathAdmin: Created page with " Find the volume of the solid obtained by rotating about the x-axis the region bounded by y=\sqr..."

MathAdmin: Created page with " Find the volume of the solid obtained by rotating about the $x$ -axis the region bounded by $y=\sqr..."$