009C Sample Final 3, Problem 8 - Revision history

MathAdmin: Created page with "A curve is given in polar coordinates by $r=4+3\sin \theta$ :: $0\leq \theta \leq 2\pi$
2017-03-19T19:00:53Z

Created page with "<span class="exam">A curve is given in polar coordinates by <math style="vertical-align: -2px">r=4+3\sin \theta</math> ::<math>0\leq \theta \leq 2\pi</math> <span class..."

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<span class="exam">A curve is given in polar coordinates by  <math style="vertical-align: -2px">r=4+3\sin \theta</math>
::<math>0\leq \theta \leq 2\pi</math>

<span class="exam">(a) Sketch the curve.

<span class="exam">(b) Find the area enclosed by the curve.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|The area under a polar curve   <math style="vertical-align: -5px">r=f(\theta)</math>  is given by
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       <math>\int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta</math>  for appropriate values of  <math>\alpha_1,\alpha_2.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a)  
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| 
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|[[File:9CSF3_8.jpg |center|400px]]
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'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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|The area enclosed by the curve,  <math style="vertical-align: 0px">A</math>  is
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        <math>\begin{array}{rcl}
\displaystyle{A} & = & \displaystyle{\int_0^{2\pi} \frac{1}{2}r^2~d\theta}\\
&&\\
& = & \displaystyle{\int_0^{2\pi} \frac{1}{2} (4+3\sin\theta)^2~d\theta.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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|Using the double angle formula for  <math style="vertical-align: -5px">\cos(2\theta),</math>  we have
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       <math>\begin{array}{rcl}
\displaystyle{A} & = & \displaystyle{\frac{1}{2}\int_0^{2\pi} (16+24\sin\theta+9\sin^2\theta)~d\theta} \\
&&\\
& = & \displaystyle{\frac{1}{2}\int_0^{2\pi} \bigg(16+24\sin\theta+\frac{9}{2}(1-\cos(2\theta))\bigg)~d\theta}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg[16\theta-24\cos\theta+\frac{9}{2}\theta-\frac{9}{4}\sin(2\theta)\bigg]\bigg|_0^{2\pi}}\\
&&\\
& = & \displaystyle{\frac{1}{2}\bigg[\frac{41}{2}\theta-24\cos\theta-\frac{9}{4}\sin(2\theta)\bigg]\bigg|_0^{2\pi}.}\\
\end{array}</math>
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|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
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|Lastly, we evaluate to get
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       <math>\begin{array}{rcl}
\displaystyle{A} & = &\displaystyle{\frac{1}{2}\bigg[\frac{41}{2}(2\pi)-24\cos(2\pi)-\frac{9}{4}\sin(4\pi)\bigg]-\frac{1}{2}\bigg[\frac{41}{2}(0)-24\cos(0)-\frac{9}{4}\sin(0)\bigg]}\\
&&\\
& = & \displaystyle{\frac{1}{2}(41\pi-24)-\frac{1}{2}(-24)}\\
&&\\
& = & \displaystyle{\frac{41\pi}{2}.}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|    '''(a)'''     See above
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|    '''(b)'''     <math>\frac{41\pi}{2}</math>
|}
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]