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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> | + | <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>
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| | <span class="exam">(a) Sketch the curve. | | <span class="exam">(a) Sketch the curve. |
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| | <span class="exam">(b) Find the area enclosed by the curve. | | <span class="exam">(b) Find the area enclosed by the curve. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 3, Problem 8 Solution|'''<u>Solution</u>''']] |
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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>
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| − | '''Solution:''' | + | [[009C Sample Final 3, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(a)
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| − | |Insert graph
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |The area of the curve, <math>A</math> is
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{A} & = & \displaystyle{\int_0^{2\pi} \frac{1}{2}r^2~d\theta}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_0^{2\pi} \frac{1}{2} (4+3\sin\theta)^2~d\theta.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !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}
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| − | \displaystyle{A} & = & \displaystyle{\frac{1}{2}\int_0^{2\pi} (16+24\sin\theta+9\sin^2\theta)~d\theta} \\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}\int_0^{2\pi} \bigg(16+24\sin\theta+\frac{9}{2}(1-\cos(2\theta))\bigg)~d\theta}\\
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| − | &&\\
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| − | & = & \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}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}\bigg[\frac{41}{2}\theta-24\cos\theta-\frac{9}{4}\sin(2\theta)\bigg]\bigg|_0^{2\pi}.}\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |Lastly, we evaluate to get
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| − | <math>\begin{array}{rcl}
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| − | \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]}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}(41\pi-24)-\frac{1}{2}(-24)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{41\pi}{2}.}\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' See above
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| − | | '''(b)''' <math>\frac{41\pi}{2}</math>
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| − | |}
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| | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
