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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 1, 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 1, 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 sketch
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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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| − | |Since the graph has symmetry (as seen in the graph), the area of the curve is
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| − | <math>2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta)^2)~d\theta.</math>
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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">\sin(2\theta),</math> we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta))^2~d\theta} & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}1+2\sin(2\theta)+\sin^2(2\theta)~d\theta} \\
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| − | &&\\
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| − | & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}1+2\sin(2\theta)+\frac{1-\cos(4\theta)}{2}~d\theta}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{3}{2}+2\sin(2\theta)-\frac{\cos(4\theta)}{2}~d\theta}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3}{2}\theta-\cos(2\theta)-\frac{\sin(4\theta)}{8}\bigg|_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}.}\\
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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{\frac{3}{2}\theta-\cos(2\theta)-\frac{\sin(4\theta)}{8}\bigg|_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}} & = &\displaystyle{\frac{3}{2}\bigg(\frac{3\pi}{4}\bigg)-\cos\bigg(\frac{3\pi}{2}\bigg)-\frac{\sin(3\pi)}{8}-\bigg[\frac{3}{2}\bigg(-\frac{\pi}{4}\bigg)-\cos\bigg(-\frac{\pi}{2}\bigg)-\frac{\sin(-\pi)}{8}\bigg]}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{9\pi}{8}+\frac{3\pi}{8}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3\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 Step 1 above.
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| − | | '''(b)''' <math>\frac{3\pi}{2}</math>
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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
