009C Sample Final 3, Problem 8

A curve is given in polar coordinates by  $r=4+3\sin \theta$

$0\leq \theta \leq 2\pi$

(a) Sketch the curve.

(b) Find the area enclosed by the curve.

Foundations:
The area under a polar curve   $r=f(\theta )$  is given by
$\int _{\alpha _{1}}^{\alpha _{2}}{\frac {1}{2}}r^{2}~d\theta$  for appropriate values of  $\alpha _{1},\alpha _{2}.$

Solution:

(a)
Insert graph

(b)

Step 1:
The area enclosed by the curve,  $A$  is
${\begin{array}{rcl}\displaystyle {A}&=&\displaystyle {\int _{0}^{2\pi }{\frac {1}{2}}r^{2}~d\theta }\\&&\\&=&\displaystyle {\int _{0}^{2\pi }{\frac {1}{2}}(4+3\sin \theta )^{2}~d\theta .}\end{array}}$

Step 2:
Using the double angle formula for  $\cos(2\theta ),$  we have
${\begin{array}{rcl}\displaystyle {A}&=&\displaystyle {{\frac {1}{2}}\int _{0}^{2\pi }(16+24\sin \theta +9\sin ^{2}\theta )~d\theta }\\&&\\&=&\displaystyle {{\frac {1}{2}}\int _{0}^{2\pi }{\bigg (}16+24\sin \theta +{\frac {9}{2}}(1-\cos(2\theta )){\bigg )}~d\theta }\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg [}16\theta -24\cos \theta +{\frac {9}{2}}\theta -{\frac {9}{4}}\sin(2\theta ){\bigg ]}{\bigg |}_{0}^{2\pi }}\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg [}{\frac {41}{2}}\theta -24\cos \theta -{\frac {9}{4}}\sin(2\theta ){\bigg ]}{\bigg |}_{0}^{2\pi }.}\\\end{array}}$

Step 3:
Lastly, we evaluate to get
${\begin{array}{rcl}\displaystyle {A}&=&\displaystyle {{\frac {1}{2}}{\bigg [}{\frac {41}{2}}(2\pi )-24\cos(2\pi )-{\frac {9}{4}}\sin(4\pi ){\bigg ]}-{\frac {1}{2}}{\bigg [}{\frac {41}{2}}(0)-24\cos(0)-{\frac {9}{4}}\sin(0){\bigg ]}}\\&&\\&=&\displaystyle {{\frac {1}{2}}(41\pi -24)-{\frac {1}{2}}(-24)}\\&&\\&=&\displaystyle {{\frac {41\pi }{2}}.}\\\end{array}}$

Final Answer:
(a)     See above
(b)     ${\frac {41\pi }{2}}$

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