009C Sample Final 1, Problem 8

A curve is given in polar coordinates by

$r=1+\sin 2\theta$

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

a) Sketch the curve.

b) Find the area enclosed by the curve.

ExpandFoundations:
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)

ExpandStep 1:
Insert sketch

(b)

ExpandStep 1:
Since the graph has symmetry (as seen in the graph), the area of the curve is
$2\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}{\frac {1}{2}}(1+\sin(2\theta )^{2})~d\theta .$

ExpandStep 2:
Using the double angle formula for $\sin(2\theta ),$ we have
${\begin{array}{rcl}\displaystyle {2\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}{\frac {1}{2}}(1+\sin(2\theta ))^{2}~d\theta }&=&\displaystyle {\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}1+2\sin(2\theta )+\sin ^{2}(2\theta )~d\theta }\\&&\\&=&\displaystyle {\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}1+2\sin(2\theta )+{\frac {1-\cos(4\theta )}{2}}~d\theta }\\&&\\&=&\displaystyle {\int _{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}{\frac {3}{2}}+2\sin(2\theta )-{\frac {\cos(4\theta )}{2}}~d\theta }\\&&\\&=&\displaystyle {{\frac {3}{2}}\theta -\cos(2\theta )-{\frac {\sin(4\theta )}{8}}{\bigg |}_{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}.}\\\end{array}}$

ExpandStep 3:
Lastly, we evaluate to get
${\begin{array}{rcl}\displaystyle {{\frac {3}{2}}\theta -\cos(2\theta )-{\frac {\sin(4\theta )}{8}}{\bigg |}_{-{\frac {\pi }{4}}}^{\frac {3\pi }{4}}}&=&\displaystyle {{\frac {3}{2}}{\bigg (}{\frac {3\pi }{4}}{\bigg )}-\cos {\bigg (}{\frac {3\pi }{2}}{\bigg )}-{\frac {\sin(3\pi )}{8}}-{\bigg [}{\frac {3}{2}}{\bigg (}-{\frac {\pi }{4}}{\bigg )}-\cos {\bigg (}-{\frac {\pi }{2}}{\bigg )}-{\frac {\sin(-\pi )}{8}}{\bigg ]}}\\&&\\&=&\displaystyle {{\frac {9\pi }{8}}+{\frac {3\pi }{8}}}\\&&\\&=&\displaystyle {{\frac {3\pi }{2}}.}\\\end{array}}$

ExpandFinal Answer:
(a) See Step 1 above.
(b) ${\frac {3\pi }{2}}$

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