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.

Foundations:
Area under a polar curve

Solution:

(a)

Step 1:
Insert sketch

(b)

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

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

Step 3:

Final Answer:
(a)
(b)

Return to Sample Exam