Find the area of the region inside
r = 11 sinθ but outside r = 1.
Please help!
Setting them equal to get the points of intersection we get
$\displaystyle 11\sin(\theta)=1 \iff \sin(\theta)=\frac{1}{11} \iff \theta =\sin^{-1}\left( \frac{1}{11}\right)$
This is the reference angle in the second quadrant so the upper limit of integration is $\displaystyle \pi-\sin(1/11)$
$\displaystyle A=\frac{1}{2}\int_{\sin^{-1}(1/11)}^{\pi-\sin^{-1}(1/11)}((11\sin(\theta))^2-1^2)d\theta$