Math Help - Polar Curves - Enclosed Area

1. Polar Curves - Enclosed Area

Find the area of the region inside
r = 11 sinθ but outside r = 1.

2. Originally Posted by Del
Find the area of the region inside
r = 11 sinθ but outside r = 1.

Setting them equal to get the points of intersection we get

$11\sin(\theta)=1 \iff \sin(\theta)=\frac{1}{11} \iff \theta =\sin^{-1}\left( \frac{1}{11}\right)$

This is the reference angle in the second quadrant so the upper limit of integration is $\pi-\sin(1/11)$

$A=\frac{1}{2}\int_{\sin^{-1}(1/11)}^{\pi-\sin^{-1}(1/11)}((11\sin(\theta))^2-1^2)d\theta$