Find the area of the region inside: $\displaystyle r = 8 \sin \theta$ but outside $\displaystyle r = 4 $
First find the points of intersection.
So $\displaystyle 8 \sin \theta = 4, \ \sin \theta = \frac{1}{2} $ or $\displaystyle \theta = \frac{\pi}{6}, \ \frac{5 \pi}{6} $.
So $\displaystyle A = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} (8 \sin \theta)^2 \ d \theta - \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} 16 \ d \theta $.