Area between $r=4\sin(\theta)$ and $r=2$ I'm trying to find the area between $r=4\sin(\theta)$ and $r=2$.

I found the points of intersections to be $\pi/6,5\pi/6$. Which implies the area is $$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta.$$
Is this correct? Or did I find the area for the following region

The area that you have calculated is for the blue region, as you suspect. Thus, the area for the red would be that of the upper circle ($\pi\cdot2^2$) minus the blue area. See if you can express that directly as a polar integral.
