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 desired red region is just the area of a circle with radius $2$ minus the area of the blue region:
$$
\pi(2)^2 - A
= 4\pi - \frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta
= 4\pi - \left( 2\sqrt 3 + \frac{4\pi}{3} \right)
= \frac{8\pi}{3} - 2\sqrt 3
$$
