polar coordinates
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
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.
The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want $$ \underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region} + \underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1. $$
Best Answer
You are just intersecting two circles with the same radius, going through the center of each other.
The area of a circle sector with radius $R=2$ and amplitude $60^\circ$ is $\frac{1}{6}\pi R^2=\frac{2\pi}{3}$, while the area of an equilateral triangle with side length $2$ is given by $\sqrt{3}$, hence the area of the circle segment by the difference of these objects is $\frac{2\pi}{3}-\sqrt{3}$.
These results are enough to solve your question without integrals:
$$\color{red}{\mathcal{A}}=2\sqrt{3}+4\left(\frac{2\pi}{3}-\sqrt{3}\right)=\color{red}{\frac{8\pi}{3}-2\sqrt{3}}.$$