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

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:

$$\color{red}{\mathcal{A}}=2\sqrt{3}+4\left(\frac{2\pi}{3}-\sqrt{3}\right)=\color{red}{\frac{8\pi}{3}-2\sqrt{3}}.$$
