Hello, clev88!
Find the area of the region that is outside $\displaystyle r \:=\:2$ and inside $\displaystyle r^2\:=\: 8\sin2\theta$ $\displaystyle r = 2$ is a circle with radius 2.
$\displaystyle r^2 = 8\sin2\theta$ is a lemniscate ("figure 8").
The graph looks like this . . . Code:

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Due to the symmetry, we can find the area in quandrant 1 and multiply by 4.
Intersection: .$\displaystyle \begin{Bmatrix} r^2 \:=\:4 \\ r^2 \:=\:8\sin2\theta \end{Bmatrix}$
. . $\displaystyle 8\sin2\theta \:=\:4 \quad\Rightarrow\quad \sin2\theta \:=\:\frac{1}{2} \quad\Rightarrow\quad 2\theta \:=\:\frac{\pi}{6} \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{12}$
Then: .$\displaystyle A \;=\;4 \times \tfrac{1}{2}\int^{\frac{\pi}{12}}_0\bigg[8\sin2\theta  4\bigg]\,d\theta $
Got it?