Answer
$$\displaystyle{\int_{\pi/6}^{5\pi/6} \int_{2}^{4\sin(\theta)} f(r,\theta) \mathrm{d}r\ \mathrm{d}\theta}$$
Work Step by Step
First let's solve $4\sin(\theta) =2$
$\implies \sin(\theta) =1/2$$ \implies \theta = \pi/6 $ or $5\pi/6$.
$\therefore$ The region is bounded above by $r = 4\sin(\theta)$ and bounded below by $r=2$ and is in between the lines $\theta = \displaystyle{\frac{\pi}{6}}$ & $\theta = \displaystyle{\frac{5\pi}{6}}$. So the answer is $$\displaystyle{\int_{\pi/6}^{5\pi/6} \int_{2}^{4\sin(\theta)} f(r,\theta) \mathrm{d}r\ \mathrm{d}\theta}$$
