Answer
$\dfrac{8}{3}$
Work Step by Step
We write the region in the polar co-ordinates as: $0 \leq r \leq 2 \sec \theta$ and $\pi/4 \leq \theta \leq \dfrac{\pi}{3}$
and $\iint_{D} f(x,y) \ dA=\int_{\pi/4}^{\frac{\pi}{3}} \int_0^{2 \sec \theta} ( r \sin \theta) (r) \ dr \ d\theta$
Now, we have:
$\int_{\pi/4}^{\frac{\pi}{3}} \int_0^{2 \sec \theta} ( r \sin \theta) (r) \ dr \ d\theta=\int_{\pi/4}^{\frac{\pi}{3}} [\dfrac{r^3}{3}]_0^{2\sec \theta} d\theta$
or. $=\dfrac {1}{3} \times \int_{\pi/4}^{\pi/3} [8 \sec^3 \theta-0] \sin \theta \ d \theta$
or. $=\dfrac{8}{3} \int_{\pi/4}^{\pi/3}[\sec^2 \theta \sec \theta] \sin \theta \ d \theta$
or, $=\dfrac{8}{3} \times [\dfrac{\tan^2 \theta}{2}]_{\pi/4}^{\pi/3}$
or, $=\dfrac{8}{3}$
