Answer
$\dfrac{1}{3}-\dfrac{\sqrt 3}{6}$
Work Step by Step
We write the region in the polar co-ordinates as: $0 \leq r \leq 1$ and $\dfrac{\pi}{3} \leq \theta \leq \dfrac{\pi}{2}$
$\iint_{D} f(x,y) \ dA=\int_{\pi/3}^{\frac{\pi}{2}} \int_0^{1} (r \cos \theta) (r) \ dr \ d\theta$
Now, we have:
$\int_{\pi/3}^{\frac{\pi}{2}} \int_0^{1} (r \cos \theta) (r) \ dr \ d\theta=\int_{\pi/3}^{\pi/2} [\dfrac{r^3}{3}]_0^{1} d\theta$
or. $=\int_{\pi/3}^{\pi/2} [\dfrac{1}{3}-0] \cos \theta \ d \theta$
or. $=\dfrac{1}{3}[\sin \theta]_{\pi/3}^{\pi/2}$
or, $=\dfrac{1}{3}[1-\dfrac{\sqrt 3}{2}]$
or, $=\dfrac{1}{3}-\dfrac{\sqrt 3}{6}$
