Answer
$\dfrac{125}{6}$
Work Step by Step
We write the region in the polar co-ordinates as: $0 \leq r \leq 5 \csc \theta$ and $\dfrac{\pi}{4} \leq \theta \leq \dfrac{\pi}{2}$
$\iint_{D} f(x,y) \ dA=\int_{\pi/4}^{\frac{\pi}{2}} \int_0^{5 \csc \theta} ( r \cos \theta) (r) \ dr \ d\theta$
Now, we have:
$\int_{\pi/4}^{\frac{\pi}{2}} \int_0^{5 \csc \theta} ( r \cos \theta) (r) \ dr \ d\theta =\int_{\pi/4}^{\pi/2} [\dfrac{r^3}{3}]_0^{5 \csc \theta} \cos \theta d\theta$
or. $=\dfrac{125}{3}\int_{\pi/4}^{\pi/2} [\csc^3 \theta-0] \cos \theta \ d \theta$
or. $=\dfrac{125}{3}[-\dfrac{\cot^2 \theta}{2}]_{\pi/4}^{\pi/2}$
or, $=\dfrac{125}{6} \times [0-(-1)]$
or, $=\dfrac{125}{6}$
