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