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