Answer
$$4\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\int_0^{\sqrt {4 - {x^2}} } {\left( {{x^2} + {y^2}} \right)} dy} dx \cr
& R{\text{ has the bounds}} \cr
& 0 \leqslant y \leqslant \sqrt {4 - {x^2}} ,{\text{ }} - 2 \leqslant x \leqslant 2 \cr
& {\text{The region }}R{\text{ in polar coordinates is}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant 2,{\text{ 0}} \leqslant \theta \leqslant \pi } \right\} \cr
& {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr
& {\text{So}},{\text{ we have}} \cr
& \int_{ - 2}^2 {\int_0^{\sqrt {4 - {x^2}} } {\left( {{x^2} + {y^2}} \right)} dy} dx = \int_0^\pi {\int_0^2 {\left( {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } \right)} rdr} d\theta \cr
& = \int_0^\pi {\int_0^2 {{r^3}} dr} d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^\pi {\left[ {\frac{{{r^4}}}{4}} \right]_0^2} d\theta \cr
& = \int_0^\pi {\left[ {\frac{{{{\left( 2 \right)}^4}}}{4} - \frac{{{{\left( 0 \right)}^4}}}{4}} \right]} d\theta \cr
& = \int_0^\pi 4 d\theta \cr
& {\text{Integrate}} \cr
& = 4\left[ \theta \right]_0^\pi \cr
& = 4\pi \cr} $$
