Answer
$$\frac{{243}}{{10}}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_0^{\sqrt {9 - {x^2}} } {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}} dy} dx \cr
& R{\text{ has the bounds}} \cr
& 0 \leqslant y \leqslant \sqrt {9 - {x^2}} ,{\text{ }}0 \leqslant x \leqslant 3 \cr
& {\text{The region }}R{\text{ in polar coordinates is}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant 3,{\text{ 0}} \leqslant \theta \leqslant \pi /2} \right\} \cr
& {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr
& {\text{So}},{\text{ we have}} \cr
& \int_0^3 {\int_0^{\sqrt {9 - {x^2}} } {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}} dy} dx \cr
& = \int_0^{\pi /2} {\int_0^3 {{{\left( {{r^2}{{\cos }^2}\theta + {r^2}{{\sin }^2}\theta } \right)}^{3/2}}} rdr} d\theta \cr
& = \int_0^{\pi /2} {\int_0^3 {{{\left( {{r^2}} \right)}^{3/2}}} rdr} d\theta \cr
& = \int_0^{\pi /2} {\int_0^3 {{r^4}} dr} d\theta \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^{\pi /2} {\left[ {\frac{{{r^5}}}{5}} \right]_0^3} d\theta \cr
& = \int_0^{\pi /2} {\left[ {\frac{{{{\left( 3 \right)}^5}}}{5} - \frac{{{{\left( 0 \right)}^5}}}{5}} \right]} d\theta \cr
& = \int_0^{\pi /2} {\frac{{243}}{5}} d\theta \cr
& {\text{Integrate}} \cr
& = \frac{{243}}{5}\left[ \theta \right]_0^{\pi /2} \cr
& = \frac{{243}}{5}\left( {\frac{\pi }{2}} \right) \cr
& = \frac{{243}}{{10}}\pi \cr} $$
