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