Answer
$$2\pi $$
Work Step by Step
$$\eqalign{
& \int_0^4 {\int_0^{\sqrt {4y - {y^2}} } {{x^2}} dxdy} \cr
& R{\text{ has the bounds}} \cr
& 0 \leqslant x \leqslant \sqrt {4y - {y^2}} ,{\text{ }}0 \leqslant y \leqslant 4 \cr
& x = \sqrt {4y - {y^2}} \cr
& {x^2} = 4y - {y^2} \cr
& {x^2} + {y^2} = 4y \cr
& {r^2}\cos \theta + {r^2}\sin \theta = 4\sin \theta \cr
& {r^2} = 4r\sin \theta \cr
& r = 4\sin \theta \cr
& {\text{The region }}R{\text{ in polar coordinates is}} \cr
& R = \left\{ {\left( {x,y} \right):0 \leqslant r \leqslant 4\sin \theta ,{\text{ }}0 \leqslant \theta \leqslant \frac{\pi }{2}} \right\} \cr
& {\text{Where: }}x = r\cos \theta ,{\text{ }}y = r\sin \theta {\text{ and }}dydx = rdrd\theta \cr
& {\text{So}},{\text{ we have}} \cr
& \int_0^4 {\int_0^{\sqrt {4y - {y^2}} } {{x^2}} dxdy} = \int_0^{\pi /2} {\int_0^{4\sin \theta } {{{\left( {r\cos \theta } \right)}^2}rdrd\theta } } \cr
& = \int_0^{\pi /2} {\int_0^{4\sin \theta } {{r^2}{{\cos }^2}\theta rdrd\theta } } \cr
& = \int_0^{\pi /2} {\int_0^{4\sin \theta } {{r^3}{{\cos }^2}\theta drd\theta } } \cr
& {\text{Integrate with respect to }}r \cr
& = \int_0^{\pi /2} {\left[ {\frac{{{r^4}}}{4}{{\cos }^2}\theta } \right]_0^{4\sin \theta }d\theta } \cr
& = \int_0^{\pi /2} {\left[ {\frac{{{{\left( {4\sin \theta } \right)}^4}}}{4}{{\cos }^2}\theta - \frac{{{{\left( 0 \right)}^4}}}{4}{{\cos }^2}\theta } \right]d\theta } \cr
& = \int_0^{\pi /2} {64{{\sin }^4}\theta {{\cos }^2}\theta d\theta } \cr
& {\text{Integrate by a graphing calculator}} \cr
& = 2\pi \cr} $$
