Need help with this.
$\displaystyle r^2 = x^2 +y^2 $
$\displaystyle y = r\sin \theta $
$\displaystyle x = r \cos \theta $
lets convert the boundaries
$\displaystyle x = \sqrt{4y-y^2 } $
$\displaystyle x^2 = 4y -y^2 \Rightarrow x^2+y^2 = 4y $
$\displaystyle r^2 = 4r\sin \theta $
$\displaystyle r = 4 \sin \theta $
r changes from 0 to $\displaystyle 4\sin \theta $
and theta from 0 to pi/2 the first quarter so
$\displaystyle \int_{0}^{4}\int_{0}^{\sqrt{4y-y^2}} x^2 \; dx dy $
becomes
$\displaystyle \int_{0}^{\frac{\pi}{2}} \int_{0}^{4\sin \theta} r^2 \cos^2 \theta (r) dr d\theta $
$\displaystyle \int_{0}^{\frac{\pi}{2}} \cos^2 \theta \left(\frac{r^4}{4}\right)\mid^{4 \sin \theta}_{0} d\theta $
$\displaystyle \int_{0}^{\frac{\pi}{2}} \cos^2\theta (4^3 \sin^4 \theta ) d\theta $
$\displaystyle 4^3 \int \cos ^2 \theta (\sin ^4 \theta ) d\theta $
$\displaystyle \sin \theta \cos \theta = \frac{\sin 2\theta }{2} $
$\displaystyle \sin ^2 \theta = \frac{1-\cos 2\theta}{2} $
$\displaystyle 4^3 \int \cos ^2 \theta \sin ^2 \theta ( \sin ^2\theta ) d \theta $
$\displaystyle 4^3 \int \frac{\sin ^2 2\theta}{4} \left(\frac{1- \cos 2\theta }{2}\right) d\theta $
$\displaystyle 4^3 \left(\int \frac{\sin ^2 2\theta}{4} d\theta - \int \frac{\sin ^2 2 \theta \cos 2\theta}{8} d\theta\right) $
first integral convert $\displaystyle \sin ^2 2\theta = \frac{1- \cos 4 \theta }{2} $
second one sub $\displaystyle u = \sin 2\theta $
can you continue ?