Answer
$$\frac{32}{9}$$
Work Step by Step
$\sqrt{x^{2}+y^{2}}=z$
$z=\sqrt{r^{2}}=r$
$2y=x^{2}+y^{2}$
$r^{2}=2 r \sin \theta$
$r(r-2 \sin \theta)=0$
$r=0$ or, $r=2 \sin \theta$
Thus, the integral is:
$\int_{0}^{\pi} \int_{0}^{2 \sin \theta} r^{2} \operatorname{drd} \theta$
$=\int_{0}^{\pi}\left[\frac{r^{3}}{3}\right]_{0}^{2 \sin \theta} \mathrm{d} \theta$
$=\frac{8}{3} \int_{0}^{\pi} \sin ^{3} \theta \mathrm{d} \theta$
$=\frac{8}{3} \int_{0}^{\pi} \frac{1}{4}(3 \sin \theta-\sin 3 \theta) \mathrm{d} \theta$
$=\frac{8}{12} \left[-3cos\theta+\frac{cos3 \theta}{3}\right]_{0}^{ \pi }$
$=\frac{32}{9}$
