Answer
$=\frac{\pi}{8}$
Work Step by Step
$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}}\left(y^{2}+x^{2}\right) d y d x$
changing to polar coordinates
$=\int_{0}^{\pi / 2} \int_{0}^{1}\left(r^{2}\right) r d r d \theta$
integrate and simplify
$=\int_{0}^{\pi / 2} \int_{0}^{1} r^{3} d r d \theta$
integrate with respect to $r$
$=\int_{0}^{\pi / 2}\left[\frac{r^{4}}{4}\right]_{0}^{1} d \theta$
$=\frac{1}{4} \int_{0}^{\pi / 2} d \theta$
integrate $=\frac{1}{4}[\theta]_{0}^{\pi / 2}$
$=\frac{\pi}{8}$
