Answer
$$\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right) d x
=\frac{2 \sqrt{1-y^{2}}}{3}\left(1+2 y^{2}\right)$$
Work Step by Step
Given $$\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right) d x$$
So,
\begin{align}
\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\left(x^{2}+y^{2}\right) d x&=\left[\frac{1}{3} x^{3}+y^{2} x\right]_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}\\
&=2\left[\frac{1}{3}\left(1-y^{2}\right)^{3 / 2}+y^{2}\left(1-y^{2}\right)^{1 / 2}\right]\\
&=\frac{2 \sqrt{1-y^{2}}}{3}\left(1+2 y^{2}\right)
\end{align}
