Answer
$$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}}(x+y) d x d y=\frac{2}{3}$$
Work Step by Step
Given $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}}(x+y) d x d y=$$
So, we get
\begin{align}
\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}}(x+y) d x d y&
=\int_{0}^{1}\left[\frac{1}{2} x^{2}+x y\right]_{0}^{\sqrt{1-y^{2}}} d y\\&=\int_{0}^{1}\left[\frac{1}{2}\left(1-y^{2}\right)+y \sqrt{1-y^{2}}\right] d y\\
&=\left[\frac{1}{2} y-\frac{1}{6} y^{3}-\frac{1}{2}\left(\frac{2}{3}\right)\left(1-y^{2}\right)^{3 / 2}\right]_{0}^{1}\\
&=\frac{2}{3}
\end{align}
