Answer
$\int_0^1\int_y^\sqrt{2-y^2}(x+y)dxdy=\frac{2\sqrt{2}}{3}$
Work Step by Step
Since the equation $x=y$ is a line through the origin with slope $1$, in polar coordinates it is written as $\theta=\frac{\pi}{4}$.
Meanwhile, for $0\leq y\leq 1$ the equation $x=\sqrt{2-y^2}$ is a right-half circle center the origin with radius $\sqrt{2}$.
Thus. the region enclosed by $x=\sqrt{2-y^2}$ and $x=y$ for $0\leq y\leq 1$ is a one-eighth circle in quadrant I center at $(0,0)$ with radius $\sqrt{2}$ and in polar coordinates it is written as $R=\{(r,\theta)|0\leq r\leq \sqrt{2},0\leq \theta\leq \pi/4\}$.
Convert the integral to polar coordinates and evaluate:
$\int_0^1\int_y^\sqrt{2-y^2}(x+y)dxdy=\int_0^\sqrt{2}\int_0^{\pi/4}(r\cos \theta+r\sin\theta)rd\theta dr$
$=\int_0^\sqrt{2}(r\sin\theta-r\cos\theta)r]_0^{\pi/4}dr$
$=\int_0^\sqrt{2}r^2dr$
$=\frac{r^3}{3}]_0^\sqrt{2}$
$=\frac{2\sqrt{2}}{3}$
