Given equation of curves
$x^{2}+y^{2}=2$ and $y=x^{2}$
For point of intersection, on solving the given curves, we get
$
\begin{array}{l}
y^{2}+y-2=0 \Rightarrow y^{2}+2 y-y-2=0 \\
\Rightarrow y(y+2)-1(y+2)=0 \\
\Rightarrow y=1[\because y>0]
\end{array}
$
So, the required area
$
\begin{array}{l}
\left.=\pi+2 \int_{0}^{1}\left(\sqrt{2-y^{2}}\right)-\sqrt{y}\right) d y \\
=\pi+2\left[\frac{y}{2} \sqrt{2-y^{2}}+\frac{2}{2} \sin ^{-1} \frac{y}{\sqrt{2}}-\frac{2}{3} y^{3 / 2}\right]_{0}^{1} \\
=\pi+2\left[\frac{1}{2}+\frac{\pi}{4}-\frac{2}{3}\right]=\frac{3 \pi}{2}+1-\frac{4}{3} \\
=\frac{3 \pi}{2}-\frac{1}{3} \text { sq. units }
\end{array}
$
