The area of the region bounded by the curves \(x^{2}+y^{2}=8\) and
\(y^{2}=2 x\) is

Given curves,
\(
\begin{aligned}
x^{2}+y^{2} &=8...(i)\\
and\quad
y^{2} &=2 x...(ii)
\end{aligned}
\)


On solving Eqs. (i) and (ii), we get
\(\therefore\)
\(
\begin{aligned}
x^{2}+2 x-8 &=0 \\
x^{2}+4 x-2 x-8 &=0 \\
x(x+4)-2(x+4) &=0 \\
(x-2)(x+4) &=0 \\
x=2 \text { and } y &=\pm 2
\end{aligned}
\)
\(\therefore\) Required area
\(
\begin{array}{l}
=2[\text { Area of } O A P+\text { Area of } P A B] \\
=2\left[\int_{0}^{2} \sqrt{2 x} d x+\int_{2}^{2 \sqrt{2}} \sqrt{8-x^{2}} d x\right] \\
=2\left[\sqrt{2}\left(x^{3 / 2} \cdot \frac{2}{3}\right)_{0}^{2}+\left(\frac{x}{2} \sqrt{8-x^{2}}\right.\right. \\
\left.\left.+\frac{8}{2} \sin ^{-1} \frac{x}{2 \sqrt{2}}\right)_{2}^{2 \sqrt{2}}\right] \\
=2\left[\frac{2 \sqrt{2}}{3}\left(2^{3 / 2}\right)+4 \times \frac{\pi}{2}-2-4 \times \frac{\pi}{4}\right]
\end{array}
\)
\(=2\left[\frac{2 \sqrt{2}}{3} \cdot 2 \sqrt{2}+2 \pi-2-\pi\right]\)
\(=2\left[\frac{8}{3}-2+\pi\right]=2\left(\frac{2}{3}+\pi\right)=2 \pi+\frac{4}{3}\)