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

Given curves,
\(
y^{2}=8x...(i)
\)
and
\(
y=x...(ii)
\)


On solving Eqs. (i) and (ii), we get
\(\begin{aligned} x^{2}-8 x &=0 \\ x(x-8) &=0 \\ \Rightarrow \quad x &=0,8 \end{aligned}\)
and \(\quad y=0,8\)
\(\begin{aligned} \therefore \text { Required area, }(O P A) &=\int_{0}^{8}(\sqrt{8 x}-x) d x \\ &=\left[2 \sqrt{2} \cdot \frac{2}{3} \cdot x^{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{8} \end{aligned}\)
\(=\frac{4 \sqrt{2}}{3} \cdot(8)^{3 / 2}-\frac{(8)^{2}}{2}\)
\(=\frac{4 \sqrt{2}}{3} \cdot(2)^{3} \cdot 2 \sqrt{2}-\frac{64}{2}\)
\(=\frac{16}{3} \times 8-\frac{64}{2}=64\left(\frac{2}{3}-\frac{1}{2}\right)\)
\(=64 \times \frac{(4-3)}{6}=64 \times \frac{1}{6}=\frac{32}{3}\)