The area of the region include between the parabolas $y^2=8 x$ and $\mathrm{x}^2=8 \mathrm{y}$, is
- $\frac{128}{3}$ sq. units
- $\frac{64}{3}$ sq. units
- $\frac{32 \sqrt{8}}{3}$ sq. units
- $\frac{16 \sqrt{8}}{3}$ sq. units
Solution
Refer figure
Required area is shaded. Point of intersection of given curves are $\mathrm{y}^2=8 \mathrm{x}$ and $\mathrm{x}^2=8 \mathrm{y}$ i.e.
$\begin{aligned}
& \left(\frac{x^2}{8}\right)^2=8 x \Rightarrow x\left(x^3-512\right)=0 \\
& \therefore O \equiv(0,0) \text { and } P \equiv(8,8) \\
& A=\int_0^8(2 \sqrt{2})(\sqrt{x}) d x-\int_0^8 \frac{x^2}{8} d x \\
& \left.=\frac{2 \sqrt{2}}{\left(\frac{3}{2}\right)} x^{\frac{3}{2}}\right]_0^8-\frac{1}{24}\left[x^3\right]_0^8=\left(\frac{4 \sqrt{2}}{3}\right)(8 \sqrt{8})-\frac{1}{24}(512)=\frac{64}{3} \text { sq. units }
\end{aligned}$Asked in: MHT CET 2021 (24 Sep Shift 1)