The area bounded by the parabola $y^2=x$, the straight line $y=4$ and $\mathrm{Y}$ axis is
- $2 \sqrt{7}$ sq. unit
- $\frac{64}{3}$ sq. units
- $\frac{16}{3}$ sq. units
- $7 \sqrt{2}$ sq. units
Solution
Required area is shaded.
Point of intersection of $y^2=x$ and $y=4$ is $A \equiv(16,4)$
$\begin{aligned}
\therefore A & =\int_0^4 y^2 d y \\
& =\left[\frac{y^3}{3}\right]_0^4=\frac{64}{3} \text { sq. units }
\end{aligned}$Asked in: MHT CET 2021 (21 Sep Shift 2)