Mathematics - Area Under Curves Question with Solution | TestHub

Refer figure
Required area is shaded. Point of intersection of $y^2=x$ and $y=4$ is $A\equiv (16,4)$ $$\begin{array}{rl}\therefore A& ={\int }_0^4{y}^{2}dy\\ & ={\left[\frac{y^3}{3}\right]}_0^4=\frac{64}{3}\text{ sq. units }\end{array}$$