JEE Main · 2012 · 29 Apr 2012 · Offline · Area Under Curves

The curves are $x^2=\frac{y}{4}$, giving $x=\frac{\sqrt{y}}{2}$, and $x^2=9y$, giving $x=3\sqrt{y}$, both symmetric about the $y$-axis. Between $y=0$ and $y=2$ each horizontal strip runs from the narrow parabola out to the wide one on each side, so the enclosed area is $A=2{\int }_0^2\left(3\sqrt{y}-\frac{\sqrt{y}}{2}\right)dy=5{\int }_0^2\sqrt{y}dy=5\cdot \frac{2}{3}[y^{3/2}{]}_0^2=\frac{10}{3}\cdot 2\sqrt{2}=\frac{20\sqrt{2}}{3}$.