The area bounded between the parabolas $x^2=\frac{y}{4}$ and $x^2=9 y$, and the straight line $y=2$ is

Question

Aicharya · Accepted Answer

Required area $\begin{aligned} & A=2\left[\int_0^2\left(3 \sqrt{y}-\frac{\sqrt{y}}{2}\right) d y\right]=2 \int_0^2 \frac{5 \sqrt{y}}{2} d y \\ & =5\left[\frac{y^{3 / 2}}{3 / 2}\right]_0^2=\frac{10}{3}\left[2^{3 / 2}-0\right]=\frac{20 \sqrt{2}}{3} \end{aligned}$

The area bounded between the parabolas $x^2=\frac{y}{4}$ and $x^2=9 y$, and the straight line $y=2$ is

Solution