The area (in sq. units) bounded between the parabolas $x^2=\frac{y}{4}$ and $x^2=9 y$ and the line $y=2$ is

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Aicharya · Accepted Answer

Required area $\begin{aligned} & =2 \int_0^2\left(3 \sqrt{y}-\frac{\sqrt{y}}{2}\right) \mathrm{d} y \\ & =2\left[3\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2-\frac{1}{2}\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2\right] \end{aligned}$ $\begin{aligned} & =2\left[2\left(2^{\frac{3}{2}}-0\right)-\frac{1}{3}\left(2^{\frac{3}{2}}-0\right)\right] \\ & =2\left[2(2 \sqrt{2})-\frac{1}{3}(2 \sqrt{2})\right] \\ & =\frac{20 \sqrt{2}}{3}\end{aligned}$

The area (in sq. units) bounded between the parabolas $x^2=\frac{y}{4}$ and $x^2=9 y$ and the line $y=2$ is

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