Mathematics - Area Under Curves Question with Solution | TestHub

The given parabolas are $x^2=\frac{y}{4}$ and $x^2=9y$.

Since both parabolas are symmetric about the y-axis, the required area is twice the area in the first quadrant.

In the first quadrant, the equations can be written in terms of $y$ as $x=\frac{\sqrt{y}}{2}$ and $x=3\sqrt{y}$.

The area bounded by these curves and the line $y=2$ is given by integrating with respect to $y$ from $0$ to $2$:

$A=2{\int }_0^2\left(3\sqrt{y}-\frac{\sqrt{y}}{2}\right)dy$

$A=2{\int }_0^2\frac{5}{2}\sqrt{y}dy$

$A=5{\int }_0^2{y}^{1/2}dy$

$A=5{\left[\frac{2}{3}{y}^{3/2}\right]}_0^2$

$A=\frac{10}{3}(2{)}^{3/2}$

$A=\frac{10}{3}(2\sqrt{2})=\frac{20\sqrt{2}}{3}$ sq. units.

Answer: $\frac{20\sqrt{2}}{3}$