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

The area is symmetric about the $y$ -axis. We can calculate for $x>0$ and multiply by 2.Curves: $x=\frac{\sqrt{y}}{2}$ and $x=3\sqrt{y}$.The line is $y=2$.Area $A=2{\int }_0^2(x_{\text{right}}-x_{\text{left}})dy$$A=2{\int }_0^2(3\sqrt{y}-\frac{\sqrt{y}}{2})dy$$A=2{\int }_0^2\frac{5}{2}\sqrt{y}dy=5{\int }_0^2{y}^{1/2}dy$$A=5{\left[\frac{y^{3/2}}{3/2}\right]}_0^2=\frac{10}{3}[2^{3/2}]=\frac{10}{3}[2\sqrt{2}]=\frac{20\sqrt{2}}{3}$.