1. Rewrite the parabolas in a convenient form

Given: $x^2=\frac y4 \quad \Rightarrow \quad y=4x^2$

and $x^2=9y \quad \Rightarrow \quad y=\frac{x^2}{9}$

So the curves are:

• $y=4x^2$ (narrower parabola)
• $y=\dfrac{x^2}{9}$ (wider parabola)
• the line $y=2$

2. Express $x$ in terms of $y$

Since the region is bounded by a horizontal line $y=2$, it is easiest to integrate with respect to $y$.

From $y=4x^2$: $x=\pm \frac{\sqrt y}{2}$

From $y=\frac{x^2}{9}$: $x=\pm 3\sqrt y$

For a fixed $y$, the horizontal width of the region between the two parabolas on the right side is $3\sqrt y-\frac{\sqrt y}{2}$ and similarly on the left side. Hence total width is $2\left(3\sqrt y-\frac{\sqrt y}{2}\right)=2\cdot \frac{5\sqrt y}{2}=5\sqrt y.$

3. Find the limits of integration

The region is enclosed below where the two parabolas meet.

Set: $4x^2=\frac{x^2}{9}$ $\left(4-\frac19\right)x^2=0$ $x=0 \Rightarrow y=0$

So the parabolas intersect at $(0,0)$.

The top boundary is the line $y=2$.

Thus the area is $A=\int_0^2 5\sqrt y\,dy.$

4. Evaluate the integral

$A=5\int_0^2 y^{1/2}\,dy$ $=5\left[\frac{2}{3}y^{3/2}\right]_0^2$ $=\frac{10}{3}\left(2^{3/2}\right)$

Now, $2^{3/2}=2\sqrt2$

Therefore, $A=\frac{10}{3}\cdot 2\sqrt2=\frac{20\sqrt2}{3}.$

5. Match with the options

$\boxed{\frac{20\sqrt2}{3}}$

So the correct option is C.