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(C) The given parabolas are $x^2 = \frac{y}{4}$ (or $x = \pm \frac{\sqrt{y}}{2}$) and $x^2 = 9y$ (or $x = \pm 3\sqrt{y}$). The line is $y = 2$. Due to

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

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(C) The given parabolas are $x^2 = \frac{y}{4}$ (or $x = \pm \frac{\sqrt{y}}{2}$) and $x^2 = 9y$ (or $x = \pm 3\sqrt{y}$). The line is $y = 2$. Due to symmetry about the $Y$-axis,the total area is twice the area in the first quadrant. In the first quadrant,the region is bounded by $x = 3\sqrt{y}$ and $x = \frac{\sqrt{y}}{2}$ from $y = 0$ to $y = 2$. Area $= 2 \int_0^2 \left( 3\sqrt{y} - \frac{\sqrt{y}}{2} \right) dy$ $= 2 \int_0^2 \frac{5}{2} \sqrt{y} \, dy = 5 \int_0^2 y^{1/2} \, dy$ $= 5 \left[ \frac{y^{3/2}}{3/2} \right]_0^2 = 5 \cdot \frac{2}{3} \left[ y^{3/2} \right]_0^2$ $= \frac{10}{3} (2^{3/2} - 0) = \frac{10}{3} (2\sqrt{2}) = \frac{20\sqrt{2}}{3}$ sq. units.

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

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