The area bounded by the parabolas y=4x^2, y=f | vedclass

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(D) Given parabolas are $y=4x^2 \implies x^2 = \frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$ and $y=\frac{x^2}{9} \implies x^2 = 9y \implies x = \p

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$\frac{20 \sqrt{2}}{3}$ sq. units

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(D) Given parabolas are $y=4x^2 \implies x^2 = \frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$ and $y=\frac{x^2}{9} \implies x^2 = 9y \implies x = \pm 3\sqrt{y}$.Since the region is symmetric about the $y$-axis, we calculate the area in the first quadrant and multiply by $2$.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 $A = 2 \int_{0}^{2} (3\sqrt{y} - \frac{\sqrt{y}}{2}) dy$.$A = 2 \int_{0}^{2} (\frac{6\sqrt{y} - \sqrt{y}}{2}) dy = 2 \int_{0}^{2} \frac{5\sqrt{y}}{2} dy$.$A = 5 \int_{0}^{2} y^{1/2} dy = 5 [\frac{y^{3/2}}{3/2}]_{0}^{2}$.$A = 5 	imes \frac{2}{3} [y^{3/2}]_{0}^{2} = \frac{10}{3} (2^{3/2} - 0)$.$A = \frac{10}{3} (2\sqrt{2}) = \frac{20\sqrt{2}}{3}$ sq. units.

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

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