The area between the parabolas x^2 = frac{y}{ | vedclass

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(C) The given parabolas are $x^2 = \frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$ and $x^2 = 9y \implies x = \pm 3\sqrt{y}$.Since the region is symm

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

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(C) The given parabolas are $x^2 = \frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$ and $x^2 = 9y \implies x = \pm 3\sqrt{y}$.Since the region is symmetric about the $y$-axis,we calculate the area in the first quadrant and multiply by $2$.The area $A$ is bounded by $x = 3\sqrt{y}$ (right curve) and $x = \frac{\sqrt{y}}{2}$ (left curve) from $y = 0$ to $y = 2$.$A = 2 \int_{0}^{2} \left( 3\sqrt{y} - \frac{\sqrt{y}}{2} ight) dy$$A = 2 \int_{0}^{2} \frac{5\sqrt{y}}{2} dy = 5 \int_{0}^{2} y^{1/2} dy$$A = 5 \left[ \frac{y^{3/2}}{3/2} ight]_{0}^{2} = 5 	imes \frac{2}{3} \left[ y^{3/2} ight]_{0}^{2}$$A = \frac{10}{3} \left( 2^{3/2} - 0 ight) = \frac{10}{3} 	imes 2\sqrt{2} = \frac{20\sqrt{2}}{3}$

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

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