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

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(A) Given the curves $y=4x^{2}$ and $y=\frac{x^{2}}{9}$, we express $x$ in terms of $y$:For $y=4x^{2}$, $x^{2}=\frac{y}{4} \implies x = \pm \frac{\sqr

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

Vedclass · Answer

(A) Given the curves $y=4x^{2}$ and $y=\frac{x^{2}}{9}$, we express $x$ in terms of $y$:For $y=4x^{2}$, $x^{2}=\frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$.For $y=\frac{x^{2}}{9}$, $x^{2}=9y \implies x = \pm 3\sqrt{y}$.The region is bounded by $y=0$ to $y=2$ and is symmetric about the $y$-axis.The area $A$ is given by:$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} 	imes (2)^{3/2}$$A = \frac{10}{3} 	imes 2\sqrt{2} = \frac{20\sqrt{2}}{3} 	ext{ sq. unit}$.

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

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