The area bounded by the parabolas y = 9x^2, y | vedclass

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(B) The given curves are $y = 9x^2$ (or $x^2 = \frac{y}{9}$) and $y = \frac{x^2}{16}$ (or $x^2 = 16y$).Since the region is symmetric about the $y$-axi

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$\frac{44}{9}$ sq. units

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(B) The given curves are $y = 9x^2$ (or $x^2 = \frac{y}{9}$) and $y = \frac{x^2}{16}$ (or $x^2 = 16y$).Since the region is symmetric about the $y$-axis, we calculate the area in the first quadrant and multiply by $2$.For $y = 9x^2$, $x = \frac{\sqrt{y}}{3}$.For $y = \frac{x^2}{16}$, $x = 4\sqrt{y}$.The area $A$ is given by $2 \int_{0}^{1} (x_{	ext{right}} - x_{	ext{left}}) dy$.$A = 2 \int_{0}^{1} (4\sqrt{y} - \frac{\sqrt{y}}{3}) dy$.$A = 2 \int_{0}^{1} (4 - \frac{1}{3}) \sqrt{y} dy = 2 	imes \frac{11}{3} \int_{0}^{1} y^{1/2} dy$.$A = \frac{22}{3} [\frac{y^{3/2}}{3/2}]_{0}^{1} = \frac{22}{3} 	imes \frac{2}{3} = \frac{44}{9}$ sq. units.

The area bounded by the parabolas $y = 9x^2$, $y = \frac{x^2}{16}$ and the line $y = 1$ is

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