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(A) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. Here,$a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$. The ellips

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$12$

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(A) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. Here,$a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$. The ellipse is symmetrical about both the $x-$axis and $y-$axis. Therefore,the total area bounded by the ellipse is $4 	imes$ (Area of the region in the first quadrant). Area in the first quadrant $= \int_{0}^{4} y \, dx$. From the equation,$y = 3 \sqrt{1 - \frac{x^2}{16}} = \frac{3}{4} \sqrt{16 - x^2}$. Area $= \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} \, dx = \frac{3}{4} \left[ \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1} \left( \frac{x}{4} ight) ight]_{0}^{4}$. $= \frac{3}{4} \left[ (0 + 8 \sin^{-1}(1)) - (0 + 8 \sin^{-1}(0)) ight] = \frac{3}{4} \left[ 8 	imes \frac{\pi}{2} ight] = \frac{3}{4} [4 \pi] = 3 \pi$. Total area $= 4 	imes 3 \pi = 12 \pi$ square units.

Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. (in $\pi$)

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