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# Find the area of the region bounded by the ellipse x^2/16+y^2/9=1.

Q : 4        Find the area of the region bounded by the ellipse  $\dpi{100} \frac{x^2}{16}+\frac{y^2}{9}=1.$

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The area bounded by the ellipse : $\dpi{100} \frac{x^2}{16}+\frac{y^2}{9}=1.$

Area will be 4 times the area of EAB.

Therefore,  $Area\ of\ EAB= \int^4_{0} y dx$

$= \int^4_{0}3\sqrt{1-\frac{x^2}{16}} dx$

$= \frac{3}{4}\int^4_{0}\sqrt{16-x^2} dx$

$= \frac{3}{4}\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]^4_{0}$

$= \frac{3}{4}\left [ 2\sqrt{16-16} +8\sin^{-1}(1)-0-8\sin^{-1}(0)\right ]$

$= \frac{3}{4}\left [ \frac{8\pi}{2} \right ]$

$= \frac{3}{4}\left [ 4\pi \right ] =3\pi$

Therefore the area bounded by the ellipse will be $= 4\times {3\pi} = 12\pi\ units.$

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