# Q : 5        Find the area of the region bounded by the ellipse  $\dpi{100} \small \frac{x^2}{4}+\frac{y^2}{9}=1$

D Divya Prakash Singh

The area bounded by the ellipse : $\dpi{100} \small \frac{x^2}{4}+\frac{y^2}{9}=1$

The area will be 4 times the area of EAB.

Therefore,  $Area\ of\ EAB= \int^2_{0} y dx$

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

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

$= \frac{3}{2}\left [ \frac{x}{2}\sqrt4-x^2 +\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{0}$

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

$= \frac{3\pi}{2}$

Therefore the area bounded by the ellipse will be $= 4\times \frac{3\pi}{2} = 6\pi\ units.$

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