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  • Find the area of the region bounded by the ellipse x^2/4+y^2/9=1.

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

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The area bounded by the ellipse : \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.

 

 

Posted by

Divya Prakash Singh

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