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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  \frac{x^2}{16}+\frac{y^2}{9}=1.

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

Ellipse 5

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.

 

Posted by

Divya Prakash Singh

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