Question

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

Answer 1

The given equation of the ellipse, $\frac{x^2}{16}+\frac{y^2}{9}=1$ ,can be represented as

It can be observed that the ellipse is symmetrical about x-axis and y-axis.

∴Area bounded by ellipse=4×Area of OAB

Area of OAB=$\int^4_0ydx$

=$\int^4_03\sqrt{1-\frac{x^2}{16}}dx$

=$\frac{3}{4}\int^4_0\sqrt{16-x^2}dx$

=$\frac{3}{4}\bigg[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4}\bigg]^4_0$

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

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

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

=3$\pi$

Therefore, area bounded by the ellipse= 4×3$\pi$=12$\pi$ units.