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,

$\frac{x^2}{16}+\frac{y^2}{9}=1$,

the standard form of ellipse is given by

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

by comparing we get,

$a=4andb=3$

It can be observed that the ellipse is symmetrical about $x$-axis and $y$-axis.
$\therefore$ Area bounded by ellipse $=4\times$ Area of $OAB$

Area of $OAB$ $={\int }_0^{a}ydx$ $={\int }_0^{4}ydx$

$={\int }_0^{4}3\sqrt{1-\frac{x^2}{16}}dx$

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

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

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

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

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

Therefore, are bounded by the ellipse $=4\times 3\pi =12\pi units$