Find the area of the region bounded by the ellipse
Question:

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

Solution:

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.

$\therefore$ Area bounded by ellipse $=4 \times$ Area of $\mathrm{OAB}$

Area of $\mathrm{OAB}=\int_{0}^{4} y d x$

$=\int_{0}^{4} 3 \sqrt{1-\frac{x^{2}}{16}} d x$

$=\frac{3}{4} \int_{0}^{1} \sqrt{16-x^{2}} d x$

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

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

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

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

$=3 \pi$

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