Find the area of the region bounded by the ellipse

The given equation of the ellipse can be represented as

$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$

$\Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}}$         ...(1)

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

∴ Area bounded by ellipse = 4 × Area OAB

$\therefore$ Area of $\mathrm{OAB}=\int_{0}^{2} y d x$

$=\int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}} d x \quad[$ Using (1) $]$

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

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

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

$=\frac{3 \pi}{2}$

Therefore, area bounded by the ellipse $=4 \times \frac{3 \pi}{2}=6 \pi$ units