Solution:

The given eq. of the ellipse can be represented as

$\begin{array}{l}
\frac{x^{2}}{4}+\frac{y^{2}}{9}=1 \\
\Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}}
\end{array}$
It can be observed that the ellipse is symmetrical about $x$-axis and $y$-axis.
$\therefore$ Area bounded by ellipse $=4 \times$ Area $\mathrm{OAB}$
$\therefore$ Area of $\mathrm{OAB}=\int_{0}^{2} y d x$
$\begin{aligned} &=\int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}} d x \quad[\mathrm{U} \operatorname{sing}(1)] \\ &=\frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} d x \\ &=\frac{3}{2}\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-} \frac{x}{2}\right]_{0}^{2} \\ &=\frac{3}{2}\left[\frac{2 \pi}{2}\right] \\ &=\frac{3 \pi}{2} \end{aligned}$
As a result, area bounded by the ellipse $=4 \times \frac{3 \pi}{2}=6 \pi$ units