Solution:
Area of the smaller region bounded by the ellipse, $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and the line, $\frac{x}{3}+\frac{y}{2}=1$ is represented by the shaded region BCAB as

$\therefore$ Area $B C A B=$ Area $(O B C A O)-$ Area (OBAO)
$\begin{array}{l}
=\int^{3} 2 \sqrt{1-\frac{x^{2}}{9}} d x-\int_{0}^{1} 2\left(1-\frac{x}{3}\right) d x \\
=\frac{2}{3}\left[\int_{0}^{3} \sqrt{9-x^{2}} d x\right]-\frac{2}{3} \int_{0}^{3}(3-x) d x \\
=\frac{2}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{0}^{3}-\frac{2}{3}\left[3 x-\frac{x^{2}}{2}\right]_{0}^{3} \\
=\frac{2}{3}\left[\frac{9}{2}\left(\frac{\pi}{2}\right)\right]-\frac{2}{3}\left[9-\frac{9}{2}\right] \\
=\frac{2}{3}\left[\frac{9 \pi}{4}-\frac{9}{2}\right] \\
=\frac{2}{3} \times \frac{9}{4}(\pi-2) \\
=\frac{3}{2}(\pi-2) \text { sq. units }
\end{array}$