# Find the area of the smaller region bounded by the ellipse

Question:

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$

Solution:

The 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

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

$=\int_{0}^{3} 2 \sqrt{1-\frac{x^{2}}{9}} d x-\int_{0}^{3} 2\left(1-\frac{x}{3}\right) d x$

$=\frac{2}{3}\left[\int_{0}^{3} \sqrt{9-x^{2}} d x\right]-\frac{2}{3} \int_{0}^{3}(3-x) d x$

$=\frac{2}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{0}^{3}-\frac{2}{3}\left[3 x-\frac{x^{2}}{2}\right]_{0}^{3}$

$=\frac{2}{3}\left[\frac{9}{2}\left(\frac{\pi}{2}\right)\right]-\frac{2}{3}\left[9-\frac{9}{2}\right]$

$=\frac{2}{3}\left[\frac{9 \pi}{4}-\frac{9}{2}\right]$

$=\frac{2}{3} \times \frac{9}{4}(\pi-2)$

$=\frac{3}{2}(\pi-2)$ units