Question

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

Answer 1

The area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$ is represented by the shaded region $BCAB$
$\therefore AreaBACB=Area\left(OBCAO\right)-Area\left(OBAO\right)$
$={\int }_0^{a}b\sqrt{1-\frac{x^2}{a^2}}dx-{\int }_0^{a}b(1-\frac{x}{a})dx$
$=\frac{b}{a}{\int }_0^a\sqrt{a^2-x^2}dx-\frac{b}{a}{\int }_0^a(a-x)dx$
$=\frac{b}{a}[{\{\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\}}_0^a-{\{ax-\frac{x^2}{2}\}}_0^a]$
$=\frac{b}{a}[\left\{\frac{a^2}{2}\left(\frac{\pi }{2}\right)\right\}-\{a^2-\frac{a^2}{2}\}]$
$=\frac{b}{a}[\frac{a^2\pi }{4}-\frac{a^2}{2}]$
$=\frac{b{a}^2}{2a}[\frac{\pi }{2}-1]$
$=\frac{ab}{2}[\frac{\pi }{2}-1]$
$=\frac{ab}{4}(\pi -2)$