Let $AOB$ be the positive quadrant of the ellipse   $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with  $OA=a$  and $OB=b$. Show that the area bounded between the chord $AB$ and the arc $AB$ of the ellipse is $\left(\frac{\pi -2}{4}\right)ab\text{ sq units }$.

Given ellipse eq is   $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow y^2/b^2=1-\frac{x^2}{a^2}$

$\Rightarrow y^2=b^2\left[\frac{a^2-x^2}{a^2}\right]$

$\Rightarrow y=\sqrt{\frac{b^2}{a^2}\left(a^2-x^2\right)}$

$\Rightarrow y=\frac{b}{a}\sqrt{\left(a^2-x^2\right)}$........(1)

The vertices of an ellipse are $A=(a,0)$

$⟺A^1=(-a,0)\text{ and }B(0,b)\mathrm{\&}{B}^1(0,-b)$

$A_1=\int \text{ area of an ellipse in }1\text{ st quadrant }$

$={\int }_0^a \frac{b}{a}\sqrt{a^2-x^2}dx=\frac{b}{a}{\int }_0^a \sqrt{a^2-x^2}dx$

$=\frac{b}{a}{\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)\right]}_0^a$

$=\frac{b}{a}\left[0+\frac{a^2}{2}{\sin }^{-1}\left(1\right)-0\right]=\frac{b}{a}\left[\frac{a^2}{2}\cdot \frac{\pi }{2}\right]$

$A_1=\frac{b}{a}\cdot \frac{\pi a^2}{4}\Rightarrow A_1=\frac{\pi ab}{4}\text{ sq.units }$

Area of  $\mathrm{△}OAB,A_2=\frac{1}{2}ab$

$\therefore$ area of between arc $AB$ and chord $AB$ is  $\frac{\pi }{4}ab-\frac{1}{2}ab=\left(\frac{\pi -2}{4}\right)ab\text{ sq.units }$