Question

Find the area bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the ordinates $x=0$ and $x=ae$, where $b^2=a^2(1-e^2)$ and $e<1$

Answer 1

Required area $A$ is given by
$A=2$ (Area of shaded region in first quadrant)
$\Rightarrow A=2{\int }_0^{ae}|y|dx=2{\int }_0^{ae}ydx$ $[\because y\ge 0\therefore |y|=y]$
$\Rightarrow A=2\frac{b}{a}\left[{\int }_0^{ae}\sqrt{a^2-x^2}dx\right]$
$\Rightarrow A=\frac{2b}{a}{[\frac{1}{2}\times \sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\frac{x}{a}]}_{ae}^0$
$\Rightarrow A=\frac{2b}{a}[\frac{ae}{2}\sqrt{a^2-a^2{e}^2}+\frac{1}{2}{a}^2{\sin }^{-1}e]$
$\Rightarrow A=\frac{b}{a}(a^{2}e\sqrt{1-e^2}+a^2{\sin }^{-1}e)=ab(e\sqrt{1-e^2}+{\sin }^{-1}e)$