Question

Find the area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Answer 1

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

Drawing ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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

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

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

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

$\Rightarrow y=\pm \frac{b}{a}\sqrt{a^2-x^2}$

Since $AOBA$ is in $1^{st}$ quadrant, the value of $y$ is positive

$\therefore y=\frac{b}{a}\sqrt{a^2-b^2}$

Since circle is symmetric about $x\text{axis and}y\text{axis}$

$\therefore \text{Area of ellipse}=4\times \text{Area of AOBA}$

$4\times {\int }_0^{a}ydx$

$=\frac{4b}{a}{\int }_0^a\sqrt{a^2-x^2}dx$

$=\frac{4b}{a}{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}]}_0^a$

$=\frac{4b}{a}[0+\frac{a^2}{2}{\sin }^{-1}(1)-0-0]$

$=\frac{4b}{a}\times \frac{a^2}{2}{\sin }^{-1}\left(1\right)$

$=2ab\times {\sin }^{-1}\left(1\right)$

$=2ab\times \frac{\pi }{2}$

$=\pi ab$

Final answer:

Therefore, required area $=\pi ab\text{square units}$.