Question

Find the area of the region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, by integration.

Answer 1

We have to find the area of the region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The area enclosed by the ellipse is shown in the following figure.

Since the ellipse is symmetric about both the axis,

Area of ellipse $=4\times$ Area of $OAB$

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

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

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

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

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

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

Since $OAB$ is in first quadrant, $y$ is positive.

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

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

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

$=\frac{4b}{a}{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\left(\frac{x}{a}\right)]}_0^a$ $\{\because \int \sqrt{a^2-x^{2}dx}=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\left(\frac{x}{a}\right)\}$

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

$=\frac{4b}{a}\left[\frac{a^2}{2}\left(\frac{\pi }{2}\right)\right]$

$=\pi ab$

Therefore the area of the region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi absq.units$