Question

Show that the area of the region bounded by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (ellipse) is $\pi ab$. Also deduce the area of the circle $x^2+y^2=a^2$.

Answer 1

We shall calculate the area in first quadrant.

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

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

So, from $x=0$ to $x=a$ area in first quadrant $={\int }_0^{a}ydx$

$={\int }_0^{a}b\sqrt{1-\frac{x^2}{a^2}}dx$

Let $\frac{x}{a}=t$

$dx=adt$

$={\int }_0^{a}b\sqrt{1-t^2}adt$

$=ab{\int }_0^a\sqrt{1-t^2}dt$

Let $t=\sin \theta$

$dt=\cos \theta d\theta$

$\sqrt{1-t^2}=\cos \theta$

$=ab{\int }_0^{\frac{\pi }{2}}\cos \theta \cos \theta d\theta$

$=\frac{ab}{2}{\int }_0^{\frac{\pi }{2}}2{\cos }^2\theta d\theta =\frac{ab}{2}{\int }_0^{\frac{\pi }{2}}(1+\cos 2\theta )d\theta$

$=\frac{ab}{2}\left({\int }_0^{\frac{\pi }{2}}1d\theta +{\int }_0^{\frac{\pi }{2}}\cos 2\theta d\theta \right)$

$=\frac{ab}{2}({\left[\theta \right]}_0^{\frac{\pi }{2}}+{\left[\frac{\sin 2\theta }{2}\right]}_0^{\frac{\pi }{2}})$

$=\frac{ab}{2}(\frac{\pi }{2}+0)=\frac{\pi ab}{4}$

Total area $=4\times$ are in $Q_1$

$=4\times \frac{\pi ab}{4}=\pi ab$

For circle $x^2+y^2=a^2$

Keep $b=a$

it becomes $\pi a^2$