Question

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

Answer 1

We know that ellipse is symmetrical about $x$ and $y$ axes, so we will find the area of one quadrant and multiply it by 4.

$$A=4{\int }_0^{a}ydx$$.....$\left(1\right)$

Lets write $x=a\sin \theta ⟹y=b\cos \theta$

The limits:

$x:0\to a$

$\theta :0\to \pi /2$

$dx=a\cos \theta d\theta$

$$\therefore \left(1\right)⟹A=4{\int }_0^{\pi /2}b\cos \theta \left(a\cos \theta d\theta \right)$$

$$⟹A=4ab{\int }_0^{\pi /2}{\cos }^2\theta d\theta$$

$$A=4ab{\int }_0^{\pi /2}\frac{(1-\cos 2\theta )}{2}d\theta$$

$$A=2ab{\int }_0^{\pi /2}d\theta -{\int }_0^{\pi /2}\cos 2\theta d\theta$$

$$A=2ab(\pi /2-0)-\frac{\sin 2\theta }{2}{|}_0^{\pi /2}$$

$$A=\pi ab-0$$

$$\therefore A=\pi ab$$