Area bounded by the curve $x{y}^2=a^2(a-x)$ and y-axis is.

The curve is symmetrical about the x-axis and lines in $0<x\le a$

$\therefore$ required $\mathrm{area}=2{\int }_0^a ydx=2{\int }_0^a a\sqrt{\frac{a-x}{x}}dx$

Put $x=a{\sin }^2\theta \Rightarrow dx=2a\sin \theta \cos \theta d\theta$

Required $\text{ Area }=2a{\int }_0^{\pi /2} \frac{a-a{\sin }^2\theta }{a{\sin }^2\theta }\cdot 2a\sin \theta \cos \theta d\theta$

$$\begin{gathered} =(2a{)}^2{\int }_0^{\pi /2} {\cos }^2\theta d\theta =2a^2{\int }_0^{\pi /2} (1+\cos 2\theta )d\theta \\ =2a^2\cdot {\left|\theta +\frac{\sin 2\theta }{2}\right|}_0^{\pi /2}=2a^2\left[\frac{\pi }{2}+0-0\right]=\pi a^2 \end{gathered}$$