Given curve is
$a y^{2}=x^{2}(a-x),(a>\,0)$
The required area
$=2 \int\limits_{0}^{a} x \sqrt{\frac{a-x}{a}} d x$
$=\frac{2}{\sqrt{a}} \int_{0}^{a} x \sqrt{a-x} d x$
Put $ a-x=t^{2}$, at $x=0, t=\sqrt{a}$ and,
at $ x=a, t=0$
and $-dx =2t\,dt$
So, required area $=-\int\limits_{\sqrt{a}}^{0} \frac{4}{\sqrt{a}} \times t^{2}\left(a-t^{2}\right) d t $
$=\frac{4}{\sqrt{a}} \int_{0}^{\sqrt{a}}\left(a t^{2}-t^{4}\right) d t=\frac{4}{\sqrt{a}}\left[\frac{a t^{3}}{3}-\frac{t^{5}}{5}\right]_{0}^{\sqrt{a}} $
$=\frac{4}{\sqrt{a}}\left(\frac{a^{2} \sqrt{a}}{3}-\frac{a^{2} \sqrt{a}}{5}\right)=\frac{8}{15} a^{2}$
