CBSE Class 12 · 2016 · Area Under CurvesLong Answer

The region lies in the first quadrant, inside the circle $x^2+y^2=2ax$, i.e. $(x-a{)}^2+y^2=a^2$ (centre $(a,0)$, radius $a$), and outside/above the parabola $y^2=ax$.
They meet where $x^2+ax=2ax\Rightarrow x^2=ax\Rightarrow x=a$ (and $x=0$); at $x=a$, $y=a$, so the point of intersection is $(a,a)$.
For $0\le x\le a$ the upper edge is the circle $y=\sqrt{a^2-(x-a{)}^2}$ and the lower edge is the parabola $y=\sqrt{ax}$:

${\int }_0^a\sqrt{a^2-(x-a{)}^2}dx=\frac{\pi a^2}{4}$ (quarter circle) and ${\int }_0^a\sqrt{ax}dx=\frac{2}{3}{a}^2$, so