CBSE Class 12 · 2008 · Area Under CurvesLong Answer

The circle $x^2+y^2=8x$ is $(x-4{)}^2+y^2=16$ (centre $(4,0)$, radius $4$); the parabola is $y^2=4x$.
Solving $x^2+4x=8x\Rightarrow x^2-4x=0\Rightarrow x=0,4$, so above the $x$-axis they meet at $O(0,0)$ and $A(4,4)$.
For $0\le x\le 4$ the upper boundary of the required region is the circle $y=\sqrt{8x-x^2}$ and the lower boundary is the parabola $y=2\sqrt{x}$:

${\int }_0^4\sqrt{16-(x-4{)}^2}dx=[\frac{x-4}{2}\sqrt{16-(x-4{)}^2}+8{\sin }^{-1}\frac{x-4}{4}{]}_0^4=0-(8{\sin }^{-1}(-1))=4\pi ,$
and ${\int }_0^{4}2\sqrt{x}dx=\frac{4}{3}\cdot 8=\frac{32}{3}.$