The area of the shaded portion is:

The correct answer is Option (1) - $\left(\pi-\frac{8}{3}\right)$ sq. units

finding intersection $y^2=2x$

$x^2+y^2=4x⇒x^2+2x=4x$

$⇒x^2=2x$

$⇒x=0,2$

so area = $\int\limits_0^2\sqrt{4x-x^2}-\sqrt{2x}dx$

$=\int\limits_0^2\sqrt{2^2-(x-2)^2}-\sqrt{2x}dx$

$=\left[\frac{(x-2)}{2}\sqrt{4x-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}-\frac{2\sqrt{2}}{3}\sqrt{x}(x)\right]_0^2$

$=\pi-\frac{8}{3}$ sq. units