The area (in sq. units) of the region $\{\left(x, y\right):y^2\ge 2x$ and $x^2+y^2\le 4x, x \ge 0, y \ge 0\}$ is

$y^2\ge 2x$ (Area outside the parabola)

$x^2+y^2\le 4x$ (Area inside the circle)


Finding point of intersection of the curves, we get, 

$x^2+y^2=4x$ and $y^2=2x$

$\Rightarrow x^2+2x=4x$

$\Rightarrow x^2=2x$

$\Rightarrow x=0, x=2$

$\mathrm{I}\mathrm{f} x=0, \mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n} y=0 \mathrm{and}\mathrm{ }\mathrm{if} x=2, \mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n} y= \pm 2.$

Coordinates of $\mathrm{A}\left(0,\mathrm{ }0\right)\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d} B\left(2, 2\right)$

As $x \ge 0 y \ge 0$ only area above $x$-axis would be considered.

Hence, area

$=\left[\underset{0}{\overset{2}{\int }}\sqrt{4x-x^2} dx- \sqrt{2} \underset{0}{\overset{2}{\int }}\sqrt{x} dx\right]$

$=\left[\underset{0}{\overset{2}{\int }}\sqrt{4-{\left(x-2\right)}^2} dx- \sqrt{2} \underset{0}{\overset{2}{\int }}\sqrt{x} dx\right]$

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

$=\left[0-2{\sin }^{-1}\left(-1\right)\right]- \sqrt{2} .\frac{2}{3}2\sqrt{2}]\mathrm{ =}$$\left[\pi -\frac{8}{3}\right]$ sq. units