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

$π-\frac{8}{3}$

The required area A is given by

$A=\int\limits_0^2|y_1-y_2|dx$

$⇒A=\int\limits_0^2(\sqrt{4x-x^2}-\sqrt{2x})dx$

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

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

$⇒A=-\frac{8}{3}-2(-\frac{π}{2})=π-\frac{8}{3}$