Given,

$x^2+{\left(y-2\right)}^2\le 4 \& x^2\ge 2y$

Now on solving, $x^2+{\left(y-2\right)}^2=4 \& x^2=2y$ we get intersecting point of the curves as $\left(\pm 2,2\right)$,

Now plotting the diagram we get,

So, from above diagram area of required region will be,

 $=2\left[\frac{1}{4}\left(\pi \times 2^2\right)-{\int }_0^2\sqrt{2}·\sqrt{y}dy\right]$

$=2\left[\pi -{\left[\frac{\sqrt{2}\cdot y^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^2\right]$

$=2\left[\pi -\frac{2\sqrt{2}}{3}·2\sqrt{2}\right]$

$=2\left[\pi -\frac{8}{3}\right]$

$=2\pi -\frac{16}{3}$