The limit of x are $x=2$ and $x=0$ (A straight line)

The limits of y are $y=2$ which is a line and $y= \sqrt {2x}$

$\therefore y^2 = 2x$ which is a parabola Intersection of parabola

And line $y=2$ putting $y=2$ in equation of parabola.

$\therefore 4 = 2x$

$\therefore x=2$ and $y=2 $

Point of intersection is $(2,2)$

Now changing the order the vertical strip is replaced by horizontal strip.

$\therefore y$ varies from 0 to 2

1) Outer limit y

$Y=0$ to $y=z$

2) Inner limit x

Upper limit is equation of parabola which is $y^2 = 2x$

$\therefore x = y^2/2$

Lower limit $x=0$

$$\therefore I=\int\limits_{y=0}^{y=2}\int\limits_{x=0}^{y^2/2}\dfrac {y^2}{\sqrt{y^4-4x^2}}dxdy$$

Dividing numerator and denominator by 2 we get

$$I=\int\limits^2_{y=0}\int\limits^{y^2/2}_{x=0}\dfrac {\frac {y^2}2}{\sqrt{\frac {y^2}4-x^2}}dxdy$$ $ I=\dfrac 12 \int\limits^2_{y=0} y^2 \int\limits_{x=0}^{\frac {y^2}2}\dfrac 1{\sqrt{\Bigg(\frac {y^2}2-x^2\Bigg)}}\\ =\dfrac 12\int\limits^2_{y=0}y^2\Bigg[\sin^{-1}\Bigg(\dfrac x{\frac {y^2}2}\Bigg)\Bigg]_0^{\frac {y^2}2}dy\\ =\dfrac 12 \int\limits_0^2 y^2\Big[\sin^{-1}\dfrac {y^2/2}{y^2/2}-\sin^{-1}0\Big]dy\\ =\dfrac 12\int\limits_0^2y^2(\dfrac \pi2)dy\\ =\dfrac \pi4\Big[\dfrac {y^3}3\Big]^2_0\\ =\dfrac \pi4\dfrac 83\\ =\dfrac {2\pi}3$