Double Integral with absolute value.

Question

I'm trying to get
$$\int_0^2\int_0^{\sqrt{2x}}y^2|x-y|dydx.$$
I'm struggling with that absolute value. I'm not sure how to "divide" the integral into two. I know that
$$|x-y|=\begin{cases} x-y &\text{ if }x\geq y, \\
y-x &\text{ if }x\leq y.\end{cases}$$
My problem is I have that $y=\sqrt{2x}$ in the top integral extreme.

Answer 1

Let draw a sketch for $\sqrt {2x}$ and $x$ then divide the region in two parts for $x-y\ge 0$ and $x-y<0$ and the integral accordingly.