How to find the new bounds for a transformed function of a random variable?

Question

I have the equation $f(x,y) = 12xy(1-y)$ with the bounds $0 < x < 1$ and $0 < y < 1$ .

We need to find the PDF of $Z = XY^2$. To do this we find the joint PDF of $Y$ and $Z$ , and then integrate out y. To integrate out $y$, the bounds were transformed to $√z < y < 1$. How do the bounds get transformed to this?

Answer 1

When you transform from the $(x,y)$-plane to the $(z,y)$-plane you get: $$0<x<1 \implies 0<z/y^2<1 \implies 0<z<y^2\implies 0<\sqrt{z}<|y|. $$ Combined with the inequality $0<y<1$, we get the region on the right hand side below.