How to find bounds of integration in finding CDF of $XY$ Suppose you had $f_{X,Y}(x,y) = \begin{cases}1/4,&0<X<2, 0 <Y<X^3 \\0,& \text{otherwise}   \end{cases}$
How would you find a CDF for $Z=XY$?
I know it's of the form
$$F_Z(z) = P(XY \leq z) = \iint_{\{(x,y):xy\leq z\}}f_{X,Y}(x,y)~dx~dy$$
I'm just a bit unclear on how to set up the bounds of the integral.
 A: Note that $Y<X^3$ and $XY<z \implies Y<z/X$; i.e., $Y < \min\{X^3,z/X\}$. 
For $X^3<z/X$, or $X < z^{1/4}$, we have
$$Y < \min\{X^3,z/X\} = X^3;$$
otherwise
$$Y < \min\{X^3,z/X\} = z/X.$$
Consequently,
$$\Pr\{XY <z\} = \int_{0}^{z^{1/4}}\int_{0}^{x^3}\frac{1}{4}dy\,dx + \int_{z^{1/4}}^{2}\int_{0}^{z/x}\frac{1}{4}dy\,dx,$$
where $0 < z < 16$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int_{0}^{2}{1 \over 4}
{\bracks{0 < z/x < x^{3}} \over \verts{x}}\,\dd x & =
{1 \over 4}\bracks{z > 0}
\int_{0}^{2}{\bracks{x > z^{1/4}} \over x}\,\dd x
\\[5mm] & =
{1 \over 4}\bracks{z > 0}\bracks{z^{1/4} < 2}\int_{z^{1/4}}^{2}
{\dd x\over x}
\\[5mm] & =
{1 \over 4}\bracks{0 < z < 16}\ln\pars{2 \over z^{1/4}}
\end{align}

