PDF of Z = XY for Jointly Uniform (X,Y) with Parabolic Region 
Suppose that $(X,Y)$ is uniformly distributed on the subset of $\; \mathbb R^2$ defined by the inequalities $0 < X < 1$ and $0 < Y < X^2$.
Determine the probability density function of the random variable $Z = XY$

The $X^2$ in the $Y$ interval is seriously throwing me off and I'm not sure where to start. I have
$$f_{X,Y}(x,y) = 3 \qquad (x,y)\ \epsilon\;(0,1)\times(0,X^2)$$
$$f_{X,Y}(x,y) = 0 \qquad \qquad \qquad \quad otherwise$$
But I do not know where to go from here. I need to be able to explain my answer so I'd like to understand how/why it works. Thanks!
 A: You have found that$f_{X,Y}(x,y)=3\cdot\mathbf 1_{0<x<1, 0<y<x^2}$ .  $\checkmark$
From there use the Jacobian Transformation Theorem.
$$\begin{align}f_{X,Z}(x,z) &= \begin{Vmatrix}\dfrac{\partial[x,z/x]}{\partial[x,z]}\end{Vmatrix}~f_{X,Y}(x,z/x)\\[1ex] &= \dfrac 1{\lvert x\rvert}\cdot 3\cdot\mathbf 1_{0< x< 1~,~ 0< z/x< x^2}\\[1ex] &= \dfrac 3{x}\cdot\mathbf 1_{0< z< 1~,~ \sqrt[3]z< x< 1}\end{align}$$
After which integration with respect to $x$ finds the marginal distribution. (Notice the support)

 $$\begin{align}f_Z(z) &= \int_\Bbb R f_{X,Z}(x,z)~\mathsf d x\\[1ex] &= \int_{\sqrt[3]z}^1\frac 3x~\mathsf d x\cdot\mathbf 1_{0<z< 1}\\[1ex] &= -3\ln(\sqrt[3]z)\cdot\mathbf 1_{0< z< 1}\\[1ex] &= -\ln(z)\cdot\mathbf 1_{0< z< 1}\end{align}$$

A: I've looked over some examples and believe I have a solution.
$$f_{X,Y}(x,y) = 
\begin{cases}
3 &&& \text{for} \quad0<X<1, & 0<Y<X^2 \\
0 &&& \text{otherwise}
\end{cases}
$$
Then,
$$F_Z(z) = \begin{cases}
0 &&& Z \leq 0 \\
\Bbb P (Z \leq z) = \Bbb P (XY \leq z) = \Bbb P(Y \leq \frac{1}{X}z) &&& 0<Z<X^2 \\
1 &&& Z \geq X^2
\end{cases}
$$
Our support region is $$A = \{(X,Y)\; |\; 0 < X < 1,\; \; 0 < Y < X^2\} \cap \{(X,Y) \;| \;Y \leq \frac{1}{X} z\}$$ 
Integrating $f_{X,Y}(x,y)$ twice over this region, where the two "curves" $Y=X^2$ and $Y = \frac{1}{X}z$ intersect at $(z^{\frac{1}{3}},z^{\frac{2}{3}})$
$$ \Bbb P(Y \leq \frac{1}{X}z) = \iint_A3 \ dx\ dy = 3\int\limits^{z^{\frac{1}{3}}}_0\int\limits^{x^{2}}_0\ dy\ dx \ + \ 3\int\limits^{1}_{z^{\frac{1}{3}}}\int\limits^{\frac{z}{x}}_{0}\ dy\ dx \\
= x^3 \bigg\rvert_{0}^{z^{\frac{1}{3}}}+3z \ln{x}\bigg\rvert^0_{z^{\frac{1}{3}}} = z - 3z\ln{z^{\frac{1}{3}}} = z(1-\ln{z})$$
So we have;
$$F_Z(z) = \begin{cases} 0 & Z\leq 0 \\
z(1-\ln{z}) & 0 < Z < X^2 \\ 
1 & Z\geq X^2
\end{cases}$$
then differentiate to obtain $f_X(x)$;
$$f_X(x) = \begin{cases}
-\ln{z} & 0 < Z < X^2 \\
0 & \text{otherwise}
\end{cases}$$
I'm still unsure about my interval for Z, but otherwise I believe this is correct?
