Uniform Distributions in Probability X, Y, and Z are independent and uniformly distributed over [0,1].
I'm trying to find the distribution of XY by using the joint transformation T = X, W = XY.  
We haven't learned transformations yet, but this is a review question that I can't seem to tackle. I was wondering how I would approach this problem or set up the initial equation in order to solve.  
I know it involves deriving the joint pdf of T and W, and integrating T out to obtain the marginal pdf of W, but I don't know how to set up the start of the problem.
 A: The transformation is $f_{T,W}(t,w) = f_{X,Y}(x,y)\;|J(x,y)|^{-1}$
Where $J$ is the Jacobian: $J(x,y) = \begin{Vmatrix}
\dfrac{\partial t}{\partial x} & \dfrac{\partial t}{\partial y}
\\
\dfrac{\partial w}{\partial x} & \dfrac{\partial w}{\partial y}
\end{Vmatrix}$
And of course: $t = x 
\\ w = xy
\\ f_{X,Y}(x.y) = \mathbf 1_{(0,1)}(x) \; \mathbf 1_{(0,1)}(y)$
So we have: $f_{T,W}(t,w) = \mathbf 1_{(0,1)}(t) \; \mathbf 1_{(0,1)}(w/t) \; |t|^{-1}$
Thus $$\begin{align}
f_W(w) & = \int\limits_{-\infty}^\infty \mathbf 1_{(0,1)}(t) \; \mathbf 1_{(0,1)}(w/t) \; |t|^{-1} \operatorname d t
\\[1ex]
 & = \int_{w}^1 \frac 1 t \mathbf 1_{(0,1)}(w) \operatorname d t
\\[1ex]
 & = - \ln w \quad \mathbf 1_{(0,1)}(w)
\end{align}$$

Alternatively.  Find the CDF and differentiate.
The CDF is determined by integrating the joint pdf over the area of the unit square beneath the hyperbola $y=1/x$.
$$\begin{align}
f_W(w) & = \frac{\mathrm d \;}{\mathrm d w} \left( \int\limits_0^w \int\limits_0^1 \operatorname d y\operatorname d x + \int\limits_w^1 \int\limits_0^{w/x} \operatorname d y \operatorname d x \right) \; \mathbf 1_{(0,1)}(w)
\\[1ex]
 & = \frac{\mathrm d \;}{\mathrm d w} \left( w - w \ln w \right) \; \mathbf 1_{(0,1)}(w)
\\[1ex]
 & = - \ln w \quad \mathbf 1_{(0,1)}(w)
\end{align}$$
