Finding density function of random variable, which is division of two other random variables. I have following 2-dimensional random variable $(x,y)$:
$$
f(x,y) = 1, \quad 0 \leq x \leq 1, \quad 0 < y \leq 1
$$
I have to find density function of random variable $Z = \frac{X}{Y}$.
I am trying to define another 2-D random variable:
$$
Z = \frac{X}{Y} \\
W = Y
$$
We can define a function:
$$
g(x,y) = (\frac{x}{y}, y)\\
g^{-1}(z,w) = (zw, w)
$$
Jacobian determinant is equal to $w$.
Then, new density function $h$ should be equal to:
$$
h(z,w) = f(zw, w) \cdot w = w
$$
If we look at these variables, we see that $z \in [0;\infty]$ and $w \in [0,1]$.
However, if we try to find marginal density for variable $z$:
$$
h_z(z) = \int_0^1 w \mathrm{d} w = \frac{1}{2}
$$
But then, obiviously, if density function is equal to $\frac{1}{2}$ and $z \in [0;\infty]$, it can't be density function, because sum of probabilities doesn't sum to $1$.
Where have I made mistake?

The joint density of $(X,Y)$ is
$$
f(x,y)=\mathbf{1}_{0<x<1}\mathbf{1}_{0<y<1}
$$
and hence the density of $(X/Y,Y)$ is
$$
h(z,w)=|w|f(zw,w)=w\mathbf{1}_{0<zw<1}\mathbf{1}_{0<w<1}=w\mathbf{1}_{0<z<w^{-1}}\mathbf{1}_{0<w<1}.
$$
The marginal density of $X/Y$ is 
$$
h(z)=\int_\mathbb{R} h(z,w)\,\mathrm dw
$$
which is zero for $z\leq 0$ and for $z>0$ equals
$$
h(z)=\int_{\mathbb{R}}w\mathbf{1}_{0<w<z^{-1}}\mathbf{1}_{0<w<1}\,\mathrm dw=\int_0^{\min(1,z^{-1})}w\,\mathrm dw=\frac12 \min(1,z^{-2}).
$$
Thus we conclude that
$$
h(z)=
\begin{cases}
0,\quad &z\leq 0,\\
\frac12, \quad &0<z<1,\\
\frac12 z^{-2},\quad &z\geq 1.
\end{cases}
$$
