Problem finding Jacobian when computing density I'm having trouble finding the Jacobian when trying to compute a distribution. If $(X,Y)$ is a point on a unit disk with radius $1$, I'd like to find the density of the distance between the point and the centre of the disk. 
So the joint density function of $X$ and $Y$ is 
$$
f_{XY}(x,y) = \begin{cases}
1/\pi, & x^2 + y^2 \leq 1,\\
0,     & \text{otherwise}
\end{cases}.
$$
I put $U = \sqrt{X^2+Y^2}$ and use an auxiliary random variable $V = \mathrm{arctan}(Y/X).$ So there is a 2-to-1 mapping.
Now I wonder what the Jacobian might be. Or, well, the book tells me it's $u$, but I don't know how to get there myself. Using wolfram alpha (admittedly I'm probably just using it wrong) gets me the answer $0$, not $u$. 
Normally, I would work with simpler auxiliary variables, which have helped me compute the Jacobian for other problems, but in this case I'm adviced to use $\mathrm{arctan}(Y/X)$ and that could be what makes this so hard for me.  
 A: Full solution, using Dirac delta:
$$
p_U(u)=\int_D dx dy \frac{1}{\pi}\delta\left(u-\sqrt{x^2+y^2}\right)\ .
$$
Making a change to polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$, whose Jacobian is $=r$, 
$$
p_U(u)=\frac{1}{\pi}\int_0^{2\pi}d\theta\int_0^1 dr\ r\ \delta(r-u)=2u\qquad 0\leq u\leq 1\ ,
$$
which is correctly normalized, $\int_0^1 du\ p_U(u)=1$.

About the Jacobian, I am not sure what the problem is [and also not sure about the statement '2-to-1' you made...you had two variables $X,Y$ and introduce two other variables $U,V$]. The Jacobian between the old variables $(X,Y)$ and the new ones $(U,V)$ is
$$
\det\begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{pmatrix}
=
\det\begin{pmatrix}
\frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\
-\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2}
\end{pmatrix}=\frac{1}{\sqrt{x^2+y^2}}=\frac{1}{u}
$$
(obviously, then, the Jacobian between the new variables and the old ones is $u$).
