Hi I am learning probability on my own, and came across this problem online, find pdf of of z given
$$f_{xy} = \frac{1}{2\pi\sigma^2}\exp(-\frac{x^2+y^2}{2\sigma^2})$$
$$Z = X^2+Y^2$$
The solution uses
$ \sin^2{\phi} +\cos^2{\phi} = 1 $
$ x = r\cos(\phi)$
$ y = r\sin(\phi) $
$ z = x^2 + y^2 = r^2\cos^2(\phi) + r^2\sin^2(\phi) = r^2$
to get
$f_{r\phi} = \frac{r}{2\pi\sigma^2}\exp(-\frac{r^2}{2\sigma^2}) $
this is where I got lost, where does the r in the $\frac{r}{2\pi\sigma^2}$ come from? This solution explains "the factor of r is from the differential area relationship $dxdy = rdrd\phi$", but where does dxdy come from?
If $R=f(X,Y)$ and $\Phi=g(X,Y)$ with $r = \sqrt{x^2+y^2}$ and $\phi=arctan(\dfrac{y}{x})$. Then the joint density function $$f_{R\Phi}(r,\phi) = |J|f_{XY}(r,\phi)$$
Where Jacobian is $|J|=$$ \begin{vmatrix} \dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial \phi}\\ \dfrac{\partial y}{\partial r}&\dfrac{\partial y}{\partial \phi}\\ \end{vmatrix} = \begin{vmatrix} cos(\phi) & -rsin(\phi) \\ sin(\phi) & rcos(\phi)\end{vmatrix} =r$.
$$f_{R\Phi}(r,\phi) = r \dfrac{1}{2\pi\sigma^2}e^{-\dfrac{(x^2+y^2)}{2\sigma^2}} = \dfrac{r}{2\pi\sigma^2}e^{-\dfrac{r^2}{2\sigma^2}} $$