The joint pdf of $(X,Y)$ is given by:
$f_{X,Y}(x,y)=4xye^{-(x^2+y^2)}$ if $0<x,y<\infty$
I have to find the distribution of $R=\sqrt{X^2+Y^2}$.
I have approached this problem in the following two ways:
First approach:
Consider the transformation $(X,Y)\rightarrow (R,Y)$ where $R=\sqrt{X^2+Y^2}$.
$x=\sqrt{r^2-y^2}$ as $x>0$
$0<x,y<\infty \implies 0<r<\infty$
Again, $r^2>y^2$.
So, $0<y<r$.
Jacobian of the transformation is:
$J=J(\frac{x,y}{r,y})=\frac{\delta x}{\delta r}=\frac{r}{\sqrt{r^2-y^2}}$.
So, joint pdf of $(R,Y)$ is:
$f_{R,Y}(r,y)=f_{X,Y}(\sqrt{r^2-y^2},y)|J|$
$=4rye^{-r^2}$, $0<y<r<\infty$
So, pdf of $R$ is:
$f_{R}(r)=\int_{0}^{r}4rye^{-r^2}dy$
$=2r^{3}e^{-r^2}$, $0<r<\infty$
Second approach:
Consider the transformation $(X,Y)\rightarrow (R,\Theta)$ where $x=r\cos\theta$ and $y=r\sin\theta$.
So, $r=\sqrt{x^2+y^2}$.
$0<x,y<\infty \implies 0<r<\infty,0<\theta<2\pi$
Jacobian of the transformation is:
$J=J(\frac{x,y}{r,\theta})=\begin{vmatrix}\cos\theta & -r\sin\theta\\ \sin\theta& r\cos\theta\end{vmatrix}=r$
So, joint pdf of $(R,\Theta)$ is:
$f_{R,\Theta}(r,\theta)=f_{X,Y}(r\cos\theta,r\sin\theta)|J|$
$=4r^3\sin\theta\cos\theta e^{-r^2}$, $0<r<\infty,0<\theta<2\pi$
So, pdf of $R$ is:
$f_{R}(r)=\int_{0}^{2\pi}4r^3\sin\theta\cos\theta e^{-r^2}d\theta$
$=2r^3e^{-r^2}\int_{0}^{2\pi}\sin 2\theta d\theta$
$=0$
I don't get it why the above two approaches give me different answers. Which one is correct? Where is the mistake? Anyone please help me to clear this confusion. Thanks in advance.
Everything looks perfect. Only you need to observe, in the second approach, that:
$X=\{(x,y), x>0, y>0\}$
gets mapped to:
$R=\{(\theta,r), 0<\theta<\pi/2, r>0\}$
Therefore the integral does not extend from $0$ to $2\pi$ but from $0$ to $\pi/2$