Find the distribution of $ R=\sqrt{X^2+Y^2}$ where $ (X,Y)$ is uniform on the unit ball (new version) $(X,Y) $ is a random vector such that
$$ f_{(X,Y)}(x,y)=\begin{cases}  \frac{1}{\pi} & \text{if} & x^2+y^2\leq 1 \\
                  0 & \text{if} & x^2+y^2> 1      
  \end{cases} $$
I want to find $ f_R(r)$ where $R=\sqrt{X^2+Y^2} $. I know that someone has already asked this question (Here is the link) but I think I have a different solution which involves more calculus. Here it is:
Define $g:\mathbb{R}^2 \to \mathbb{R}^2 $ by $g(x,y)=(\sqrt{x^2+y^2},x) $, this function is inyective in the set $ S_o=\{ (x,y)|y\geq 0\}$ and has inverse $ g^{-1}(u,v)=\left( \sqrt{v^2-u^2},v\right)$. The jacobian:
$$J_{g^{-1}}(u,v)= \begin{pmatrix} 
\frac{-u}{\sqrt{v^2-u^2}} & \frac{v}{\sqrt{v^2-u^2}} \\
0 & 1 
\end{pmatrix}\rightarrow |\det(J_{g^{-1}}(u,v))|=\frac{|u|}{\sqrt{v^2-u^2}}$$
Therefore
$$ f_{(U,V)}=\frac{f_{(X,Y)}(\sqrt{v^2-u^2})|u|}{\sqrt{v^2-u^2}} $$
Since we want to find $f_{R}=f_{U} $ it suffies to calculate
$$ f_{U}(u)=\int_{\mathbb{R}} \frac{f_{(X,Y)}(\sqrt{v^2-u^2})|u|}{\sqrt{v^2-u^2}} \;dv$$
Am I right in this idea? 
Thank you all in advance
 A: I'm not sure if what I asked is right (I think it is) but now I have a diferent and probably easier solution for the problem, thanks to a hint given in Probability Essentials by Jean Jacod Philip Protter:
Consider the following function:
 $$ g:\mathbb{R}^2\to \mathbb{R}^2 \quad g(x,y)=\left(\sqrt{x^2+y^2},\arctan\left(\frac{x}{y}\right)\right) $$
It is inyective and has inverse $ g^{-1}(r,s) =(r\sin(s),r\cos(s)) $ then 
$$ f_{g(X,Y)}(r,s)=f_{(X,Y)}(g^{-1}(r,s))|\det(J_{g^{-1}})(r,s)| =\frac{r}{\pi}1_{[0,1]}(r^2)=\frac{r}{\pi}1_{[-1,1]}(r) $$
Recall that we want to know $f_R $ which is the marginal distribution of $  f_{g(X,Y)}$:
$$ f_R(r)= \int_{\mathbb{R}} f_{g(X,Y)}(r,s) \; ds = \int_{0}^{2\pi} \frac{r}{\pi}1_{[-1,1]}(r) \; ds =2r 1_{[-1,1]}(r)   $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{f}_{R}\pars{r} = \int_{0}^{2\pi}{1 \over \pi}\bracks{0 < r < 1}r\,\dd\theta =
\bbx{\bracks{0 < r < 1}\pars{2r}}
\end{align}
