The task is to calculate two-dimentional probability density of $[X,Y]$ if we know that: $$ R \sim U(0,1), \quad \Phi \sim U(0, 2\pi), $$ $R$ and $\Phi$ are independent and $$ X := R \cos(\Phi), \quad Y:= R \sin(\Phi). $$
We have that: $$ f_{[R, \Phi]}(r, \phi) = \mathbf{1}_{[0,1]}(r) \frac{1}{2 \pi} \mathbf{1}_{[0,2\pi]}(\phi) $$
$$ \begin{cases} X = R \cos(\Phi) \\ Y = R \sin(\Phi) \end{cases} \quad \Rightarrow \quad \begin{cases} \Phi = \arcsin \Big{(} \frac{Y}{ \sqrt{X^2 + Y^2} } \Big{)}\\ R = \sqrt{X^2 + Y^2} \end{cases} $$
$$ |J| = R = \sqrt{X^2 + Y^2} $$
$$ g_{[X,Y]}(x,y) = f( \Phi(x,y), R(x,y) ) |J| = $$
$$ \mathbf{1}_{[0,1]}\Big{(} \sqrt{x^2 + y^2} \Big{)} \frac{1}{2 \pi} \mathbf{1}_{[0, 2\pi]} \Big{(} \arcsin \Big{(} \frac{y}{\sqrt{x^2+y^2}} \Big{)}\Big{)} \sqrt{x^2+y^2} $$
Is this solution correct so far?
Now I have a problem solving:
$$ \begin{cases} 0 \le \sqrt{x^2 + y^2} \le 1 \\ 0 \le \arcsin \Big{(} \frac{y}{\sqrt{x^2+y^2}} \Big{)} \le 2\pi \end{cases} $$
The jacobian determinant is with respect to $(x,y)$, so it should be
$$J=J\left(\frac{r,\phi}{x,y}\right)=\frac1{J\left(\frac{x,y}{r,\phi}\right)}=\frac1r=\frac1{\sqrt{x^2+y^2}}$$
Note that $$0<r<1\implies r^2<1\implies x^2+y^2<1$$
And $$0<\phi<2\pi\implies \tan\phi=\frac{y}{x}\in \mathbb R\implies (x,y)\in \mathbb R^2$$
So joint density of $(X,Y)$ is $$f_{X,Y}(x,y)=f_{R,\Phi}(r,\phi)|J|=\cdots$$