Given: $$f(x,y) = 2x^2y + \sqrt{y}, \quad 0<x<1, 0<y<1$$ $$U=\min{\{X,Y\}}, V = \max{\{X,Y\}}$$
I am interested in the joint distribution of $U$ and $V$.
Thus: $$\begin{cases} U=X \\ V=Y \end{cases}\rightarrow\begin{cases} X=U \\ Y=V \end{cases}$$
In this case: $ J_1=\begin{vmatrix} 1 & 0 \\ 0& 1 \end{vmatrix}=1 $.
The transformation could also be: $$\begin{cases} U=Y \\ V=X \end{cases}\rightarrow\begin{cases} X=V \\ Y=U \end{cases} $$
In this case: $ J_2=\begin{vmatrix} 0 & 1 \\ 1& 0 \end{vmatrix} =-1 $.
Hence: $$f(u,v) = (2x^2y + \sqrt{y}) |J_1| \mathbf{1}_{x=u,y=v} + (2x^2y + \sqrt{y}) |J_2| \mathbf{1}_{x=v,y=u} $$ $$= 2uv(u + v) + (\sqrt{v} + \sqrt{u}) $$ Thus: $$ f(u,v) = 2uv(u + v) + (\sqrt{v} + \sqrt{u}) \quad \text{where} \quad 0<u<1, 0<v<1, v>u$$ Is there anyting wrong with this ? is there anyway to simplify the approach