If $X$ and $Y$ are iid with $U(0,2)$, $Z=2X+Y$ and $V=e^X$
What is the joint density of $(Z,V)$?
$\begin{align} V=e^X &\Rightarrow X=lnV\\ Z=2X+Y &\Rightarrow Y=Z-2lnV \end{align}$
$ J= \begin{vmatrix} \frac{\partial x}{\partial z} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial z} & \frac{\partial y}{\partial v} \\ \end{vmatrix} = \begin{vmatrix} 0 & \frac{1}{v} \\ 1 & -\frac{2}{v} \end{vmatrix} =-\frac{1}{v} $
$\begin{align} f_{Z,V}(z,v)&=f_{X,Y}(lnV,Z-2lnV)*|J| \\ &=f_X(lnV)f_Y(Z-2lnV)*|-\frac{1}{v}| \end{align}$
$\begin{align} 0 \le lnV \le 2 &\Rightarrow 1 \le V \le e^x \\ 0 \le Z-2lnV \le 2 &\Rightarrow 2lnV \le Z \le 2+2lnV \\ &\Rightarrow 0 \le Z \le 6 \end{align}$
For $1 \le v \le e^x$ and $0 \le z \le 6$,
$\begin{align}
f_{Z,V}(z,v)&=\frac{1}{2}*\frac{1}{2}*\frac{1}{v} \\
&=\frac{1}{4v}
\end{align}$
However, when I check the answer by using double integration
$\begin{align}
\int_{1}^{e^2} \int_{0}^{6}\frac{1}{4v} \,dz\,dv &= \frac{3}{2}\int_{1}^{e^2}\frac{1}{v} \,dv\\
&=\frac{3}{2}*(2-0) \\
&=3
\end{align}$
The integral of density by the support should instead be 1, that means my answer is wrong, what have I done wrong?
Indeed $Z\in [0,6]$, but if $\varphi $ is s.t. $\varphi (X,Y)=(U,V)$, then it's not true that $$\varphi(\{(X,Y)\in (0,2)^2\})= [1,e^2]\times [0,6].$$ What you nevertheless have is $$\varphi (\{(X,Y)\in (0,2)^2\}=\{(V,Z)\mid V\in (1,e^2), 2\ln(V)\leq Z\leq 2\ln(V)+2\}).$$