$\iint_D y^3 dA$, where $D$ is bounded by $y = 2 e^{-x}$, $y = 3 e^{-x}$, $y = e^x$, $y = 4 e^x$ I'm trying to solve

Calculate $\iint_D y^3 dA$, where $D$ is bounded by $y = 2 e^{-x}$, $y= 3 e^{-x}$, $y = e^x$, $y = 4 e^x$

This question is similar to this question, i.e. we need to transform our coordinate axis to a one in which $D$ will look nicer.
If I do 
$$u = e^{-x},$$
the region look nicer, but still ugly, and the boundary of the integral is still problematic, so I need to transform $y$ also, but I couldn't figure out how to transform $y$.
So, how can we transform $y$ in this case ? 
Any help or hint is appreciated.

Let $u=ye^x$ and $v=ye^{-x}$, then $x=\frac{1}{2}(\ln u-\ln v)$ and $y=\sqrt{uv}$. Also, $D$ is bounded by $u=2$, $u=3$, $v=1$, $v=4$. Compute the Jacobian:
$$
\frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix}\frac{1}{2u} &\frac{1}{2}\sqrt{\frac{v}{u}} \\ -\frac{1}{2v}&\frac{1}{2}\sqrt{\frac{u}{v}}\end{vmatrix}=\frac{1}{2\sqrt{uv}}.
$$
Thus
\begin{align}
\iint_D y^3 dA &= \int_1^4\int_2^3 u^{3/2}v^{3/2}\cdot \frac{1}{2u^{1/2}v^{1/2}} dudv\\
&=\int_1^4\int_2^3 \frac{uv}{2}dudv.
\end{align}
