Change of variables for triple integrals Question: Consider the one-to-one transformation $(u; v;w) \to (x; y; z)$ defined by the equations $u = x + y + z; uv = y + z; uvw = z;$ which maps the unit cube $U$ defined by $0 \lt u \lt 1, 0 < v < 1, 0 < w < 1$ onto the tetrahedron $T$ defined by $x > 0, y > 0, z > 0, x + y + z < 1.$
I need to evaluate the integral $\int \int \int e^{-(x+y+z)^3} \;dz \;dy \;dz$ changing the variables.
For the Jacobian I got $u^2v(1-2v),$ then the integral would be
$\int_0^1 \int_0^1 \int_0^1 e^{-u^3} |u^2  v (1-2v) | \; du \; dv \; dw$
Am I correct so far?
I'm struggling to integrate $e^{-u^3}$ from here. 
 A: This is a community-wiki answer trying to remove this question from the unanswered queue.


For the Jacobian I got $u^2v(1-2v)$

This is not correct. As Christian Blatter pointed out in the comments, the Jacobian should be $u^2 v$. For 
$$
x = u(1-v), \;y= uv(1-w),\; z = uvw.
$$
Hence the Jacobian (w/o absolute value) is:
$$
\det\begin{vmatrix} 1-v & -u & 0
\\ 
v(1-w) & u(1-w) & -uv
\\
vw & uw & uv \end{vmatrix} 
= (1-v)\begin{vmatrix} 
 u(1-w) & -uv
\\
 uw & uv \end{vmatrix} + u  \begin{vmatrix}  
v(1-w)   & -uv
\\
vw  & uv \end{vmatrix}
= u^2v,
$$
and ${|\det J|} = u^2v$ as well.


I'm struggling to integrate $e^{-u^3}$ from here.

Even in the integral you gave with the incorrect Jacobian, you could do
$$\int_0^1 \int_0^1 |v (1-2v) | \left(\int_0^1 e^{-u^3} u^2   \; du\right) \, dv \, dw$$
like Ross Millikan said. The inner most integral about $u$ yields a constant, for the rest you can do: 
$$
\int_0^1 \int_0^1 |v (1-2v) |  \, dv \, dw = 
\int_0^1 \int_0^{1/2}  v (1-2v)   \, dv \, dw 
+ \int_0^1 \int_{1/2}^1  v (2v-1)    \, dv \, dw.
$$

Finally, with the correct Jacobian, the integral
$$
\iiint_T e^{-(x+y+z)^3}dxdydz = \iiint_U e^{-u^3} u^2 v\,dudvdw = \frac{e-1}{6e}.
$$
A: You can integrate $\int e^{-u^3}u^2du$ by the substitution $p=u^3, dp=3u^2\; du$, giving $\int e^{-u^3}u^2du=\frac {1}3\int e^{-p}\; dp$
