Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I think your Jacobian is wrong. I got $J(u,v,w)=u^2v$. – Christian Blatter Jun 19 '13 at 12:48

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
$$ \iiint_T e^{-(x+y+z)^3}dxdydz = \iiint_U e^{-u^3} u^2 v\,dudvdw = \frac{e-1}{6e}. $$

share|cite|improve this answer

You can integrate $\int e^{-u^3}u^2du$ by the substitution $p=u^3, dp=3u^2\; du$, giving $\int e^{-u^3}u^2du=\frac {1}3\int e^{-p}\; dp$

share|cite|improve this answer
thanks! That was fairly simple.. But is the approach correct though? And how should I deal with the modulus, since when v is more than 1/2, it will be negative.. – Keksainis Oct 24 '12 at 21:19
@Keksainis: the approach is fine. I didn't check the Jacobian and am surprised that it changes sign with $v$. – Ross Millikan Oct 24 '12 at 21:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.