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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.

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

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

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

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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

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