Change of variables in $\int_{\Omega}(x+y+z)^2xyz \: dxdydz$

Question

I have the triple integral $\int_{\Omega}(x+y+z)^2xyz \: dxdydz$ , with $\Omega=\lbrace (x,y,z)\in \mathbb{R}^3:0\leq x+y+z\leq 1 , \:1\geq x\geq0, \: \: 1\geq y\geq 0, \: \:1\geq z\geq 0 \rbrace$. I set

$\begin{cases}x=u\\y=v\\ z=w(1-u-v)\end{cases}$ but the resut isn't correct.

The result must be $1/960$.

Answer 1

Put \begin{align*} u &= z & v &= y+z & w &=x+y+z \end{align*} Then $\Omega$ is described by \begin{align*} 0\leq u\leq 1 && u\leq v\leq 1 && v\leq w\leq 1 \end{align*} Moreover, we have $$ \frac{\partial(u,v,z)}{\partial(x,y,z)}= \det \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} = -1 $$ This gives \begin{align*} \iiint_\Omega (x+y+z)^2xyz\,dV &= \int_0^1\int_u^1\int_v^1 w^2(w-v)(v-u)u\left\lvert\frac{\partial(x,y,z)}{\partial(u,v,w)}\right\rvert\,dw\,dv\,du \\ &= \int_0^1\int_u^1\int_v^1 w^2(w-v)(v-u)u\,dw\,dv\,du \\ &\vdots \\ &=\frac{1}{960} \end{align*}