Here is the statement :
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Let\ \varphi : [0,1]^{3} \rightarrow S\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (u,v,w)\mapsto (u(1-v) , uv(1-w) , uvw)$
where $S=\left \{ (x,y,z)\in \mathbb{R}^{3} : x\geqslant 0, y \geqslant 0,z \geqslant 0, x+y+z\leqslant 1 \right \}$
a) Show that $ \varphi$ restricted to $]0,1[^{3}$ is a $C^{1}$ diffeomorphism from $]0,1[^{3}$ to $Š=\left \{ (x,y,z)\in \mathbb{R}^{3} : x> 0, y> 0,z >0, x+y+z<1 \right \}$
$\textit{Indication : Show that $ \varphi$ makes a bijection and then use inverse function theorem}$.
b) Deduce the value of $\int \int \int_{S}\frac{1}{(1+x+y+z)^2}dxdydz$.
$$\underline{Solution}$$
a) Let $(x,y,z) ∈ S$ $\Rightarrow$ $\left\{ \begin{matrix}x=u(1-v)) \\ y=uv(1-w)) \\ z=uvw \end{matrix}\right.$
So :
$\cdot x+y+z = u-uv+uv-uvw+uvw = u ∈ ]0,1[$
$\cdot u(1-v)=x \Rightarrow v=1-\frac{x}{x+y+z} = \frac{y+z}{x+y+z} ∈ ]0,1[$
$\cdot uv(1-w)=y \Rightarrow w=1-\frac{y}{uv} = 1 -\frac{y}{y+z} ∈ ]0,1[$
And then we see that we can write in a unique way every u,v,w so there is a bijection from $]0,1[^{3}$ to Š.
Now let's compute the Jacobian to see if it's a local diffeomorphism :
$Jacobian(\varphi ')$= $\begin{pmatrix}
1-v \ \ \ \ \ -u\ \ \ \ \ \ \ \ \ \ \ 0 \\ \ \ \
v(1-w) \ \ u(1-w))\ \ -uv \\
vw \ \ \ \ \ \ \ \ \ uw\ \ \ \ \ \ \ \ \ uv \\
\end{pmatrix}$ $\rightarrow$ det=$u^2v > 0 \ \forall \ (u,v,w) \ \in \ ]0,1[^3$.
So we have a local diffeomorphism + bijection $\rightarrow$ we have a global diffeomorphism !
$\\ $
b) : $\int \int \int_{S}\frac{1}{(1+x+y+z)^2}dxdydz$ = $\int \int \int_{[0,1]^3}\frac{1}{(1+u)^2}*det(Jacobian(\varphi '))dudvdw$
The determinant is $u^2v$ and then :
$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+u)^2}*u^2vdudvdw$ = $\int_{0}^{1}1dw\int_{0}^{1}vdv\int_{0}^{1}\frac{u^2}{(1+u)^2}du$ = $-ln(2) + \frac{3}{4}$.