My textbook defines simplex coordinates in $\mathbb{R}^n$ by:
$\Phi: \mathbb{R}^n\to \mathbb{R}^n , (x_1,\ldots,x_n)=\Phi(u_1,\ldots,u_n)$ where:
\begin{align*} x_1 & =u_1(1-u_2), & u_1 & =x_1+x_2+\cdots + x_n \\[6pt] x_2 & =u_1u_2(1-u_3), & u_1 & =\frac{\phantom{x_1 + {}} x_2+\cdots +x_n}{x_1+x_2+\cdots +x_n} \\ &\,\,\, \vdots & & \,\,\, \vdots \\ x_k & =u_1u_2\cdots u_k(1-u_{k+1}), & u_k & =\frac{\phantom{x_{k-1}+{}}x_k+\cdots +x_n}{x_{k-1} + x_k+\cdots + x_n} \\ & \,\,\, \vdots & & \,\,\, \vdots \\ x_n & =u_1u_2\cdots u_n, & u_n & =\frac{\phantom{x_{n-1}+{}}x_n}{x_{n-1}+x_n} \end{align*}
Problem:
A tetrahedron ($3$-simplex) is given by the vertices $(0,0,0)$, $(1,0,0)$, $(0,2,0)$, $(0,0,3)$. Find $\int _T xz+y^2 \, \mathrm{d}V$.
I know that I have to use the transformation formula. But how do I describe the tetrahedron with the new coordinate system?
Hint:
Your tetrahedron is given by $$T = \{(x,2y,3z) : x,y,z \ge 0, x+y+z \le 1\}.$$ New coordinates $(u,v,w)$ are given by $$\begin{cases} u = x+y+z, \\ v = \frac{y+z}{x+y+z}, \\ w = \frac{z}{y+z}\end{cases} \qquad \begin{cases} x = u(1-v), \\ y=uv(1-w), \\ z=uvw\end{cases}$$ so check that $$T=\{(x,2y,3z) : (x,y,z) = \Phi(u,v,w), 0 \le u,v,w \le 1\}) = \Psi([0,1]^3)$$
where $$\Psi(u,v,w) = (u(1-v),2uv(1-w),3uvw).$$