Consider the following surface:
$$A=\{(x,y,z,t)\in\mathbb{R^4}:0\leq x,y,z,t\leq1,\quad x+y+z+t=1\}$$
I need to parametrize it to be able to calculate its volume. Of course, I thought on seeing it as the graph of an $\mathbb{R^3}$ function, namely:
$\phi:\mathbb{R^3}\longrightarrow\mathbb{R^4} \quad\phi(x,y,z)=1-x-y-z$
The problem is, where to define this function? If I consider the domain $[0,1]^3$, then the fourth variable can go down to $-2$, which is not what I want. If, to try to solve this, I consider $[0,\frac{1}{3}]^3$, then this time the first 3 variables can only go up to a third, and hence I'm missing most points, like $(1,0,0,0)\in A$, for instance.
How to find the proper domain to define this function? Or maybe this is not the proper parametrization?
Your parametrization is fine with $t=\phi(x,y,z)$. You're now integrating over $0 \leq x+y+z \leq 1$ which you can also describe with $0 \leq x \leq 1$ and $0 \leq y \leq 1-x$ and $0 \leq z \leq 1-x-y$.
So after you compute your volume element your integral is $$ \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 2 dz dy dx $$