Parameterize the following surface: $x +y +z=1$, $x,y,z>0$.

Question

I need to parameterize the following surface: $x +y +z = 1$, $x,y,z>0$.

I tried to put: $\sigma(u,v)=(u+v,u-v,-2u+1)$, but does it solve the case of $x,y,z>0$?

Answer 1

Your suggestion is a parameterisation once you restrict to $u$ and $v$ such that $u + v > 0$, $u - v > 0$, and $-2u + 1 > 0$ (e.g. $|v| < u < \frac{1}{2}$).

Alternatively, note that we can rearrange the equation $x + y + z = 1$ as $z = 1 - x - y$. Now we can write the surface as the graph of the function $f: \mathbb{R}^2 \to \mathbb{R}$, $f(x, y) = 1 - x - y$. As $z = 1 - x - y$, the inequality $z > 0$ is equivalent to $x + y < 1$. Therefore, a parameterisation of the surface is given by $\sigma : T \to \mathbb{R}^3$, $\sigma(u, v) = (u, v, 1 - u - v)$ where $T = \{(u, v) \in \mathbb{R}^2 \mid u, v > 0, u + v < 1\}$; geometrically, $T$ is the interior of a right-angled triangle with vertices $(0, 0)$, $(1, 0)$, $(0, 1)$.