Parametrization of a tetrahedron?

Question

I need to do a surface integral over the surface bounded by $x + 2y + z = 4$ and the coordinate planes. How do you go about parametrizing this surface? My first thought was with $x = x$, $y = y$, and $z = 4 - x - 2y$, but there are no numerical bounds for x and y which will only capture the surface within the first octant. Is there another parametrization? Also, I realize it would be way easier to use the divergence theorem, but the question here is to confirm the theorem by evaluating it as a surface integral and with the theorem. Any help is appreciated, thanks.

Answer 1

No, it is right, you parameterize it as $$x=u$$ $$y=v$$ $$z=4-u-2v$$ Where $$0\leq u \leq 4$$ $$0\leq v \leq \frac{4-u}{2}$$ I have just renamed the parameterizing variables.

Answer 2

The complete boundary surface $\partial T$ of your full tetrahedron $T$ is a so-called $2$-chain consisting of four triangles. The total surface integral $\int_{\partial T}$ is then the sum of four integrals computed individually.

A parametrization of the "roof" of $T$ is given by $$(x,y)\mapsto(x,\ y,\ 4-x-2y)\qquad \left(0\leq y\leq {4-x\over2},\ \ 0\leq x\leq4\right)\ ,$$ whereas a parametrization of the "floor" of $T$ is given by $$(x,y)\mapsto(x,\ y,\ 0)\qquad \left(0\leq y\leq {4-x\over2},\ \ 0\leq x\leq4\right)\ .$$ In both cases you have to check that the outward normal comes out right. I leave the parametrization of the two "walls" to you.