How do I do this....and how many ways are there to do this?
I think dwsmith is misinterpreting the question. The "surface of the tetrahedron" includes all four faces.
One method, of course, would be to integrate on all four faces separately.
a) For example, the face with vertices (0,0,0), (1,0,0), and (0,1,0) is the triangle with x= 0, y= 0 and x+ y= 1 as boundaries. To integrate over that, let x range from 0 to 1. For each x, then, y ranges from 0 to 1- x:
$\displaystyle \int_{x= 0}^1\int_{y= 0}^{1- x} (x^2+ y^2) dy dx$
(0f course, z= 0 on this face.)
b) To integrate on the face with vertices (0,0,0), (1,0,0), and (0,0,1), do the same in the xz-plane- just replace "y" with "z".
c) To integrate on the face with vertices (0,0,0), (0,1,0), and (0,0,1) do the same in the yz-plane- just replace "x" with "z".
It should not be hard to see, from symmetry, that those three integrals will give the same result- so just do the first one and multiply by 3.
d) The face with vertices (1,0,0), (0,1,0) and (0,0,1) is a little harder. Projecting down to the xy-plane, we have the same triangle as in (a) but now, since the plane is x+ y+ z= 1, z= 1- x- y:
$\displaystyle \int_{x=0}^1\int_{y=0}^{1- x} \sqrt{x^2+ y^2+ (1- x- y)^2}dy dx$.