# Calculating the Surface of a Tetrahedron

• Dec 13th 2010, 08:31 PM
spruancejr
Calculating the Surface of a Tetrahedron
I'm having trouble setting up the problem:

Compute $\int\int_s xy dS$, where S is the surface of the tetrahedron with sides z = 0, y = 0, x + z = 1, and x = y.

How should I approach this?
• Dec 13th 2010, 09:45 PM
snowtea
First draw a picture of the problem and the tetrahedron.

The straight forward way is to just perform the surface integral for each face of the tetrahedron and sum them. Could you try to do this, and progress a bit further?

There may be a way to use the Divergence Theorem if you can find a vector field that would give you that surface integral, but seeing the problem stated in this form, I would probably not go this route.
• Dec 13th 2010, 10:06 PM
spruancejr
I'm trying to figure out how to integrate over each face (i.e. finding the equations to use).
• Dec 14th 2010, 04:16 AM
HallsofIvy
Well, that was the purpose of drawing the figure! Each face is a triangle so you are asked for the total area of four triangles. You don't really need to integrate at all! The three faces, y= 0, z= 0, x+ z= 1, meet at (1, 0, 0). The three faces, y= 0, z= 0, y= x, meet at (0, 0, 0). The three faces, z= 0, x+ z= 1, y= x, meet at (1, 1, 0), and the three faces, y= 0, x+ z= 1, y= x, meet at (0, 0, 1). Those are the four vertices of the tetrahedron. The four faces have vertices (0, 0, 0), (1, 0, 0), (0, 0, 1); (0, 0, 0), (1, 0, 0), (1, 1, 0); (0, 0, 0)(0, 0, 1), (1, 1 0; and (1, 0, 0), (0, 0, 1), (1, 1, 0). You could, for example, use Heron's formula, Heron's formula - Wikipedia, the free encyclopedia, for the area of each of those.

If you do want to integrate, the triangle with vertices (0, 0, 0), (1, 0, 0), and (0, 0, 1) is in the xz-plane which has "differential of surface area" dxdz. The triangle with vertices (0, 0, 0), (1, 0, 0), and (1, 1, 0) is in the xy plane which has differential of surface area dxdy. The triangle with vertices at (0, 0, 0), (0, 0, 1), and (1, 1, 0) is in the plane y= x which has differential of surface area $\sqrt{2}dxdz$. The triangle with vertices at (1, 0, 0), (0, 0, 1), and (1, 1, 0) is in the plane x+ z= 1 which has differential of surface area $\sqrt{2}dxdy$.