A seemingly simple triple integral has me stumped

1. Jan 29, 2009

Seda

1. The problem statement, all variables and given/known data

Essentially, do the volume integral of z^2 over the tetrahedron with vetices at (0,0,0) (1,0,0) (0,1,0) (0,0,1)

3. The attempt at a solution

There seems to be a ton(!) of brute-force algebra involved. Enough to make me question if I'm doing the problem right.
I set up the triple integral of z^2 in the order dzdydx with the following limits of integration.

z=0 to z= 1-x-y
y=0 to y= 1-x
x=0 to x=1

It didn't take to long for me to end up with trying to integrate a humongous polynomial in the second interval.

Evaluating z^3 / 3 at z = 1-x-y was fun enough.

But now after integrating again, I have to evaluate y/3 -xy-y^2/2+xy^2+x^2*y + y^3/3 + 1/3*x^3*y et cetera at y = 1-x seems to be a nightmare.

Am I tackling this the wrong way?

2. Jan 29, 2009

Dick

No, I think your approach is correct. Though I'm not sure I agree with your xy polynomial. You just have to work through the mess and be careful. If don't have to use a triple integral you can use a shortcut. If A(z) is the area of a cross section of the tetrahedron at a constant value of z, and you can use geometry to get a formula for that area in terms of z, then the triple integral is the integral of A(z)*z^2*dz. Hmm. That sort of suggests that you change the order of integration so you integrate over z last, it might be easier.

3. Jan 29, 2009

Seda

Yeah not only have I found a few mistakes, but i typed some parts in wrong, I was really just writing the polynomial to show how ugly.

Ill try integrating over x last then.

4. Jan 29, 2009

Dick

Z last, I think. You already did X last.

5. Jan 29, 2009

Seda

sorry, i meant z

6. Jan 29, 2009

Seda

Ok i have an answer now, thanks for the help.

Yes, integrating in the different order helped alot. Much cleaner algebra there.

7. Jan 29, 2009

Dick

Right. If your integrand depends on a subset of the variables, integrate over those variables last. It keeps them constants for as long as possible.