# 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.