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A seemingly simple triple integral has me stumped

  1. Jan 29, 2009 #1
    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. jcsd
  3. Jan 29, 2009 #2


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    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.
  4. Jan 29, 2009 #3
    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.
  5. Jan 29, 2009 #4


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    Z last, I think. You already did X last.
  6. Jan 29, 2009 #5
    sorry, i meant z
  7. Jan 29, 2009 #6
    Ok i have an answer now, thanks for the help.

    Yes, integrating in the different order helped alot. Much cleaner algebra there.
  8. Jan 29, 2009 #7


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    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.
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