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Volume of a tetrahedron of a function

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the volume integral of the function T = z^2 over the tetrahedron with corners at
    (0,0,0), (1,0,0) , (0,1,0), and (0,0,1)

    3. The attempt at a solution

    z to x (1,0,-1)
    z to y (0,1,-1)

    Then i crossed them to get (1,1,1)

    Found the plane n dot (x-1, y , z) = x+y+z-1=0

    Normally i'd have no problem from here, but i'm not sure how i'm supposed to incorporate the T=z^2
     
  2. jcsd
  3. Sep 14, 2011 #2

    Char. Limit

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    Gold Member

    So you're dealing with the volume integral:

    [tex]\int \int \int z^2 dz dy dx[/tex]

    First thing is to find the bounds. You know that this is the volume contained in the first octant and bound by x+y+z-1=0. This can be rearranged to z=1-x-y, and the lower bound of z is zero (because it's in the first octant). So we can get our first bounds:

    [tex]\int \int \int_0^{1-x-y} z^2 dz dy dx[/tex]

    From there can you see the other two sets of bounds?
     
  4. Sep 14, 2011 #3
    okay, so when z is 0 y= 1-x and x goes from 0 to 1. If that is right i did this initially and i need to have more confidence in my answers...

    thank you for the clarification
     
  5. Sep 14, 2011 #4

    Char. Limit

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    Gold Member

    No worries. Have a great day!
     
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