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Centroid of a Solid (triple integral)

  1. Jul 14, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the centroid of the solid:
    the tetrahedron in the first octant enclosed by the coordinate planes and the plane x+y+z=1.


    2. Relevant equations

    xcenter = [tex]\frac{\int\int\int_G x dV}{V}[/tex]

    ycenter = [tex]\frac{\int\int\int_G y dV}{V}[/tex]

    zcenter = [tex]\frac{\int\int\int_G z dV}{V}[/tex]

    3. The attempt at a solution

    I have shown my attempt for xcenter, as the same problem arises for each one.

    [tex]\frac{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} x dzdydx}{\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV} [/tex]

    but [tex]\int_{x=0}^1\int_{y=0}^1\int_{z=0}^{1-y-x} dV[/tex] is equal to zero,
    so the above expression is undefined.

    According to my text, the answer should be (1/4,1/4,1/4). Could someone point out what I did wrong? (Perhaps my bounds of integration?)
     
  2. jcsd
  3. Jul 14, 2008 #2

    tiny-tim

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    Hi Knissp! :smile:

    Try [tex]\int_{x=0}^1\int_{y=0}^ {1-x}\int_{z=0}^{1-y-x}[/tex] :wink:
     
  4. Jul 14, 2008 #3
    Oh, I see now! Thank you!
     
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