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Equation of the plane, keep getting z=0 but can't be

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    ok so the problem is to evaluate (triple integral)x^2 dV of the solid tetrahedron with vertices (0,0,0) (1,0,0) (0,1,0) and (1,0,1)



    2. Relevant equations



    3. The attempt at a solution
    so (triple integral)x^2 dzdydx
    i made the x-y plane be the base with 0<x<1 and 0<y<1-x
    for the z limits, i know they should be 0 to the plane but i am having trouble getting the equation of the plane. i need to use the points (1,0,0) (0,1,0) and (1,0,1) right?
    but when i do the cross products of the two vectors <-1,1,0> and <0,0,-1> i get that the k component is 0. which makes no sense looking at the picture because z is not always zero. what am i doing wrong?
     
  2. jcsd
  3. Dec 3, 2008 #2

    Dick

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    You aren't drawing the right picture. The plane defined by those three points is perpendicular to the x-y plane. I think you want to use (0,0,0), (0,1,0) and (1,0,1) to define the plane that lies above the region in the x-y plane. Try the picture again.
     
  4. Dec 3, 2008 #3
    now i'm getting a plane equation of z=x which makes no sense when i integrate
     
  5. Dec 3, 2008 #4

    Dick

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    That's correct. But how can that make no sense when you integrate? It's a perfectly reasonable limiting plane. One variable in your integral is a dummy, one is the actual limit.
     
  6. Dec 3, 2008 #5
    got it, thanks!
     
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