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Flux of F over the surface of a cube

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the flux of F over the surface of the cube with vertices [tex]( \pm 1, \pm 1, \pm 1)[/tex] using the outer normal.
    F(x,y,z)=(x+y)i+zj+xzk

    2. Relevant equations
    Flux of F over S is
    [tex]\iint F \cdot n dS[/tex]


    3. The attempt at a solution
    I think the normal should be 1 in the respective directions for each side of the cube, and I keep getting 0 when I try to work out the math.
     
  2. jcsd
  3. Nov 23, 2008 #2

    Dick

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    I used the divergence theorem, and I don't get zero. How did you get zero?
     
  4. Nov 23, 2008 #3
    Well that would make sense, the divergence theorem is the next section, haha. I was just following the example in the book multiplying the normal and F and finding the integral? (for each side, since it isn't continuous)

    Do you know any more elementary ways to do it?
     
  5. Nov 23, 2008 #4

    Dick

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    If you don't have the divergence theorem, then you just have to do what you said. Dot the normal with F over each face integrate and sum them. You should get the same answer. It's not that hard and you can do it that way. But since you didn't tell us how you got zero out of it, it's hard to say what you are doing wrong.
     
  6. Nov 24, 2008 #5
    I did these three integrals
    [tex]\iint (1+y) dydz[/tex]
    [tex]\iint z dxdz[/tex]
    [tex]\iint x dxdy[/tex]

    With -1 to 1 as the limits for all three. Do I need to split those up into two each for each side? I feel like I do but I'm not sure how to do that.
     
  7. Nov 24, 2008 #6

    HallsofIvy

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    Yes, [tex]\int_{-1}^1\int_{-1}^1 z dxdz= 0[/tex] and [tex]\int_{-1}^1\int_{-1}^1 x dydx= 0[/itex], but [tex]\int_{-1}^1\int_{-1}^1(y+1) dydz[/itex] is NOT 0.
     
  8. Nov 24, 2008 #7
    Alright, that was my problem, I figured it out now.

    Thanks!
     
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