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Gaussian theorem question

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose the one-dimensional field A = Kx * ax exists in a region. Illustrate the validity of the Gaussian theorem by evaluating its volume and surface integrals inside and on the rectangular parallelepiped bounded by the surfaces: x=1,x=4,y=2,y=-2,z=0 and z=3, for a given A.


    2. Relevant equations

    (Right) [itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz ay + (left) [itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz -(ay) + (top) [itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy az + (bottom) [itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy -(az) + (front) [itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz (ax) + (back) [itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz -(ax)

    Direction on the left is applied to the integral on its right.
    3. The attempt at a solution

    For the Right side
    [itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz ay
    My answer to this integral is 45/2

    For the left side
    [itex]\int_0^3[/itex][itex]\int_1^4x[/itex]dxdz -(ay)
    My answer to this integral is -45/2

    For the top side
    [itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy az
    My answer is 30

    For the bottom
    [itex]\int_{-2}^2[/itex][itex]\int_1^4x[/itex]dxdy -(az)

    For the front
    [itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz (ax)

    For the back
    [itex]\int_0^3[/itex][itex]\int_{-2}^2[/itex]dydz -(ax)

    When I add these up, I get zero... however, when I use the divergence theorem I get 36.

    This answer is suppose to equal the answer I get via the divergence theorem formula. I'm confused :(
  2. jcsd
  3. Feb 4, 2013 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    There are only 2 sides that "see" the field A head-on. Mathematically, there are only two sides for which ∫A*ds ≠ 0 where ds is a vector element of area on any side. Which sides are those?

    Now integrate A over those two sides, remembering that the dot-product A*ds will be positive for one side and negative for the other. In other words, the normal to any closed surface always points out of the surface.
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