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Stokes theorem over a cube

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data

    I have to use stokes' theorem and calculate the surface integral, where the function F = <xy,2yz,3zx> and the surface is the cube bounded by the points (2,0,0), (0,2,0),(0,0,2),(0,2,2),(2,0,2),(2,2,0),(2,2,2). The back side of the cube is open.
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    2. Relevant equations



    3. The attempt at a solution

    I found the curlF = <-2y,-3z,-x>.

    I know next I have to find a unit normal vector. But exactly which side of the cube am I supposed to find the normal vector to? Or would I HAVE to take the line integrals for the bottom side of the cube?
     
    Last edited: Jun 2, 2009
  2. jcsd
  3. Jun 2, 2009 #2
    If I understand correctly, you want to calculate

    Int{F.dA}

    I think that divergence theorem could be useful

    Int{F.dA} = Int{(div F)dV}

    where "div F" is gradient operator (Del) dotted with F.

    Stokes/curl theorem helps if you need

    Int{F.dl}

    then you use

    Int{F.dl} = Int{(curl F).dA}

    and now "curl F" is cross product of Del and F.

    Since the back side is open I think that either you need to do each of five sides in turn or maybe do the whole cube with divergence theorem and then subtract the integral over the back side from it.

    ---
     
    Last edited by a moderator: Aug 6, 2009
  4. Jun 2, 2009 #3

    HallsofIvy

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    You don't find the normal vector to any side. To use Stokes theorem you find the line integral around the boundary of the surface. Here that means around the four edges of the missing back, not bottom.

    If you did want to do this by integrating over the surface, you would find the unit normal vector of all of the five faces and integrate each one separately.
     
  5. Jun 2, 2009 #4
    Ahh I get it now. Thanks!
     
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