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Intuition with applying Stoke's theorem to a cube.

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    F(x,y,z) = xyzi+xyj+x^2yzk

    Surface is the top and four sides of cube with vertices at <+/-1,+/-1,+/-1>


    2. Relevant equations

    ∫∫curlF * ds = ∫F*dr

    3. The attempt at a solution

    At Z = 1, I broke up the surface into 4 lines, parameterized them and combined everything but it seems like the book may have been trying to get a simpler idea across in the case of a cube. I can't quite figure that out.

    Is there a simpler intuitive approach to finding the surface integral of a cube?
     
  2. jcsd
  3. Apr 30, 2013 #2
    Hrm, seems like if I apply Green's Theorem it's faster.
     
  4. Apr 30, 2013 #3

    LCKurtz

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    The four lines at z = 1 are not the boundary of that open cube. The boundary of that cube surface is the square in the plane z = -1. The interior of that square has the same boundary as your open cube. So the integral over the 5 faces of the cube can be calculated by doing the integral over the square in the z = -1 plane since they share the same boundary. Just one easy surface.
     
  5. Apr 30, 2013 #4

    Dick

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    I'd second that opinion. And I'd also think about the divergence theorem. What's the flux over the whole cube? That should reduce the problem to just computing the flux over the missing face.
     
  6. May 1, 2013 #5
    It says that S consists of the top and the four sides "(but not the bottom)"

    Doesn't that mean, if we ignored the fact that all faces will be equal, that we can not apply Stoke's since there is no surface at the intersecting plane of z=-1?
     
  7. May 1, 2013 #6
    Thanks, seems like calculating the flux works out pretty well here too.
     
  8. May 1, 2013 #7

    LCKurtz

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    No. I don't know what you mean by the other faces being equal. Think of a hollow cube sitting on the plane z = -1. The edge resting on the plane is the boundary of the cube that you would use for Stokes theorem. The square that edge describes is the missing face sharing the same boundary. Both flux integrals would be equal to the circuit integral around that edge so they are equal. It is similar to Dick's idea.
     
  9. May 1, 2013 #8
    According to Stokes' the ∫∫s1 curlF * ds = ∫∫s2 curlF * ds.

    So I'm basically saying that technically we are given that there is no surface at z = -1 but we can assume that because it's a cube that it would be the same surface as any other side.

    Right?

    I'm getting a little too technical when it comes to the wording of the question but I'm just trying to really lock in vector analysis.
     
  10. May 1, 2013 #9

    LCKurtz

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    I think you are understanding it. The above theorem is what you want to use, where the surface ##S_2## is the flat square ##-1\le x \le 1,\, -1\le y \le 1,\, z=-1##, oriented upward. (Assuming the original cube was directed outward, which you never specified, and should have). So integrate curl F over that instead of using the other five sides. You can do that because ##S_2## and the other five sides share the same boundary.
     
    Last edited: May 1, 2013
  11. May 1, 2013 #10
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