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What does this symbol mean (Countour integration over a boundary)?

  1. Dec 2, 2011 #1
    I keep seeing this symbol, something like [itex]\oint[/itex][itex]_{{\partial}\omega}[/itex][itex]\omega[/itex], I know it is a contour integral and read that [itex]{\partial}\omega[/itex] is called a boundary, but I don't know what it means or why there isn't a differential at the end of it. Can someone please answer these questions and explain how to evaluate it? Thanks in advance.
     
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  3. Dec 3, 2011 #2

    HallsofIvy

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    Well, you would not see what you have given because if [itex]\omega[/itex] is appropriate in one of those uses, it is inappropriate in the other. You might well see [itex]\int_{\partial S} \omega[/itex]. Here, S is a set and [itex]\partial S[/itex] is the boundary of that set. There is no "dx" because [itex]\omega[/itex] is already a differential possibly not in terms of "x".

    If that is a path integral in three dimensional Euclidean space, it might be [itex]\int_{\partial S} f(x, y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz[/itex] where we are taking the integral over the boundary of some two dimensional set S in R3 and the differential to be integrated is [itex]\omega= f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z)dz[/itex].

    On the other hand, if S is a three-dimensional subset of R3, it boundary, [itex]\partial S[/itex] would be a surface and an integral on it would be [itex]\int_{\partial S} f(x,y,z)dxdy+ g(x,y,z)dydz+ h(x,y,z)dxdz[/itex]. Here the differential to be integrated is [itex]\omega= f(x,y,z)dxdy+ g(x,y,z)dydz+ h(x,y,z)dxdz[/itex].
     
  4. Dec 3, 2011 #3
    This clears up most of my questions, but what does the boundary mean? Also, how do I know what s,ω, and [itex]\partial S[/itex] are; would they be given to me in a real problem? If I knew that [itex]\partial S[/itex] was a surface, path, or contour, then could I just replace it with another letter (say L) and integrate accordingly, and would I need to use Stokes' theorem, kelvin-stokes theorem, or the divergence theorem to evaluate it? How would I go about doing this? Basically, what I'm asking is, if S is a path or surface, then how would integrating over [itex]\partial S[/itex] be different than integration over S? Sorry I am asking so many questions but I can't seem to find anything that defines this in detail.
     
  5. Dec 3, 2011 #4

    I like Serena

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    Suppose S is a circle disk.

    Integrating over S means integrating over the surface of the disk (surface integral).
    Integrating over ∂S means integrating over the perimeter of the circle (contour integral).

    Note that ∂S is always closed (a closed curve in this case), whereas S is not closed.


    Now suppose S is a sphere volume.
    Can you think up what ∂S is?


    And yes, in a real problem you would have more information about all your stuff.
     
  6. Dec 3, 2011 #5
    ∂S would be the volume of the surface area (simply the surface area of a sphere). But how would I evaluate an integral over ∂S? Would I have to convert it to an integral over S? If so, how would I go about doing this (I think I read on wikipedia that I can use stokes theorem by taking the exterior derivative, which is just the directional derivative in Euclidean Space, and integrating over S)?
     
    Last edited: Dec 3, 2011
  7. Dec 3, 2011 #6

    I like Serena

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    Yup.

    An integral over a curve is done like this:
    http://en.wikipedia.org/wiki/Line_integral#Line_integral_of_a_scalar_field


    There are 2 major theorems about this.
    Gauss's theorem and Stokes theorem (which is the same as Green's theorem except that Green is for 2 dimensions).
    They show identities between an integral over an area, and an integral over the boundary of that area.

    Sometimes it is easier to apply one of those theorems to get what you want, but the integrals are properly defined without using them.
     
  8. Dec 3, 2011 #7
    Thanks, I know how to evaluate a line integral, but didn't realize that an integral over a boundary was just a line integral.

    I'm pretty familiar with using Green's, Gauss', and the Curl Theorem (which is also applicable in this case unless I'm mistaken), but how would I apply Stokes' theorem? Would I just have to take the exterior derivative of the differential ω? If so, what does the directional derivative of a differential look like?

    Now that I think about it, I probably should have put this in a differential geometry forum instead.
     
  9. Dec 3, 2011 #8

    I like Serena

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    You can find your generalized form here:
    http://en.wikipedia.org/wiki/Stokes'_theorem#General_formulation
    [tex]\int_\Omega \textrm{ d}\omega = \oint_{\partial \Omega} \omega[/tex]
    ω is defined here as an (n-1)-differential-form (generalization of a line integral).
    dω is defined as the exterior derivative of ω (generalization of a surface integral of the curl).

    Note that the surface Ω uses a different symbol and indeed is different from the differential-form ω.


    And how would you calculate one or the other?
    Well, typically by using a pullback chart φ:Ω→ℝn:
    [tex]\int_\Omega \alpha = \int_{\phi(\Omega)} (\phi^{-1})^* ~ \alpha[/tex]
    This matches of course a line integral, or a surface integral.
     
    Last edited: Dec 3, 2011
  10. Dec 3, 2011 #9
    So would I be correct in saying
    [itex]\oint_{\partial \Omega} \omega[/itex]=[itex]\int_\Omega dω = \int_{\phi(\Omega)} (\psi)* d \omega[/itex]= [itex] \int_{\phi} d (\psi* \omega)[/itex]

    Where [itex]\phi[/itex] is the function that maps the surface Ω to ℝ^n (pullback) where ω is an n-form and ψ is the inverse of [itex]\phi[/itex] that maps ℝ^n to Ω (pushforward)? Also, d is the directional derivative, which I now have to take of the product of a function (ψ) and the n-form ω. How would I take the directional derivative of this product? Would the next step be using Green's/Stokes theorem to convert the line integral into definite integrals based off of the extrema of [itex]\phi[/itex] in ℝ^n?

    Can someone move this to the differential geometry forum, or is there some way for me to do it myself?
     
    Last edited: Dec 3, 2011
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