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Thread: Stokes's Theorem

  1. #1 Stokes's Theorem 
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    I'm having issues comprehending what it means, and what it is useful for. if anyone could give me an explaination on it, I would be real grateful


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  3. #2 Re: Stokes's Theorem 
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    Quote Originally Posted by Arcane_Mathamatition
    I'm having issues comprehending what it means, and what it is useful for. if anyone could give me an explaination on it, I would be real grateful
    You may need to be a bit more precise in your question.

    Generally Stokes theorem says that a the integral of a differential operator applied to a function taken over some open region is equal to the integral of the original function over the boundary of the region.

    This is a generalization of the fundamental theorem of calculus. Consider an interval [a,b]. The boundary is the set of two points (a,b}. So now, the fundamental theorem of calculus says



    f(b)-f(a) can be interepreted as the integral of f over the oriented boundary of [a,b]

    This basic logic applies to the Divergence Theorem and what you are probably calling Stokes Theorem. In the manner in which I described it the theorem is called Stokes Theorem or the generalized Stokes Theorem.

    This Wiki article explains this notion in more detail http://en.wikipedia.org/wiki/Stokes'_theorem


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  4. #3  
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    it's not the divergence theorem, the one I'm having problems with is that the integral over the boundary of some surface (peicewise-smooth curve) such that where F is a vector feild with continuous partial derivatives in all of
    the problem i have is telling me to verify this is true, but I don't really understand it. simple exlaination is all I need, I hope. on the article you provided, it was the special case of Kelvin-Stokes theorem, if that helps you to see where I'm stuck
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  5. #4  
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    To get an idea of what's going on, work out the case where S is the surface of a box, with the bottom panel removed.

    Another, actually easier example: Let S be a square in the plane, and C the boundary.

    These example may sound too simple. But I kid you not, they pretty much capture everything that's going on. In other words, with some care, you can use examples of this nature to construct a proof of the general case.
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  6. #5  
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    Quote Originally Posted by Arcane_Mathamatition
    it's not the divergence theorem, the one I'm having problems with is that the integral over the boundary of some surface (peicewise-smooth curve) such that where F is a vector feild with continuous partial derivatives in all of
    the problem i have is telling me to verify this is true, but I don't really understand it. simple exlaination is all I need, I hope. on the article you provided, it was the special case of Kelvin-Stokes theorem, if that helps you to see where I'm stuck
    I'm not sure where you are stuck. Salsaonline's suggestion is a good one.

    If that is not doing the trick for you, then perhaps if you state quite explicitly the exercise that you are trying to do it would help us to understand your issue.

    Are you trying to work a specific example ? Prove the general theorem in the special case of a particular sort of surface ? Prove the general theorem itself ?
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  7. #6 Re: Stokes's Theorem 
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    Quote Originally Posted by Arcane_Mathamatition
    I'm having issues comprehending what it means, and what it is useful for. if anyone could give me an explaination on it, I would be real grateful
    Well lookin at it and geting rid of the maths, is it not saying for example:-
    If you walk over a hill for example or over some hills from A to B then the difference in height etween A and B will be the same as difference if you add up all the differences in height of all the steps you take.
    Basically stating the blindlingly obvious!!
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  8. #7 Re: Stokes's Theorem 
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    Quote Originally Posted by esbo
    Quote Originally Posted by Arcane_Mathamatition
    I'm having issues comprehending what it means, and what it is useful for. if anyone could give me an explaination on it, I would be real grateful
    Well lookin at it and geting rid of the maths, is it not saying for example:-
    If you walk over a hill for example or over some hills from A to B then the difference in height etween A and B will be the same as difference if you add up all the differences in height of all the steps you take.
    Basically stating the blindlingly obvious!!
    It is saying no such thing.

    What you have stated is indeed blindingly obvious. But it has nothing to do with Stokes Theorem, which is not so obvious.

    It is saying that if you integrate the curl of the flux of a vector field through a disc that gives the same thiing as the line integral of the flux around the loop that bounds the disc. Combined with Maxwell's equations it gives you Faraday's law of magnetic induction.

    It is not particularly obvious, which is why it has earned a name and is a well-known and important result in vector analysis and in more modern terms in differential geometry. It can be made "obvious" when treated in terms of differential forms and intetgration over chains, but what is actually done there is to incorporate the hard work into the definitions. The definitions are not so obvious or intuitive, at least initially. When stated in those terms Stokes theorem is seen to be a generalization of the fundamental theorem of calculus, quite a significant generalization.
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  9. #8  
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    I was given a paraboloid surface, and told to integrate the vector field over it. I think I understand it now, I was just stubborn and wanted to solve the integral over the entire surface, rather than just the boundary. I got it now, I hope. We just want to show that the integral over the boundary itself is equivalent to the integral of the curl of the vector field over the surface, which is true. both go to 0.

    the Curl of F does it rather obviously, while the line integral becomes (with polar coordinates)

    all of those in the integral, the trig functions, are periodic repeating that integrate to zero over every period 2\pi and since our integral is over a 'distance' of 2\pi, I feel confident that I can say that they will be 0, without having to solve them (for right now)
    Wise men speak because they have something to say; Fools, because they have to say something.
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