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Math Help - Stokes and Guass theorm Integral Problem

  1. #1
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    Stokes and Guass theorm Integral Problem

    In my attached working I try and use stokes theorm but end up getting dr/dt = 0 . and cant get any further this way

    I used the guass therom and got pi.

    If anybody can see what the problem is?
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  2. #2
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    For those who don't like to open pdf files (not as prone to viruses as word files, but still)
    The problem is to find \oint_S \nabla\times F\cdot dS where F= zi+ xj+ yk and S is the circle x^2+ y^2= 1 in the plane z= 1 using Stokes theorem or Gauss's divergence theorem.

    Stoke's theorem says \oint_S \nabla\times F\cdot dS= \int\int_R F\cdot ds
    Since that has " \nabla\cdot F", that is the obvious one to use.

    Taking x= cos(\t), y= sin(\t), z= 1
    (heatly, you don't include " z= 1", which I consider an important part of the parameterization. Of course, here, because z is constant, it doesn't matter).

    You then say "r(t)= cos^2t+ sin^2t" which is non-sense! That is the same as saying that r(t)= 1 so of course dr= 0. In order that " F\cdot ds",where F is a vector, make sense, ds must be a vector, not a scalar.

    What you want is that r(t) is the position vector xi+ yj+ zk= cos(t)i+ sin(t)j+ k. Then dr= dx i+ dy j+ dz k= -sin(t)dt i+ cos(t)dt j. (dz= 0 so there is no k component.)
    F\cdot ds= (zi+ xj+ yk)\cdot(-sin(t)i+ cos(t)j)dt= (i+ cos(t)j+ sin(t)k)\cdot(-sin(t)i+ cos(t)j)dt= (-sin(t)+ cos^2(t))dt.

    \oint\int \nabla F\cdot dS= \int_{t= 0}^{2\pi} (-sin(t)+ cos^2(t))dt

    That is not 0 because of the cos^2(t).

    You then proceed to "use" Gauss's divergence theorem which doesn't apply here because there is no " \nabla F" in it and because it requires integration over a solid region and there is no solid region given in this problem.
    Last edited by HallsofIvy; May 23rd 2011 at 04:32 AM.
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  3. #3
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    Thanks a lot for that,really helpful.
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