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Prove a Statement about the Line Integral of a Vector Field

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the vector field [itex]\vec{v} = (-y\hat{x} + x\hat{y})/(x^2+y^2)[/itex]
    Show that [itex]\oint \vec{dl}\cdot\vec{v} = 2\pi\oint dl[/itex] for any closed path, where [itex]dl[/itex] is the line integral around the path.


    2. Relevant equations
    Stokes' Theorem: [itex]\oint_{\delta R} \vec{dl}\cdot\vec{v} = \int_R \vec{dA}\cdot(\vec{\nabla}\times\vec{v})[/itex]


    3. The attempt at a solution
    I tried finding the curl of v, and got that the x and y components were zero, and the z component is [itex]\frac{\delta}{\delta x}(x/(x^2+y^2)) + \frac{\delta}{\delta y}(y/(x^2+y^2))[/itex], which becomes [itex](y^2-x^2+x^2-y^2)/(x^2+y^2)^2[/itex], which is also zero. But this can't be right, because that would imply that [itex]\oint \vec{dl}\cdot\vec{v} = \int \vec{dA}\cdot0 = 0[/itex] by Stokes' Theorem, which would mean the thing I'm trying to show is false, since [itex]2\pi\oint dl[/itex] is not always zero.
     
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  3. Feb 14, 2012 #2

    SammyS

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    Could it be that [itex]2\pi\oint dl[/itex] is always zero ?
     
  4. Feb 14, 2012 #3
    I don't think it could, because [itex]\oint dl[/itex] represents the length of the line being integrated over, doesn't it? And there's no reason why the length of a closed loop must always be zero.
     
  5. Feb 14, 2012 #4

    Dick

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    Yes, it does. Your Stoke's theorem argument doesn't apply to curves that wind around the origin since the vector field is singular at the origin. But it does to curves that don't. And they don't all have arclength 0. It think the question got garbled somehow. I think your vector integral really computes a winding number around the origin. It's not related to arclength.
     
  6. Feb 14, 2012 #5
    What exactly do you mean by a "winding number around the origin?"
     
  7. Feb 14, 2012 #6

    Dick

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    http://en.wikipedia.org/wiki/Winding_number Your vector integral is zero if the curve doesn't contain the origin. As you checked with Stokes's theorem. It's 2pi if the curve winds once around the origin counterclockwise. It's -2pi if it winds once clockwise. Etc. Look down in the wikipedia reference to the 'Differential Geometry' section. You'll see your vector integral. I think the question they gave you is messed up.
     
  8. Feb 14, 2012 #7
    I played around with the vector field and found that it does seem to hold that [itex]\oint \vec{dl}\cdot\vec{v}[/itex] is equal to 2*pi times the winding number of any given closed loop, so I'm going to assume that's what my professor wanted me to prove. I can prove this is true when the winding number is zero by the divergence theorem (as you stated), but how do I go about proving it when the loop circles around the origin?
     
  9. Feb 15, 2012 #8

    Dick

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    You take a contour where the winding number is easy to compute. Say a contour along the unit circle. Now take your hard to compute contour. Draw a segment connecting the two contours. Figure out how you can run around the two contours using the segment bridging them and only enclose a region that does not contain the origin.
     
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