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Stokes' theorem and unit vector

  1. Aug 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Use Stokes' theorem to show that

    [tex]\oint\ \hat{t}*ds = 0[/tex]

    Integration is done closed curve C and [tex]\hat{t}[/tex] is a unit tangent vector to the curve C
    2. Relevant equations
    Stokes' theorem

    [tex]\oint F* \hat{t}*ds = \int\int \hat{n}*curl(F)*ds[/tex]

    3. The attempt at a solution

    Ok, this is really teasing me because I know that is probably simple. Could someone help please?
  2. jcsd
  3. Aug 13, 2009 #2


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    Your notation is pretty sloppy. If t=(tx,ty,tz) the result of your first integral is the vector whose first coordinate is the integral of tx*ds, second is integral ty*ds, and third integral tz*ds. You want to show all of those are zero. Apply Stokes to the constant vector field F=(1,0,0). What does that tell you? What other vector fields would be good to use?
  4. Aug 14, 2009 #3
    Yes, sorry about the notation, it is direct copy from the book that I'm studying. t is a unit vector with components (dx/ds, dy/ds, dz/ds) so after the multiplication the integral is taken from vector (dx, dy, dz)
  5. Aug 14, 2009 #4
    Ok, now I think I got it. I should use Stokes' theorem with F=1 (scalar). What bothered me was that I was all the time looking at the cross product in curl which is not defined for scalars. Of course the cross product is only a notation, not actual vector cross product.
  6. Aug 14, 2009 #5


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    No! You can't use Stokes theorem on a scalar. Use it on the vector F=(1,0,0)=1*i+0*j+0*k.
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