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Stokes' Theorem formula question

  1. Jun 19, 2011 #1
    eq0002M.gif

    I was wondering, how you break down dS to something with dA? I know that dS is equal to ndS. The n is equal to grad f / (magnitude of grad f) and the dS is equal to the same as the magnitude of grad f right? So is the formula the same as double integral of region D (curl F * grad f) dA?
     
  2. jcsd
  3. Jun 19, 2011 #2
    I don't know what f is (as opposed to F), but yes, that should be it.
    (A comes from area, while S comes from surface - it is basically the same)
     
  4. Jun 19, 2011 #3

    LCKurtz

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    Let's say you parameterize your surface S in terms of u and v as

    [tex]\vec R = \vec R(u,v) [/tex]

    Now since

    [tex]dS = |\vec R_u\times \vec R_v|dudv[/tex]
    and
    [tex]\hat n =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|}[/tex]
    with the sign chosen to agree with the orientation of the surface, you have
    [tex]d\vec S =\hat n dS =\pm\frac{\vec R_u\times\vec R_v}{|\vec R_u\times\vec R_v|}
    |\vec R_u\times\vec R_v|dudv=\pm\vec R_u\times \vec R_vdudv[/tex]
    This allows you to express everything in terms of u and v with appropriate uv limits.
     
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