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Stokes Theorem in cylindrical coordinates

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A vector field A is in cylindrical coordinates is given.

    A circle S of radius ρ is defined.

    The line integral [tex]\int[/tex]A∙dl and the surface integral [tex]\int[/tex]∇×A.dS are different.

    2. Relevant equations

    Field: A = ρcos(φ/2)uρ2 sin(φ/4) uφ+(1+z)uz (1)

    3. The attempt at a solution

    The line integal of A over the circumference of the circle S is

    [tex]\int[/tex]Adl =[tex]\int_0^{2\pi}[/tex]ρ3sin(φ/4)dφ = 4ρ3 (2)

    The surface integral over the area of the circle is

    [tex]\int[/tex]∇×A.dS = [tex]\int_0^{2\pi} \int_0^\rho[/tex](3ρsin(φ/4)+(1/2)sin(φ/2))ρdρdφ = [tex]\int_0^{2\pi}[/tex](ρ3sin(φ/4)+ρ2/4 sin(φ/2))dφ=4ρ32 (3)

    Observing more closely, in the surface integral (3), we get an additional term [tex]\int_0^{2\pi}[/tex](ρ2/4sin(φ/2))dφ which is not present in the line integral (2).
    This gives rise to the second term ρ2 in (3) after the integration.
    What is the reason this new term appears in the surface integral, but not the line integral?
     
    Last edited: Sep 28, 2009
  2. jcsd
  3. Sep 28, 2009 #2

    gabbagabbahey

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    Hi SDas, welcome to PF!:smile:

    Hint: what are the conditions [itex]\textbf{A}[/itex] must satisfy for Stokes theorem to be true on a given region?:wink:
     
  4. Sep 29, 2009 #3
    Hi gabbagabbahey, thanks for the welcome.

    I've been trying to check the conditions for Stoke's theorem to hold. I could not find them in any textbook on electromagnetics. Intuitively, the field A should
    (i) be bounded, and
    (ii) have continuous, first order partial derivatives w.r.t. ρ, φ, and z.
    Am I missing something?
     
  5. Sep 29, 2009 #4

    Dick

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    Is your field even continuous? What happens when the angular coordinate passes through 2pi?
     
  6. Sep 29, 2009 #5
    Got it! Of course it isn't. I should not have missed that. The sin(φ/4) and the cos(φ/2) should have been sin(4φ) and cos(2φ) to make A be the same at φ=0 and φ=2π. That was the mistake in what I was trying to do.
     
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