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Gauss' and Stokes' Theorems

  1. Jan 28, 2007 #1
    Please check my work for the following problem:

    The problem statement, all variables and given/known data

    By subsituting A(r) = c [tex]\phi[/tex](r) in Gauss's and Stokes theorems, where c is an arbitrary constant vector, find these two other "fundamental theorems":

    a) [tex] \int_{\tau} \nabla \phi d \tau = \int_{S} \phi ds[/tex]
    b) [tex]- \int_{S} \nabla \phi \times ds = \int_{C} \phi dl[/tex]

    The attempt at a solution

    So I start with 'a' and I'll subsitute: A(r) = c [tex]\phi[/tex](r)

    Original equation:
    [tex] \int_{\tau} (\nabla \cdot A) d \tau = \int_{S} A \cdot ds[/tex]
    Subsitution:
    [tex] \int_{\tau} (\nabla \cdot c \phi) d \tau = \int_{S} c \phi \cdot ds[/tex]
    [tex]c \int_{\tau} (\nabla \cdot \phi) d \tau = c \int_{S} \phi \cdot ds[/tex]
    this leads us back to the equation that we want:
    [tex] \int_{\tau} \nabla \phi d \tau = \int_{S} \phi ds[/tex]

    right?

    So I start with 'b' and I'll subsitute: A(r) = c [tex]\phi[/tex](r)


    Original equation:
    [tex] \int_{s} (\nabla \times A) \cdot ds = \int_{C} A \cdot dl[/tex]
    Subsitution:
    [tex] \int_{s} (\nabla \times c \phi) \cdot ds = \int_{C} c \phi \cdot dl[/tex]
    [tex] \int_{s} \nabla \cdot ( c \phi \times ds) = c \int_{C} \phi \cdot dl[/tex]

    i am stuck here on what to do for part b.
     
  2. jcsd
  3. Jan 29, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Are you aware that
    [tex]\nabla \times c\phi= c (\nabla \times \phi)[/tex]?
     
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