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Div, Curl and all that jazz: Proving identities

  1. Apr 10, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\left(\nabla \times \vec{F}\right)\cdot d\vec{V} = \oint \left(\vec{\hat{n}} \times \vec{F} \right) dS [/tex]

    2. Relevant equations
    In the previous part of the question, we proved that:

    [tex]\nabla \cdot \left( \vec{F} \times \vec{d} \right) = \vec{d} \cdot \nabla \times \vec {F} [/tex]
    (where d is a constant vector)
    And also, it looks like we'll need to use the Divergence theorem.

    3. The attempt at a solution

    OK, so, here I go!

    [tex] \int\left(\nabla \times \vec{F} \right)\cdot \vec{\hat{n}}dV \\

    = \int \nabla \cdot \left( \vec{F} \times \vec{\hat{n}} \right) dV [/tex]

    By the relation above proved from the previous part of the question. Next, I used the divergence theorem:


    \int \nabla \cdot \left( \vec{F} \times \vec{\hat{n}} \right) dV = \oint \left( \vec{F} \times \vec{\hat{n}} \right) \cdot d\vec{S} [/tex]

    My question is...Is [tex] \oint \left( \vec{F} \times \vec{\hat{n}} \right) \cdot d\vec{S} = \oint \left(\vec{\hat{n}} \times \vec{F} \right) \cdot d \vec{S} [/tex]?

    My initial thought is that it isn't, as the cross product isn't commutative. If that is thecase, where else have I gone wrong?

  2. jcsd
  3. Apr 10, 2008 #2


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    What is the orientation of the surface in each integral?
  4. Apr 10, 2008 #3
    Erm, I'm don't know what an the orientation of a surface is, but I've had an idea.

    I know that:

    [tex] \vec{F} \times \vec{\hat{n}} = -\vec{\hat{n}} \times \vec{F} [/tex]

    and that:

    [tex] \int_{a}^{b} f(x)dx = - \int_{b}^{a} f(x)dx [/tex]

    But, since it's a closed integral, I guess if we reverse the 'order' of integration, it doesn't matter if we do: the start and end points are the same...and the minuses cancel. Seems a bit of mathematical trickery to me, though. :frown:
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