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Math Help - Stokes' Theorem

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    Super Member craig's Avatar
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    Stokes' Theorem

    Show, using Stokes' Theorm, that

    \oint_C \phi \nabla \phi \cdot \mathbf{dl} = 0

    Firstly Stokes gives us:

    \oint_C \phi \nabla \phi \cdot \mathbf{dl} = \int_S \nabla \times [\phi \nabla \phi] \cdot \mathbf{dS}

    Expanding \nabla \times [\phi (\nabla \phi)] gives:

    \phi \nabla \times \nabla \phi + \nabla \phi \times \nabla \phi

    The second part of this is zero for any vector, so this leaves the integral as:

    \int_S \phi \nabla \times \nabla \phi \cdot \mathbf{dS}

    Anyone any ideas?

    Cheers in advance!
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  2. #2
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    Quote Originally Posted by craig View Post
    Show, using Stokes' Theorm, that

    \oint_C \phi \nabla \phi \cdot \mathbf{dl} = 0

    Firstly Stokes gives us:

    \oint_C \phi \nabla \phi \cdot \mathbf{dl} = \int_S \nabla \times [\phi \nabla \phi] \cdot \mathbf{dS}

    Expanding \nabla \times [\phi (\nabla \phi)] gives:

    \phi \nabla \times \nabla \phi + \nabla \phi \times \nabla \phi

    The second part of this is zero for any vector, so this leaves the integral as:

    \int_S \phi \nabla \times \nabla \phi \cdot \mathbf{dS}

    Anyone any ideas?

    Cheers in advance!
    Remember that \phi(x,y,z) is a scalar function!

    \displaystyle \nabla \phi =\frac{\partial \phi}{\partial x}\mathbf{i}+\frac{\partial \phi}{\partial y}\mathbf{j}+\frac{\partial \phi}{\partial z}\mathbf{k}

    \displaystyle \phi \nabla \times \nabla \phi=\phi \left(\nabla \times \nabla \phi \right)

    Expand out what is inside the parenthesis and you will get zero (As long as \phi is twice continuously differentiable)

    Remember 2nd continuous partial derivatives commute Clairaut's theorem

    Symmetry of second derivatives - Wikipedia, the free encyclopedia
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Remember that \phi(x,y,z) is a scalar function!

    \displaystyle \phi \nabla \times \nabla \phi=\phi \left(\nabla \times \nabla \phi \right)
    Ahh of course! I was thinking of it as a vector function.

    If we take \displaystyle \left(\nabla \times \nabla \phi \right), we get,

    \displaystyle \left(\frac{\partial}{\partial y} \frac{\partial \phi}{\partial z} - \frac{\partial}{\partial z} \frac{\partial \phi}{\partial y},... \right) = \left(0,0,0 \right)

    Hence the integral is equal to zero!

    Many thanks for that
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