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Let $\partial \mathcal{S}$ be a simple closed curve and $\mathcal{S}$ the region enclosed by it, with $\mathcal{S} \subset \mathbf{R}^3$. Let $\vec{F}$ be a vector field in $\mathbf{R}^3$.

If $\mathcal{S}$ is contained in the $x$-$y$ plane and $\vec{F}$ has only $x$ and $y$ components, one can check that \begin{equation} \int_{\partial \mathcal{S}} \vec{F} \times \mathrm{d} \vec{l} = \hat{n}\int_\mathcal{S} \nabla \cdot \vec{F} \, \mathrm{d}x \, \mathrm{d}y \, , \end{equation}

where $\hat{n}$ is given by the right-hand-screw rule in relation to the sense of the line integral around $\partial \mathcal{S}$. This is just another application of Stokes' Theorem.

I have tried to find the appropriate generalization for $\mathcal{S}$ and $\vec{F}$ not constrained to live in a plane, with $\partial \mathcal{S}$ still a simple closed curve. When I try to work the result using the standard form of Stokes' theorem and vector identities I get \begin{equation} \int_{\partial \mathcal{S}} \vec{F} \times \mathrm{d} \vec{l} = \int_\mathcal{S}\left[ ( \nabla \cdot \vec{F} ) \hat{n}- \left( \nabla F_i \right) n_i \right]\, \mathrm{d}s \, , \end{equation} with implicit summation over $i$. This seems to be consistent with the 2D case and I think it works but I could not find it elsewhere and I wanted to make sure if this is right. I would be very grateful if you could let me know if you agree.

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    $\begingroup$ I got the same result starting from Stoke's theorem. $\endgroup$ – mastrok Feb 24 '18 at 13:30
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Probably your way of deriving it is the same as below. I just find convenient to use index form.

Stoke's Theorem: $\oint_{\partial S} E_idl_i = \int_S \epsilon_{ijk}\partial_jE_k dS_i $ for arbitrary surface $S$ with the same boundary.

We have \begin{align} \left(\oint_{\partial S} \vec{F}\times d\vec{l}\right)_i &= \oint_{\partial S} \epsilon_{ijk}F_j dl_k \\ &= \int_{ S} \epsilon_{lmn} \partial_m(\epsilon_{ijn}F_j) dS_l \\ &= \int_{ S} (\delta_{li}\delta_{mj}-\delta_{lj}\delta_{mi})\partial_mF_j dS_l \\ &=\int_{ S} (\partial_kF_k dS_i - \partial_i F_l dS_l), \end{align} which is exactly your expression.

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    $\begingroup$ Thanks @mastrok. I did use this. I'll wait and see if someone comes up with something else before the bounty expires. Otherwise you'll get it. $\endgroup$ – secavara Feb 25 '18 at 1:28

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