The Vector Integral page on the Wolfram mathworld website lists as eq.(4) the following vector integral identity:

If $$\mathbf{F}:=\mathbf{c}\,F,$$ then $$\int_{C}F\,ds=\int_{S}d\mathbf{a}\times\nabla\mathbf{F}.$$

It seems clear from context that $\mathbf{c}$ is supposed to be a constant vector, though the page doesn't explicitly list this as a condition.

Try as I might, I can't wrap my head around this integral. In fact, it looks like patent nonsense to me, since on the LHS we have a scalar line integral (that is, the line-element $ds$ is a scalar) of a scalar field resulting in a scalar value, while the integral on the RHS looks to me like it must be a 2nd rank tensor.

I'd be surprised if this is an error on Wolfram's part that's somehow gone unnoticed all this time, so I can only assume that I have a fundamental misunderstanding somewhere that's preventing me from parsing this formula correctly. Can someone please set me straight?


My thanks to @TylerHG for helping me confirm that this is indeed a typo on Wolfram's part. Instead of closing this question, I'd like to ask a follow-up question prompted by the falsehood of the identity proposed above. Given a scalar field $f(\mathbf{r})$ and a surface $\Sigma$ with boundary $\partial\Sigma$, does there exist any kind of Stokes' theorem analogue for the scalar line integral,

$$\oint_{\partial\Sigma}f\,\mathrm{d}\ell =\,???$$

  • 1
    $\begingroup$ Yea it is a typo. That F should be a scalar and the ds should be a vector. Look at this mathworld.wolfram.com/CurlTheorem.html $\endgroup$ – ClassicStyle Jun 13 '14 at 20:23
  • $\begingroup$ @TylerHG Well I'll be... Thank you for restoring my sanity. $\endgroup$ – David H Jun 13 '14 at 20:36
  • 1
    $\begingroup$ The cross product of a vector and the gradient of a vector...nooooooo! @DavidH $\endgroup$ – ClassicStyle Jun 13 '14 at 20:38
  • $\begingroup$ @TylerHG You can imagine upsetting this was to me. :o $\endgroup$ – David H Jun 13 '14 at 20:58

For any type of Stokes-like result to hold, you have to be able to interpret $\int_{\partial\Sigma} f ds$ as the line integral of a vector field over the curve $\partial\Sigma$, and indeed, you can write $$ \int_{\partial \Sigma} f ds = \int_{\partial \Sigma} (f \mathbf{t}) \cdot d\mathbf{r}, $$ where $\mathbf{t}$ denotes the unit tangent vector field on $\partial\Sigma$. For Stokes's theorem to apply, you have to be able to extend $f \mathbf{t}$ to a vector field on all of $\Sigma$; since $f$ is (presumably) defined on some neighbourhood of $\Sigma$ in $\mathbb{R}^3$ anyway, the entire problem is in extending the unit tangent vector field $\mathbf{t}$ to a vector field $\tilde{\mathbf{t}}$ on all of $\Sigma$. If such a vector field $\tilde{\mathbf{t}}$ exists, then, and only then, can you apply Stokes's theorem to conclude that $$ \int_{\partial \Sigma} f ds = \int_{\partial \Sigma} (f \tilde{\mathbf{t}}) \cdot d\mathbf{r} = \int_\Sigma \nabla \times (f\tilde{\mathbf{t}}) \cdot d\mathbf{S} = \int_\Sigma \left( \nabla f \times \tilde{\mathbf{t}} + f \nabla \times \tilde{\mathbf{t}} \right) \cdot d\mathbf{S}, $$ i.e., that $$ \int_{\partial \Sigma} f ds = \int_\Sigma \left( \nabla f \times \tilde{\mathbf{t}} \right) \cdot d\mathbf{S} + \int_\Sigma f \left(\nabla \times \tilde{\mathbf{t}}\right) \cdot d\mathbf{S}. $$ If $\mathbf{t}$ cannot be extended to a vector field on $\Sigma$, or equivalently, if $\mathbf{t}$ isn't the restriction to $\partial\Sigma$ of some vector field defined on all of $\Sigma$, then you're out of luck. However, in any reasonable situation, $\mathbf{t}$ can be extended to a vector field on an open neighbourhood of $\partial\Sigma$ in $\mathbb{R}^3$, in which case you can use bump functions to get an extension $\tilde{\mathbf{t}}$ of $\mathbf{t}$ that vanishes outside that neighbourhood.

There is exactly one other possibility, which is to use the divergence theorem for surfaces: $$ \int_{\partial \Sigma} \mathbf{F} \cdot (\mathbf{t}\times\mathbf{n}) ds = \int_\Sigma \operatorname{div}_\Sigma(\mathbf{F}) dS, $$ where $\mathbf{t}$, as before, is the unit tangent vector field on $\partial\Sigma$, $\mathbf{n}$ is the unit normal vector field on $\Sigma$, and $\operatorname{div}_\Sigma$ denotes the divergence on $S$, i.e., $$ \operatorname{div}_\Sigma(\mathbf{F}) := \mathbf{n} \cdot \nabla \times (\mathbf{n} \times \mathbf{F}). $$ Then, since $$ \int_{\partial\Sigma} fds = \int_{\partial\Sigma} f(\mathbf{t}\times\mathbf{n}) \cdot (\mathbf{t}\times\mathbf{n}) ds, $$ it follows that if $\mathbf{t}\times\mathbf{n}$ extends to a vector field $\mathbf{v}$ on $\Sigma$, then you can apply the divergence theorem for surfaces to conclude that $$ \int_{\partial\Sigma} fds = \int_{\partial\Sigma} f\mathbf{v} \cdot (\mathbf{t}\times\mathbf{n}) ds = \int_\Sigma \operatorname{div}_\Sigma(f\mathbf{v}) dS. $$ This, however, gives you nothing new; on the one hand, if $\mathbf{t}$ does indeed extend to $\tilde{\mathbf{t}}$, then you can set $\mathbf{v} = \tilde{\mathbf{t}} \times \mathbf{n}$, whilst on the other, if $\mathbf{v}$ exists, then you can check, using the BAC-CAB formula, that $$ \tilde{\mathbf{t}} := \mathbf{n} \times \mathbf{v} $$ correctly extends $\mathbf{t}$.

Since Stokes's theorem and the divergence theorem for surfaces are the only two ways to rewrite, in the language of vector analysis, the abstract Stokes's theorem for $1$-forms and $2$-dimensional submanifolds with boundary, i.e., $$ \int_{\partial\Sigma} \omega = \int_\Sigma d\omega, $$ in the special case where the ambient space is $\mathbb{R}^3$, this really now exhausts all the obvious possibilities for a Stokes-like result.

  • $\begingroup$ Very informative! This is actually a question that's nagged at me on and off for at least a few years. Good to finally have my answer. :) $\endgroup$ – David H Jun 13 '14 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.