I've seen Stokes' theorem written as $\oint \mathbf{F} \cdot \ d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \hat{n} \ dS$ and $\oint \mathbf{F} \cdot \ d\mathbf{r} = \iint_S ( \nabla \times \mathbf{F} ) \cdot \ dS$. I was under the impression that the first one, $\oint \mathbf{F} \cdot \ d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \hat{n} \ dS$, is the correct one, but I've seen the other one multiple times now, so I'm wondering if I misunderstood something?

Are these two equivalent? If so, then where is the normal vector $\hat{n}$ in the second one? If not, then why, and what am I misunderstanding?

I would greatly appreciate it if people could please take the time to clarify this.

  • $\begingroup$ It might just be a question of notation, as some authors prefer to include the normal in the surface element $dS$. But another possibility that comes to mind is the dimension. You can write down Stokes theorem in 2 and 3 dimensions. In 3 dimensions, you have a surface $S$ with a corresponding normal. In 2 dimensions, there is no normal any more, as $S$ is just a region of space. This works as $\nabla \times F$ can be considered a scalar in $\mathbb{R}^2$ anyway. $\endgroup$
    – mlk
    Aug 16, 2017 at 14:01
  • $\begingroup$ @mlk thanks for the response. I've seen $\iint_S ( \nabla \times \mathbf{F} ) \cdot \ dS$ used in 3 dimensions, so I don't think that's where the difference lies. As you said, it could just be a matter of notation. Either way, I'd appreciate it if anyone could provide confirmation (for my peace of mind). $\endgroup$ Aug 16, 2017 at 14:09
  • $\begingroup$ I would just write it as $\oint_{\partial \Omega} F= \iint_{\Omega} \nabla\times F$ to be honest, no need for all the decoration. $\endgroup$ Aug 16, 2017 at 17:58

1 Answer 1


Physicists like to use the notation $\operatorname d \vec S$, i.e. they assign a direction to the area element, so it becomes a vector and the scalar product is "well-defined". This direction is simply the unit normal. Hence $\operatorname d \vec S= \hat n \operatorname d S$. I guesse some authors simply ommit the arrows as usual in the mathematics literature.

In summary, this are equivalent notations.

  • $\begingroup$ Good to know. Thank you for the assistance! $\endgroup$ Aug 16, 2017 at 14:18
  • $\begingroup$ Glad I could help $\endgroup$
    – klirk
    Aug 16, 2017 at 14:18

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