# Confusion about integration theorem (stokes?)

While studying electrodynamics, I got confused by one specific step that as been used in the book when calculating the magnetic momentum under a closed curve (current wire) $\partial F$:

$$\int_{\partial F} \textbf{r}\times\mathrm{d}\textbf{r} = e_\textbf{z}\mathrm{vol}\:F$$

This seems like some kind of stokes theorem, but when expanding the surface, it would say

$$\int_{\partial F} \textbf{r}\times\mathrm{d}\textbf{r} = \int_F \textbf{r} \cdot \textrm{d}\textbf{n}$$

But isn't Stoke's theorem the other way around, having the cross product (or exterior derivative) on the side of the surface?

Q1: What is the confusion here?

EDIT:

F is a surface in $\mathbb{R}^2 \times \{0\}$, which makes $\partial F$ a closed curve in said space. $\mathbf{r}$ denotes the position vector. $\mathbf{e_z}$ is the third vector of the standard basis of $\mathbb{R}^3$.

The first part will probably be trivial to answer, but since I often get confused by such details, I'd like to take the approach using differential forms and the exterior derivative.

So I tried to calculate, $r$ being a 1-form:

$$\mathrm{d}(\textbf{r}\wedge\mathrm{d}\textbf{r}) = \mathrm{d}\textbf{r} \wedge \mathrm{d}\textbf{r} + (-1)^1 \textbf{r}\wedge\mathrm{d}\mathrm{d}\textbf{r} = 0 + 0 = 0$$ Using that $\wedge$ is bilinear and $\mathrm{d}^2=0$. That cannot be true though, since the integral is nonzero.

Q2: where does the mistake of that approach lie?

• What is $r$? What is $e_z$? $F$ is a curve in $\mathbb{R}^2$? Jan 10, 2016 at 21:11