# Surface integral of normal vector

1. Aug 28, 2011

### daudaudaudau

Hi. Does anyone know how to prove that
$$\int \int dS \hat \mathbf n = \int \mathbf r \times d\mathbf r$$

i.e., the surface integral of the unit normal vector equals the line integral on the r.h.s. ?

2. Aug 28, 2011

### I like Serena

Hi daudaudaudau!

This looks a bit like a special case of the Kelvin–Stokes theorem.
(Fixed your latex.)

But I suspect you mixed up the operations.

The Kelvin-Stokes theorem says:

Last edited: Aug 28, 2011
3. Aug 28, 2011

### daudaudaudau

No, I really do want to integrate the normal vector over a surface, i.e. the result should be a vector.

4. Aug 28, 2011

### I like Serena

I've been stumped at this for awhile, and Googling hasn't helped much.

I did think up a proof:

$$\oint \mathbf r \times d\mathbf r = \oint \begin{pmatrix}ydz - zdy \\ zdx - xdz \\ xdy - ydx \end{pmatrix} = \hat {\mathbf x} \oint \begin{pmatrix}0 \\ -z \\ y\end{pmatrix} \cdot \begin{pmatrix}dx \\ dy \\ dz\end{pmatrix} + \hat {\mathbf y} \oint \begin{pmatrix}z \\ 0 \\ -x\end{pmatrix} \cdot \begin{pmatrix}dx \\ dy \\ dz\end{pmatrix} + \hat {\mathbf z} \oint \begin{pmatrix}-y \\ x \\ 0\end{pmatrix} \cdot \begin{pmatrix}dx \\ dy \\ dz\end{pmatrix}$$
$$\oint \mathbf r \times d\mathbf r = \hat {\mathbf x} \oint \begin{pmatrix}0 \\ -z \\ y\end{pmatrix} \cdot d\mathbf r + \hat {\mathbf y} \oint \begin{pmatrix}z \\ 0 \\ -x\end{pmatrix} \cdot d\mathbf r + \hat {\mathbf z} \oint \begin{pmatrix}-y \\ x \\ 0\end{pmatrix} \cdot d\mathbf r$$

Applying the Kelvin-Stokes theorem we get:
$$\oint \mathbf r \times d\mathbf r = \hat {\mathbf x} \iint (\nabla \times \begin{pmatrix}0 \\ -z \\ y\end{pmatrix}) \cdot d\mathbf S + \hat {\mathbf y} \iint (\nabla \times \begin{pmatrix}z \\ 0 \\ -x\end{pmatrix}) \cdot d\mathbf S + \hat {\mathbf z} \iint (\nabla \times \begin{pmatrix}-y \\ x \\ 0\end{pmatrix}) \cdot d\mathbf S$$
$$\oint \mathbf r \times d\mathbf r = \hat {\mathbf x} \iint 2 \hat {\mathbf x} \cdot d\mathbf S + \hat {\mathbf y} \iint 2 \hat {\mathbf y} \cdot d\mathbf S + \hat {\mathbf z} \iint 2 \hat {\mathbf z} \cdot d\mathbf S$$
$$\oint \mathbf r \times d\mathbf r = 2 \iint d\mathbf S = 2 \iint dS \hat {\mathbf n}$$

However, this is a factor 2 different from the equation you proposed.

Last edited: Aug 28, 2011
5. Aug 29, 2011

### daudaudaudau

Sorry, I was missing a factor of two in the equation i posted, so your proof is correct :)

6. Sep 1, 2011

### Anthony

Alternatively, for arbitrary fixed vector $\mathbf{a} \in \mathbf{R}^3$
$$\mathbf{a} \cdot \oint \mathbf{r} \wedge \mathrm{d}\mathbf{r} = \oint \mathbf{a} \cdot (\mathbf{r} \wedge \mathrm{d}\mathbf{r}) = \oint (\mathbf{a} \wedge \mathbf{r}) \cdot \mathrm{d}\mathbf{r} = \iint \nabla \wedge ( \mathbf{a} \wedge \mathbf{r})\cdot \mathbf{n}\, \mathrm{d}S = \iint 2\mathbf{a}\cdot \mathbf{n}\, \mathrm{d}S = \mathbf{a} \cdot \iint 2\mathbf{n}\, \mathrm{d}S$$
and the result follows since $\mathbf{a}$ was arbitrary.

7. Sep 1, 2011

### daudaudaudau

Thanks, Anthony!

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