Surface integral of normal vector

[daudaudaudau]

Hi. Does anyone know how to prove that
<br /> \int \int dS \hat \mathbf n = \int \mathbf r \times d\mathbf r<br />

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

 

[I like Serena]

Hi daudaudaudau!

Hi. Does anyone know how to prove that
<br /> \int \int dS \hat {\mathbf n} = \int \mathbf r \times d\mathbf r<br />

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

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:

 

[daudaudaudau]

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

 

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

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

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

 

[daudaudaudau]

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

 

[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.

 

[daudaudaudau]

Thanks, Anthony!