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Surface integral of normal vector

  1. Aug 28, 2011 #1
    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. jcsd
  3. Aug 28, 2011 #2

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    Hi daudaudaudau! :smile:

    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
  4. Aug 28, 2011 #3
    No, I really do want to integrate the normal vector over a surface, i.e. the result should be a vector.
  5. Aug 28, 2011 #4

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

    I did think up a proof:

    [tex]\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}
    [tex]\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:
    [tex]\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
    [tex]\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
    [tex]\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
  6. Aug 29, 2011 #5
    Sorry, I was missing a factor of two in the equation i posted, so your proof is correct :)
  7. Sep 1, 2011 #6
    Alternatively, for arbitrary fixed vector [itex]\mathbf{a} \in \mathbf{R}^3[/itex]
    [tex] \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 [/tex]
    and the result follows since [itex]\mathbf{a}[/itex] was arbitrary.
  8. Sep 1, 2011 #7
    Thanks, Anthony!
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