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$$ \int_C\mathbf{r}(\mathbf{r}\cdot d\mathbf{r})=\iint_S\mathbf{r}\wedge d\mathbf{S} $$

With $\mathbf{r} = (x,y,z)$ being a 3-dimensional vector.

How do you get this result using Stokes' theorem?

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  • $\begingroup$ explain what is $r$ $\endgroup$ – Emanuele Paolini Jun 20 '13 at 19:15
  • $\begingroup$ r = (x,y,z) $\endgroup$ – AronK Jun 20 '13 at 19:16
  • $\begingroup$ Hi, @AronK I have edited the title to make it more detailed and added a differential geometry tag. $\endgroup$ – Shuhao Cao Jun 20 '13 at 22:22
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Starting from the standard classical Kelvin-Stokes relation, $$ \iint_S \nabla \times \mathbf{F} \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}, $$

let $\mathbf{F} = \phi \mathbf{c} $, where $\mathbf{c}$ is an arbitrary constant vector and $\phi$ is a scalar field. Next we turn the vector identity crank a few times until something nice pops out. On the LHS,

$$\nabla\times\mathbf{F}= \nabla \times (\phi\mathbf{c})= \nabla\phi\times\mathbf{c} + \phi\nabla\times\mathbf{c}= \nabla\phi\times\mathbf{c},$$

$$\nabla\times\mathbf{F}\cdot d\mathbf{S} =(\nabla\phi\times\mathbf{c}) \cdot d\mathbf{S} = \mathbf{c} \cdot (d\mathbf{S} \times \nabla\phi),$$

$$\iint_S\nabla\times\mathbf{F}\cdot d\mathbf{S} =\iint_S \mathbf{c} \cdot (d\mathbf{S} \times \nabla\phi) = \mathbf{c} \cdot \iint_S (d\mathbf{S} \times \nabla\phi).$$

On the RHS, $$ \mathbf{F} \cdot d\mathbf{r} =(\phi\mathbf{c})\cdot d\mathbf{r} = \mathbf{c} \cdot (\phi d\mathbf{r}),$$

$$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C \mathbf{c} \cdot (\phi d\mathbf{r}) = \mathbf{c} \cdot \oint_C \phi d\mathbf{r}.$$

Now since the equality we now have holds for any arbitrary constant vector $\mathbf{c}$, we can arrive at the following corollary to KST:

$$ \iint_S (d\mathbf{S} \times \nabla\phi) = \oint_C \phi d\mathbf{r}. $$

The above intermediate result takes care of most the work for this problem. The last step is to think of a scalar field with the right gradient, which is fairly straightforward for the problem at hand. I'll leave this as an exercise at least for now.

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  • $\begingroup$ Nice answer +1, also you might wanna point out that $\phi = \mathbf{r}\cdot \mathbf{r}/2$. $\endgroup$ – Shuhao Cao Jun 21 '13 at 0:30
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Your identity should have a factor of half in it.

$$ \newcommand{\b}{\mathbf}\iint_S\b{r}\wedge d\b{S} \\ = \iint_S (x,y,z)\wedge (dy\wedge dz, dz\wedge dx, dx\wedge dy) \\ =\begin{pmatrix} \iint_S y \wedge dx\wedge dy - z\wedge dz\wedge dx \\ \iint_S z\wedge dy\wedge dz- x\wedge dx\wedge dy \\ \iint_S x\wedge dx\wedge dy- y \wedge dy\wedge dz \end{pmatrix} \\ =\frac{1}{2}\begin{pmatrix} \iint_S d(y^2\wedge dx) + d(z^2\wedge dx) \\ \iint_S d(z^2\wedge dy)+ d(x^2\wedge dy) \\ \iint_S d(x^2\wedge dz)+ d(y^2\wedge dz). \end{pmatrix} \\ = \frac{1}{2}\begin{pmatrix} \iint_S d(y^2\wedge dx) + d(z^2\wedge dx) + \color{red}{d(x^2\wedge dx)} \\ \iint_S d(z^2\wedge dy) + d(x^2\wedge dy) + \color{red}{d(y^2\wedge dy)} \\ \iint_S d(x^2\wedge dz) + d(y^2\wedge dz) + \color{red}{d(z^2\wedge dz)}. \end{pmatrix} $$ The red terms are artificially added, for $d(x^2\wedge dx) = 0$, same for $y$, $z$ terms. Generalized Stokes theorem reads: $$ \iint_S d\omega = \oint_{\partial S} \omega.\tag{1} $$ Hence: $$ \iint_S\b{r}\wedge d\b{S} = \frac{1}{2}\begin{pmatrix} \oint_{\partial S} y^2\wedge dx + z^2\wedge dx + x^2\wedge dx \\ \oint_{\partial S} z^2\wedge dy + x^2\wedge dy + y^2\wedge dy \\ \oint_{\partial S} x^2\wedge dz + y^2\wedge dz + z^2\wedge dz. \end{pmatrix} \\ =\frac12 \oint_{\partial S} (x^2+y^2+z^2) \wedge (dx,dy,dz) = \frac12 \oint_{\partial S} \b{r}\cdot \b{r} \wedge d\b{r}. $$ For a $0$-form like $x^2+y^2+z^2$ wedge with a $1$-form like $dx$ is just multiplication, therefore: $$ \iint_S\b{r}\wedge d\b{S} = \frac12 \oint_{\partial S}\b{r}\cdot \b{r} d\b{r}. $$


Notice that in David H's answer, his $\phi = \b{r}\cdot\b{r}/2$. And the Kelvin-Stokes theorem is (1) in disguise. In his proof, $\b{c}$ is a constant vector, actually any irrotational vector field will do, for example $\b{c} = (x,y,z).$

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