Prove an identity about $\iint_S\mathbf{r}\wedge d\mathbf{S}$ using Stokes' theorem $$
\int_C\mathbf{r}(\mathbf{r}\cdot d\mathbf{r})=\iint_S\mathbf{r}\wedge d\mathbf{S}
$$
With $\mathbf{r} = (x,y,z)$ being a 3-dimensional vector.
How do you get this result using Stokes' theorem?
 A: 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.
A: 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 $\b{c} = (x,y,z).$
