Surface integral with vector integrand identity On page 155 of G.E. Hay Vector and Tensor Analysis, he asks the reader to prove that
$$\int_S \mathbf{n}\times\mathbf{x}\ dS = 0$$
where $\mathbf{n}$ is the unit outer normal and $\mathbf{x}$ is the position vector of a general point X on the closed surface S.
The vector integrand will always be perpendicular to the unit normal, so there isn't any way to rewrite this as the dot product of some function with the unit normal. Working with components, this problem is equivalent to proving that
$$\mathbf{i}\int_S n_yz-n_zy\ dS+\mathbf{j}\int...dS+\mathbf{k}\int...dS = 0$$
My next thought was to show that each of these integrals could be expressed as the dot product of some field with the normal, which I did (as I write this).
For the first term:
$$\mathbf{i}\int_S n_yz-n_zy\ dS=\mathbf{i}\int_S\left<0,z,-y\right>\cdot \mathbf{n}\ dS$$
And since the vector $\left<0,z,-y\right>$ has divergence $0$, we can use Stokes theorem to prove that the surface integral over any closed surface $S$ will be zero. This can be repeated for the other terms.
I don't know if this is actually a valid proof, or if I've overlooked something, and it's also remarkably inelegant. I'd appreciate any thoughts on the attempt, or some alternative approach that isn't so ugly. I am also curious if the method of this proof implies that Stokes' and Gauss' Theorem hold in general for surface integrals with vector integrands. The vector field in question, $\mathbf{n}\times\mathbf{x}$, also has zero divergence, and it seems possible that the Divergence Theorem could be extended to this class of integrals.
Thanks
 A: We can apply the version of the Divergence Theorem that states 
$$\int_V \nabla \times A(\vec r)\,dV=\oint_S \hat n \times \vec A(\vec r)\,dS \tag 1$$
for suitable $S$ and $\vec A$.  Here, we have $\vec A(\vec r)=\vec r$ so that $\nabla \times \vec A(\vec r)=\nabla \times \vec r=0$.  And the coveted result follows immediately.

NOTE:
To show that $(1)$ is a special case of the Divergence Theorem, let $\vec A(\vec r)= \vec B(\vec r)\times \hat x_i$.  Then, applying the identity 
$$\nabla \cdot (\vec F\times \vec G)=(\nabla \times \vec F)\cdot \vec G-\vec F \cdot(\nabla \times G)$$
with $\vec F=\vec B$ and $\vec G=\hat x_i$, we find that
$$\nabla \cdot (\vec B\times \hat x_i)=(\nabla \times \vec B)\cdot \hat x_i \tag 2$$
Applying the Divergence Theorem to $(2)$ reveals
$$\begin{align}
\hat x_i \cdot \int_V \nabla \times \vec B\,dV&=\int_V \nabla \cdot (\vec B\times \hat x_i)\,dV\\\\
&=\oint_S (\vec B\times \hat x_i)\cdot \hat n\,dS\\\\
&=\hat x_i \cdot \oint_S \hat n \times \vec B\,dS \tag 3
\end{align}$$
Since $\hat x_i$ is any Cartesian unit vector, we have from $(3)$
$$\int_V \nabla \times \vec B\,dV = \oint_S \hat n \times \vec B\,dS$$
as was to be shown!
