Showing symmetry of the stress tensor by applying divergence theorem to $\int\int_{\delta V(t)} \vec{x}\times \vec{t} dS$ I'm currently working through the symmetry of the stress tensor, in relation to viscous flow. I am looking at this by examining the conservation of angular momentum equation for a material volume $V(t)$ with unit normal $\vec{n}=(n_1,n_2,n_3)$. I am having issue with applying the divergence theorem to this term
$$\int\int_{\delta V(t)} \vec{x}\times \vec{t} dS$$
Where $\vec{x}=(x_1,x_2,x_3)$ and $\vec{t}$ is the stress vector where $\vec{t}=\vec{e}_i\sigma_{ij}n_j$, using the summation convenction, where $\sigma_{ij}$ is stress vector.
If I can extract a normal from this expression I can use the divergence theorem to convert to a volume integral and combine with the other terms of the conservation of angular momentum equation, which are volume integrals, this will lead to showing $\sigma_{ij}=\sigma_{ji}$.
Many thanks to anyone who could help.
EDIT: Under angular momentum on this page is basically doing what i'm looking for, but can't for the life of me see how they do it -or what their notation relates to
http://bobbyness.net/NerdyStuff/Navier%20Stokes%20Equations/Navier%20Stokes.html
EDIT2: Here is a link to the notes i'm learning from, page 14
http://www.maths.ox.ac.uk/system/files/coursematerial/2012/2386/9/B6aLectureNotes_img.pdf
 A: Sorry, this is a bit too late but here goes. 
The first thing is to rewrite the cross product using indices and the summation convention
$\mathbf{x} \times \mathbf{t}|_r = \epsilon_{rmn} x_m t_n$
where $\epsilon_{rmn}$ is the Levi-Civita symbol.
Now rewrite the traction vector $t_n$ as $\sigma_{jn} n_j$ which is how you sneak a normal into the equation.
Before you can apply the Divergence theorem to the resulting integral, ie.
$\int_{S} \epsilon_{rmn} x_m \sigma_{jn} n_j dS$ 
you have to understand that the integrand without the $n_j$ is an object $T$ with two free indices. So you can think of this integrand as 
$\int_{S} T_{rj} n_j dS$
By the Divergence theorem this is
$$
\int_{S} T_{rj} n_j dS = \int_{V} \frac{\partial T_{rj}}{\partial x_j} dV
$$
Now you can resub for $T_{rj}$, so the integral is
$$
\int_{V} \frac{\partial ({\epsilon_{rmn} x_m \sigma_{jn}})}{\partial x_j} dV
$$
Take the constant $\epsilon$ out, apply partial derivatives and simplify; you get
$$
\int_{V} \epsilon_{rmn} \left ( \sigma_{mn} + x_m \frac{\partial {\sigma_{jn}}}{\partial x_j} \right) dV
$$
You can proceed from here. You'll see the momentum balance appear if you group the other terms, and you'll finally just have
$$
\int_{V} \epsilon_{rmn} \sigma_{mn} dV = 0
$$
which yields the result because $V$ is an arbitrary fluid volume and the Levi-Civita symbol is skew-symmetric.
