# 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

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Is it fair to say that $\vec{e}_i$ are tangent to the surface $\delta V(t)$? I need a few more details on the conventions of your notation. Interesting problem. I have a derivation of the inertia tensor from KE of a rigid body, the inertia tensor naturally arises a symmetric tensor which gives the components of the quadratic form in the angular velocity... but your problem is fluid physics so perhaps my starting point is wrong. –  James S. Cook Dec 6 '12 at 18:25
@JamesS.Cook: At sorry, I tried to include as many possible but I see that now could be confusing, $\vec{e_1}=\vec{i}, \vec{e_2}=\vec{j}, \vec{e_3}=\vec{k}$ –  Freeman Dec 6 '12 at 19:02