# Understanding tensor divergence notation in an integral

Given a smooth tensor valued function $\sigma:R^2\rightarrow R^{2\times2}$, I'm trying to show that

$\int_\Omega \nabla\cdot\sigma=\int_{\partial\Omega}\sigma n$,

where $\Omega$ is a connected simple region in $R^2$, $\partial\Omega$ is a closed simple curve along boundary of $\Omega$, and $n$ is a unit vector.

I'm struggling with several points in this problem:

1. How can I represent $\sigma$ explicitly?
2. How can I explicitly represent $\sigma n$?
3. Given that $\nabla\cdot\sigma$ is a vector, would the differential of the left hand side of the equation still be $dxdy$?
4. I am told that I can use divergence theorem explicitly, but the only version I'm familiar with applies to integrals whose integrands are scalar quantities? What is the tensor-divergence theorem equivalent?

Any help would be greatly appreciated! :) Thanks.

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$$\int_\Omega\sum_{i=1}^2\frac{\partial}{\partial x_i}\sigma_{ij}dxdy=\int_{\partial\Omega}\sum_{i=1}^2 n_i\sigma_{ij}ds\qquad j=1,2$$
in practice you apply the usual divergence theorem to each vector $\boldsymbol{\sigma}_j=(\sigma_{1j},\sigma_{2j})$.
So, you mean "the usual divergence theorem" for each column vector of $\sigma$? – Paul Sep 14 '12 at 20:56
In the definition of $\sigma$, it maps a vector to a matrix. But in your notation of the right hand side, wouldn't $\sigma n$ be a vector? Isn't this a contradiction? – Paul Sep 14 '12 at 21:29
No, the notation $\sigma:R^2\rightarrow R^{2\times2}$ means that in each point of $\Omega$ (two coordinates) is defined a matrix (four numbers in a square). Like each $2\times2$ matrix, $\sigma$ can be considered, in each point, as a linear map from $R^2$ to $R^2$, i.e. it maps vectors to vectors (by a matrix multiplication). – enzotib Sep 14 '12 at 21:35
Oh... ok... So, at each point, a matrix is assigned. Could I also consider it as if $\sigma$ were a matrix whose elements are functions of variables in $R^2$? – Paul Sep 14 '12 at 21:40