How does one integrate an equation with a tensor? I was looking at the equation for the differential force $dF$ caused from a stress tensor $\sigma$ acting on a differential surface $dS$. In mathematical terms,
$dF=\sigma \cdot dS$
Taking the integral of both sides yields.
$$F=\int \sigma \cdot dS$$
What is the nature of this integral? How does on go about calculating it? I would like to see an example given a non-trivial $\sigma$.
Can this be generalized to $F=\int \sigma \cdot dV$?
My idea was to integrate with respect to the variables $x$ and $y$ and represent $dS$ as a normal vector multiplied by the $dx$ and $dy$ differentials. Taking the product of the right hand side, you then integrate with respect to the differentials.
 A: You can imagine three equations one for each component of the force and then the tensor can be thought of as three vectors.
For instance $F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$ where $d\vec S=\hat n dA.$ So now it is a regular flux surface integral. You just have three of them:
$F_x=\int \sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_zdA$,
$F_y=\int \sigma_{yx}n_x+\sigma_{yy}n_y+\sigma_{yz}n_zdA,$ and $F_z=\int \sigma_{zx}n_x+\sigma_{zy}n_y+\sigma_{zz}n_zdA.$
See how it almost looks like a vector on the left and a matrix times a vector on the right? It's not more complicated than that.
$$F=\int \sigma dS=\int \sigma n dA.$$
A: Seems as though my idea is the correct one. Here's an example.
Since, $dS$ is normal to the surface and has a magnitude equal to the area of a differential. $dS$ can be represented by a cross product of the differentials of the relevant surface.
Given,
$$\sigma= \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} $$
We want to find the force over the surface of a square parallel to the x-y plane. The cross product of the differentials yields $[0,0,dx \cdot dy]$. Taking the transpose of this we get $dS$. Putting this into the formula yields,
$$\sigma \cdot dS=[dx \cdot dy,0, dx \ dy]^t$$ 
Taking the integral over the square gives,
$$F=[s^2,0,s^2]^t$$
Where $s$ is the side length of the square. This method can be generalized for other surfaces of course.
