is the stress tensor (elasticity theory) actually a pseudo tensor? Please, is the stress tensor (elasticity theory) actually a pseudo tensor? It seems to me it must change its sign when coordinate system changes its orientation.
The argument is as follows:
We must integrate something over a small two dimensional surface to obtain the force vector which is applied to this surface. Thus this "something" must be a tensor of the form $\sum_{j<k}\omega^i_{jk}dx^j\wedge dx^k$ All indexes run 1,2,3
The tensor $\omega^i_{jk}dx^j$ is skew symmetric in the subscripts ,therefore it has 9 independent components. So it corresponds to a 3*3 matrix. This matrix is a stress tensor. But the correspondence can be only pseudo (axial) tensored. There is no pure tensor correspondence that takes three indexed tensor to two indexed one.
 A: No.  I think the stress tensor is a rank two tensor that converts the outward pointing unit normal to the plane into a force: $F_i = \sum_{j} w_{ij} \nu^j$.  In effect, $\nu$ is the Hodge dual to $dx^j\wedge dx^k$.  Think about what it would be in $n$ dimensions - you would have to wedge $(n-1)$ 1-forms to describe a plane, and it is so much easier to consider a single vector perpendicular to the plane.
A: If I'm understanding your question correctly then I believe you're right. The transformation from a pair of antisymmetric indices to a single index is (as Stephen pointed out) the three-dimensional case of the Hodge dual, which you may be more familiar with as contraction with the Levi-Civita "pseudotensor" $\epsilon$. 
In DG we typically make a choice of preferred orientation, after which the Hodge dual really sends tensors to tensors and $\epsilon$ is a genuine tensor. One just has to keep in mind that the components of $\epsilon$ will have signs depending on the orientation of your chosen frame - you cannot know the components of $\epsilon$ based on the coordinate system and the metric alone.
It seems in physics that it is more conventional to have $\epsilon^1{}_{23} = 1$ in any orthonormal frame, so that it is instead a "pseudotensor" satisfying a transformation law that flips the sign when you flip orientation. When using this latter formalism, the $(1,1)$-valency stress-tensor $T^i_l = \frac12 \omega^i_{jk}\epsilon^{jk}{}_l$ is a pseudotensor due to the presence of the pseudotensor $\epsilon$. (In fact I believe it's a "mixed" tensor of sorts - the upper index transforms tensorially while the lower one transforms pseudotensorially.)
A: to obtain the force acting on a surface $S$ one must compute an integral
$$F^i=\int_S\sqrt g(w_{i1}dx^2\wedge dx^3+w_{i2}dx^3\wedge dx^1+w_{i3}dx^1\wedge dx^2)$$ ($w_{ij}$ is the stress tensor as above)
suppose that $w_{ij}$ is a true tensor then the form $\sqrt g(w_{i1}dx^2\wedge dx^3+w_{i2}dx^3\wedge dx^1+w_{i3}dx^1\wedge dx^2)$ is a pseudo tensor and thus the force F is also a pseudo tensor. Thats not good
On the other hand , a density $\rho$ is also a pseudo scalar Indeed,
$$m=\int_V\rho\sqrt g dx^1\wedge dx^2\wedge dx^3$$
the mass m is a true scalar, the form $\sqrt g dx^1\wedge dx^2\wedge dx^3$ is a pseudo tensor, so that $\rho$ is a pseudo tensor 
Consider the mechanical equation
$$\rho a^k=\rho F^k+\nabla_iw^{ki}$$
The left side is a pseudo vector (due to $\rho$) and  the right one is also a pseudo vector. The are no contradictions
