# Interpretation of stokes theorem

I recently solved the following task:

Let $$A = [0,1]^3$$ and $$\omega = \dfrac{x_1^2 x_2^3}{1+x_3^2} \ dx_1 \wedge dx_3$$ Show that this fulfills stokes theorem by showing that $$\displaystyle \int_A \ d\omega \ = \ \displaystyle \int_{\partial A} \ \omega$$.

That worked out really well. As the solution I got for both sides $$-\dfrac{\pi}{12}$$. The question is how to interpret this (maybe physically). I just learned about diffrential forms a few days ago and Im not sure how to interpret them. I think that integrating a 1-form over a curve gives the work that the vectorfield applies on a particle that walks along this curve. Is $$-\dfrac{\pi}{12}$$ something like the work that the vectorfield applies on the cube or something like that? Please keep in mind that Im not a physicist at all.

• I assume that $[0,1^3]$ should be $[0,1]^3$. – md2perpe Jan 27 at 20:35
• Oh yes. Im going to change that immediately. – Arjihad Jan 27 at 20:36
• Why do you feel a need to interpret this physically? – md2perpe Jan 27 at 20:42
• I just wonder what this result tells me. What does it mean to get a negative value? Using the standard integrals the value would stand for the area under a curve or a volume but what does this value stand for? If this would just be a random number I guess the whole chapter would not be interesting. – Arjihad Jan 27 at 20:47
• This exercise was just an example to have you test the theorem. It doesn't need an interpretation just like $\int_0^1 3x^2 + 7x + 5 \, dx$ doesn't need an interpretation. It could represent work, but it could as well just represent the area under a function. – md2perpe Jan 27 at 20:55

OK, so the physical interpretation is that you're finding the flux of the vector field $$\vec F = \begin{bmatrix} 0 \\ -x_1^2x_2^3/(1+x_3^2) \\ 0\end{bmatrix}$$ outwards across $$\partial A$$. In this setting, $$d\omega = (\text{div}\, \vec F) dx_1\wedge dx_2\wedge dx_3$$, and $$\text{div}\,\vec F$$ tells you infinitesimally whether you have a source or a sink (or neither) at each point; adding these up all over $$A$$ gives the net flux across the boundary.