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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.

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  • $\begingroup$ I assume that $[0,1^3]$ should be $[0,1]^3$. $\endgroup$ – md2perpe Jan 27 at 20:35
  • $\begingroup$ Oh yes. Im going to change that immediately. $\endgroup$ – Arjihad Jan 27 at 20:36
  • $\begingroup$ Why do you feel a need to interpret this physically? $\endgroup$ – md2perpe Jan 27 at 20:42
  • $\begingroup$ 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. $\endgroup$ – Arjihad Jan 27 at 20:47
  • $\begingroup$ 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. $\endgroup$ – md2perpe Jan 27 at 20:55
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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.

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