Stokes' theorem( here I am only talking about the special $\mathbb{R}^3$ case) contains a line integral $\int_{\partial S} \langle f, \tau \rangle ds$.
(Actually, I would be confident if somebody could explain me the meaning of this integral and how it is defined ) If you can do this, then you do not even have to read the rest of the question :-)
The problem is I have never seen this $ds$ before. Why don't people write on both sides of the integral the surface measure $dS$? Wouldn't this be more mathematical precise( sure in that case you need to look at $\partial S$ as a 1-dimensional submanifold as we are talking about a curve)?
As I was told in the comment section, $ds$ is also a surface measure but (as far as I understood it now) a different one than $dS$, which is why people use different notation, probably in order to stress that we are talking about two different submanifolds(the curve and the surface).
But in that case ( I assume ) if $\gamma(t)$ is a chart of the curve, by definition of the surface measure we should have $\int_{\partial S} \langle f, \tau \rangle ds = \int_0^{t'}\langle f(\gamma(t)), \tau(\gamma(t)) \rangle ||\gamma'(t)||_2 dt$ is this correct? Am I interpreting right, what $ds$ does?
Thank you very much for your help. Unfortunately, no one has confirmed my result so far, so I would highly appreciate it if anybody could approve what I have written down so far.
$s$ represents arclength along the curve $\partial S$. This is a line integral along the boundary, not a surface integral. Your formula in terms of a parametrization (not chart) is correct.