Accepting Stokes' theorem (101) as a premise, how do I prove the fundamental theorem of calculus (102)? I know that the FTC is a straightforward specialization of Stokes' theorem and so proving $\text{(101)} \implies \text{(102)}$ should be trivial. I'm using the triviality of this implication to try to improve my understanding of Stokes' theorem. I vaguely remember Stokes' theorem from school, but didn't really understand it then and don't really understand it now. I wrote both theorems with the boundary-related part on the left and the interior-related part on the right.

Basically, what values do I pick for $\Omega$ and $\omega$ to turn (101) into (102)?

$$ \int_{\partial \Omega} \omega = \int_{\Omega} d \omega \tag{101} $$

$$ F(b) - F(a) =\int_{a}^{b} f(x)dx \tag{102} $$

Explaining the notation, $\omega$ is a differential form and and $\Omega$ is an orientable manifold. I think this means that $\Omega$ is not inherently equipped with an orientation but is capable of receiving one. So, $\Omega$ cannot be something like a Möbius strip.

I'm trying to get the left sides to match first, but I'm stuck.

So it seems like the most straightforward way to prove the implication is to have $\Omega$ be a closed interval on the real line $[a, b]$.

However, the boundary of $[a,b]$ is a set of two points $\{a, b\}$. I'm trying to understand what $\omega$ would have to be to make $\int_{\{a, b\}} \omega$ equal to $F(b)-F(a)$ rather than $F(b)+F(a)$.

What values to pick for $\Omega$ and $\omega$?


You should take $\omega = F$ (which is a $0$-form), and you should take $\Omega = [a,b]$ as you say, with the natural orientation (given by choosing the identity charts to be positively oriented, or to say it a different way, by choosing the nonvanishing $1$-form $dx$ on $[a,b]$, where $x$ is the identity coordinate function). This induces an orientation on $\partial[a,b] = \{a,b\}$, giving $\{a\}$ negative local orientation and $\{b\}$ positive local orientation, so that integrating $F$ on it gives $F(b)-F(a)$ instead of $F(b)+F(a)$. The statement of Stokes' theorem includes the assumption that $\partial \Omega$ is also oriented, in this specific way induced by the orientation on $\Omega$. Look here for more details in the 1-dimensional case.


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