I am reading Analysis on Manifolds by Munkres, and I am at the part about Stokes' Theorem right now. I am trying to understand the material in this section, but at the moment it seems contradictory.

(1D version of) Stokes' Theorem: Let M be a compact oriented 1-manifold in $\mathbb{R}^n$, $A \subseteq \mathbb{R}^n$ open s.t. $M \subseteq A$.

  • If $\partial M \neq \emptyset$, give it the induced orientation, then $\int_M df =\int_{ \partial M} f$
  • If $\partial M = \emptyset$, then $\int_M df$=0.

Yet we have the following (where $\exists$ a $C^\infty$ bijection $\alpha:[a,b] \to M$ s.t. $D \alpha(t) \neq 0$ $\forall t \in [a,b]$, $p=\alpha(a), q=\alpha(b)$)
enter image description here
Stokes' Thm agrees with Thm 37.3, since $\partial M= \{ p,q \}$. However, consider the following:

  • $M= \alpha(a,b]$ is the image of $(a,b]$ under $\alpha$, so $\partial M= \{ q \}$, and $\int_M df=\int_{ \partial M} f=f(q)$
  • $M= \alpha [a,b)$, so $\partial M =\{ p \}$, and $\int_M df=\int_{ \partial M} f=-f(p)$
  • $M= \alpha (a,b)$, so $\partial M = \emptyset$, and $\int_M df$=0

Note the proof of Thm 37.3 starts by noting $\alpha (a,b)$ is M, except for a set of measure zero (which is true in all 3 cases above), so it seems by the same reasoning we should have $\int_M df=f(q)-f(p)$ in all 3 cases above.
I am very confused at the moment and would greatly appreciate any help to resolve this apparent contradiction. I am new to this subject and my only exposure is the material before and including this in Munkres.

  • $\begingroup$ @xpaul: Your understanding is wrong, as well. Jason: Stokes's Theorem applies to a compact manifold with boundary. You need to parametrize by the closed interval. You don't get a choice. $\endgroup$ – Ted Shifrin Jan 20 '17 at 23:37
  • $\begingroup$ @TedShifrin Thank you so much for pointing that out, I see my error $\endgroup$ – Jason Jan 21 '17 at 17:54

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