Question on Computation of Integral of a Form

Again:

I'm trying to understand the result of a certain integral of a form in a paper I'm reading (for which I do not, unfortunately, have a link):

We start with a surface S that is oriented, compact, and with non-empty boundary, and we construct its mapping torus $S_{\phi}$by $\phi$.

We then define a set S of forms by:

$$S:= \{\omega \in \Omega^1(S) \mid \text{(1) } \omega=(1+s)d\theta \text{ near \partial S, and (2) } d\omega \text{ is a volume form on S }\}$$

and, near each boundary component of $S$ , we use coordinates $(s,\theta)$ in $[0,1]x\times S^1$.

Then we set out to show $S$ is non-empty, by assuming there is a form $\gamma$ on $S$

that satisfies condition (1) above ,near $\partial S$, i.e., $\gamma=(1+s)d\theta$

Now :

This is stumping me:

Note that $$\int_S d\gamma = \int_{\partial S}\gamma = 2\pi|\partial S|$$ where $|\partial S|$ is the number of boundary components.

Now, I understand the first equality is just Stoke's theorem, but I don't understand why the area is necessarily equal to $2\pi|\partial S|$. I'd appreciate your suggestions.

Note that near $\partial S$ we have $\gamma = (1+s)d\theta$, hence on $\partial S$, $\gamma = d\theta$, so on each component $C$ of $\partial S$, the integral over $\gamma$ gives $\int_{S^1} d\theta = 2\pi$.
• @RickyBobby As far as I understand, on $\partial S$, we have $s = 0$. Commented Jul 26, 2013 at 12:26