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Let first state the version of Stokes theorem I am looking at

For any smooth $(n-1)$-form $\omega$ with compact support on the oriented $n$-dimensional manifold $M,$ $\int_M d\omega = \int_{\partial M} \omega$.

I never learn to know how to use this properly. One specific problem I have at hand is the following:

$f$ is a smooth function defined on the unit disk $\Delta,$ $$ \eta = \frac{1}{2 \pi i} \frac{f(w)\,dw}{w-z}, $$ $$ d\eta = - \frac{1}{2\pi i} \frac{\partial f(w)}{\partial\bar w} \frac{d \omega \wedge d \bar \omega}{w-z}, $$ let $\Delta_{\varepsilon}$ be the disc of radius $\varepsilon$ around $z$. Then by stokes theorem:

$$ \frac{1}{2\pi i}\int_{\partial \Delta_{\varepsilon}} \frac{f(w) \, dw}{w-z} = \frac{1}{2\pi i} \int_{\partial \Delta}\frac{f(w) \, dw}{w-z} + \frac{1}{2\pi i} \int_{ \Delta - \Delta_{\varepsilon}} \frac{\partial f(w)}{\partial\bar w} \frac{d \omega \wedge d \bar \omega}{w-z}.$$

Can someone explain to me in detail how does the Stokes theorem apply here?

The orientation is important in the proof of the stokes theorem since we want to choose a set of oriented atlas. But I do not see how do we see the orientation in real applications.

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  • $\begingroup$ This question is a little imprecise, IMHO. Where exactly are you stuck? Note that the manifold we're integrating over is an annulus: it's the unit disk, with a smaller disk cut out of it. That's what we're integrating over in the integral on the far right. If you bring the rest of the integrals to the left side, you'll see we're integrating over -(outer boundary of annulus) + (inner boundary of annulus). This is the oriented boundary, from the induced orientation of the unit disk (like, always keep the interior on your left). $\endgroup$
    – Steve D
    Sep 25, 2017 at 21:05
  • $\begingroup$ Exactly I do not see why the induced orientation on the boundary should have opposite signs $\endgroup$
    – Keith
    Sep 25, 2017 at 21:07
  • $\begingroup$ To induce an orientation, you pick a vector in the tangent space that points "inward" (that is, not in the subspace spanned by boundary tangent vectors). The vector pointing inward from the outer boundary circle is going the opposite way of the one pointing inward from the inner boundary circle. $\endgroup$
    – Steve D
    Sep 25, 2017 at 21:11

1 Answer 1

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Consider applying Stokes theorem on the annulus $\Delta-\Delta_{\epsilon}$, using your differential form $\eta$:

$\int_{\Delta-\Delta_{\epsilon}} d\eta = \int_{\partial(\Delta-\Delta_{\epsilon})} \eta$ .

The boundary $\partial(\Delta-\Delta_{\epsilon})$ consists of 2 circles, the outer one being the circumference of $\Delta$ and the inner one being the circumference of $\Delta_{\epsilon}$. Like Steve have commented, these 2 circles are oppositely oriented and so the above equation becomes

$\int_{\Delta-\Delta_{\epsilon}} d\eta = \int_{\partial\Delta} \eta -\int_{\partial\Delta_{\epsilon}} \eta$ .

Then plug in your $\eta$ and $d\eta$ to get what you want to show.

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