Closed integration of exact form

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SUMMARY

The discussion centers on the application of Stokes' theorem in the context of exact forms in differential geometry. It establishes that if ##\omega## is an exact form represented as ##\omega = d\eta##, then the integral of ##\omega## over the boundary ##\partial \Omega## of a region ##\Omega## is zero, expressed mathematically as $$\oint_{\partial \Omega} \omega = 0$$. This conclusion follows from the property that the exterior derivative of an exact form is zero, leading to the equation $$\int_{\Omega} d\omega = \int_{\Omega} dd\eta = 0$$.

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Jhenrique
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If ##\omega## is an exact form ##( \omega = d\eta )## and ##\Omega## is the region of integration and ##\partial \Omega## represents the boundary of integration, so the following equation is correct:
$$\\ \oint_{\partial \Omega} \omega = 0$$?
 
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By Stoke's theorem:

$$\oint_{\partial\Omega} \omega =\int_{\Omega} d\omega=\int_{\Omega} dd\eta=0$$
 
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