Stoke's Theorem: Difference between revisions

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: $$\int_M d\omega = \int_{\partial M}\omega$$
<math>\int_M d\omega = \int_{\partial M}\omega</math>


== Resources: ==
== Resources: ==
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== Discussion: ==
== Discussion: ==
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Latest revision as of 16:39, 19 February 2023

[math]\displaystyle{ \int_M d\omega = \int_{\partial M}\omega }[/math]

Resources:[edit]

Discussion:[edit]