# Stoke's Theorem Question

**URL:** <https://boards.straightdope.com/t/stokes-theorem-question/946850>\
**Category:** Factual Questions\
**Created:** [July 19, 2021, 8:01pm UTC](https://boards.straightdope.com/t/stokes-theorem-question/946850 "2021-07-19T20:01:00Z")\
**Posts on this page:** 5\
**Page:** 1

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**Author:** ![Someonewastaken](https://avatars.discourse-cdn.com/v4/letter/s/ba8739/32.png) [@Someonewastaken](https://boards.straightdope.com/u/Someonewastaken)\
**Post date:** [July 19, 2021, 8:01pm UTC](https://boards.straightdope.com/t/stokes-theorem-question/946850/1 "2021-07-19T20:01:00Z")

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Why do we take the dot product of n and f. f is the vector field and n is the unit vector normal to the plane being analyzed.

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**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\
**Post date:** [July 19, 2021, 8:11pm UTC](https://boards.straightdope.com/t/stokes-theorem-question/946850/2 "2021-07-19T20:11:17Z")

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If you integrate that over the surface, you get the flux of the vector field through that surface.

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**Author:** ![Someonewastaken](https://avatars.discourse-cdn.com/v4/letter/s/ba8739/32.png) [@Someonewastaken](https://boards.straightdope.com/u/Someonewastaken)\
**Post date:** [July 19, 2021, 11:24pm UTC](https://boards.straightdope.com/t/stokes-theorem-question/946850/3 "2021-07-19T23:24:21Z")

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Oh! Duh, thanks.

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**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [July 19, 2021, 11:41pm UTC](https://boards.straightdope.com/t/stokes-theorem-question/946850/4 "2021-07-19T23:41:39Z")

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If the flow is perpendicular to the surface, then n \cdot f will just be the magnitude of f, since all of the flow is going through the surface. On the other hand, if the flow is parallel, then n \cdot f will be zero–again, expected since none of the flow is going through the surface (it’s all going _along_ the surface). And intermediate angles will have intermediate values (proportional to \cos(\theta)).

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**Author:** ![leahcim](https://avatars.discourse-cdn.com/v4/letter/l/b4bc9f/32.png) [@leahcim](https://boards.straightdope.com/u/leahcim)\
**Post date:** [July 20, 2021, 12:03am UTC](https://boards.straightdope.com/t/stokes-theorem-question/946850/5 "2021-07-20T00:03:36Z")

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Nit: The dude’s name was “George Stokes”, so it is “Stokes’ Theorem”, not “Stoke’s Theorem”.