My Math Forum what is Stokes’s theorem?

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 March 27th, 2018, 11:22 PM #1 Banned Camp   Joined: Nov 2017 From: india Posts: 204 Thanks: 2 what is Stokes’s theorem? hello what is application of Stokes’s theorem?
 March 27th, 2018, 11:54 PM #2 Senior Member   Joined: Oct 2009 Posts: 772 Thanks: 279 Please stop these threads with one line questions. I don't mind if you post them, but I really doubt that you're learning anything useful here. Please tell us what math and physics you do know, why you want to know about Stokes and what you already know about Stokes.
 March 28th, 2018, 02:44 AM #3 Banned Camp   Joined: Nov 2017 From: india Posts: 204 Thanks: 2 i know nothing about stokres i want to learn it.
 March 28th, 2018, 03:43 AM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,791 Thanks: 630 Math Focus: Yet to find out. Maybe focus on odd an even numbers first... Thanks from SDK and integration
March 28th, 2018, 04:24 AM   #5
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 Originally Posted by integration i know nothing about stokres i want to learn it.
Sure nice, you want to learn Stokes. You don't tell me why or what you know, so I have limited information. But get any multivariable calculus book and work through it. You'll get to Stokes at the end.

 March 28th, 2018, 04:31 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 First, Stoke' theorem has nothing to do with 'Linear Algebra'. You can, however, find it in any Calculus text. In it's most general form it is a generalization of the fundamental theorem of Calculus- it says that $\displaystyle \int_{\partial \Omega} \omega= \int_\Omega d\omega$. Here $\displaystyle \Omega$ is some n-dimensional differentiable manifold and $\displaystyle \partial \Omega$ is its n-1 dimensional boundary; $\displaystyle \omega$ is an n-1 dimensional differential form and $\displaystyle d\omega$ is its differential. In one dimension, $\displaystyle \Omega$ is, say, the interval [a, b] and $\displaystyle \partial \Omega$ is its boundary, the two points a, and b. $\displaystyle \omega$ is some function of one variable, f(x), and $\displaystyle d\omega$ is its differential, dx. In those terms, Stoke's theorem says that $\displaystyle \int_a^b df= f(b)- f(a)$, the "Fundamental Theorem of Calculus". "Stoke's Theorem" in two dimensions is called, in most Calculus texts, "Green's theorem" $\displaystyle \oint_C \left(Ldx+ Mdy\right)$$\displaystyle = \int_D \left(\frac{\partial M}{\partial x}+ \frac{\partial L}{\partial y}\right) dxdy$. The form most often given as "Stoke's Theorem" in Calculus texts is three dimensional: $\displaystyle \int_\Gamma F\cdot d\Gamma= \int_S \nabla\times F \cdot dS$. Now, does any of that make sense to you? Have you taken a Calculus class? I get the impression, from what you have said that you are a young person, with a good healthy curiosity, who comes across these words in reading but have not taken much mathematics classes. You have to understand that all of these things involve deep background knowledge. You are welcome to ask questions but please let us know where you ran acros these words and what you do understand about them so we will know what kind of explanations will make sense to you. Thanks from studiot
 March 28th, 2018, 05:00 AM #7 Senior Member   Joined: Jun 2015 From: England Posts: 905 Thanks: 271 Thanks Country Boy, that Alabama Approach is particularly clear (mathematically).

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