# Talk:Kelvin–Stokes theorem

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## Contents

Oh Wikipedia... Why such a complicated proof? Why can't you show an elementary derivation meant for human beings? i feel it is very complicated proof can you just elaborate simply — Preceding unsigned comment added by Adk.jaggu (talkcontribs) 02:36, 1 May 2016 (UTC)

Both [1] and Kelvin–Stokes theorem work for me (cut and pasted on a Mac). Do you have trouble with the dash? —Kusma (t·c) 09:27, 28 January 2014 (UTC)

## Merging into Stokes' Theorem

This page should be merged in Stokes' Theorem as they're the same topic. If there's anything worth salvaging in this article it should be moved, but I'm not convinced there is anything. 131.111.184.8 (talk) 00:34, 6 November 2014 (UTC)

I think a separate article is appropriate. The article Stokes' theorem is meant to treat the general case concerning differential forms and manifolds. This article is about the special case that concerns the flux of a curl through a surface. We have separate articles for Green's theorem and the Divergence theorem, which are also important special cases of Stokes' theorem. Sławomir Biały (talk) 12:13, 6 November 2014 (UTC)
Unlike the other special cases of Stokes' theorem, Kelvin-Stokes is normally referred to as just Stokes' theorem. Even if you want to split the article then the current page here needs to be wiped and started from scratch. It is completely irredeemable. 131.111.184.8 (talk) 13:30, 14 December 2014 (UTC)
I do not agree that the article is completely irredeemable. The main issue is the impenetrable formal style, not the actual content (although I'm don't really understand what's going on in the last section). Sławomir Biały (talk) 14:03, 14 December 2014 (UTC)
I see no way of fixing the style without a total rewrite: hence, irredeemable. I'm also not convinced there's anything that needs adding to the existing material at Stokes' theorem#Kelvin–Stokes theorem. A nice proof might be useful, but remember that Wikipedia isn't meant to be a textbook. 131.111.184.8 (talk) 20:27, 11 February 2015 (UTC)
I coincide with . What is nowadays called "Stokes' theorem" in higher mathematics is a generalization of the curl theorem, the divergence theorem, Green's theorem and even the fundamental theorem of calculus, and the formulation of the general theorem is not easily recognisable as a generalization encompassing these theorems by a nonspecialist. So it is far better to have separate articles for those theorems, and a different, and necessarily of higher level, general Stokes' theorem. --Txebixev (talk) 15:56, 3 March 2015 (UTC)
And I partly agree with 131.111.184.8 in that the page almost needs a total rewriting, from the very first section, with a misleading statement of the theorem. --Txebixev (talk) 16:05, 3 March 2015 (UTC)
I oppose the merge per above, but I also do agree that it needs a total rewriting. In my opinion the use of bra-ket notation is needlessly confusing for a reader who likely has little knowledge of linear algebra.--Jasper Deng (talk) 20:32, 12 August 2015 (UTC)

This article barely mentions differential forms. The article Stokes' theorem is about the differential forms formulation. So, if anything, a merge would result in an article that is much less "elementary" than the present one. Granted, this article is pretty far from ideal. But the solution is to improve it, not to merge it with another, ostensibly more advanced one. 14:55, 10 October 2015 (UTC)

## Simpler but less general proof

This proof, used in introductory multivariable calculus courses, is far more elementary and requires no knowledge of anything beyond Green's theorem, Fubini's theorem and the multivariable chain rule, though it only works when the image of S on the xy plane allows expression of the surface integral as a single double integral over its image on the xy plane, which we shall call R. Let g: R2R be such that if (x, y, z) is on S, then z = g(x, y). Further assume without loss of generality that S is positively oriented such that the normal ${\displaystyle {\frac {\partial }{\partial x}}(x,y,g(x,y))\times {\frac {\partial }{\partial y}}(x,y,g(x,y))}$ points upwards (if it does not, we can reverse the cross product). Let F have its x, y, and z components be respectively the scalar functions L, M, and N of three variables.

Recall that then ${\displaystyle \iint _{S}(\nabla \times \mathbf {F} )\cdot d\mathbf {S} =\iint _{R}(\nabla \times \mathbf {F} )\cdot ({\frac {\partial }{\partial x}}(x,y,g(x,y))\times {\frac {\partial }{\partial y}}(x,y,g(x,y)))d(x,y)}$ ${\displaystyle =\iint _{R}(\nabla \times \mathbf {F} (x,y,g(x,y)))\cdot (-g_{x}(x,y),-g_{y}(x,y),1)d(x,y)}$.

Decompose the integral into (for brevity, L(x, y, g(x, y)) etc. will be denoted simply as L etc.):

${\displaystyle \iint _{R}(\nabla \times \mathbf {F} )\cdot (-g_{x},-g_{y},1)d(x,y)=\iint _{R}-(N_{y}-M_{z})g_{x}-(L_{z}-N_{x})g_{y}+(M_{x}-L_{y})d(x,y)}$

Similarly, decompose the line integral over the boundary of S:

${\displaystyle \oint _{\partial S}\mathbf {F} \cdot d\mathbf {x} =\oint _{\partial S}Ldx+\oint _{\partial S}Mdy+\oint _{\partial S}Ndz}$

Express the line integral of the z component in terms of x and y, with dz being replaced by expressions in x and y using the chain rule:

${\displaystyle \oint _{\partial S}Ndz=\oint _{\partial S}N(x,y,g(x,y))(g_{x}(x,y)dx+g_{y}(x,y)dy}$.

Since the dependence on z is now gone from the equation, it is now just

${\displaystyle \oint _{\partial R}N(x,y,g(x,y))(g_{x}(x,y)dx+g_{y}(x,y)dy)}$.

Then the whole line integral is ${\displaystyle \oint _{\partial S}\mathbf {F} \cdot d\mathbf {x} =\oint _{\partial R}(L(x,y,g(x,y))+N(x,y,g(x,y))g_{x}(x,y),M(x,y,g(x,y))+N(x,y,g(x,y))g_{y}(x,y),0)\cdot d\mathbf {x} }$

It is now reduced to a 2D line integral, where we can apply Green's theorem:

${\displaystyle \oint _{\partial R}(L(x,y,g(x,y))+N(x,y,g(x,y))g_{x}(x,y),M(x,y,g(x,y))+N(x,y,g(x,y))g_{y}(x,y))\cdot d\mathbf {x} }$ ${\displaystyle =\iint _{R}(M_{x}(x,y,g(x,y))+M_{z}(x,y,g(x,y))g_{x}(x,y)+(N_{x}(x,y,g(x,y))+N_{z}(x,y,g(x,y))g_{x}(x,y))g_{y}(x,y)+N(x,y,g(x,y)))g_{yx}(x,y))d(x,y)}$ ${\displaystyle -\iint _{R}(L_{y}(x,y,g(x,y))+L_{z}(x,y,g(x,y))g_{y}(x,y)+(N_{y}(x,y,g(x,y))+N_{z}(x,y,g(x,y))g_{y}(x,y))g_{x}(x,y)+N(x,y,g(x,y))g_{xy}(x,y)d(x,y)}$

where I have split the resulting double integral to allow it to be written over two lines. The whole integral, when like terms are cancelled (including those equal by symmetry of second partial derivatives), is equal to the double integral originally derived. This completes the proof.--Jasper Deng (talk) 09:52, 14 August 2015 (UTC)

Both proofs just use vector calculus, but the one in the article may be in unfamiliar notation that makes it difficult to read. The one you gave is for the special case when the surface is a graph, but it can easily be made very general by a partition of unity argument (basically, breaking up the surface into little pieces, each of which is a graph). I'd say include both proofs. 12:34, 14 August 2015 (UTC)
Yes, why not. Try also to find a reference. I vaguely recall Marsden & Tromba Vector Calculus having a similar proof, but may recall wrong. I can find out for sure later. YohanN7 (talk) 18:41, 14 August 2015 (UTC)

## Proof for a general triangle

Does anybody know of a reference for a proof on a general triangle using the change of variables formula from the unit triangle? It is very easy to show that this proof applies to any triangle mesh with manifold topology and a single boundary curve. This would also prove the theorem for any piecewise smooth surface with a single boundary curve because every piecewise smooth surface is the limit of a sequence of triangle meshes.

This proof actually proves Green's theorem rather than relying on it. Another benefit of this proof is that the same principle could be used to prove the divergence theorem (using change of variables formula to general tetrahedron from the unit tetrahedron) and also the general dimensional case. Jrheller1 (talk) 18:23, 6 August 2016 (UTC)

## Jordan

Why citing the Jordan curve in the statement of the theorem? Isn't this just an embedding of a disc in three-space? Ylebru (talk) 10:30, 6 May 2017 (UTC)

The domain D is not a disc. It is a region bounded by a piecewise smooth Jordan curve. So it could have corners and cusps, for example. Sławomir Biały (talk) 10:35, 6 May 2017 (UTC)
I see. Wouldn't it be better to write directly that S is a surface in space with piecewise smooth boundary? I don't see the point of invoking Jordan's theorem here. Moreover, it seems to me that the theorem holds for any surface in 3-space, not only simply connected ones (or am I missing something?) Ylebru (talk) 16:22, 6 May 2017 (UTC)
No you're quite right that Stokes theorem holds for any surface with rectifiable boundary, but I'm uncertain whether that would still be considered the (historical) "Kelvin-Stokes theorem". The theorem that is the subject of this article reduces it to the two dimensional Green's theorem, which arguably does require the Jordan curve theorem to be formulated properly. Sławomir Biały (talk) 21:51, 6 May 2017 (UTC)