Talk:Stokes' theorem

This article has previously been nominated to be moved. Please review the prior discussions if you are considering re-nomination.

Discussions:

RM, Stokes theorem → Stokes' theorem, Moved, 24 January 2006, discussion
RM, Stokes' theorem → Stokes's theorem, Not moved, 4 December 2018, discussion
- MRV, Endorsed, 23 December 2018, discussion
RM, Stokes' theorem → Stokes's theorem, Not moved, 23 August 2023, discussion

Older comments

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 (talk • contribs) 02:36, 1 May 2016 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

I coincide with @Sławomir Biały:. 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)[reply]

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)[reply]

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)[reply]

Why is this page written in such an advanced manner? This is a topic of introductory multivariable calculus. Why in the world would the definition include one-forms and exterior derivatives? This article is absolutely terrible. — Preceding unsigned comment added by 76.108.236.176 (talk) 17:18, 9 October 2015 (UTC)[reply]

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. Sławomir
Biały 14:55, 10 October 2015 (UTC)[reply]

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: R²→R 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 ${\frac {\partial }{\partial x}}(x,y,g(x,y))\times {\frac {\partial }{\partial y}}(x,y,g(x,y))$ 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 $\iint _{S}(\nabla \times \mathbf {F} )\cdot d\mathbf {S} =\iint _{R}(\nabla \times \mathbf {F} )\cdot ({\frac {\partial }{\partial x}}(x,y,g(x,y))\times {\frac {\partial }{\partial y}}(x,y,g(x,y)))d(x,y)$ $=\iint _{R}(\nabla \times \mathbf {F} (x,y,g(x,y)))\cdot (-g_{x}(x,y),-g_{y}(x,y),1)d(x,y)$ .

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

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

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

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

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:

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

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

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

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

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

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

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)[reply]

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. Sławomir
Biały 12:34, 14 August 2015 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Merge

I think Stokes' theorem article should be merged into this article because both articles basically talk about the same thing. Brian Everlasting (talk) 18:25, 27 January 2018 (UTC)[reply]

I disagree (strongly). Stokes' theorem is a theorem about integration of differential forms in higher dimensions and on arbitrary manifolds. Kelvin-Stokes is a special case only. Merging the two will create a mess, and / or, an extremely long article. Jakob.scholbach (talk) 08:09, 29 January 2018 (UTC)[reply]

I also disagree. A merge is not appropriate. A rewrite of this article is what is needed. Sławomir Biały (talk) 10:56, 29 January 2018 (UTC)[reply]

I abandoned the merge proposal. Both articles still need a lot of work. Brian Everlasting (talk) 22:17, 3 February 2018 (UTC)[reply]

Question about long complicated Proof

I'm wondering if we should just delete or move the proof because it's very long and complicated and doesn't add anything to the article. Brian Everlasting (talk) 19:42, 27 January 2018 (UTC)[reply]

Just a thought: this article might be more appropriate for Wikibooks. Sławomir Biały (talk) 19:58, 27 January 2018 (UTC)[reply]

Indeed this seems a bit beyond the proofs and proof description usually considered acceptable in WP and outsourcing it Wikibooks or Proofwiki seems the best solution.--Kmhkmh (talk) 02:47, 29 January 2018 (UTC)[reply]

Yeah, we should just include a short proof of a special case. Most calculus students will not find the page in its current form useful. Nerd271 (talk) 18:53, 4 April 2020 (UTC)[reply]

First step of proof confusing

It refers to psi and gamma and parameterizations, but offers no description of what they are. The only reference is to a Wikipedia page on "theorems." Very sloppy. Come on, people, we can do better here. Sheesh. — Preceding unsigned comment added by 146.142.1.10 (talk) 17:20, 5 February 2020 (UTC)[reply]

Agreed. Leaving note here to hopefully address this later. briardew (talk) 17:05, 20 January 2023 (UTC)[reply]

What is Γ in the first equation? Please define it, describe what this equation is saying. Also, please number all the equations. — Preceding unsigned comment added by Suneecat (talk • contribs) 15:25, 15 February 2023 (UTC)[reply]

Redirect links differing in case of one letter only

There are numerous redirect links to this article and the one on the Generalized Stokes' Theorem. Or should I say Generalized Stokes theorem? Or one of the many other variants, some of which point here, some there. Forgive me, I do not want to insult anyone, but I can't but wonder how people who write and edit these articles, and presumably must be reasonably smart mathematicians themselves, can make such an ugly mess of things. 5.186.55.135 (talk) 14:52, 14 July 2023 (UTC)[reply]

Move discussion in progress

There is a move discussion in progress on Talk:Bayes' theorem which affects this page. Please participate on that page and not in this talk page section. Thank you. —RMCD bot 17:01, 23 August 2023 (UTC)[reply]