Talk:Stokes' theorem

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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

Latest comment: 9 years ago2 comments2 people in discussion

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

Latest comment: 9 years ago10 comments6 people in discussion

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

Latest comment: 1 day ago5 comments4 people in discussion

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 ${\displaystyle {\frac {\partial }{\partial x}}(x,y,g(x,y))\times {\frac {\partial }{\partial y}}(x,y,g(x,y))}$  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)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

The proof in the main article is unusual, but is perfect in its rigor. It can be understood as the pullback of a differential form written using a vector field.

Your proof, on the other hand, is popular, but many people mistakenly believe it proves Stokes' theorem. However, as someone pointed out before, it only proves the extremely limited case of graph surfaces.

When considering applications to physics, the disadvantage of limiting the proof to graph surfaces is enormous. If your proof is to be considered a proof, then the only time we can mathematically guarantee that a conservative force field has the properties of a conservative force field is when a curve borders a graph surface.

As other comments have said, by introducing partition of unity, we can upgrade the proof to the level necessary to explain conservative force fields, but is it really reasonable to explain partition of unity to someone who cannot even understand this proof?

Mentaro Ikeuchi (talk) 00:09, 30 August 2025 (UTC)Reply

In the general proof, for a piecewise smooth parameterization r(u,v), let P(u,v),Q(u,v)be

${\displaystyle P(u,v):=\mathbf {F} \!{\bigl (}r(u,v){\bigr )}\cdot r_{u}(u,v)}$ 

${\displaystyle Q(u,v):=\mathbf {F} (r(u,v))\cdot r_{v}(u,v)}$ 

and apply green's theorem.

In your case,

${\displaystyle r(x,y)=(x,y,g(x,y))}$ 

then,

${\displaystyle r_{x}=(1,0,g_{x}),}$ 

${\displaystyle r_{y}=(0,1,g_{y})}$ 

So,

${\displaystyle \,P(x,y)=\mathbf {F} {\bigl (}x,y,g(x,y){\bigr )}\cdot (1,0,g_{x}(x,y))=F_{1}(x,y,g(x,y))+g_{x}(x,y)\,F_{3}(x,y,g(x,y))}$ 

${\displaystyle \,Q(x,y)=\mathbf {F} {\bigl (}x,y,g(x,y){\bigr )}\cdot (0,1,g_{x}(x,y))=F_{2}(x,y,g(x,y))+g_{y}(x,y)\,F_{3}(x,y,g(x,y))}$ 

As such, the proof in the main article is not only general, but is also essentially very simple and systematic, with the advantage that the pullback operation and the proof of Green's theorem can be performed separately. This makes the essence much clearer than popular proofs that mix these two operations together, and it would make a good bridge to differential form theory.

On the other hand, it is also true that the proof in the main article appears complicated, even though its essence is much simpler and more structured than your proof. This is because the definitions of P and Q are postponed.

The proof in the main article can be considered good in that it "heuristically" derives what P and Q should be in Step 1, but personally I think it would be easier to understand intuitively if Steps 1 and 2 were swapped, i.e., it would be clearer that it is more systematic and simple than the classical, mixed-up pseudo-proof.

Furthermore, the lengthy calculation that qualifies as a "simple" proof merely demonstrates a specific example of the algebraic formula for vector analysis in Step 3, but the formula in Step 3 is certainly not intuitive. In order to provide a more structured explanation, it seems better to postpone the proof of this formula until later.

With these two improvements, the simplicity and structured features of the proof in the main article will become more apparent. Mentaro Ikeuchi (talk) 11:32, 30 August 2025 (UTC)Reply

Proof for a general triangle

Latest comment: 1 day ago2 comments2 people in discussion

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

Proving that Green's Theorem holds on a triangle in two-dimensional space should probably be done in an article on Green's Theorem.

To extend Green's Theorem on a triangle in two-dimensional space to three-dimensional space, essentially the same operations as the proof in this article are required. This operation is usually written in terms of pullback of differential forms, but there is no problem in describing the same thing using vector fields, as in this article. Mentaro Ikeuchi (talk) 00:30, 30 August 2025 (UTC)Reply

Jordan

Latest comment: 1 day ago6 comments3 people in discussion

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

The statement in this article can be understood as a strict translation of the pullback of differential 2-forms and 1-forms into the language of vector fields.

Thus, as in the theory of differential forms, the statement in this article also makes Stokes' theorem valid for Möbius strips, provided that the form depends on the pullback.

Furthermore, if ∂Σ = φ(γ(t)), then even in extremely typical cases such as 2D polar coordinates, ∂Σ contains an interior point of Σ.

To "solve" this problem, one possible strategy would be to restrict D to planar triangles and then make the statement for a PL-manifold K.

In the case of a PL manifold,

A: = ∂Σ \cap Σ^o (Σ^o is the interior of Σ)

B: = Σ^f (essentially, B is the topological boundary of Σ)

When we set these,

∂Σ = A \cap B

and we can say that a single simplicial element τ belonging to A is an edge simplicial element of two different 2-dimensional simplicial elements σ1 and σ2, but is not an edge simplicial element of any other two simplicial elements.

However, we cannot say anything more without imposing the condition that the PL manifold is orientable, which means that the orientation of τ determined by σ1 is opposite to the orientation determined by σ2. Therefore, specifying such detailed conditions would make everything seem tautological.

Furthermore, the pullback calculation essentially requires the same proof as in this article, and introducing a PL manifold would make it even more difficult. Mentaro Ikeuchi (talk) 13:09, 30 August 2025 (UTC)Reply

I think this is simply because the Wikipedia article on Green's Theorem makes a statement about a ___domain D enclosed by a simple closed curve C. I can't think of any reasonable reason to restrict the construction of the pullback to a specific case of Green's Theorem. Mentaro Ikeuchi (talk) 12:08, 30 August 2025 (UTC)Reply

Merge

Latest comment: 7 years ago4 comments3 people in discussion

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

Latest comment: 1 day ago5 comments5 people in discussion

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

In fact, pseudo-proofs that are said to be "simple" are merely special cases that conflate several types of calculations that are not so simple. "Proofs" that are said to be "simple" are not as simple as they claim. This is because traditional "proofs" conflate pullback and Green's theorem. Furthermore, proving that 2 + 3 = 3 + 2 does not prove that a + b = b + a. Mentaro Ikeuchi (talk) 12:13, 30 August 2025 (UTC)Reply

First step of proof confusing

Latest comment: 1 day ago5 comments4 people in discussion

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

Although it dates back to a fairly old version, the theorem statement clearly states that Γ = φ(γ) and that γ is a simple closed curve in R^2, so it seems you, not the author, are the one being careless.

I understand that if you don't have a good at mathematics, you might not understand why we need to be so careful and treat Γ and ∂Σ as separate entities. But this gap isn't serious enough to warrant mentioning in the main article, and it depends on how you define the boundary (whether the boundary can include interior points), so it would be better to include a warning to avoid the hopeless misunderstanding you mentioned. This is a problem you'll likely encounter immediately if you try to perform specific calculations in 2D polar coordinates.

Essentially, the chain boundary and the topological boundary don't necessarily coincide, but the line integrals at interior points cancel out. Mentaro Ikeuchi (talk) 12:38, 30 August 2025 (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

The author who wrote Γ = φ(γ(t)) instead of ∂Σ seems to be an extremely careful and rigorous thinker, but popular statements in vector analysis are so careless that many people fail to appreciate his thoughtfulness.

Well,

Understanding Γ = ∂Σ

is a bit careless, but it's not a big problem. It could even be said that the author who wrote Γ = φ(γ(t)) was too strict.

If we take Γ = ∂Σ, then clearly ∂Σ can contain interior points of Σ. This occurs in very simple concrete examples (such as two-dimensional polar coordinates). However, even if ∂Σ contains points that are not boundaries, the line integrals over the interior points cancel out, so there's no problem in the end.

So, although a bit careless, it's fine to think of Γ = ∂Σ. For more precision, it's better to distinguish between chains (usually symbolized as C) and topological boundaries. Mentaro Ikeuchi (talk) 12:24, 30 August 2025 (UTC)Reply

Redirect links differing in case of one letter only

Latest comment: 2 years ago1 comment1 person in discussion

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

Latest comment: 2 years ago1 comment1 person in discussion

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

Unclear statement

Latest comment: 1 day ago2 comments2 people in discussion

The section Theorem contains this sentence:

"Suppose ${\displaystyle \psi :D\to \mathbb {R} ^{3}}$  is piecewise smooth at the neighborhood of ${\displaystyle D}$ , with ${\displaystyle \Sigma =\psi (D)}$ ."

It is enrtirely unclear what the phrase "at the neighborhood of ${\displaystyle D}$ " might mean. Especially because D is the ___domain of the mapping ${\displaystyle \psi }$ , so its only "neighborhood" in this statement can be itself.

(Also: If this statement made sense, then in English it should be "in the neighborhood of ${\displaystyle D}$ ".)

(Also: Since no "neighborhood of ${\displaystyle D}$ " has ever been referred to previously, it is entirely inappropriate to speak of "the neighborhood of ${\displaystyle D}$ ".)

2601:204:F181:9410:A5B6:1F84:4490:A95D (talk) 19:33, 26 May 2025 (UTC)Reply

As stated in the previous paragraph, D is the region enclosed by the Jordan curve in R^2. This means that D is a closed set in R^2.

To be more clear, we can say that a "neighborhood of D" is an "open set that contains D."　This condition seems to be a stronger condition than the condition for differentiability at interior points, or in other words, the most elementary statement of differentiability at the boundary. Mentaro Ikeuchi (talk) 23:27, 29 August 2025 (UTC)Reply