Stokes' Theorem - Revision history

Lila at 13:54, 3 November 2021

2021-11-03T13:54:50Z

Lila: /* First step of the proof (parametrization of integral) */

2021-11-03T13:53:47Z

First step of the proof (parametrization of integral)

Lila: /* Elementary proof */

2021-11-03T13:53:10Z

Elementary proof

Lila: /* Theorem */

2021-11-03T13:51:42Z

Theorem

Lila at 13:50, 3 November 2021

2021-11-03T13:50:58Z

Lila at 13:50, 3 November 2021

2021-11-03T13:50:18Z

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Lila: Created page with "n}}. '''Stok..."

2021-11-03T13:43:53Z

Created page with "thumb|right|An illustration of Stokes' theorem, with surface {{math|Σ}}, its boundary {{math|∂Σ}} and the normal vector {{mvar|n}}. '''Stok..."

New page

[[Image:Stokes' Theorem.svg|thumb|right|An illustration of Stokes' theorem, with surface {{math|Σ}}, its boundary {{math|∂Σ}} and the normal vector {{mvar|n}}.]]

'''Stokes' theorem''', also known as '''Kelvin–Stokes theorem''' after [[Lord Kelvin]] and [[Sir George Stokes, 1st Baronet|George Stokes]], the '''fundamental theorem for curls''' or simply the '''curl theorem''', is a [[theorem]] in [[vector calculus]] on <math>\mathbb{R}^3</math>. Given a [[vector field]], the theorem relates the [[Surface integral|integral]] of the [[Curl (mathematics)|curl]] of the vector field over some surface, to the [[line integral]] of the vector field around the boundary of the surface. The classical Stokes' theorem can be stated in one sentence: The [[line integral]] of a vector field over a loop is equal to the ''[[flux]] of its curl'' through the enclosed surface.

Stokes' theorem is a special case of the [[generalized Stokes' theorem]]. In particular, a vector field on <math>\mathbb{R}^3</math> can be considered as a [[differential form|1-form]] in which case its curl is its [[exterior derivative]], a 2-form.

==Theorem==

Let <math>\Sigma</math> be a smooth oriented surface in {{math|'''R'''3}} with boundary <math>\partial \Sigma</math>. If a vector field <math>\mathbf{A} = (P(x, y, z), Q(x, y, z), R(x, y, z))</math> is defined and has continuous first order [[partial derivatives]] in a region containing <math>\Sigma</math>, then

<math display=block>
\iint_\Sigma (\nabla \times \mathbf{A}) \cdot \mathrm{d}\mathbf{a} = \oint_{\partial\Sigma} \mathbf{A} \cdot
\mathrm{d}\mathbf{l}.
</math>
More explicitly, the equality says that
<math display=block>
\begin{align}
&\iint_\Sigma \left(\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z} \right)\,\mathrm{d}y\, \mathrm{d}z +\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\, \mathrm{d}z\, \mathrm{d}x +\left (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\, \mathrm{d}x\, \mathrm{d}y\right) \\
& = \oint_{\partial\Sigma} \Bigl(P\, \mathrm{d}x+Q\, \mathrm{d}y+R\, \mathrm{d}z\Bigr).
\end{align}
</math>

The main challenge in a precise statement of Stokes' theorem is in defining the notion of a boundary. Surfaces such as the [[Koch snowflake]], for example, are well-known not to exhibit a Riemann-integrable boundary, and the notion of surface measure in [[Lebesgue integration|Lebesgue theory]] cannot be defined for a non-[[Lipschitz function|Lipschitz]] surface. One (advanced) technique is to pass to a [[weak formulation]] and then apply the machinery of [[geometric measure theory]]; for that approach see the [[coarea formula]]. In this article, we instead use a more elementary definition, based on the fact that a boundary can be discerned for full-dimensional subsets of {{math|'''R'''{{sup|2}}}}.

Let {{math|''γ'': [''a'', ''b''] → '''R'''2}} be a [[piecewise]] smooth [[Jordan curve|Jordan plane curve]]. The [[Jordan curve theorem]] implies that {{mvar|γ}} divides {{math|'''R'''2}} into two components, a [[compact space|compact]] one and another that is non-compact. Let {{mvar|D}} denote the compact part; then {{mvar|D}} is bounded by {{mvar|γ}}. It now suffices to transfer this notion of boundary along a continuous map to our surface in {{math|'''R'''{{sup|3}}}}. But we already have such a map: the [[Parametrization (geometry)|parametrization]] of {{math|Σ}}.

Suppose {{math|''ψ'': ''D'' → '''R'''3}} is smooth, with {{math|1=Σ = ''ψ''(''D'')}}. If {{math|Γ}} is the [[space curve]] defined by {{math|1=Γ(''t'') = ''ψ''(''γ''(''t''))}},<ref group="note" name=cgamma>{{math|Γ}} may not be a [[Jordan curve]], if the loop {{mvar|γ}} interacts poorly with {{mvar|ψ}}. Nonetheless, {{math|Γ}} is always a [[loop (topology)|loop]], and topologically a [[connected sum]] of [[countable set|countably-many]] Jordan curves, so that the integrals are well-defined.</ref> then we call {{math|Γ}} the boundary of {{math|Σ}}, written {{math|∂Σ}}.

With the above notation, if {{math|'''F'''}} is any smooth vector field on {{math|'''R'''3}}, then<ref name="Jame">{{cite book|url={{Google books |plainurl=yes |id=btIhvKZCkTsC |page=786 }}|title=Essential Calculus: Early Transcendentals|last=Stewart|first=James|publisher=Cole|year=2010}}</ref><ref name="bath">Robert Scheichl, lecture notes for [[University of Bath]] mathematics course [http://www.maths.bath.ac.uk/~masrs/ma20010/stokesproofs.pdf]</ref><math display="block">\oint_{\partial\Sigma} \mathbf{F}\, \cdot\, \mathrm{d}{\mathbf{\Gamma}} = \iint_{\Sigma} \nabla\times\mathbf{F}\, \cdot\, \mathrm{d}\mathbf{S}. </math>

==Proof==
The proof of the theorem consists of 4 steps. We assume [[Green's theorem]], so what is of concern is how to boil down the three-dimensional complicated problem (Stokes' theorem) to a two-dimensional rudimentary problem (Green's theorem).<ref>{{cite book|title=Vector Calculus|last=Colley|first=Susan Jane|edition=4th|publisher=Pearson|year=2002|location=Boston|pages=500–3}}</ref> When proving this theorem, mathematicians normally deduce it as a special case of a [[Generalized Stokes' theorem|more general result]], which is stated in terms of [[differential form]]s, and proved using more sophisticated machinery. While powerful, these techniques require substantial background, so the proof below avoids them, and does not presuppose any knowledge beyond a familiarity with basic vector calculus.<ref name="bath"/> At the end of this section, a short alternate proof of Stokes' theorem is given, as a corollary of the generalized Stokes' Theorem.

===Elementary proof===
====First step of the proof (parametrization of integral)====
As in {{slink|#Theorem}}, we reduce the dimension by using the natural parametrization of the surface. Let {{math|'''''ψ'''''}} and {{mvar|γ}} be as in that section, and note that by change of variables

<math display=block>\oint_{\partial\Sigma}{\mathbf{F}(\mathbf{x})\cdot\,\mathrm{d}\mathbf{l}}
= \oint_{\gamma}{\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot\,\mathrm{d}\boldsymbol{\psi}(\mathbf{y})}
= \oint_{\gamma}{\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))J_{\mathbf{y}}(\boldsymbol{\psi})\,\mathrm{d}\mathbf{y}}</math>

where {{mvar|Jψ}} stands for the [[Jacobian matrix and determinant|Jacobian matrix]] of {{mvar|ψ}}.

Now let {{math|{'''e'''''u'', '''e'''''v''}<nowiki/>}} be an orthonormal basis in the coordinate directions of {{math|'''R'''2}}. Recognizing that the columns of {{math|''J'''''y''''''''ψ'''''}} are precisely the partial derivatives of {{math|'''''ψ'''''}} at {{math|'''y'''}}, we can expand the previous equation in coordinates as

<math display=block>\begin{align}
\oint_{\partial\Sigma}{\mathbf{F}(\mathbf{x})\cdot\,\mathrm{d}\mathbf{l}}
&= \oint_{\gamma}{\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))J_{\mathbf{y}}(\boldsymbol{\psi})\mathbf{e}_u(\mathbf{e}_u\cdot\,\mathrm{d}\mathbf{y}) + \mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))J_{\mathbf{y}}(\boldsymbol{\psi})\mathbf{e}_v(\mathbf{e}_v\cdot\,\mathrm{d}\mathbf{y})} \\
&=\oint_{\gamma}{\left(\left(\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot\frac{\partial\boldsymbol{\psi}}{\partial u}(\mathbf{y})\right)\mathbf{e}_u + \left(\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot\frac{\partial\boldsymbol{\psi}}{\partial v}(\mathbf{y})\right)\mathbf{e}_v\right)\cdot\,\mathrm{d}\mathbf{y}}
\end{align}</math>

====Second step in the proof (defining the pullback)====
The previous step suggests we define the function

<math display=block>\mathbf{P}(u,v) = \left(\mathbf{F}(\boldsymbol{\psi}(u,v))\cdot\frac{\partial\boldsymbol{\psi}}{\partial u}(u,v)\right)\mathbf{e}_u + \left(\mathbf{F}(\boldsymbol{\psi}(u,v))\cdot\frac{\partial\boldsymbol{\psi}}{\partial v} \right)\mathbf{e}_v</math>

This is the [[Pullback (differential geometry)|pullback]] of {{math|'''F'''}} along {{math|'''''ψ'''''}}, and, by the above, it satisfies

<math display=block>\oint_{\partial\Sigma}{\mathbf{F}(\mathbf{x})\cdot\,\mathrm{d}\mathbf{l}}=\oint_{\gamma}{\mathbf{P}(\mathbf{y})\cdot\,\mathrm{d}\mathbf{l}}</math>

We have successfully reduced one side of Stokes' theorem to a 2-dimensional formula; we now turn to the other side.

====Third step of the proof (second equation)====
First, calculate the partial derivatives appearing in [[Green's theorem]], via the [[General Leibniz rule|product rule]]:

<math display=block>\begin{align}
\frac{\partial P_1}{\partial v} &= \frac{\partial (\mathbf{F}\circ \boldsymbol{\psi})}{\partial v}\cdot\frac{\partial \boldsymbol\psi}{\partial u} + (\mathbf{F}\circ \boldsymbol\psi) \cdot\frac{\partial^2 \boldsymbol\psi}{\partial v \, \partial u} \\[5pt]
\frac{\partial P_2}{\partial u} &= \frac{\partial (\mathbf{F}\circ \boldsymbol{\psi})}{\partial u}\cdot\frac{\partial \boldsymbol\psi}{\partial v} + (\mathbf{F}\circ \boldsymbol\psi) \cdot\frac{\partial^2 \boldsymbol\psi}{\partial u \, \partial v}
\end{align}</math>

Conveniently, the second term vanishes in the difference, by [[equality of mixed partials]]. So,

<math display=block>\begin{align}
\frac{\partial P_1}{\partial v} - \frac{\partial P_2}{\partial u}
&= \frac{\partial (\mathbf{F}\circ \boldsymbol\psi)}{\partial v}\cdot\frac{\partial \boldsymbol\psi}{\partial u} - \frac{\partial (\mathbf{F}\circ \boldsymbol\psi)}{\partial u}\cdot\frac{\partial \boldsymbol\psi}{\partial v} \\[5pt]
&= \frac{\partial \boldsymbol\psi}{\partial u}(J_{\boldsymbol\psi(u,v)}\mathbf{F})\frac{\partial \boldsymbol\psi}{\partial v} - \frac{\partial \boldsymbol\psi}{\partial v}(J_{\boldsymbol\psi(u,v)}\mathbf{F})\frac{\partial \boldsymbol\psi}{\partial u} && \text{(chain rule)}\\[5pt]
&= \frac{\partial \boldsymbol\psi}{\partial u}\left(J_{\boldsymbol\psi(u,v)}\mathbf{F}-{(J_{\boldsymbol\psi(u,v)}\mathbf{F})}^{\mathsf{T}}\right)\frac{\partial \boldsymbol\psi}{\partial v}
\end{align}
</math>

But now consider the matrix in that quadratic form—that is, <math>J_{\boldsymbol\psi(u,v)}\mathbf{F}-(J_{\boldsymbol\psi(u,v)}\mathbf{F})^{\mathsf{T}}</math>. We claim this matrix in fact describes a cross product.

To be precise, let <math>A=(A_{ij})_{ij}</math> be an arbitrary {{math|3 × 3}} matrix and let

<math display=block>\mathbf{a} = \begin{bmatrix}A_{32}-A_{23} \\ A_{13}-A_{31} \\ A_{21}-A_{12}\end{bmatrix}</math>

Note that {{math|'''x''' ↦ '''a''' × '''x'''}} is linear, so it is determined by its action on basis elements. But by direct calculation
<math display="block">\begin{align}
\left(A-A^{\mathsf{T}}\right)\mathbf{e}_1 &= \begin{bmatrix} 0 \\ a_3 \\ -a_2 \end{bmatrix} = \mathbf{a}\times\mathbf{e}_1\\
\left(A-A^{\mathsf{T}}\right)\mathbf{e}_2 &= \begin{bmatrix} -a_3 \\ 0 \\ a_1 \end{bmatrix} = \mathbf{a}\times\mathbf{e}_2\\
\left(A-A^{\mathsf{T}}\right)\mathbf{e}_3 &= \begin{bmatrix} a_2 \\ -a_1 \\ 0 \end{bmatrix} = \mathbf{a}\times\mathbf{e}_3
\end{align}</math>

Thus {{math|1=(''A'' − ''A''{{sup|T}})'''x''' = '''a''' × '''x'''}} for any {{math|'''x'''}}. Substituting {{math|''J'' '''F'''}} for {{mvar|A}}, we obtain

<math display=block>\left({(J_{\boldsymbol\psi(u,v)}\mathbf{F})}_{\psi(u,v)} - {(J_{\boldsymbol\psi(u,v)}\mathbf{F})}^{\mathsf{T}} \right) \mathbf{x} =(\nabla\times\mathbf{F})\times \mathbf{x}, \quad \text{for all}\, \mathbf{x}\in\R^{3}</math>

We can now recognize the difference of partials as a [[Triple product|(scalar) triple product]]:

<math display=block>\begin{align}
\frac{\partial P_1}{\partial v} - \frac{\partial P_2}{\partial u}
&= \frac{\partial \boldsymbol\psi}{\partial u}\cdot(\nabla\times\mathbf{F}) \times \frac{\partial \boldsymbol\psi}{\partial v} \\
&= \det \begin{bmatrix} (\nabla\times\mathbf{F})(\boldsymbol\psi(u,v)) & \frac{\partial \boldsymbol\psi}{\partial u}(u,v) & \frac{\partial \boldsymbol\psi}{\partial v}(u,v) \end{bmatrix}
\end{align}</math>

On the other hand, the definition of a [[surface integral]] also includes a triple product—the very same one!

<math display=block>\begin{align}
\iint_S (\nabla\times\mathbf{F})\cdot \, d^2\mathbf{S}
&=\iint_D {(\nabla\times\mathbf{F})(\boldsymbol\psi(u,v))\cdot\left(\frac{\partial \boldsymbol\psi}{\partial u}(u,v)\times \frac{\partial \boldsymbol\psi}{\partial v}(u,v)\,\mathrm{d}u\,\mathrm{d}v\right)}\\
&= \iint_D \det \begin{bmatrix} (\nabla\times\mathbf{F})(\boldsymbol\psi(u,v)) & \frac{\partial \boldsymbol\psi}{\partial u}(u,v) & \frac{\partial \boldsymbol\psi}{\partial v}(u,v) \end{bmatrix} \,\mathrm{d}u \,\mathrm{d}v
\end{align}</math>

So, we obtain

<math display=block> \iint_S (\nabla\times\mathbf{F})\cdot \,\mathrm{d}^2\mathbf{S} = \iint_D \left( \frac{\partial P_2}{\partial u} - \frac{\partial P_1}{\partial v} \right) \,\mathrm{d}u\,\mathrm{d}v </math>

====Fourth step of the proof (reduction to Green's theorem)====
Combining the second and third steps, and then applying [[Green's theorem]] completes the proof.

===Proof via differential forms===
{{math|'''R''' → '''R'''{{sup|3}}}} can be identified with the differential 1-forms on {{math|'''R'''{{sup|3}}}} via the map

<math display=block>F_1\mathbf{e}_1+F_2\mathbf{e}_2+F_3\mathbf{e}_3 \mapsto F_1\,\mathrm{d}x+F_2\,\mathrm{d}y+F_3\mathrm{d}z .</math>

Write the differential 1-form associated to a function {{math|'''F'''}} as {{math|''ω'''''F'''}}. Then one can calculate that

<math display=block>\star\omega_{\nabla\times\mathbf{F}}=\mathrm{d}\omega_{\mathbf{F}}</math>

where {{math|★}} is the [[Hodge star]] and <math>\mathrm{d}</math> is the [[exterior derivative]]. Thus, by generalized Stokes' theorem,<ref>{{cite book |last=Edwards |first=Harold M. |title=Advanced Calculus: A Differential Forms Approach |publisher=Birkhäuser |year=1994 |isbn=0-8176-3707-9 }}</ref>

<math display=block>\oint_{\partial\Sigma}{\mathbf{F}\cdot\,\mathrm{d}\mathbf{l}}
=\oint_{\partial\Sigma}{\omega_{\mathbf{F}}}
=\int_{\Sigma}{\mathrm{d}\omega_{\mathbf{F}}}
=\int_{\Sigma}{\star\omega_{\nabla\times\mathbf{F}}}
=\iint_{\Sigma}{\nabla\times\mathbf{F}\cdot\,\mathrm{d}^2\mathbf{S}}
</math>

==Licensing==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Stokes%27_theorem Stokes' Theorem] under a CC BY-SA license

@@ Line 22: / Line 22: @@
 </math>
-The main challenge in a precise statement of Stokes' theorem is in defining the notion of a boundary.  Surfaces such as the Koch snowflake, for example, are well-known not to exhibit a Riemann-integrable boundary, and the notion of surface measure in Lebesgue theory cannot be defined for a non-Lipschitz surface.  One (advanced) technique is to pass to a weak formulation and then apply the machinery of geometric measure theory; for that approach see the coarea formula.  In this article, we instead use a more elementary definition, based on the fact that a boundary can be discerned for full-dimensional subsets of {{math|'''R'''{{sup|2}}}}.
+The main challenge in a precise statement of Stokes' theorem is in defining the notion of a boundary.  Surfaces such as the Koch snowflake, for example, are well-known not to exhibit a Riemann-integrable boundary, and the notion of surface measure in Lebesgue theory cannot be defined for a non-Lipschitz surface.  One (advanced) technique is to pass to a weak formulation and then apply the machinery of geometric measure theory; for that approach see the coarea formula.  In this article, we instead use a more elementary definition, based on the fact that a boundary can be discerned for full-dimensional subsets of {{math|'''R'''<sup>2</sup>}}.
-Let {{math|''γ'': [''a'', ''b''] → '''R'''<sup>2</sup>}} be a piecewise smooth Jordan plane curve. The Jordan curve theorem implies that {{mvar|γ}} divides {{math|'''R'''<sup>2</sup>}} into two components, a compact one and another that is non-compact. Let {{mvar|D}} denote the compact part; then {{mvar|D}} is bounded by {{mvar|γ}}.  It now suffices to transfer this notion of boundary along a continuous map to our surface in {{math|'''R'''{{sup|3}}}}.  But we already have such a map: the parametrization of {{math|Σ}}.
+Let {{math|''γ'': [''a'', ''b''] → '''R'''<sup>2</sup>}} be a piecewise smooth Jordan plane curve. The Jordan curve theorem implies that {{mvar|γ}} divides {{math|'''R'''<sup>2</sup>}} into two components, a compact one and another that is non-compact. Let {{mvar|D}} denote the compact part; then {{mvar|D}} is bounded by {{mvar|γ}}.  It now suffices to transfer this notion of boundary along a continuous map to our surface in {{math|'''R'''<sup>3</sup>}}.  But we already have such a map: the parametrization of {{math|Σ}}.
 Suppose {{math|''ψ'': ''D'' → '''R'''<sup>3</sup>}} is smooth, with {{math|1=Σ = ''ψ''(''D'')}}. If {{math|Γ}} is the space curve defined by {{math|1=Γ(''t'') = ''ψ''(''γ''(''t''))}}, then we call {{math|Γ}} the boundary of {{math|Σ}}, written {{math|∂Σ}}.

@@ Line 26: / Line 26: @@
 Let {{math|''γ'': [''a'', ''b''] → '''R'''<sup>2</sup>}} be a piecewise smooth Jordan plane curve. The Jordan curve theorem implies that {{mvar|γ}} divides {{math|'''R'''<sup>2</sup>}} into two components, a compact one and another that is non-compact. Let {{mvar|D}} denote the compact part; then {{mvar|D}} is bounded by {{mvar|γ}}.  It now suffices to transfer this notion of boundary along a continuous map to our surface in {{math|'''R'''{{sup|3}}}}.  But we already have such a map: the parametrization of {{math|Σ}}.
-Suppose {{math|''ψ'': ''D'' → '''R'''<sup>3</sup>}} is smooth, with {{math|1=Σ = ''ψ''(''D'')}}. If {{math|Γ}} is the space curve defined by {{math|1=Γ(''t'') = ''ψ''(''γ''(''t''))}}, {{math|Γ}} may not be a Jordan curve, if the loop {{mvar|γ}} interacts poorly with {{mvar|ψ}}.  Nonetheless, {{math|Γ}} is always a loop, and topologically a connected sum of countably-many Jordan curves, so that the integrals are well-defined.</ref> then we call {{math|Γ}} the boundary of {{math|Σ}}, written {{math|∂Σ}}.
+Suppose {{math|''ψ'': ''D'' → '''R'''<sup>3</sup>}} is smooth, with {{math|1=Σ = ''ψ''(''D'')}}. If {{math|Γ}} is the space curve defined by {{math|1=Γ(''t'') = ''ψ''(''γ''(''t''))}}, then we call {{math|Γ}} the boundary of {{math|Σ}}, written {{math|∂Σ}}.
 With the above notation, if {{math|'''F'''}} is any smooth vector field on {{math|'''R'''<sup>3</sup>}}, then
@@ Line 33: / Line 33: @@
 ==Proof==
-The proof of the theorem consists of 4 steps. We assume Green's theorem, so what is of concern is how to boil down the three-dimensional complicated problem (Stokes' theorem) to a two-dimensional rudimentary problem (Green's theorem). When proving this theorem, mathematicians normally deduce it as a special case of a more general result, which is stated in terms of differential forms, and proved using more sophisticated machinery. While powerful, these techniques require substantial background, so the proof below avoids them, and does not presuppose any knowledge beyond a familiarity with basic vector calculus.<ref name="bath"/> At the end of this section, a short alternate proof of Stokes' theorem is given, as a corollary of the generalized Stokes' Theorem.
+The proof of the theorem consists of 4 steps. We assume Green's theorem, so what is of concern is how to boil down the three-dimensional complicated problem (Stokes' theorem) to a two-dimensional rudimentary problem (Green's theorem). When proving this theorem, mathematicians normally deduce it as a special case of a more general result, which is stated in terms of differential forms, and proved using more sophisticated machinery. While powerful, these techniques require substantial background, so the proof below avoids them, and does not presuppose any knowledge beyond a familiarity with basic vector calculus. At the end of this section, a short alternate proof of Stokes' theorem is given, as a corollary of the generalized Stokes' Theorem.
 ===Elementary proof===

@@ Line 95: / Line 95: @@
 \end{align}</math>
-Thus {{math|1=(''A'' − ''A''{{sup|T}})'''x''' = '''a''' × '''x'''}} for any {{math|'''x'''}}.  Substituting {{math|''J'' '''F'''}} for {{mvar|A}}, we obtain
+Thus {{math|1=(''A'' − ''A''<sup>''T''</sup>)'''x''' = '''a''' × '''x'''}} for any {{math|'''x'''}}.  Substituting {{math|''J'' '''F'''}} for {{mvar|A}}, we obtain
 <math display=block>\left({(J_{\boldsymbol\psi(u,v)}\mathbf{F})}_{\psi(u,v)} - {(J_{\boldsymbol\psi(u,v)}\mathbf{F})}^{\mathsf{T}} \right) \mathbf{x} =(\nabla\times\mathbf{F})\times \mathbf{x}, \quad \text{for all}\, \mathbf{x}\in\R^{3}</math>
@@ Line 123: / Line 123: @@
 ===Proof via differential forms===
-{{math|'''R''' → '''R'''{{sup|3}}}} can be identified with the differential 1-forms on {{math|'''R'''{{sup|3}}}} via the map
+{{math|'''R''' → '''R'''<sup>3</sup>}} can be identified with the differential 1-forms on {{math|'''R'''<sup>3</sup>}} via the map
 <math display=block>F_1\mathbf{e}_1+F_2\mathbf{e}_2+F_3\mathbf{e}_3 \mapsto F_1\,\mathrm{d}x+F_2\,\mathrm{d}y+F_3\mathrm{d}z .</math>

@@ Line 37: / Line 37: @@
 ===Elementary proof===
 ====First step of the proof (parametrization of integral)====
-As in <math>\s</math>Theorem, we reduce the dimension by using the natural parametrization of the surface.  Let {{math|'''''ψ'''''}} and {{mvar|γ}} be as in that section, and note that by change of variables
+As in <math>\S</math>Theorem, we reduce the dimension by using the natural parametrization of the surface.  Let {{math|'''''ψ'''''}} and {{mvar|γ}} be as in that section, and note that by change of variables
 <math display=block>\oint_{\partial\Sigma}{\mathbf{F}(\mathbf{x})\cdot\,\mathrm{d}\mathbf{l}}