thumb|right|An illustration of Stokes' theorem, with surface , its boundary  and the normal vector .
Stokes' theorem, also known as Kelvin–Stokes theorem Nagayoshi Iwahori, et al.:"Bi-Bun-Seki-Bun-Gaku" Sho-Ka-Bou(jp) 1983/12  (Written in Japanese)Atsuo Fujimoto;"Vector-Kai-Seki Gendai su-gaku rekucha zu.
C(1)" Bai-Fu-Kan(jp)(1979/01)  [] (Written in Japanese) after Lord Kelvin and George Stokes, the fundamental theorem for curls or simply the curl theorem, is a theorem in vector calculus on \mathbb{R}^3.
Given a vector field, the theorem relates the integral of the 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 \mathbb{R}^3 can be considered as a 1-form in which case its curl is its exterior derivative, a 2-form.
Theorem
Let \Sigma be a smooth oriented surface in  with boundary \partial \Sigma.
If a vector field \mathbf{A} = (P(x, y, z), Q(x, y, z), R(x, y, z)) is defined and has continuous first order partial derivatives in a region containing \Sigma, then
\iint_\Sigma (\nabla \times \mathbf{A}) \cdot \mathrm{d}\mathbf{a}  =  \oint_{\partial\Sigma} \mathbf{A} \cdot  \mathrm{d}\mathbf{l}.
More explicitly, the equality says that
\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}
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 .
Let  be a piecewise smooth Jordan plane curve.
The Jordan curve theorem implies that  divides  into two components, a compact one and another that is non-compact.
Let  denote the compact part; then  is bounded by .
It now suffices to transfer this notion of boundary along a continuous map to our surface in .
But we already have such a map: the parametrization of .
Suppose  is smooth, with .
If  is the space curve defined by , may not be a Jordan curve, if the loop  interacts poorly with .
Nonetheless,  is always a loop, and topologically a connected sum of countably-many Jordan curves, so that the integrals are well-defined.
then we call  the boundary of , written .
With the above notation, if  is any smooth vector field on , thenRobert Scheichl, lecture notes for University of Bath mathematics course  \oint_{\partial\Sigma} \mathbf{F}\, \cdot\, \mathrm{d}{\mathbf{\Gamma}}  = \iint_{\Sigma} \nabla\times\mathbf{F}\, \cdot\, \mathrm{d}\mathbf{S}.
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 and linear algebra.
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 , we reduce the dimension by using the natural parametrization of the surface.
Let  and  be as in that section, and note that by change of variables
\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}}
where  stands for the Jacobian matrix of .
Now let  be an orthonormal basis in the coordinate directions of .
Recognizing that the columns of  are precisely the partial derivatives of  at , we can expand the previous equation in coordinates as
\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} Second step in the proof (defining the pullback)
The previous step suggests we define the function
\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
This is the pullback of  along , and, by the above, it satisfies
\oint_{\partial\Sigma}{\mathbf{F}(\mathbf{x})\cdot\,\mathrm{d}\mathbf{l}}=\oint_{\gamma}{\mathbf{P}(\mathbf{y})\cdot\,\mathrm{d}\mathbf{l}}
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 product rule:
\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}
Conveniently, the second term vanishes in the difference, by equality of mixed partials.
So,
\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}
But now consider the matrix in that quadratic form—that is, J_{\boldsymbol\psi(u,v)}\mathbf{F}-(J_{\boldsymbol\psi(u,v)}\mathbf{F})^{\mathsf{T}}.
We claim this matrix in fact describes a cross product.
To be precise, let A=(A_{ij})_{ij} be an arbitrary  matrix and let
\mathbf{a} = \begin{bmatrix}A_{32}-A_{23} \\ A_{13}-A_{31} \\ A_{21}-A_{12}\end{bmatrix}
Note that  is linear, so it is determined by its action on basis elements.
But by direct calculation \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}
Thus  for any .
Substituting  for , we obtain
\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}
We can now recognize the difference of partials as a (scalar) triple product:
\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}
On the other hand, the definition of a surface integral also includes a triple product—the very same one!
\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}
So, we obtain
\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 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
The functions  can be identified with the differential 1-forms on  via the map
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 .
Write the differential 1-form associated to a function  as .
Then one can calculate that
\star\omega_{\nabla\times\mathbf{F}}=\mathrm{d}\omega_{\mathbf{F}}
where  is the Hodge star and \mathrm{d} is the exterior derivative.
Thus, by generalized Stokes' theorem,
\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}} Applications
In fluid dynamics
In this section, we will discuss the lamellar vector field based on Stokes' theorem.
Irrotational fields
Definition 2-1 (irrotational field).
A smooth vector field  on an open  is irrotational if .
If the domain of  is simply connected, then  is a conservative vector field.
Helmholtz's theorems
In this section, we will introduce a theorem that is derived from Stokes' theorem and characterizes vortex-free vector fields.
In fluid dynamics it is called Helmholtz's theorems.
Theorem 2-1 (Helmholtz's theorem in fluid dynamics).
Let  be an open subset with a lamellar vector field  and let  be piecewise smooth loops.
If there is a function  such that
[TLH0]  is piecewise smooth,
[TLH1]  for all ,
[TLH2]  for all ,
[TLH3]   for all .
Then, \int_{c_0} \mathbf{F} \, \mathrm{d}c_0=\int_{c_1} \mathbf{F} \, \mathrm{d}c_1
Some textbooks such as Lawrence call the relationship between  and  stated in theorem 2-1 as "homotopic" and the function  as "homotopy between  and ".
However, "homotopic" or "homotopy" in above-mentioned sense are different (stronger than) typical definitions of "homotopic" or "homotopy"; the latter omit condition [TLH3].
So from now on we refer to homotopy (homotope) in the sense of theorem 2-1 as a tubular homotopy (resp. tubular-homotopic).
Proof of the theorem
thumb|The definitions of
In what follows, we abuse notation and use "" for concatenation of paths in the fundamental groupoid and "" for reversing the orientation of a path.
Let , and split  into four line segments .
\begin{align} \gamma_1:[0,1] \to D;\quad&\gamma_1(t) = (t, 0) \\ \gamma_2:[0,1] \to D;\quad&\gamma_2(s) = (1, s) \\ \gamma_3:[0,1] \to D;\quad&\gamma_3(t) = (1-t, 1) \\ \gamma_4:[0,1] \to D;\quad&\gamma_4(s) = (0, 1-s) \end{align} so that \partial D = \gamma_1+\gamma_2+\gamma_3+\gamma_4
By our assumption that  and  are piecewise smooth homotopic, there is a piecewise smooth homotopy
\begin{align} \Gamma_i(t) &= H(\gamma_{i}(t)) && i=1, 2, 3, 4 \\ \Gamma(t) &= H(\gamma(t)) =(\Gamma_1 \oplus \Gamma_2 \oplus \Gamma_3 \oplus \Gamma_4)(t) \end{align}
Let  be the image of  under .
That
\iint_S \nabla\times\mathbf{F}\, \mathrm{d}S = \oint_\Gamma \mathbf{F}\, \mathrm{d}\Gamma
follows immediately from Stokes' theorem.
is lamellar, so the left side vanishes, i.e.
0=\oint_\Gamma \mathbf{F}\, \mathrm{d}\Gamma = \sum_{i=1}^4 \oint_{\Gamma_i} \mathbf{F} \, \mathrm{d}\Gamma
As  is tubular, .
Thus the line integrals along  and  cancel, leaving
0=\oint_{\Gamma_1} \mathbf{F} \, \mathrm{d}\Gamma +\oint_{\Gamma_3} \mathbf{F} \, \mathrm{d}\Gamma
On the other hand,  and , so that the desired equality follows almost immediately.
Conservative forces
Helmholtz's theorem gives an explanation as to why the work done by a conservative force in changing an object's position is path independent.
First, we introduce the Lemma 2-2, which is a corollary of and a special case of Helmholtz's theorem.
Lemma 2-2.
Let  be an open subset, with a Lamellar vector field  and a piecewise smooth loop .
Fix a point , if there is a homotopy (tube-like-homotopy)  such that
[SC0]  is piecewise smooth,
[SC1]  for all ,
[SC2]  for all ,
[SC3]  for all .
Then, \int_{c_0} \mathbf{F} \, \mathrm{d}c_0=0
Lemma 2-2 follows from theorem 2–1.
In Lemma 2-2, the existence of  satisfying [SC0] to [SC3] is crucial.
If  is simply connected, such  exists.
The definition of simply connected space follows:
Definition 2-2 (simply connected space).
Let  be non-empty and path-connected.
is called simply connected if and only if for any continuous loop,  there exists a continuous tubular homotopy  from  to a fixed point ; that is,
[SC0']  is continuous,
[SC1]  for all ,
[SC2]  for all ,
[SC3]   for all .
The claim that "for a conservative force, the work done in changing an object's position is path independent" might seem to follow immediately.
But recall that simple-connection only guarantees the existence of a continuous homotopy satisfying [SC1-3]; we seek a piecewise smooth homotopy satisfying those conditions instead.
However, the gap in regularity is resolved by the Whitney approximation theorem.L. S. Pontryagin, Smooth manifolds and their applications in homotopy theory, American Mathematical Society Translations, Ser.
2, Vol. 11, American Mathematical Society, Providence, R.I., 1959, pp.
1–114.
(22 #5980 ).
See theorems 7 & 8.
We thus obtain the following theorem.
Theorem 2-2.
Let  be open and simply connected with an irrotational vector field .
For all piecewise smooth loops  \int_{c_0} \mathbf{F} \, \mathrm{d}c_0 = 0 Maxwell's equations
In the physics of electromagnetism, Stokes' theorem provides the justification for the equivalence of the differential form of the Maxwell–Faraday equation and the Maxwell–Ampère equation and the integral form of these equations.
For Faraday's law, Stokes' theorem is applied to the electric field, \mathbf{E}:
\oint_{\partial\Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{l}= \iint_\Sigma \mathbf{\nabla}\times \mathbf{E} \cdot \mathrm{d} \mathbf{S} .
For Ampère's law, Stokes' theorem is applied to the magnetic field, \mathbf{B}:
\oint_{\partial\Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{l}= \iint_\Sigma \mathbf{\nabla}\times \mathbf{B} \cdot \mathrm{d} \mathbf{S} .
Notes
References
