Stokes' theorem

 ( Electromagnetism ) 

Stokes' theorem,

also known as the Kelvin–Stokes theorem

Iwahori Nagayoshi (1983). 9784785310394, Shokabo. . ISBN 9784785310394

after Lord Kelvin and George Stokes, the fundamental theorem for curls, or simply the curl theorem,

David J. Griffiths (2026). 9780321856562, Pearson. ISBN 9780321856562

or rotor theorem is a theorem in vector calculus on three-dimensional Euclidean space and real coordinate space, $\backslash R^3$ $\backslash mathbb\{R\}^n$ (where n is a natural number) is the set of all ordered n-tuples of real numbers, regarded as column vectors. By defining vector addition and scalar multiplication componentwise, $\backslash mathbb\{R\}^n$ becomes a real vector space equipped with the usual operations of linear algebra. From the viewpoint of matrix operations, $\backslash mathbb\{R\}^n$ can be identified with $\backslash mathcal\{M\}(n,1,\backslash mathbb\{R\})$, the set of real $n\backslash times\; 1$ matrices. Any affine linear function on $\backslash mathbb\{R\}^n$ can then be written in the form $A\; x\; +\; b$, where A is a matrix and b is a constant vector. . Given a vector field, the theorem relates the Surface 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 theorem of Stokes can be stated in one sentence:

The line integral of a vector field over a loop is equal to the surface integral of its curl over the enclosed surface.

Stokes' theorem is a special case of the generalized Stokes theorem.

John M. Lee (2026). 9781441999825, Springer. ISBN 9781441999825

In particular, a vector field on $\backslash R^3$ can be considered as a 1-form in which case its curl is its exterior derivative, a 2-form.

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Theorem Let $\backslash Sigma$ be a smooth oriented surface in $\backslash R^3$, parametrized by $\backslash mathbf\backslash Sigma(u,v)$, with boundary $\backslash partial\; \backslash Sigma\; \backslash equiv\; \backslash Gamma$, parametrized by $\backslash mathbf\backslash Gamma(t)$. If a vector field

$\backslash mathbf\{F\}(x,y,z)\; =\; (F\_x(x,\; y,\; z),\; F\_y(x,\; y,\; z),\; F\_z(x,\; y,\; z))$

has continuous first-order partial derivatives in $\backslash Sigma$, then $$\backslash iint\_\backslash Sigma\; (\backslash nabla\; \backslash times\; \backslash mathbf\{F\})\; \backslash cdot\; d\backslash mathbf\{\backslash Sigma\}\; =\; \backslash oint\_\{\backslash partial\backslash Sigma\}\; \backslash mathbf\{F\}\; \backslash cdot\; d\backslash mathbf\{\backslash Gamma\}$$ with the shorthands for the line elment $d\backslash mathbf\backslash Gamma\; =\; \backslash frac\{d\backslash mathbf\backslash Gamma\}\{dt\}dt$ and the surface element $$d\backslash mathbf\{\backslash Sigma\}\; =\; \backslash mathbf\{n\}\; d\backslash Sigma\; =\; \backslash left(\; \backslash frac\{\backslash partial\; \backslash mathbf\backslash Sigma\}\{\backslash partial\; u\}\; \backslash times\; \backslash frac\{\backslash partial\; \backslash mathbf\backslash Sigma\}\{\backslash partial\; v\}\; \backslash right)\; du\; dv$$ where $\backslash mathbf\{n\}(u,v)$ is the vector orthogonal to the surface at the point $\backslash mathbf\backslash Sigma(u,v)$.

More explicitly, with $\backslash wedge$ being the wedge product, the equality says that

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 $\backslash R^2$.

A more detailed statement will be given for subsequent discussions. Let $\backslash gamma:a,b\backslash to\backslash R^2$ be a piecewise smooth Jordan curve: a simple closed curve in the plane. The Jordan curve theorem implies that $\backslash gamma$ divides $\backslash R^2$ into two components, a compact space one and another that is non-compact. Let $D$ denote the compact part; then $D$ is bounded by $\backslash gamma$. It now suffices to transfer this notion of boundary along a continuous map to our surface in $\backslash R^3$. But we already have such a map: the parametrization of $\backslash Sigma$.

Suppose $\backslash psi:D\backslash to\backslash R^3$ is piecewise smooth at the neighborhood of $D$From the definition of $D$, $D$ is obviously a bounded closed set in $\backslash R^2$. "A neighborhood of D" means "an open set in $\backslash R^2$ that contains D." 　, with $\backslash Sigma=\backslash psi(D)$.$\backslash Sigma=\backslash psi(D)$ represents the image set of $D$ by $\backslash psi$ If $\backslash Gamma$ is the space curve defined by $\backslash Gamma(t)=\backslash psi(\backslash gamma(t))$$\backslash Gamma$ may not be a Jordan curve if the loop $\backslash gamma$ interacts poorly with $\backslash psi$. Nonetheless, $\backslash Gamma$ is always a loop or loops, and topologically a connected sum of countable set Jordan curves, so that the integrals are well-defined. then we call $\backslash Gamma$ the boundary of $\backslash Sigma$, written $\backslash partial\backslash Sigma$ Even if we consider two-dimensional polar coordinates, then if we let ∂Σ = Γ, then ∂Σ can clearly contain a topological interior point of Σ. However, in orientable combinatorial manifolds, this is not such a serious problem, since the line integrals over the interior points cancel out: in an orientable manifold, the line integral over ∂Σ coincides with the line integral over the true boundary.An even more important example is that this theorem also applies to manifolds without a topological boundary, such as the sphere or the torus. In such cases, if ∂Σ = Γ, then ∂Σ is completely contained within Σ. However, in such cases, the resulting line integrals over Γ completely cancel each other out. This allows us to assert that the surface integral of the curl of a vector field on a manifold without a boundary is zero.

With the above notation, if $\backslash mathbf\{F\}$ is any smooth vector field on $\backslash R^3$, then

Robert Scheichl, lecture notes for University of Bath mathematics course$$\backslash oint\_\{\backslash partial\backslash Sigma\}\; \backslash mathbf\{F\}\backslash ,\; \backslash cdot\backslash ,\; d\{\backslash mathbf\{\backslash Gamma\}\}\; =\; \backslash iint\_\{\backslash Sigma\}\; \backslash nabla\backslash times\backslash mathbf\{F\}\backslash ,\; \backslash cdot\backslash ,\; d\backslash mathbf\{\backslash Sigma\}.$$

Here, the "$\backslash cdot$" represents the dot product in $\backslash R^3$.

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Special case of a more general theorem Stokes' theorem can be viewed as a special case of the following identity: $$\backslash oint\_\{\backslash partial\backslash Sigma\}\; (\backslash mathbf\{F\}\backslash ,\; \backslash cdot\backslash ,\; d\{\backslash mathbf\{\backslash Gamma\}\})\backslash ,\backslash mathbf\{g\}\; =\; \backslash iint\_\{\backslash Sigma\}\backslash left\backslash mathbf\{g\},$$ where $\backslash mathbf\{g\}$ is any smooth vector or scalar field in $\backslash mathbb\{R\}^3$. When $\backslash mathbf\{g\}$ is a uniform scalar field, the standard Stokes' theorem is recovered.

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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).

Susan Jane Colley (2026). 9780321780652, Pearson. . ISBN 9780321780652

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 alternative proof of Stokes' theorem is given, as a corollary of the generalized Stokes' theorem.

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Elementary proof

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First step of the elementary 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_{\gamma}{\mathbf{F}(\boldsymbol{\psi}(\mathbf{\gamma}))\cdot\,\mathrm{d}\boldsymbol{\psi}(\mathbf{\gamma})} = \oint_{\gamma}{\mathbf{F}(\boldsymbol{\psi}(\mathbf{y}))\cdot J_{\mathbf{y}}(\boldsymbol{\psi})\,\mathrm{d}\gamma} where stands for the Jacobian matrix of at .

Now let be an orthonormal basis in the coordinate directions of .In this article, $$\backslash mathbf\{e\}\_u=\; \backslash begin\{bmatrix\}1\; \backslash \backslash \; 0\; \backslash end\{bmatrix\}\; ,\; \backslash mathbf\{e\}\_v\; =\; \backslash begin\{bmatrix\}0\; \backslash \backslash \; 1\; \backslash end\{bmatrix\}\; .$$ Note that, in some textbooks on vector analysis, these are assigned to different things. For example, in some text book's notation, can mean the following respectively. In this article, however, these are two completely different things. $$\backslash mathbf\{t\}\_\{u\}\; =\backslash frac\; \{1\}\{h\_u\}\backslash frac\; \{\backslash partial\; \backslash varphi\}\{\backslash partial\; u\}\; \backslash ,\; ,\; \backslash mathbf\{t\}\_\{v\}\; =\backslash frac\; \{1\}\{h\_v\}\backslash frac\; \{\backslash partial\; \backslash varphi\}\{\backslash partial\; v\}.$$ Here, $$h\_u\; =\; \backslash left\backslash |\backslash frac\; \{\backslash partial\; \backslash varphi\}\{\backslash partial\; u\}\backslash right\backslash |\; ,\; h\_v\; =\; \backslash left\backslash |\backslash frac\; \{\backslash partial\; \backslash varphi\}\{\backslash partial\; v\}\; \backslash right\backslash |,$$ and the "$\backslash |\; \backslash cdot\; \backslash |$" represents Euclidean norm.

Recognizing that the columns of are precisely the partial derivatives of at , we can expand the previous equation in coordinates as

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Second step in the elementary proof (defining the pullback)

The previous step suggests we define the function $$\backslash mathbf\{P\}(u,v)\; =\; \backslash left(\backslash mathbf\{F\}(\backslash boldsymbol\{\backslash psi\}(u,v))\backslash cdot\backslash frac\{\backslash partial\backslash boldsymbol\{\backslash psi\}\}\{\backslash partial\; u\}(u,v)\backslash right)\backslash mathbf\{e\}\_u\; +\; \backslash left(\backslash mathbf\{F\}(\backslash boldsymbol\{\backslash psi\}(u,v))\backslash cdot\backslash frac\{\backslash partial\backslash boldsymbol\{\backslash psi\}\}\{\backslash partial\; v\}(u,v)\; \backslash right)\backslash mathbf\{e\}\_v$$

Now, if the scalar value functions $P\_u$ and $P\_v$ are defined as follows, $$\{P\_u\}(u,v)\; =\; \backslash left(\backslash mathbf\{F\}(\backslash boldsymbol\{\backslash psi\}(u,v))\backslash cdot\backslash frac\{\backslash partial\backslash boldsymbol\{\backslash psi\}\}\{\backslash partial\; u\}(u,v)\backslash right)$$ $$\{P\_v\}(u,v)\; =\backslash left(\backslash mathbf\{F\}(\backslash boldsymbol\{\backslash psi\}(u,v))\backslash cdot\backslash frac\{\backslash partial\backslash boldsymbol\{\backslash psi\}\}\{\backslash partial\; v\}(u,v)\; \backslash right)$$ then, $$\backslash mathbf\{P\}(u,v)\; =\; \{P\_u\}(u,v)\; \backslash mathbf\{e\}\_u\; +\; \{P\_v\}(u,v)\; \backslash mathbf\{e\}\_v\; .$$

This is the pullback of along , and, by the above, it satisfies

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

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Third step of the elementary proof (second equation)

First, calculate the partial derivatives appearing in Green's theorem, via the product rule:

Conveniently, the second term vanishes in the difference, by equality of mixed partials. So, For all $\backslash textbf\{a\},\; \backslash textbf\{b\}\; \backslash in\; \backslash mathbb\{R\}^\{n\}$, for all $A\; ;\; n\backslash times\; n$ square matrix, $\backslash textbf\{a\}\backslash cdot\; A\; \backslash textbf\{b\}\; =\; \backslash textbf\{a\}^\backslash mathsf\{T\}A\; \backslash textbf\{b\}$ and therefore $\backslash textbf\{a\}\backslash cdot\; A\; \backslash textbf\{b\}\; =\; \backslash textbf\{b\}\; \backslash cdot\; A^\backslash mathsf\{T\}\; \backslash textbf\{a\}$.

But now consider the matrix in that quadratic form—that is, $J\_\{\backslash boldsymbol\backslash psi(u,v)\}\backslash mathbf\{F\}-(J\_\{\backslash boldsymbol\backslash psi(u,v)\}\backslash mathbf\{F\})^\{\backslash mathsf\{T\}\}$. We claim this matrix in fact describes a cross product. Here the superscript "$\{\}^\{\backslash mathsf\{T\}\}$" represents the Transpose.

To be precise, let $A=(A\_\{ij\})\_\{ij\}$ be an arbitrary matrix and let $$\backslash mathbf\{a\}=\; \backslash begin\{bmatrix\}a\_1\; \backslash \backslash \; a\_2\; \backslash \backslash \; a\_3\backslash end\{bmatrix\}\; =\; \backslash begin\{bmatrix\}A\_\{32\}-A\_\{23\}\; \backslash \backslash \; A\_\{13\}-A\_\{31\}\; \backslash \backslash \; A\_\{21\}-A\_\{12\}\backslash end\{bmatrix\}$$

Note that is linear, so it is determined by its action on basis elements. But by direct calculation Here, represents an orthonormal basis in the coordinate directions of $\backslash R^3$.In this article, $$\backslash mathbf\{e\}\_1=\; \backslash begin\{bmatrix\}1\; \backslash \backslash \; 0\; \backslash \backslash \; 0\backslash end\{bmatrix\}\; ,\; \backslash mathbf\{e\}\_2\; =\; \backslash begin\{bmatrix\}0\; \backslash \backslash \; 1\; \backslash \backslash \; 0\backslash end\{bmatrix\}\; ,\; \backslash mathbf\{e\}\_3\; =\; \backslash begin\{bmatrix\}0\; \backslash \backslash \; 0\; \backslash \backslash \; 1\backslash end\{bmatrix\}\; .$$ Note that, in some textbooks on vector analysis, these are assigned to different things.

Thus for any .

Substituting $\{(J\_\{\backslash boldsymbol\backslash psi(u,v)\}\backslash mathbf\{F\})\}$ for , we obtain $$\backslash left(\{(J\_\{\backslash boldsymbol\backslash psi(u,v)\}\backslash mathbf\{F\})\}\; -\; \{(J\_\{\backslash boldsymbol\backslash psi(u,v)\}\backslash mathbf\{F\})\}^\{\backslash mathsf\{T\}\}\; \backslash right)\; \backslash mathbf\{x\}\; =(\backslash nabla\backslash times\backslash mathbf\{F\})\backslash times\; \backslash mathbf\{x\},\; \backslash quad\; \backslash text\{for\; all\}\backslash ,\; \backslash mathbf\{x\}\backslash in\backslash R^\{3\}$$

We can now recognize the difference of partials as a Triple product:

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

So, we obtain $$\backslash iint\_\backslash Sigma\; (\backslash nabla\backslash times\backslash mathbf\{F\})\backslash cdot\; \backslash ,\backslash mathrm\{d\}\backslash mathbf\{\backslash Sigma\; \}\; =\; \backslash iint\_D\; \backslash left(\; \backslash frac\{\backslash partial\; P\_v\}\{\backslash partial\; u\}\; -\; \backslash frac\{\backslash partial\; P\_u\}\{\backslash partial\; v\}\; \backslash right)\; \backslash ,\backslash mathrm\{d\}u\backslash ,\backslash mathrm\{d\}v$$

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Fourth step of the elementary proof (reduction to Green's theorem)

Combining the second and third steps and then applying Green's theorem completes the proof. Green's theorem asserts the following: for any region D bounded by the Jordans closed curve γ and two scalar-valued smooth functions $P\_u(u,v),\; P\_v(u,v)$ defined on D;

= \iint_D \left( \frac{\partial P_v}{\partial u} - \frac{\partial P_u}{\partial v} \right) \,\mathrm{d}u\,\mathrm{d}v

We can substitute the conclusion of STEP2 into the left-hand side of Green's theorem above, and substitute the conclusion of STEP3 into the right-hand side. Q.E.D.

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Proof via differential forms

The functions $\backslash R^3\backslash to\backslash R^3$ can be identified with the differential 1-forms on $\backslash R^3$ via the map $$F\_x\backslash mathbf\{e\}\_1+F\_y\backslash mathbf\{e\}\_2+F\_z\backslash mathbf\{e\}\_3\; \backslash mapsto\; F\_x\backslash ,\backslash mathrm\{d\}x\; +\; F\_y\backslash ,\backslash mathrm\{d\}y\; +\; F\_z\backslash ,\backslash mathrm\{d\}z\; .$$

Write the differential 1-form associated to a function as . Then one can calculate that $$\backslash star\backslash omega\_\{\backslash nabla\backslash times\backslash mathbf\{F\}\}=\backslash mathrm\{d\}\backslash omega\_\{\backslash mathbf\{F\}\},$$ where is the Hodge star and $\backslash mathrm\{d\}$ is the exterior derivative. Thus, by generalized Stokes' theorem,

Harold M. Edwards (1994). 9780817637071, Birkhäuser. ISBN 9780817637071

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Applications

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Irrotational fields In this section, we will discuss the irrotational field (lamellar vector field) based on Stokes' theorem.

Definition 2-1 (irrotational field). A smooth vector field on an open set $U\backslash subseteq\backslash R^3$ is irrotational (lamellar vector field) if .

This concept is very fundamental in mechanics; as we'll prove later, if is irrotational and the domain of is simply connected, then is a conservative vector field.

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Helmholtz's theorem In this section, we will introduce a theorem that is derived from Stokes' theorem and characterizes vortex-free vector fields. In classical mechanics and fluid dynamics it is called Helmholtz's theorem.

Theorem 2-1 (Helmholtz's theorem in fluid dynamics). Let $U\backslash subseteq\backslash R^3$ be an open set 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, $$\backslash int\_\{c\_0\}\; \backslash mathbf\{F\}\; \backslash ,\; \backslash mathrm\{d\}c\_0=\backslash int\_\{c\_1\}\; \backslash mathbf\{F\}\; \backslash ,\; \backslash mathrm\{d\}c\_1$$

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) Homotopy 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).

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Proof of Helmholtz's theorem In what follows, we abuse notation and use "$\backslash oplus$" for concatenation of paths in the fundamental groupoid and "$\backslash ominus$" for reversing the orientation of a path.

Let , and split into four line segments . so that $$\backslash partial\; D\; =\; \backslash gamma\_1\; \backslash oplus\; \backslash gamma\_2\; \backslash oplus\; \backslash gamma\_3\; \backslash oplus\; \backslash gamma\_4$$

By our assumption that and are piecewise smooth homotopic, there is a piecewise smooth homotopy

Let be the image of under . That $$\backslash iint\_S\; \backslash nabla\backslash times\backslash mathbf\{F\}\backslash ,\; \backslash mathrm\{d\}S\; =\; \backslash oint\_\backslash Gamma\; \backslash mathbf\{F\}\backslash ,\; \backslash mathrm\{d\}\backslash Gamma$$ follows immediately from Stokes' theorem. is lamellar, so the left side vanishes, i.e. $$0=\backslash oint\_\backslash Gamma\; \backslash mathbf\{F\}\backslash ,\; \backslash mathrm\{d\}\backslash Gamma\; =\; \backslash sum\_\{i=1\}^4\; \backslash oint\_\{\backslash Gamma\_i\}\; \backslash mathbf\{F\}\; \backslash ,\; \backslash mathrm\{d\}\backslash Gamma$$

As is tubular(satisfying TLH3),$\backslash Gamma\_2\; =\; \backslash ominus\; \backslash Gamma\_4$ and $\backslash Gamma\_2\; =\; \backslash ominus\; \backslash Gamma\_4$. Thus the line integrals along and cancel, leaving $$0=\backslash oint\_\{\backslash Gamma\_1\}\; \backslash mathbf\{F\}\; \backslash ,\; \backslash mathrm\{d\}\backslash Gamma\; +\backslash oint\_\{\backslash Gamma\_3\}\; \backslash mathbf\{F\}\; \backslash ,\; \backslash mathrm\{d\}\backslash Gamma$$

On the other hand, , $c\_3\; =\; \backslash ominus\; \backslash Gamma\_3$, so that the desired equality follows almost immediately.

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Conservative forces Above 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 $U\backslash subseteq\backslash R^3$ be an open set subset, with a Lamellar vector field and a piecewise smooth loop . Fix a point , if there is a homotopy such that

• SC0 is piecewise smooth,
• SC1 for all ,
• SC2 for all ,
• SC3 for all .

Then, $$\backslash int\_\{c\_0\}\; \backslash mathbf\{F\}\; \backslash ,\; \backslash mathrm\{d\}c\_0=0$$

Above Lemma 2-2 follows from theorem 2–1. In Lemma 2-2, the existence of satisfying SC0 to SC3 is crucial;the question is whether such a homotopy can be taken for arbitrary loops. If is simply connected, such exists. The definition of simply connected space follows:

Definition 2-2 (simply connected space). Let $M\backslash subseteq\backslash R^n$ 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 if the M is simply connected.　However, 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.

Fortunately, the gap in regularity is resolved by the Whitney's approximation theorem.

See theorems 7 & 8. In other words, the possibility of finding a continuous homotopy, but not being able to integrate over it, is actually eliminated with the benefit of higher mathematics. We thus obtain the following theorem.

Theorem 2-2. Let $U\backslash subseteq\backslash R^3$ be open set and simply connected with an irrotational vector field . For all piecewise smooth loops $$\backslash int\_\{c\_0\}\; \backslash mathbf\{F\}\; \backslash ,\; \backslash mathrm\{d\}c\_0\; =\; 0$$

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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, $\backslash mathbf\{E\}$: $$\backslash oint\_\{\backslash partial\backslash Sigma\}\; \backslash mathbf\{E\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{l\}=\; \backslash iint\_\backslash Sigma\; \backslash mathbf\{\backslash nabla\}\backslash times\; \backslash mathbf\{E\}\; \backslash cdot\; \backslash mathrm\{d\}\; \backslash mathbf\{S\}\; .$$

For Ampère's law, Stokes' theorem is applied to the magnetic field, $\backslash mathbf\{B\}$: $$\backslash oint\_\{\backslash partial\backslash Sigma\}\; \backslash mathbf\{B\}\; \backslash cdot\; \backslash mathrm\{d\}\backslash boldsymbol\{l\}=\; \backslash iint\_\backslash Sigma\; \backslash mathbf\{\backslash nabla\}\backslash times\; \backslash mathbf\{B\}\; \backslash cdot\; \backslash mathrm\{d\}\; \backslash mathbf\{S\}\; .$$

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Notes

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