Stokes' theorem, also known as the Kelvin–Stokes theorem, is a theorem in vector calculus that relates the behavior of a vector field along the edge of a surface to the behavior of its curl on the surface itself. In its usual three-dimensional form, it says that the total circulation of a vector field around a closed curve is equal to the total curl of the field through a surface bounded by that curve. If Σ is an oriented surface with boundary ∂Σ, Stokes' theorem is commonly written as
∮ ∂ Σ F ⋅ d r = ∬ Σ ( ∇ × F ) ⋅ n d S . {\displaystyle \oint _{\partial \Sigma }\mathbf {F} \cdot d\mathbf {r} =\iint _{\Sigma }(\nabla \times \mathbf {F} )\cdot \mathbf {n} \,dS.}
Here the left side is the line integral of the vector field around the boundary curve, while the right side is the surface integral of its curl over the surface. Informally, the theorem says that adding up the local rotation of a vector field across a surface gives the net circulation around its edge. The theorem is also called the fundamental theorem for curls, the curl theorem, or the rotor theorem. It is a special case of the generalized Stokes theorem. In the language of differential forms, the vector field corresponds to a 1-form and its curl corresponds to the exterior derivative of that form.
Theorem Let Σ {\displaystyle \Sigma } be a smooth oriented surface in R 3 {\displaystyle \mathbb {R} ^{3}} , parametrized by Σ ( u , v ) {\displaystyle \mathbf {\Sigma } (u,v)} , with boundary ∂ Σ ≡ Γ {\displaystyle \partial \Sigma \equiv \Gamma } , parametrized by Γ ( t ) {\displaystyle \mathbf {\Gamma } (t)} . If a vector field
F ( x , y , z ) = ( F x ( x , y , z ) , F y ( x , y , z ) , F z ( x , y , z ) ) {\displaystyle \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 Σ {\displaystyle \Sigma } , then
∬ Σ ( ∇ × F ) ⋅ d Σ = ∮ ∂ Σ F ⋅ d Γ {\displaystyle \iint _{\Sigma }(\nabla \times \mathbf {F} )\cdot d\mathbf {\Sigma } =\oint _{\partial \Sigma }\mathbf {F} \cdot d\mathbf {\Gamma } }
with the shorthands for the line element d Γ = d Γ d t d t {\displaystyle d\mathbf {\Gamma } ={\frac {d\mathbf {\Gamma } }{dt}}dt} and the surface element
d Σ = n d Σ = ( ∂ Σ ∂ u × ∂ Σ ∂ v ) d u d v {\displaystyle d\mathbf {\Sigma } =\mathbf {n} d\Sigma =\left({\frac {\partial \mathbf {\Sigma } }{\partial u}}\times {\frac {\partial \mathbf {\Sigma } }{\partial v}}\right)dudv}
where n ( u , v ) {\displaystyle \mathbf {n} (u,v)} is the vector orthogonal to the surface at the point Σ ( u , v ) {\displaystyle \mathbf {\Sigma } (u,v)} . The equality can be expressed in terms of differential forms, with ∧ {\displaystyle \wedge } being the wedge product and d {\displaystyle {\text{d}}} the exterior derivative:
∬ Σ ( ( ∂ F z ∂ y − ∂ F y ∂ z ) d y ∧ d z + ( ∂ F x ∂ z − ∂ F z ∂ x ) d z ∧ d x + ( ∂ F y ∂ x − ∂ F x ∂ y ) d x ∧ d y ) = ∮ ∂ Σ ( F x d x + F y d y + F z d z ) . {\displaystyle {\begin{aligned}&\iint _{\Sigma }\left(\left({\frac {\partial F_{z}}{\partial y}}-{\frac {\partial F_{y}}{\partial z}}\right)\,\mathrm {d} y\wedge \mathrm {d} z+\left({\frac {\partial F_{x}}{\partial z}}-{\frac {\partial F_{z}}{\partial x}}\right)\,\mathrm {d} z\wedge \mathrm {d} x+\left({\frac {\partial F_{y}}{\partial x}}-{\frac {\partial F_{x}}{\partial y}}\right)\,\mathrm {d} x\wedge \mathrm {d} y\right)\\&=\oint _{\partial \Sigma }{\Bigl (}F_{x}\,\mathrm {d} x+F_{y}\,\mathrm {d} y+F_{z}\,\mathrm {d} z{\Bigr )}.\end{aligned}}}
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 R 2 {\displaystyle \mathbb {R} ^{2}} . A more detailed statement will be given for subsequent discussions. Let γ : [ a , b ] → R 2 {\displaystyle \gamma :[a,b]\to \mathbb {R} ^{2}} be a piecewise smooth Jordan plane curve: a simple closed curve in the plane. The Jordan curve theorem implies that γ {\displaystyle \gamma } divides R 2 {\displaystyle \mathbb {R} ^{2}} into two components, a compact one and another that is non-compact. Let D {\displaystyle D} denote the compact part; then D {\displaystyle D} is bounded by γ {\displaystyle \gamma } . It now suffices to transfer this notion of boundary along a continuous map to our surface in R 3 {\displaystyle \mathbb {R} ^{3}} . But we already have such a map: the parametrization of Σ {\displaystyle \Sigma } . Suppose ψ : D → R 3 {\displaystyle \psi :D\to \mathbb {R} ^{3}} is piecewise smooth at the neighborhood of D {\displaystyle D} , with Σ = ψ ( D ) {\displaystyle \Sigma =\psi (D)} . If Γ {\displaystyle \Gamma } is the space curve defined by Γ ( t ) = ψ ( γ ( t ) ) {\displaystyle \Gamma (t)=\psi (\gamma (t))} then we call Γ {\displaystyle \Gamma } the boundary of Σ {\displaystyle \Sigma } , written ∂ Σ {\displaystyle \partial \Sigma }
With the above notation, if F {\displaystyle \mathbf {F} } is any smooth vector field on R 3 {\displaystyle \mathbb {R} ^{3}} , then ∮ ∂ Σ F ⋅ d Γ = ∬ Σ ∇ × F ⋅ d Σ . {\displaystyle \oint _{\partial \Sigma }\mathbf {F} \,\cdot \,d{\mathbf {\Gamma } }=\iint _{\Sigma }\nabla \times \mathbf {F} \,\cdot \,d\mathbf {\Sigma } .}
Here, the " ⋅ {\displaystyle \cdot } " represents the dot product in R 3 {\displaystyle \mathbb {R} ^{3}} .
Special case of a more general theorem Stokes' theorem can be viewed as a special case of the following identity:
∮ ∂ Σ ( F ⋅ d Γ ) g = ∬ Σ [ d Σ ⋅ ( ∇ × F − F × ∇ ) ] g , {\displaystyle \oint _{\partial \Sigma }(\mathbf {F} \,\cdot \,d{\mathbf {\Gamma } })\,\mathbf {g} =\iint _{\Sigma }\left[d\mathbf {\Sigma } \cdot \left(\nabla \times \mathbf {F} -\mathbf {F} \times \nabla \right)\right]\mathbf {g} ,}
where g {\displaystyle \mathbf {g} } is any smooth vector or scalar field in R 3 {\displaystyle \mathbb {R} ^{3}} . When g {\displaystyle \mathbf {g} } is a uniform scalar field, the standard Stokes' theorem is recovered.
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 alternative proof of Stokes' theorem is given, as a corollary of the generalized Stokes' theorem.
Elementary proof
First step of the elementary proof (parametrization of integral) As in § Theorem, we reduce the dimension by using the natural parametrization of the surface. Let ψ and γ be as in that section, and note that by change of variables
∮ ∂ Σ F ( x ) ⋅ d Γ = ∮ γ F ( ψ ( γ ) ) ⋅ d ψ ( γ ) = ∮ γ F ( ψ ( y ) ) ⋅ J y ( ψ ) d γ {\displaystyle \oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,\mathrm {d} \mathbf {\Gamma } }=\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 Jyψ stands for the Jacobian matrix of ψ at y = γ(t). Now let {eu, ev} be an orthonormal basis in the coordinate directions of R2. Recognizing that the columns of Jyψ are precisely the partial derivatives of ψ at y, we can expand the previous equation in coordinates as
∮ ∂ Σ F ( x ) ⋅ d Γ = ∮ γ F ( ψ ( y ) ) ⋅ J y ( ψ ) e u ( e u ⋅ d y ) + F ( ψ ( y ) ) ⋅ J y ( ψ ) e v ( e v ⋅ d y ) = ∮ γ ( ( F ( ψ ( y ) ) ⋅ ∂ ψ ∂ u ( y ) ) e u + ( F ( ψ ( y ) ) ⋅ ∂ ψ ∂ v ( y ) ) e v ) ⋅ d y {\displaystyle {\begin{aligned}\oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,\mathrm {d} \mathbf {\Gamma } }&=\oint _{\gamma }{\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))\cdot J_{\mathbf {y} }({\boldsymbol {\psi }})\mathbf {e} _{u}(\mathbf {e} _{u}\cdot \,\mathrm {d} \mathbf {y} )+\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))\cdot 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{aligned}}}
Second step in the elementary proof (defining the pullback) The previous step suggests we define the function
P ( u , v ) = ( F ( ψ ( u , v ) ) ⋅ ∂ ψ ∂ u ( u , v ) ) e u + ( F ( ψ ( u , v ) ) ⋅ ∂ ψ ∂ v ( u , v ) ) e v {\displaystyle \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}}(u,v)\right)\mathbf {e} _{v}}
Now, if the scalar value functions P u {\displaystyle P_{u}} and P v {\displaystyle P_{v}} are defined as follows,
P u ( u , v ) = ( F ( ψ ( u , v ) ) ⋅ ∂ ψ ∂ u ( u , v ) ) {\displaystyle {P_{u}}(u,v)=\left(\mathbf {F} ({\boldsymbol {\psi }}(u,v))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(u,v)\right)}
P v ( u , v ) = ( F ( ψ ( u , v ) ) ⋅ ∂ ψ ∂ v ( u , v ) ) {\displaystyle {P_{v}}(u,v)=\left(\mathbf {F} ({\boldsymbol {\psi }}(u,v))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(u,v)\right)}
then,
P ( u , v ) = P u ( u , v ) e u + P v ( u , v ) e v . {\displaystyle \mathbf {P} (u,v)={P_{u}}(u,v)\mathbf {e} _{u}+{P_{v}}(u,v)\mathbf {e} _{v}.}
This is the pullback of F along ψ, and, by the above, it satisfies
∮ ∂ Σ F ( x ) ⋅ d l = ∮ γ P ( y ) ⋅ d l = ∮ γ ( P u ( u , v ) e u + P v ( u , v ) e v ) ⋅ d l {\displaystyle \oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,\mathrm {d} \mathbf {l} }=\oint _{\gamma }{\mathbf {P} (\mathbf {y} )\cdot \,\mathrm {d} \mathbf {l} }=\oint _{\gamma }{({P_{u}}(u,v)\mathbf {e} _{u}+{P_{v}}(u,v)\mathbf {e} _{v})\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 elementary proof (second equation) First, calculate the partial derivatives appearing in Green's theorem, via the product rule:
∂ P u ∂ v = ∂ ( F ∘ ψ ) ∂ v ⋅ ∂ ψ ∂ u + ( F ∘ ψ ) ⋅ ∂ 2 ψ ∂ v ∂ u ∂ P v ∂ u = ∂ ( F ∘ ψ ) ∂ u ⋅ ∂ ψ ∂ v + ( F ∘ ψ ) ⋅ ∂ 2 ψ ∂ u ∂ v {\displaystyle {\begin{aligned}{\frac {\partial P_{u}}{\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_{v}}{\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{aligned}}}
Conveniently, the second term vanishes in the difference, by equality of mixed partials. So,
∂ P v ∂ u