Stokes' Theorem

An illustration of Stokes' theorem, with surface Σ, its boundary ∂Σ and the normal vector n.

Stokes' theorem, also known as Kelvin–Stokes theorem 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 R³ 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 ${\begin{aligned}&\iint _{\Sigma }\left(\left({\frac {\partial R}{\partial y}}-{\frac {\partial Q}{\partial z}}\right)\,\mathrm {d} y\,\mathrm {d} z+\left({\frac {\partial P}{\partial z}}-{\frac {\partial R}{\partial x}}\right)\,\mathrm {d} z\,\mathrm {d} x+\left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\,\mathrm {d} x\,\mathrm {d} y\right)\\&=\oint _{\partial \Sigma }{\Bigl (}P\,\mathrm {d} x+Q\,\mathrm {d} y+R\,\mathrm {d} z{\Bigr )}.\end{aligned}}$

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 R².

Let γ: [a, b] → R² be a piecewise smooth Jordan plane curve. The Jordan curve theorem implies that γ divides R² into two components, a compact one and another that is non-compact. Let D denote the compact part; then D is bounded by γ. It now suffices to transfer this notion of boundary along a continuous map to our surface in R³. But we already have such a map: the parametrization of Σ.

Suppose ψ: D → R³ is smooth, with Σ = ψ(D). If Γ is the space curve defined by Γ(t) = ψ(γ(t)), then we call Γ the boundary of Σ, written ∂Σ.

With the above notation, if F is any smooth vector field on R³, then

$\oint _{\partial \Sigma }\mathbf {F} \,\cdot \,\mathrm {d} {\mathbf {\Gamma } }=\iint _{\Sigma }\nabla \times \mathbf {F} \,\cdot \,\mathrm {d} \mathbf {S} .$

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. 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 $\S$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

$$\oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,\mathrm {d} \mathbf {l} }=\oint _{\gamma }{\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))\cdot \,\mathrm {d} {\boldsymbol {\psi }}(\mathbf {y} )}=\oint _{\gamma }{\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))J_{\mathbf {y} }({\boldsymbol {\psi }})\,\mathrm {d} \mathbf {y} }$$

where Jψ stands for the Jacobian matrix of ψ.

Now let {e_u, e_v} be an orthonormal basis in the coordinate directions of R₂. Recognizing that the columns of J_yψ are precisely the partial derivatives of ψ at y, we can expand the previous equation in coordinates as

${\begin{aligned}\oint _{\partial \Sigma }{\mathbf {F} (\mathbf {x} )\cdot \,\mathrm {d} \mathbf {l} }&=\oint _{\gamma }{\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))J_{\mathbf {y} }({\boldsymbol {\psi }})\mathbf {e} _{u}(\mathbf {e} _{u}\cdot \,\mathrm {d} \mathbf {y} )+\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))J_{\mathbf {y} }({\boldsymbol {\psi }})\mathbf {e} _{v}(\mathbf {e} _{v}\cdot \,\mathrm {d} \mathbf {y} )}\\&=\oint _{\gamma }{\left(\left(\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(\mathbf {y} )\right)\mathbf {e} _{u}+\left(\mathbf {F} ({\boldsymbol {\psi }}(\mathbf {y} ))\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(\mathbf {y} )\right)\mathbf {e} _{v}\right)\cdot \,\mathrm {d} \mathbf {y} }\end{aligned}}$

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 F along ψ, 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:

${\begin{aligned}{\frac {\partial P_{1}}{\partial v}}&={\frac {\partial (\mathbf {F} \circ {\boldsymbol {\psi }})}{\partial v}}\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial u}}+(\mathbf {F} \circ {\boldsymbol {\psi }})\cdot {\frac {\partial ^{2}{\boldsymbol {\psi }}}{\partial v\,\partial u}}\\[5pt]{\frac {\partial P_{2}}{\partial u}}&={\frac {\partial (\mathbf {F} \circ {\boldsymbol {\psi }})}{\partial u}}\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}+(\mathbf {F} \circ {\boldsymbol {\psi }})\cdot {\frac {\partial ^{2}{\boldsymbol {\psi }}}{\partial u\,\partial v}}\end{aligned}}$

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

$${\begin{aligned}{\frac {\partial P_{1}}{\partial v}}-{\frac {\partial P_{2}}{\partial u}}&={\frac {\partial (\mathbf {F} \circ {\boldsymbol {\psi }})}{\partial v}}\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial u}}-{\frac {\partial (\mathbf {F} \circ {\boldsymbol {\psi }})}{\partial u}}\cdot {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}\\[5pt]&={\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(J_{{\boldsymbol {\psi }}(u,v)}\mathbf {F} ){\frac {\partial {\boldsymbol {\psi }}}{\partial v}}-{\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(J_{{\boldsymbol {\psi }}(u,v)}\mathbf {F} ){\frac {\partial {\boldsymbol {\psi }}}{\partial u}}&&{\text{(chain rule)}}\\[5pt]&={\frac {\partial {\boldsymbol {\psi }}}{\partial u}}\left(J_{{\boldsymbol {\psi }}(u,v)}\mathbf {F} -{(J_{{\boldsymbol {\psi }}(u,v)}\mathbf {F} )}^{\mathsf {T}}\right){\frac {\partial {\boldsymbol {\psi }}}{\partial v}}\end{aligned}}$$

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 3 × 3 matrix and let

$$\mathbf {a} ={\begin{bmatrix}A_{32}-A_{23}\\A_{13}-A_{31}\\A_{21}-A_{12}\end{bmatrix}}$$

Note that x ↦ a × x is linear, so it is determined by its action on basis elements. But by direct calculation

$${\begin{aligned}\left(A-A^{\mathsf {T}}\right)\mathbf {e} _{1}&={\begin{bmatrix}0\\a_{3}\\-a_{2}\end{bmatrix}}=\mathbf {a} \times \mathbf {e} _{1}\\\left(A-A^{\mathsf {T}}\right)\mathbf {e} _{2}&={\begin{bmatrix}-a_{3}\\0\\a_{1}\end{bmatrix}}=\mathbf {a} \times \mathbf {e} _{2}\\\left(A-A^{\mathsf {T}}\right)\mathbf {e} _{3}&={\begin{bmatrix}a_{2}\\-a_{1}\\0\end{bmatrix}}=\mathbf {a} \times \mathbf {e} _{3}\end{aligned}}$$

Thus (A − A^T)x = a × x for any x. Substituting J F for A, we obtain

$$\left({(J_{{\boldsymbol {\psi }}(u,v)}\mathbf {F} )}_{\psi (u,v)}-{(J_{{\boldsymbol {\psi }}(u,v)}\mathbf {F} )}^{\mathsf {T}}\right)\mathbf {x} =(\nabla \times \mathbf {F} )\times \mathbf {x} ,\quad {\text{for all}}\,\mathbf {x} \in \mathbb {R} ^{3}$$

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

${\begin{aligned}{\frac {\partial P_{1}}{\partial v}}-{\frac {\partial P_{2}}{\partial u}}&={\frac {\partial {\boldsymbol {\psi }}}{\partial u}}\cdot (\nabla \times \mathbf {F} )\times {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}\\&=\det {\begin{bmatrix}(\nabla \times \mathbf {F} )({\boldsymbol {\psi }}(u,v))&{\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(u,v)&{\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(u,v)\end{bmatrix}}\end{aligned}}$

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

${\begin{aligned}\iint _{S}(\nabla \times \mathbf {F} )\cdot \,d^{2}\mathbf {S} &=\iint _{D}{(\nabla \times \mathbf {F} )({\boldsymbol {\psi }}(u,v))\cdot \left({\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(u,v)\times {\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(u,v)\,\mathrm {d} u\,\mathrm {d} v\right)}\\&=\iint _{D}\det {\begin{bmatrix}(\nabla \times \mathbf {F} )({\boldsymbol {\psi }}(u,v))&{\frac {\partial {\boldsymbol {\psi }}}{\partial u}}(u,v)&{\frac {\partial {\boldsymbol {\psi }}}{\partial v}}(u,v)\end{bmatrix}}\,\mathrm {d} u\,\mathrm {d} v\end{aligned}}$

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

R → R³ can be identified with the differential 1-forms on R³ 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 F as ω_F. 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,

$$\oint _{\partial \Sigma }{\mathbf {F} \cdot \,\mathrm {d} \mathbf {l} }=\oint _{\partial \Sigma }{\omega _{\mathbf {F} }}=\int _{\Sigma }{\mathrm {d} \omega _{\mathbf {F} }}=\int _{\Sigma }{\star \omega _{\nabla \times \mathbf {F} }}=\iint _{\Sigma }{\nabla \times \mathbf {F} \cdot \,\mathrm {d} ^{2}\mathbf {S} }$$

Licensing

Content obtained and/or adapted from:

• Stokes' Theorem under a CC BY-SA license