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Stokes' Theorem

Stokes' theorem extends Green's theorem from regions in the plane to surfaces in three-dimensional space. It relates the circulation of a vector field around a boundary curve to the flux of its curl through the surface. The boundary may be a space curve; it need not lie in a plane.
STOKES' THEOREM
Let SS be a compact, oriented, piecewise-smooth surface in R3\mathbb{R}^3 with chosen unit normal n\mathbf{n}. Suppose its boundary C=∂SC=\partial S is a simple, closed, piecewise-smooth curve with the positive orientation induced by n\mathbf{n}. If F\mathbf{F} has continuous first partial derivatives on an open set containing SS, then
∮CF⋅dr=∬S(∇×F)⋅n dS.\begin{equation*}\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S (\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS.\end{equation*}

Motivation

Stokes' theorem is named after Sir George Stokes (1819–1903), an Irish mathematical physicist known for his studies of fluid flow and light. At Cambridge University, Stokes held the Lucasian Professorship of Mathematics, the same chair once held by Isaac Newton.
The theorem's discovery is credited to the Scottish physicist Sir William Thomson (1824–1907), later known as Lord Kelvin. Thomson described the result to Stokes in a letter in 1850. In 1854, Stokes asked students to prove it in an examination at Cambridge University. It is not known whether any of those students succeeded.

Curl and grad

The surface integral which appears in Stokes’ theorem can be expressed more simply in terms of the curl of a vector field. Let F \mathbf{F} be a differentiable vector field given by
F(x,y,z)=P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k.\begin{equation*}\mathbf{F}(x,y,z)=P(x,y,z)\mathbf{i}+Q(x,y,z)\mathbf{j}+R(x,y,z)\mathbf{k}.\end{equation*}
CURL
The curl of F is another vector field defined by the equation
curl⁡F=(∂R∂y−∂Q∂z)i+(∂P∂z−∂R∂x)j+(∂Q∂x−∂P∂y)k.\begin{equation*}\operatorname{curl}\mathbf{F}=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\mathbf{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathbf{k}.\end{equation*}
To remember this expression, introduce the vector differential operator ∇\nabla, pronounced ``del'':
∇=i∂∂x+j∂∂y+k∂∂z.\begin{equation*}\nabla=\mathbf{i}\frac{\partial}{\partial x}+\mathbf{j}\frac{\partial}{\partial y}+\mathbf{k}\frac{\partial}{\partial z}.\end{equation*}
Applying it to a differentiable scalar function f(x,y,z)f(x,y,z) produces the gradient of ff:
grad⁡f=∇f=∂f∂xi+∂f∂yj+∂f∂zk.\begin{equation*}\operatorname{grad}f=\nabla f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}.\end{equation*}
We can also form a symbolic cross product of ∇\nabla with F\mathbf{F}. The determinant below is a mnemonic: each differential operator acts on the component function to its right in the expansion. Expanding along the first row gives
∇×F=∣ijk∂∂x∂∂y∂∂zPQR∣=(∂R∂y−∂Q∂z)i+(∂P∂z−∂R∂x)j+(∂Q∂x−∂P∂y)k=curl⁡F.\begin{align*}\nabla\times\mathbf{F}&=\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\P & Q & R\end{vmatrix} \\&=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\mathbf{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\mathbf{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathbf{k} \\&=\operatorname{curl}\mathbf{F}.\end{align*}
Thus the curl can be remembered through the compact identity
curl⁡F=∇×F,\begin{equation*}\operatorname{curl}\mathbf{F}=\nabla\times\mathbf{F},\end{equation*}
which is the notation used in Stokes' theorem.

Proof

We prove the theorem for the special case of a surface given by a graph z=g(x,y)z=g(x,y). Vertical projection sends the surface to a plane region and its boundary to the boundary of that region. This lets us apply Green's theorem in the plane.
Let DD be a bounded plane region with a simple, closed, piecewise-smooth boundary C1C_1, and suppose
S={(x,y,g(x,y)):(x,y)∈D},\begin{equation*}S=\{(x,y,g(x,y)):(x,y)\in D\},\end{equation*}
where gg has continuous second partial derivatives on a neighborhood of DD. Choose the upward orientation of SS. Then the positive direction around C=∂SC=\partial S projects to the counterclockwise direction around C1=∂DC_1=\partial D.
Write F=(P,Q,R)\mathbf{F}=(P,Q,R). For the graph parametrization r(x,y)=(x,y,g(x,y))\mathbf{r}(x,y)=(x,y,g(x,y)), the upward oriented area element is
n dS=(∂r∂x×∂r∂y) dx dy=(−∂g∂x,−∂g∂y,1) dx dy.\begin{equation*}\mathbf{n}\,dS=\left(\frac{\partial\mathbf{r}}{\partial x}\times\frac{\partial\mathbf{r}}{\partial y}\right)\,dx\,dy=\left(-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial y},1\right)\,dx\,dy.\end{equation*}
Taking its dot product with ∇×F\nabla\times\mathbf{F}, we obtain
∬S(∇×F)⋅n dS=∬D[−(∂R∂y−∂Q∂z)∂g∂x−(∂P∂z−∂R∂x)∂g∂y+(∂Q∂x−∂P∂y)] dx dy.\begin{align*}\iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS&=\iint_D\Bigl[-\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\frac{\partial g}{\partial x} \\&\qquad-\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\frac{\partial g}{\partial y} \\&\qquad+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\Bigr]\,dx\,dy.\end{align*}
Here and in the plane integrals below, P,Q,RP,Q,R and their partial derivatives are evaluated at (x,y,g(x,y))(x,y,g(x,y)).
Let γ(t)=(x(t),y(t))\gamma(t)=(x(t),y(t)), a≤t≤ba\leq t\leq b, parametrize C1C_1 in the positive direction. Its lift to SS is the corresponding positive parametrization of CC:
α(t)=(x(t),y(t),z(t)),z(t)=g(x(t),y(t)),a≤t≤b.\begin{equation*}\boldsymbol{\alpha}(t)=(x(t),y(t),z(t)),\qquad z(t)=g(x(t),y(t)),\qquad a\leq t\leq b.\end{equation*}
The chain rule gives
dzdt=∂g∂xdxdt+∂g∂ydydt.\begin{equation*}\frac{dz}{dt}=\frac{\partial g}{\partial x}\frac{dx}{dt}+\frac{\partial g}{\partial y}\frac{dy}{dt}.\end{equation*}
We can therefore evaluate the boundary integral as follows, with P,Q,RP,Q,R evaluated at α(t)\boldsymbol{\alpha}(t) in the integrals over [a,b][a,b]:
∮CF⋅dr=∫ab(Pdxdt+Qdydt+Rdzdt) dt=∫ab[Pdxdt+Qdydt+R(∂g∂xdxdt+∂g∂ydydt)] dt=∫ab[(P+R∂g∂x)dxdt+(Q+R∂g∂y)dydt] dt=∮C1(P+R∂g∂x) dx+(Q+R∂g∂y) dy=Green’s theorem∬D[∂∂x(Q+R∂g∂y)−∂∂y(P+R∂g∂x)] dx dy.\begin{align*}\oint_C\mathbf{F}\cdot d\mathbf{r}&=\int_a^b\left(P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}\right)\,dt \\&=\int_a^b\left[P\frac{dx}{dt}+Q\frac{dy}{dt}+R\left(\frac{\partial g}{\partial x}\frac{dx}{dt}+\frac{\partial g}{\partial y}\frac{dy}{dt}\right)\right]\,dt \\&=\int_a^b\left[\left(P+R\frac{\partial g}{\partial x}\right)\frac{dx}{dt}+\left(Q+R\frac{\partial g}{\partial y}\right)\frac{dy}{dt}\right]\,dt \\&=\oint_{C_1}\left(P+R\frac{\partial g}{\partial x}\right)\,dx+\left(Q+R\frac{\partial g}{\partial y}\right)\,dy \\&\overset{\text{Green's theorem}}{=}\iint_D\left[\frac{\partial}{\partial x}\left(Q+R\frac{\partial g}{\partial y}\right)-\frac{\partial}{\partial y}\left(P+R\frac{\partial g}{\partial x}\right)\right]\,dx\,dy.\end{align*}
The last step applies Green's theorem to the plane region DD. Expanding with the chain rule yields
∮CF⋅dr=∬D[(∂Q∂x+∂Q∂z∂z∂x+∂R∂x∂z∂y+∂R∂z∂z∂x∂z∂y+R∂2z∂x ∂y)−(∂P∂y+∂P∂z∂z∂y+∂R∂y∂z∂x+∂R∂z∂z∂y∂z∂x+R∂2z∂y ∂x)] dx dy.\begin{align*}\oint_C\mathbf{F}\cdot d\mathbf{r}&=\iint_D\Biggl[\Bigl(\frac{\partial Q}{\partial x}+\frac{\partial Q}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial R}{\partial x}\frac{\partial z}{\partial y}+\cancel{\frac{\partial R}{\partial z}\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}}+\cancel{R\frac{\partial^2 z}{\partial x\,\partial y}}\Bigr) \\&\qquad-\Bigl(\frac{\partial P}{\partial y}+\frac{\partial P}{\partial z}\frac{\partial z}{\partial y}+\frac{\partial R}{\partial y}\frac{\partial z}{\partial x}+\cancel{\frac{\partial R}{\partial z}\frac{\partial z}{\partial y}\frac{\partial z}{\partial x}}+\cancel{R\frac{\partial^2 z}{\partial y\,\partial x}}\Bigr)\Biggr]\,dx\,dy.\end{align*}
The two terms containing ∂R∂z\frac{\partial R}{\partial z} cancel. The two terms containing mixed second derivatives also cancel, since gg has continuous second partial derivatives and hence
∂2g∂x ∂y=∂2g∂y ∂x.\begin{equation*}\frac{\partial^2 g}{\partial x\,\partial y}=\frac{\partial^2 g}{\partial y\,\partial x}.\end{equation*}
The remaining six terms regroup into the integrand in the surface-integral formula above. Thus
∮CF⋅dr=∬S(∇×F)⋅n dS.\begin{equation*}\oint_C\mathbf{F}\cdot d\mathbf{r}=\iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS.\end{equation*}
This proves the theorem for the upward oriented graph SS.

References

  • [StewartEtAl2021]Stewart, James, Daniel K. Clegg, and Saleem Watson. Calculus: Early Transcendentals, Metric Edition. Ninth edition. Cengage, 2021.