Stokes' Theorem

Stokes' theorem is the statement that equates a vector field's circulation along a closed curve with the net curl over its bounded surface. On one side sits the flux of the curl through an oriented surface $S$ — a surface integral that sums the microscopic circulation of $\mathbf{F}$ over every patch of a membrane. On the other sits the circulation of $\mathbf{F}$ once around the rim $\partial S$ — a single line integral around the boundary loop. The theorem asserts these are equal for every $C^1$ field and every reasonable capping surface: $\iint_S(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ . The bridge between the two is the curl, which is defined as circulation per unit area. Pave $S$ with tiny tiles; the rim circulation of each tile is the local curl times its area, and when the tiles are summed every interior edge is walked twice in opposite directions and cancels. Only the outer rim, whose edges belong to a single tile, is left standing — and that is the circulation around $\partial S$ . The same picture explains a remarkable fact: the answer depends only on the boundary loop, not on how the membrane bulges. This surface-independence is the integral shadow of the differential identity $\nabla\cdot(\nabla\times\mathbf{F})=0$ . Seen from higher ground, this is one face of a single master statement, $\int_M\mathrm{d}\omega=\int_{\partial M}\omega$ , the generalized Stokes' theorem of differential forms: choosing the dimension and degree recovers the fundamental theorem of calculus, Green's theorem, the curl theorem here, and the divergence theorem — four results that turn out to be one. The classical version is also the machinery that converts Maxwell's differential curl equations into their integral Ampère and Faraday forms.

Key takeaways

Stokes' theorem equates the flux of the curl through an oriented surface to the circulation of the field around its boundary: $\iint_S(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ .
The mechanism is edge cancellation: curl is circulation per unit area, tile rims cancel on shared interior edges, and only the outer rim $\partial S$ survives.
The result depends only on the boundary curve, not on the capping surface — the integral form of the identity $\nabla\cdot(\nabla\times\mathbf{F})=0$ .
Orientations of $S$ and $\partial S$ are locked by the right-hand rule; reversing either flips a sign, so consistency is not optional.
It is the curl-degree member of the family $\int_M\mathrm{d}\omega=\int_{\partial M}\omega$ , alongside Green's theorem (flat case) and the divergence theorem (3D flux).
Applied to Maxwell's curl equations it yields Faraday's EMF law and the Ampère–Maxwell law; applied to velocity it yields the circulation–vorticity relation of fluid dynamics.
Key idea: microscopic circulation measured by $\nabla\times\mathbf{F}$ accumulates additively over a surface, interior edges cancel pairwise, and the net is the circulation around the boundary — so flux of curl equals boundary circulation and depends only on $\partial S$ .

The core insight

Microscopic circulation, measured by $\nabla\times\mathbf{F}$ , accumulates additively over a surface; interior edges cancel pairwise, leaving the net circulation around the boundary. Hence flux of curl equals boundary circulation, and the result depends only on $\partial S$ .

Where this lives today

Stokes' theorem is a daily tool wherever circulation and flux meet. In computational electromagnetics the finite-element and finite-difference methods that design antennas, chips and MRI coils are built on discrete exterior calculus — Whitney forms and the Yee lattice encode Stokes' theorem exactly on a mesh, so the discretized Faraday and Ampère laws conserve flux to machine precision by construction. In superconductivity, integrating around a loop threading a ring turns the circulation of the vector potential into the enclosed magnetic flux, which quantum mechanics forces into units of $h/2e$ — the flux quantization SQUID magnetometers exploit to sense fields a hundred billion times weaker than Earth's. The same boundary-versus-interior bookkeeping is the Aharonov–Bohm phase $\oint\mathbf{A}\cdot\mathrm{d}\mathbf{r}$ , picked up around a region the particle never enters. In aerodynamics, Kelvin's circulation theorem and the Kutta–Joukowski law express a wing's lift as the circulation around it, equal by Stokes to the vorticity flux of the shed wake. And across modern physics the generalized form $\int_M\mathrm{d}\omega=\int_{\partial M}\omega$ underlies gauge theory, topological invariants, and the index theorems tying a space's geometry to the equations living on it.

The physical picture

Think of the curl as a dense field of microscopic paddle-wheels, one at every point, each spinning at a rate set by $(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}$ . To measure the total twist a field delivers to a membrane stretched across a wire loop, you could tally every paddle-wheel on the membrane (the surface integral) or walk once around the wire adding up how hard the field pushes along your path (the line integral). Stokes' theorem says these tallies are the same number. The reason is combinatorial: when you build the surface tally out of tiny cells, adjacent cells share edges, and along every shared edge the two neighbours push in opposite directions and delete each other. What cannot delete is the outermost boundary — and that leftover is exactly the walk around the rim.

Figure

A surface paved with cells — each cell's rim circulation equals curl times area, and every interior edge is shared by two neighbours and walked in opposite senses, so it cancels; only the outer rim survives, giving Σᵢ ∮ over Cᵢ of F·dr = ∮ over ∂S of F·dr

The blue paddle-wheels above all turn the same way, forcing neighbouring cells to disagree on their shared wall: one walks that edge upward, the other downward, and the two contributions cancel. That is the whole engine of the theorem, and it also makes visible why the cap is irrelevant — two membranes spanning the same loop together enclose a solid region, and the flux of curl out of a closed region is zero, so both carry the same flux. The second figure puts real numbers on the equality using the rotational field $\mathbf{F}=(-y,x,0)$ over a hemispherical cap, where both sides land on $2\pi$ .

Figure

Stokes checked on the upper unit hemisphere for F = (−y, x, 0): the curl is the constant vector ∇×F = (0, 0, 2), so the flux ∬ over S of (∇×F)·dS = 2 × (area of the unit disk) = 2 × π = 2π, while the boundary circulation ∮ over ∂S of F·dr = ∫ from 0 to 2π of dt = 2π — both sides equal 2π, and the flat disk gives the same value

The core equations

The headline identity is the classical curl theorem: the flux of $\nabla\times\mathbf{F}$ through an oriented surface equals the circulation of $\mathbf{F}$ around its boundary. Everything below either specializes this, proves it, or applies it. The orientation of $S$ (which way $\hat{\mathbf{n}}$ points) and the traversal direction of $\partial S$ must be tied by the right-hand rule for the signs to match.

\iint_S (\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r} \qquad (1)

where: $S$ — oriented piecewise-smooth surface; $\partial S$ — its boundary curve, traversed per the right-hand rule; $\mathbf{F}$ — a $C^1$ vector field on a neighbourhood of $S$ ; $\nabla\times\mathbf{F}$ — the curl, the local circulation density; $\mathrm{d}\mathbf{S}=\hat{\mathbf{n}}\,\mathrm{d}S$ — the vector area element; $\mathrm{d}\mathbf{r}$ — the directed line element. Reading: sum the microscopic twist over the membrane on the left; walk the rim once on the right; they agree. To evaluate the left side you need the oriented area element of a parametrized surface $\mathbf{r}(u,v)$ : the cross product of the two tangent vectors gives both the parallelogram's area and, through its direction, the normal that fixes the orientation.

\mathrm{d}\mathbf{S}=\left(\frac{\partial\mathbf{r}}{\partial u}\times\frac{\partial\mathbf{r}}{\partial v}\right)\mathrm{d}u\,\mathrm{d}v,\qquad \hat{\mathbf{n}}=\frac{\partial_u\mathbf{r}\times\partial_v\mathbf{r}}{|\partial_u\mathbf{r}\times\partial_v\mathbf{r}|} \qquad (2)

where: $\mathbf{r}(u,v)$ — a parametrization of $S$ ; $u,v$ — surface coordinates; $\partial_u\mathbf{r},\partial_v\mathbf{r}$ — tangent vectors along the coordinate curves. Reading: the magnitude $|\partial_u\mathbf{r}\times\partial_v\mathbf{r}|$ is the area density and the direction is the outward normal — swapping $u\leftrightarrow v$ flips $\hat{\mathbf{n}}$ and the orientation. The right-hand side is built from the curl, best understood as the very quantity the theorem globalizes: circulation per unit area in the infinitesimal limit. This is the definition that makes the tiling proof work.

(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}=\lim_{A\to0}\frac{1}{A}\oint_{C}\mathbf{F}\cdot\mathrm{d}\mathbf{r} \qquad (3)

where: $A$ — area of a small patch bounded by the loop $C$ ; $C$ — the patch boundary, oriented by the right-hand rule about $\hat{\mathbf{n}}$ . Reading: the component of curl along $\hat{\mathbf{n}}$ is the circulation of $\mathbf{F}$ around a shrinking loop divided by the loop's area — Stokes' theorem in the small. A curl field is automatically divergence-free — the one-line identity that makes the flux in $(1)$ independent of the capping surface, since any two caps sharing a boundary differ by a closed surface through which the flux of a curl must vanish.

\nabla\cdot(\nabla\times\mathbf{F})=0 \qquad (4)

where: $\nabla\cdot$ — the divergence operator. Reading: written in components $\partial_i\varepsilon_{ijk}\partial_j F_k$ , the symmetric mixed partial $\partial_i\partial_j$ contracts against the antisymmetric $\varepsilon_{ijk}$ and vanishes identically — so the flux of any curl through a closed surface is zero. Lifting to differential forms, all four integral theorems collapse into one statement. With $\omega$ a $1$ -form on a surface and $\mathrm{d}\omega$ its exterior derivative (the curl in disguise), one recovers $(1)$ .

\int_M \mathrm{d}\omega=\int_{\partial M}\omega \qquad (5)

where: $\omega$ — a differential $k$ -form; $\mathrm{d}\omega$ — its exterior derivative, a $(k{+}1)$ -form; $M$ — an oriented $(k{+}1)$ -manifold with boundary $\partial M$ . Reading: integrating the derivative over a region equals integrating the form over its boundary — the fundamental theorem of calculus ( $k=0$ ), Green's ( $M$ planar), classical Stokes' ( $M$ a surface), and Gauss' ( $M$ a solid) are all one line. Feeding Maxwell's differential curl equations through $(1)$ produces their integral forms — the Ampère–Maxwell law from $\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\,\partial_t\mathbf{E}$ , and Faraday's law from $\nabla\times\mathbf{E}=-\partial_t\mathbf{B}$ .

\oint_{\partial S}\mathbf{B}\cdot\mathrm{d}\mathbf{r}=\mu_0\iint_S\mathbf{J}\cdot\mathrm{d}\mathbf{S}+\mu_0\varepsilon_0\frac{\mathrm{d}}{\mathrm{d}t}\iint_S\mathbf{E}\cdot\mathrm{d}\mathbf{S},\qquad \oint_{\partial S}\mathbf{E}\cdot\mathrm{d}\mathbf{r}=-\frac{\mathrm{d}}{\mathrm{d}t}\iint_S\mathbf{B}\cdot\mathrm{d}\mathbf{S} \qquad (6)

where: $\mathbf{B},\mathbf{E}$ — magnetic and electric fields; $\mathbf{J}$ — current density; $\mu_0,\varepsilon_0$ — vacuum permeability and permittivity; $\Phi_B=\iint_S\mathbf{B}\cdot\mathrm{d}\mathbf{S}$ — magnetic flux. Reading: the circulation of $\mathbf{B}$ counts the enclosed current plus the displacement term; the circulation of $\mathbf{E}$ is minus the rate of change of enclosed flux — the induced EMF.

Derivation AFrom local circulation to the global theorem

The goal is to build Stokes' theorem from the ground up: subdivide the capping surface into tiny cells, replace each cell's rim circulation by the local curl times the cell area using the definition in equation $(3)$ , and watch the shared interior edges annihilate in pairs so only the outer rim survives. This turns the microscopic definition of curl into a global statement relating the surface to its boundary, and makes the cancellation mechanism — the same one behind Green's and the divergence theorems — explicit.

Step 1. Cover $S$ with $N$ small cells $S_i$ , each bounded by a loop $C_i$ oriented consistently by the right-hand rule about $\hat{\mathbf{n}}_i$ . By equation $(3)$ , each cell contributes $\oint_{C_i}\mathbf{F}\cdot\mathrm{d}\mathbf{r}\approx(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}_i\,\Delta S_i$ . Why: curl is circulation per unit area, so to leading order the rim integral of a cell is fixed by the curl there.

Step 2. Sum over all cells: $\sum_i\oint_{C_i}\mathbf{F}\cdot\mathrm{d}\mathbf{r}\approx\sum_i(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}_i\,\Delta S_i$ , a Riemann sum converging to $\iint_S(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}$ as the mesh refines. Why: adding the cell fluxes is how a surface integral is assembled in the limit of vanishing cell size.

Step 3. Now read the left sum edge by edge. Every interior edge is shared by two adjacent cells, and consistent orientation forces the neighbours to traverse it in opposite directions, so their $\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ contributions cancel. Why: the same field along the same edge, integrated once each way, sums to zero — the combinatorial heart of the theorem.

Step 4. Only edges on the outer rim $\partial S$ belong to a single cell and survive, summing to $\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ ; so $\sum_i\oint_{C_i}\mathbf{F}\cdot\mathrm{d}\mathbf{r}=\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ . Equating with Step 2 and refining the mesh — where the leading-order approximation becomes exact for a $C^1$ field — gives equation $(1)$ . Why: the rim has no outer neighbour to delete it, and the per-cell errors vanish faster than the cell count grows.

Go deeper · Derivation B · 1 minReduction to Green's theorem on a coordinate patch

Derivation BReduction to Green's theorem on a coordinate patch

The goal is to make the tiling argument rigorous on a single smooth patch by writing the surface as a graph $z=h(x,y)$ over a planar region $D$ , projecting every integral to the plane, and invoking Green's theorem, taken as established for planar regions. We track only the $P$ -component of $\mathbf{F}=(P,Q,R)$ ; the $Q$ and $R$ parts follow identically, and summing the three reproduces the full theorem on the patch, after which general surfaces are covered by gluing patches with the previous derivation's cancellation.

Step 1. The upward-oriented area element of the graph is $\mathrm{d}\mathbf{S}=(-h_x,-h_y,1)\,\mathrm{d}x\,\mathrm{d}y$ , obtained from equation $(2)$ with $\mathbf{r}=(x,y,h(x,y))$ . Why: the tangent vectors are $(1,0,h_x)$ and $(0,1,h_y)$ , whose cross product is exactly $(-h_x,-h_y,1)$ — the gradient of $z-h$ , hence normal to the surface.

Step 2. The curl is $\nabla\times\mathbf{F}=(R_y-Q_z,\;P_z-R_x,\;Q_x-P_y)$ . Keeping only the pieces containing $P$ — $P_z$ in the $y$ -slot and $-P_y$ in the $z$ -slot — and dotting with $\mathrm{d}\mathbf{S}$ gives $[(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}]_P=(-h_y P_z-P_y)\,\mathrm{d}x\,\mathrm{d}y$ . Why: isolating the $P$ -terms lets us match them against the $P\,\mathrm{d}x$ part of the line integral independently.

Step 3. On $\partial S$ we have $z=h(x,y)$ and $\mathrm{d}x$ is common to the curve and its shadow $\partial D$ , so $\oint_{\partial S}P\,\mathrm{d}x=\oint_{\partial D}P(x,y,h(x,y))\,\mathrm{d}x$ . Why: the projection of $\partial S$ is exactly $\partial D$ , and only the $x$ -displacement matters here.

Step 4. Apply planar Green's theorem to $\tilde P\equiv P(x,y,h(x,y))$ : $\oint_{\partial D}\tilde P\,\mathrm{d}x=-\iint_D\partial_y\tilde P\,\mathrm{d}x\,\mathrm{d}y$ , with $\partial_y\tilde P=P_y+P_z h_y$ by the chain rule. Why: differentiating through the height $h$ is where the cap's shape enters — the extra $P_z h_y$ is absent for a flat surface.

Step 5. Steps 3 and 4 give $\oint_{\partial S}P\,\mathrm{d}x=-\iint_D(P_y+P_z h_y)\,\mathrm{d}x\,\mathrm{d}y$ , exactly the $P$ -part of Step 2. The $Q$ and $R$ parts follow by symmetry, so summing proves equation $(1)$ on the patch. Why: the two sides agree term by term, and the theorem is the sum of three identical computations.

Go deeper · Derivation C · 1 minSurface independence from the vanishing of the curl's divergence

Derivation CSurface independence from the vanishing of the curl's divergence

The goal is to establish the property that makes Faraday's and Ampère's laws well posed: the flux of a curl through $S$ depends only on the boundary $\partial S$ , never on which membrane you stretch across it. The engine is equation $(4)$ , $\nabla\cdot(\nabla\times\mathbf{F})=0$ ; combined with the divergence theorem it forces any two caps sharing a rim to carry identical flux — which is why a physicist may deform a surface freely to whatever shape makes an integral easy.

Step 1. Take two surfaces $S_1$ and $S_2$ sharing the same boundary curve $C$ . Reversing $S_2$ and gluing it to $S_1$ produces a single closed surface $\Sigma=S_1\cup(-S_2)$ sealing a solid region $V$ . Why: a common rim plus one flipped orientation is what closes the two caps into a boundaryless surface.

Step 2. Apply the divergence theorem to $\nabla\times\mathbf{F}$ over $V$ : $\oiint_\Sigma(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=\iiint_V\nabla\cdot(\nabla\times\mathbf{F})\,\mathrm{d}V$ . Why: Gauss converts the closed-surface flux into a volume integral of the divergence.

Step 3. By equation $(4)$ the integrand vanishes identically, so the closed flux is zero. Why: the divergence of any curl is zero everywhere, killing the volume integral regardless of field or region.

Step 4. Split $\Sigma$ back apart: the flux through $S_1$ minus that through $S_2$ is zero, hence $\iint_{S_1}(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=\iint_{S_2}(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}$ . Why: with $S_2$ reversed inside $\Sigma$ , the closed integral is the difference of the two outward fluxes, forcing their equality — only $C$ matters.

Limiting cases

Stokes' theorem contains several familiar results as special geometries or special fields. Each row is obtained by restricting the surface, the normal, or the field in equation $(1)$ .

Case	Result	Meaning
Flat surface in the $xy$ -plane	$\oint_C (P\,\mathrm{d}x+Q\,\mathrm{d}y)=\iint_S\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\mathrm{d}x\,\mathrm{d}y$	With $\hat{\mathbf{n}}=\hat{\mathbf{z}}$ only the $z$ -component of curl survives — this is Green's theorem in the plane.
Irrotational field	$\nabla\times\mathbf{F}=\mathbf{0}\Rightarrow\oint_C\mathbf{F}\cdot\mathrm{d}\mathbf{r}=0$	On a simply connected domain every loop has zero circulation, so $\mathbf{F}=\nabla\phi$ is conservative and line integrals are path-independent.
Closed surface	$\oiint_S(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=0$	A closed surface has no boundary ( $\partial S=\varnothing$ ), so the flux of any curl vanishes — the integral form of $\nabla\cdot(\nabla\times\mathbf{F})=0$ .
Shrinking loop	$(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}=\lim_{\Delta S\to0}\dfrac{1}{\Delta S}\oint_{\partial(\Delta S)}\mathbf{F}\cdot\mathrm{d}\mathbf{r}$	In the infinitesimal limit the theorem collapses back to the definition of curl as circulation per unit area.

Common trapStokes' theorem "always" makes a curl-free field circulate to zero

It does not — the theorem requires $\mathbf{F}$ to be $C^1$ on a neighbourhood of every point of the surface, and a hidden singularity breaks that hypothesis dramatically. The classic trap is the point-vortex $\mathbf{F}=(-y,x,0)/(x^2+y^2)$ , whose curl is zero everywhere it is defined, yet $\oint\mathbf{F}\cdot\mathrm{d}\mathbf{r}=2\pi$ around any loop encircling the origin. No contradiction: any cap of such a loop is pierced by the singular line at the origin, so $\mathbf{F}$ is not $C^1$ there and the theorem simply does not apply. The same subtlety underlies a current-carrying wire, the Aharonov–Bohm effect, and every non-simply-connected domain. The lesson: "curl-free" implies "zero circulation" only on a simply connected region, and orientation must stay consistent — the right-hand rule ties $\hat{\mathbf{n}}$ to the traversal of $\partial S$ , and reversing one flips a sign.

Worked examples

1 · The rotational field on a hemisphere (confidence builder)

Let $\mathbf{F}=(-y,x,0)$ and let $S$ be the upper unit hemisphere $x^2+y^2+z^2=1$ , $z\ge0$ , oriented with the outward (upward) normal. Its boundary $\partial S$ is the unit circle in the $xy$ -plane, traversed counterclockwise. Verify Stokes' theorem, then confirm surface-independence.

Step 1. Compute the curl. For $\mathbf{F}=(-y,x,0)$ the only surviving component is the $z$ -component, $\partial_x(x)-\partial_y(-y)=1-(-1)=2$ , so $\nabla\times\mathbf{F}=(0,0,2)$ , a constant vector pointing straight up.

Step 2. The surface integral. Because the curl is constant and along $\hat{\mathbf{z}}$ , its flux through the dome equals $2$ times the projected area of the dome onto the $xy$ -plane. That projection is the unit disk $D$ , of area $\pi$ , so $\iint_S(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=2\iint_D\mathrm{d}x\,\mathrm{d}y=2\pi$ . The flux of the curl is $2\pi$ .

Step 3. The line integral. Parametrize $\partial S$ as $(\cos t,\sin t,0)$ for $t\in[0,2\pi]$ , so $\mathrm{d}\mathbf{r}=(-\sin t,\cos t,0)\,\mathrm{d}t$ and $\mathbf{F}=(-\sin t,\cos t,0)$ on the circle. Then $\mathbf{F}\cdot\mathrm{d}\mathbf{r}=(\sin^2 t+\cos^2 t)\,\mathrm{d}t=\mathrm{d}t$ , and $\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r}=\int_0^{2\pi}\mathrm{d}t=2\pi$ . The circulation is $2\pi$ — equal to the flux, as Stokes' theorem demands.

Step 4. Surface-independence. Replace the dome with the flat disk $D$ (same boundary, normal $\hat{\mathbf{z}}$ ): the flux is again $2\times\text{area}=2\pi$ . Replace it with the paraboloid $z=1-x^2-y^2$ over the same circle: its projection is still the unit disk, so the flux is once more $2\pi$ . Every cap sharing the unit circle gives $2\pi$ — the answer is a property of the boundary alone.

2 · Ampère's law and why the displacement term is not optional

A long straight wire carries a steady current $I=10\ \mathrm{A}$ . Use Stokes' theorem to find the magnetic field at radius $r=5\ \mathrm{cm}$ , then examine what happens when the wire feeds a charging capacitor.

Step 1. Apply equation $(6)$ with no changing flux to a circular Amperian loop of radius $r$ coaxial with the wire: $\oint_{\partial S}\mathbf{B}\cdot\mathrm{d}\mathbf{r}=\mu_0 I_{\text{enc}}$ . By symmetry $\mathbf{B}$ is tangential and constant in magnitude on the loop, so the left side is $B\,(2\pi r)$ .

Step 2. Solve for the field: $B=\dfrac{\mu_0 I}{2\pi r}=\dfrac{(4\pi\times10^{-7})(10)}{2\pi(0.05)}=\dfrac{4\pi\times10^{-6}}{0.1\pi}=4\times10^{-5}\ \mathrm{T}$ . $B=40\ \mu\mathrm{T}$ at $5\ \mathrm{cm}$ — about the strength of Earth's field, and independent of which membrane caps the loop as long as the same current pierces it.

Step 3. Now insert a capacitor. Cap the loop with a flat disk cut by the wire — enclosed current $I$ ; cap it instead with a bulging surface slipping between the plates, where no conduction current flows — enclosed current $0$ . Without a second term the two caps disagree, contradicting surface-independence. Why: Stokes' theorem guarantees both caps give the same circulation, so the conduction-only Ampère law must be incomplete.

Step 4. The fix is Maxwell's displacement term $\mu_0\varepsilon_0\,\mathrm{d}_t\!\iint_S\mathbf{E}\cdot\mathrm{d}\mathbf{S}$ : between the plates the growing electric flux supplies exactly the missing $\mu_0 I$ . Surface-independence, demanded by Stokes' theorem, is the argument that forced the discovery of the displacement current — and hence electromagnetic waves.

Physics of the portrait

The theorem carries the name of the man who examined generations of Cambridge students on it, not the man who first wrote it down — a historical accident that says something about how vector calculus grew up inside nineteenth-century British physics.

Origins · 1 minGeorge Gabriel Stokes and how this was discovered

George Gabriel Stokes

1819–1903 · Cambridge mathematical physics; curl, circulation and the integral theorems of vector calculus

George Gabriel Stokes held the Lucasian Chair at Cambridge for over half a century and left his name across physics — Stokes' law of viscous drag, the Navier–Stokes equations, the Stokes parameters of polarization. Yet the theorem now universally called his was not his discovery. William Thomson, the future Lord Kelvin, stated the curl–circulation identity in a letter to Stokes dated 2 July 1850. Stokes set it as question eight on the Smith's Prize examination he administered at Cambridge in February 1854 — an exam whose candidates that year included the young James Clerk Maxwell, who would soon weave the result into the foundations of electromagnetism. Because every ambitious Cambridge mathematician met the problem through Stokes' examining, the attribution stuck. The deeper truth is that the classical theorem is one shadow of a far more general statement, $\int_M\mathrm{d}\omega=\int_{\partial M}\omega$ , unifying the fundamental theorem of calculus, Green's theorem, the divergence theorem and the curl theorem into a single line of exterior calculus — a synthesis that took another eighty years and the differential forms of Volterra, Cartan and de Rham to complete. Stokes gave the result a name and a home in physics; the mathematicians who followed showed it was really a theorem about the boundary of anything.

History

1850

— William Thomson (Kelvin) states the curl–circulation identity in a letter to Stokes

1854

— Stokes sets it as a Smith's Prize examination question at Cambridge, with Maxwell among the candidates

1889

— Volterra generalizes the integral theorems to higher-dimensional regions

1899

— Cartan's exterior calculus unifies grad, curl and div through the exterior derivative

1931

— de Rham's cohomology reads the master formula as a duality between forms and boundaries

Going further

Texts. Stone & Goldbart, Mathematics for Physics, §11 (vector calculus and the integral theorems) · Arfken & Weber, Mathematical Methods for Physicists, §3.8 (Stokes' theorem) · Hassani, Mathematical Physics, Ch. 15 (vector analysis) · Nakahara, Geometry, Topology and Physics, §5.4 (Stokes' theorem for forms) · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, §11.9 · Spivak, Calculus on Manifolds, Ch. 5 (the generalized theorem). Key checkpoints for the results in this article: both sides of the hemisphere example equal $2\pi$ and the projected disk area is $\pi$ ; the identity $\nabla\cdot(\nabla\times\mathbf{F})=0$ holds in components; the single-patch reduction turns on the chain rule $\partial_y\tilde P=P_y+P_z h_y$ and planar Green's theorem; and the Ampère result gives $B=\mu_0 I/2\pi r=40\ \mu\mathrm{T}$ for $I=10\ \mathrm{A}$ , $r=5\ \mathrm{cm}$ . The numbers $2\pi$ and $\pi$ are exact geometric values; the only physical constant invoked is $\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m/A}$ .

Prerequisites. Curl of a Vector Field · Line Integrals · Surface Integrals and Flux · The Del Operator and Vector Differential Identities · Green's Theorem in the Plane. Leads to. The Divergence (Gauss) Theorem · The Del Operator and Vector Differential Identities · Scalar and Vector Fields.

Ask the tutor: Want to see Stokes' theorem verified on a surface of your choice, or watch it collapse into Green's theorem, the divergence theorem, or a Maxwell equation? Ask and we will work it line by line.

Does it really not matter which surface I stretch across the loop?

For the flux of a curl, no — any two caps sharing the same boundary give the same answer, so choose whichever is easiest to integrate. The reason is $\nabla\cdot(\nabla\times\mathbf{F})=0$ : two caps with a common rim enclose a solid region, and the flux of a divergence-free field out of a closed region is zero, so the fluxes match. This is why the dome, the flat disk and the paraboloid all return $2\pi$ in the hemisphere example. The caveat: the field must be $C^1$ throughout every cap you consider; if a singularity or a current-carrying wire pierces one cap but not another, the caps stop being equivalent and you must track what threads them.

What is the difference between Stokes', Green's, and the divergence theorems?

They are three faces of one statement, $\int_M\mathrm{d}\omega=\int_{\partial M}\omega$ , distinguished by the dimension of the region. Green's theorem relates a line integral around a planar loop to a double integral over the flat region it bounds — the two-dimensional case. Classical Stokes' lifts that loop into three dimensions: the region becomes a curved surface and the flat curl $\partial_x Q-\partial_y P$ becomes $\nabla\times\mathbf{F}$ dotted with the normal. The divergence theorem is the sibling where the region is a solid volume and the boundary a closed surface, relating $\iiint\nabla\cdot\mathbf{F}$ to the outward flux. Green's is the flat special case of Stokes', and all three are the generalized theorem with $M$ of dimension two, two and three.

How can a field with zero curl have a nonzero circulation?

Only when the domain is not simply connected — when it has a hole the loop can encircle. The standard example is the vortex $\mathbf{F}=(-y,x,0)/(x^2+y^2)$ , whose curl is identically zero wherever it is defined, yet whose circulation around any loop enclosing the origin is $2\pi$ . The catch is that $\mathbf{F}$ blows up at the origin, so any cap of such a loop is punctured by the singularity and Stokes' theorem cannot be applied. Restrict to loops that avoid the hole and the circulation is indeed zero. Physically this is the mathematics behind a current-carrying wire and the Aharonov–Bohm effect: the action lives in a region the loop surrounds but does not enter.

Why can't I ignore the right-hand rule as a mere convention?

Because the two sides of the theorem carry signs and the right-hand rule keeps them consistent. The surface integral's sign depends on which way $\hat{\mathbf{n}}$ points; the line integral's sign depends on which way you walk the boundary. The rule — curl the right hand's fingers along $\partial S$ , the thumb gives $\hat{\mathbf{n}}$ — locks the two choices so both integrals flip in step. Pick the normal and the traversal independently and you can make the two sides disagree by an overall sign, so the theorem appears to fail. Reversing both orientations leaves the equality intact; reversing just one breaks it. A convention, but an unavoidable one.

Is the differential-forms version really the same theorem?

Yes, in a language that makes its universality obvious. In differential forms a vector field $\mathbf{F}$ corresponds to a $1$ -form $\omega=F_x\,\mathrm{d}x+F_y\,\mathrm{d}y+F_z\,\mathrm{d}z$ , and its exterior derivative $\mathrm{d}\omega$ is the curl repackaged as a $2$ -form. The abstract statement $\int_M\mathrm{d}\omega=\int_{\partial M}\omega$ , with $M$ a surface, unpacks term by term into $\iint_S(\nabla\times\mathbf{F})\cdot\mathrm{d}\mathbf{S}=\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ . The power of the abstraction is that the same line, with $M$ of other dimensions, also gives the fundamental theorem of calculus, Green's and the divergence theorems — and generalizes to any dimension, where no cross-product curl exists.

What does "flux of curl" actually mean physically?

The curl at a point measures the local tendency of the field to spin a tiny paddle-wheel; its component along a normal is the circulation per unit area of a shrinking loop about that axis. The flux of curl through a surface adds up this local rotation over the whole membrane, weighting each patch by how well its spin axis lines up with the normal. Stokes' theorem says this accumulated microscopic rotation equals the net push you feel dragging a test particle once around the rim. In fluids the curl of the velocity is the vorticity, and the theorem reads: the total vorticity threading a loop equals the circulation around it — the heart of Kelvin's circulation theorem and the theory of lift.