Unit 5, lesson 41

Stokes’ Theorem

How the circulation of a field around the edge of a surface equals the flux of its curl through the surface, with the orientation, smoothness and domain conditions that make this true.

  • Theory part 81
  • Practice part 82
  • The Theorems of Stokes and Divergence

Hold a soap film on a bent wire loop and blow across it. How much the air swirls around the wire can be found in two ways, by walking along the wire or by adding up the tiny swirls on the film. Stokes’ theorem says the two answers agree, provided the film and the wire are oriented to match and the field has no bad points on the film. It is Green’s theorem lifted off the flat plane onto curved surfaces.

Key terms

Oriented piecewise smooth surface

Definition A piecewise smooth surface SS is a finite union of smooth pieces S1,…,SkS_1, \dots, S_k that meet only along their edges. Each piece has a regular parametrization r\mathbf{r} on a closed bounded region DD with piecewise smooth boundary, meaning r\mathbf{r} has continuous first partial derivatives on DD, is one-to-one on the interior of DD, and has

ru×rv≠0\mathbf{r}_u \times \mathbf{r}_v \ne \mathbf{0}

on the interior of DD. The surface is oriented when a continuous unit normal n\mathbf{n} is chosen on each piece, consistently across shared edges. Precisely, on every shared edge the two pieces must induce opposite directions, in the sense of the next section. Throughout, SS is bounded and contains its boundary ∂S\partial S, which is made of finitely many simple closed piecewise smooth curves.

In plain words A few smooth patches stitched together along seams, with one side chosen as the front and kept as the front across every seam.

Example A cup made of the cylinder side x2+y2=1x^2 + y^2 = 1, 0≤z≤10 \le z \le 1, and the bottom disk z=0z = 0, x2+y2≤1x^2 + y^2 \le 1, is piecewise smooth with one seam, the bottom circle. Its boundary is the top rim. One orientation uses the normals pointing out of the cup, which are ⟨x,y,0⟩\langle x, y, 0 \rangle on the side and ⟨0,0,−1⟩\langle 0, 0, -1 \rangle on the bottom.

Induced boundary orientation

Definition Let SS be oriented by n\mathbf{n}. The induced (positive) orientation of ∂S\partial S is the direction of travel such that, walking along ∂S\partial S upright with your head pointing along n\mathbf{n}, the surface is on your left. Precisely, at a smooth point of ∂S\partial S with unit tangent T\mathbf{T} in the direction of travel, the vector n×T\mathbf{n} \times \mathbf{T} is tangent to SS and points into SS, away from the edge.

In plain words Curl the fingers of your right hand along the boundary and your thumb should point along the chosen normal. It is the same rule as counterclockwise for a flat region in the plane with normal k\mathbf{k}.

Example For the upper hemisphere x2+y2+z2=1x^2 + y^2 + z^2 = 1, z≥0z \ge 0, with outward normal, the boundary is the unit circle in the xyxy-plane. At (1,0,0)(1, 0, 0) the normal is i\mathbf{i}. Try T=j\mathbf{T} = \mathbf{j}, which gives

n×T=i×j=k,\mathbf{n} \times \mathbf{T} = \mathbf{i} \times \mathbf{j} = \mathbf{k},

pointing up into the hemisphere. So the circle runs counterclockwise when seen from above.

Compatible normal and boundary direction

Definition A normal n\mathbf{n} on SS and a direction along ∂S\partial S are compatible when the direction is the one induced by n\mathbf{n}. Reversing either one alone breaks compatibility. Reversing both keeps it. If ∂S\partial S has several closed curves, each one gets its own induced direction, and these need not look alike from a fixed viewpoint.

In plain words Every boundary loop is walked with the surface on your left. For a surface with a hole, that makes the inner loop go the opposite way to the outer one.

Example The flat annulus 1≤x2+y2≤41 \le x^2 + y^2 \le 4 in the plane z=0z = 0 with n=k\mathbf{n} = \mathbf{k} has an outer circle traversed counterclockwise and an inner circle traversed clockwise, as seen from above. For F=⟨−y,x,0⟩\mathbf{F} = \langle -y, x, 0 \rangle the circulations are 8π8\pi on the outer circle and −2π-2\pi on the inner one, a total of 6π6\pi. The curl is ⟨0,0,2⟩\langle 0, 0, 2 \rangle, so the curl flux is 22 times the area 3π3\pi, which is also 6π6\pi.

Circulation and curl flux

Definition For an oriented closed piecewise smooth curve CC and a continuous field F\mathbf{F}, the circulation of F\mathbf{F} around CC is

∮CF⋅dr=∮CF⋅T ds.\oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C \mathbf{F} \cdot \mathbf{T} \, ds.

For an oriented surface SS and a field with continuous first partial derivatives, the curl flux is

∬S(∇×F)⋅n dS.\iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS.

In plain words Circulation measures how much the field pushes you along the loop. Curl flux adds up the local spinning of the field, measured around the normal at each point of the surface.

Example Take F=⟨−y,x,0⟩\mathbf{F} = \langle -y, x, 0 \rangle and the unit circle r(t)=⟨cos⁡t,sin⁡t,0⟩\mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle, 0≤t≤2π0 \le t \le 2\pi. Then

F⋅r′(t)=sin⁡2t+cos⁡2t=1,\mathbf{F} \cdot \mathbf{r}'(t) = \sin^2 t + \cos^2 t = 1,

so the circulation is 2π2\pi. The curl is ⟨0,0,2⟩\langle 0, 0, 2 \rangle, and its flux through the unit disk with n=k\mathbf{n} = \mathbf{k} is 2⋅π=2π2 \cdot \pi = 2\pi.

Field of class C1 near the surface

Definition A vector field F=⟨P,Q,R⟩\mathbf{F} = \langle P, Q, R \rangle is of class C1C^1 on a neighborhood of SS if there is an open set UU in space containing all of SS, its boundary included, such that PP, QQ and RR have continuous first partial derivatives on UU.

In plain words The field must be smooth not just on the surface but a little beyond it in every direction, with no blowups, jumps or missing points anywhere on SS.

Example The field

F=⟨−yx2+y2, xx2+y2, 0⟩\mathbf{F} = \left\langle \frac{-y}{x^2 + y^2},\ \frac{x}{x^2 + y^2},\ 0 \right\rangle

is C1C^1 everywhere except on the zz-axis, where it is undefined, and its curl is 0\mathbf{0} wherever it is defined. Around the unit circle counterclockwise its circulation is 2π2\pi, not 00. There is no contradiction, because every surface with that circle as boundary meets the zz-axis, where F\mathbf{F} is undefined. A surface that avoids the axis works fine. The cylinder side x2+y2=1x^2 + y^2 = 1, 0≤z≤10 \le z \le 1, with outward normal, has two boundary circles. The bottom one runs counterclockwise and the top one clockwise seen from above, so the circulations are 2π2\pi and −2π-2\pi, and their sum 00 equals the curl flux.

Stokes’ theorem

Definition Let SS be a bounded oriented piecewise smooth surface whose boundary ∂S\partial S consists of finitely many simple closed piecewise smooth curves, given the induced orientation. Let F\mathbf{F} be a vector field of class C1C^1 on an open set containing SS. Then

∮∂SF⋅dr=∬S(∇×F)⋅n dS.\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS.

When ∂S\partial S has several curves, the left side is the sum of the circulations around all of them.

In plain words Total circulation around the edge equals total spin through the surface. Inside the surface, neighboring small loops cancel along shared edges, and only the outer edge survives.

Example Let F=⟨2z,x,y2⟩\mathbf{F} = \langle 2z, x, y^2 \rangle and let SS be the part of z=4−x2−y2z = 4 - x^2 - y^2 with z≥0z \ge 0, oriented upward. The boundary is the circle of radius 22 in the xyxy-plane, counterclockwise from above. Along r(t)=⟨2cos⁡t,2sin⁡t,0⟩\mathbf{r}(t) = \langle 2\cos t, 2\sin t, 0 \rangle we have z=0z = 0, so

∮∂SF⋅dr=∫02π4cos⁡2t dt=4π.\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} 4\cos^2 t \, dt = 4\pi.

On the surface side the curl is ⟨2y,2,1⟩\langle 2y, 2, 1 \rangle, the upward normal vector is ⟨2x,2y,1⟩\langle 2x, 2y, 1 \rangle, and

∬D(4xy+4y+1) dA=0+0+area(D)=4π,\begin{aligned} &\iint_D (4xy + 4y + 1) \, dA \\[0.4em] &= 0 + 0 + \text{area}(D) = 4\pi, \end{aligned}

where DD is the disk of radius 22 and the odd terms vanish by symmetry. Both sides give 4π4\pi.

Choosing a convenient surface

Definition If S1S_1 and S2S_2 are two oriented piecewise smooth surfaces with the same boundary curves, carrying the same induced orientation, and F\mathbf{F} is C1C^1 on an open set containing both, then

∬S1(∇×F)⋅dS=∬S2(∇×F)⋅dS,\begin{aligned} &\iint_{S_1} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \\[0.4em] &= \iint_{S_2} (\nabla \times \mathbf{F}) \cdot d\mathbf{S}, \end{aligned}

since both equal the circulation around the common boundary.

In plain words Curl flux depends only on the edge, so you may swap a hard surface for an easy one, as long as the orientations match and the field is smooth on the new surface too.

Example In the previous example replace the paraboloid by the flat disk x2+y2≤4x^2 + y^2 \le 4, z=0z = 0, with n=k\mathbf{n} = \mathbf{k}, which induces the same counterclockwise circle. Then (∇×F)⋅k=1(\nabla \times \mathbf{F}) \cdot \mathbf{k} = 1 and the curl flux is the area of the disk, 4π4\pi, with no parametrization needed.

Check yourself

  1. Let F=⟨z,x,y⟩\mathbf{F} = \langle z, x, y \rangle and let SS be the upper unit hemisphere with outward normal. Find the flux of ∇×F\nabla \times \mathbf{F} through SS.
Show the answer

By Stokes’ theorem it equals the circulation around the unit circle, counterclockwise from above. With r(t)=⟨cos⁡t,sin⁡t,0⟩\mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle the integrand is

z x′+x y′+y z′=cos⁡2t,z \, x' + x \, y' + y \, z' = \cos^2 t,

so the answer is π\pi. As a check, the curl is ⟨1,1,1⟩\langle 1, 1, 1 \rangle and its flux through the unit disk with normal k\mathbf{k} is π\pi.

  1. Orient the paraboloid z=x2+y2z = x^2 + y^2, z≤1z \le 1, with the normal pointing away from the zz-axis. In which direction, seen from above, is its boundary circle traversed, and what is the circulation of ⟨−y,x,0⟩\langle -y, x, 0 \rangle in that direction?
Show the answer

The normal away from the axis is a positive multiple of ⟨2x,2y,−1⟩\langle 2x, 2y, -1 \rangle, which points downward. A downward normal induces the clockwise direction seen from above, so the circulation is −2π-2\pi. The curl flux agrees, since ⟨0,0,2⟩⋅⟨2x,2y,−1⟩=−2\langle 0, 0, 2 \rangle \cdot \langle 2x, 2y, -1 \rangle = -2 integrated over the unit disk gives −2π-2\pi.

  1. The field F=⟨−yx2+y2,xx2+y2,0⟩\mathbf{F} = \left\langle \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}, 0 \right\rangle has zero curl wherever it is defined. Why can Stokes’ theorem not show that its circulation around the circle x2+y2=4x^2 + y^2 = 4, z=3z = 3, is zero?
Show the answer

Every surface bounded by that circle crosses the zz-axis, where F\mathbf{F} is not defined, so the field is not C1C^1 on an open set containing the surface. In fact the circulation counterclockwise is 2π2\pi.

How this lesson connects

It also uses Vector line integrals, curl, oriented surface flux, Green's theorem and regular parameterized surfaces.