Mathematics.

vector calculus

Stokes' Theorem

Calculus III25 minDifficulty8 out of 10

You should know: surface integrals, greens theorem

Overview

Stokes' theorem generalizes Green's theorem from a flat region in the plane to a curved surface in 3D space: it relates the circulation of a vector field around the boundary curve ∂S of an oriented surface S to the flux of the field's curl through S itself. It's the 3D statement that 'total spin on the boundary equals total local rotation summed over the surface.'

Interactive Graph

A vector field whose curl relates to a boundary integral

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Formal Definition

Definition

Let S be an oriented, piecewise-smooth surface bounded by a simple, closed, piecewise-smooth curve ∂S with orientation induced by S's normal (right-hand rule):

SFdr=S(×F)dS\oint_{\partial S} \mathbf F \cdot d\mathbf r = \iint_S (\nabla \times \mathbf F) \cdot d\mathbf S
Stokes' Theorem

Notation

NotationMeaning
S\partial SThe boundary curve of surface S, oriented consistently with S's chosen normal via the right-hand rule

Properties

Surface-independence

If S1,S2 share the same boundary S (with matching orientation), S1(×F)dS=S2(×F)dS\text{If } S_1, S_2 \text{ share the same boundary } \partial S \text{ (with matching orientation), } \iint_{S_1}(\nabla\times\mathbf F)\cdot d\mathbf S = \iint_{S_2}(\nabla\times\mathbf F)\cdot d\mathbf S

Condition: The flux of curl F depends only on the boundary curve, not the specific surface spanning it

Theorems

Theorem 1: Stokes' Theorem
SFdr=S(×F)dS for oriented surface S with compatibly oriented boundary S\oint_{\partial S} \mathbf F\cdot d\mathbf r = \iint_S(\nabla\times\mathbf F)\cdot d\mathbf S \text{ for oriented surface } S \text{ with compatibly oriented boundary } \partial S

Corollaries

Follows from Stokes' Theorem

Green’s theorem is the special case where S is a flat region in the xy-plane with upward normal, so the surface integral of curl reduces to the 2D scalar curl QxPy\text{Green's theorem is the special case where } S \text{ is a flat region in the } xy\text{-plane with upward normal, so the surface integral of curl reduces to the 2D scalar curl } \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}

Follows from Stokes' Theorem

If S is a closed surface (no boundary, S=), then S(×F)dS=0\text{If } S \text{ is a closed surface (no boundary, } \partial S = \emptyset\text{), then } \iint_S (\nabla\times\mathbf F)\cdot d\mathbf S = 0

Applications

Faraday's law of induction, ∮_C E·dr = −d/dt∬_S B·dS, is essentially Stokes' theorem applied to relate circulating electric field to changing magnetic flux.

Worked Examples

  1. First compute curl F.

    ×F=(00,00,1(1))=(0,0,2)\nabla \times \mathbf F = (0-0, 0-0, 1-(-1)) = (0,0,2)
  2. Compute the flux of curl F through S. Since curl F=(0,0,2) is constant, and the hemisphere's flux of a constant vertical field equals that of the flat disk it bounds (same boundary, by surface-independence), use the disk x²+y²≤1, z=0, normal=(0,0,1).

    S(×F)dS=disk(0,0,2)(0,0,1)dA=2π(1)2=2π\iint_S (\nabla\times\mathbf F)\cdot d\mathbf S = \iint_{\text{disk}} (0,0,2)\cdot(0,0,1)\, dA = 2 \cdot \pi(1)^2 = 2\pi
  3. Now compute the boundary line integral directly: ∂S is r(t)=(cos t, sin t, 0), t∈[0,2π].

    SFdr=02π(sint,cost,0)(sint,cost,0)dt=02π1dt=2π\oint_{\partial S} \mathbf F \cdot d\mathbf r = \int_0^{2\pi} (-\sin t, \cos t, 0)\cdot(-\sin t,\cos t,0)\,dt = \int_0^{2\pi} 1\, dt = 2\pi

Answer: Both sides equal 2π, verifying Stokes' theorem

Practice Problems

Difficulty 8/10

Use Stokes' theorem to evaluate ∬_S (∇×F)·dS where F=⟨z,x,y⟩ and S is the disk x²+y²≤4 in the plane z=0 (upward normal).

Difficulty 6/10

A vector field F has curl ∇×F = 0 everywhere (irrotational). Using Stokes' theorem, what is the circulation ∮_C F·dr around any closed loop C, and what does this imply physically?

Common Mistakes

Common Mistake

Assuming curl F for F=⟨z,x,y⟩ is zero without computing it carefully.

curl⟨z,x,y⟩ = (∂y/∂y − ∂x/∂z, ∂z/∂z − ∂y/∂x, ∂x/∂x − ∂z/∂y) = (1,1,1), NOT zero — a common error is assuming any 'cyclic-looking' field is curl-free.

Common Mistake

Forgetting that ∂S's orientation must be consistent with S's chosen normal via the right-hand rule.

If S's normal points 'up and out', ∂S must be traversed counterclockwise when viewed from the side the normal points toward — reversing this convention flips the sign of one side of the equation.

Quiz

Stokes' theorem equates the surface integral of curl F over S to:
The curl ∇×F of a vector field measures:

Historical Background

The theorem is named after Sir George Gabriel Stokes, who set it as an examination question at Cambridge in 1854, though it had been communicated to him earlier by Lord Kelvin (William Thomson) in an 1850 letter — the theorem was known to Kelvin before Stokes popularized it via the exam, a case (like several 'Stigler's law' examples in mathematics) of a result named for someone other than its original discoverer.

  1. 1850

    Kelvin communicates the result to Stokes in a letter

    Lord Kelvin

  2. 1854

    Stokes poses the theorem as a Smith's Prize examination question at Cambridge, and it becomes known by his name

    George Gabriel Stokes

Summary

  • Stokes' theorem: ∮_{∂S} F·dr = ∬_S (∇×F)·dS, relating boundary circulation to surface flux of curl.
  • Green's theorem is the flat, xy-plane special case of Stokes' theorem.
  • The flux of curl F through a surface depends only on the surface's boundary curve, not the specific surface (as long as they share the same boundary and orientation).
  • Orientation of ∂S must match S's normal via the right-hand rule.

References