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Series and multivariable theorems

Stokes' Theorem: Statement, Orientation and Worked Examples

Stokes' theorem: the circulation of a field around a closed curve equals the flux of its curl through any surface the curve bounds. Statement and examples.

Series and multivariable theorems Updated 4 min read

At a glance

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}
Hypotheses
S is an oriented, piecewise smooth surface bounded by a simple, closed, piecewise smooth curve C with positive orientation, and \mathbf{F} has continuous partial derivatives on an open region containing S
Positive orientation
Right-hand rule: with your thumb along the normal of S, your fingers curl in the direction of C
Curl
\nabla\times\mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle
Key consequence
The curl flux depends only on the boundary curve, not on which surface you pick
Special case
Green's theorem, when S is a flat region in the xy-plane
Course
Calculus 3
On this page
  1. Statement of Stokes' theorem
  2. What Stokes' theorem means
  3. Why each hypothesis matters
  4. Worked examples
  5. Common mistakes
  6. Related theorems and guides

Stokes' theorem says that the circulation of a vector field around a closed curve in space equals the total curl flowing through any surface that has that curve as its edge. It is Green's theorem lifted off the flat plane and onto curved surfaces.

Statement of Stokes' theorem

Let

  1. S be an oriented, piecewise smooth surface with unit normal \mathbf{n},
  2. C be the boundary of S, a simple, closed, piecewise smooth curve with positive orientation (the right-hand rule relative to \mathbf{n}), and
  3. \mathbf{F} = \langle P, Q, R \rangle have continuous partial derivatives on an open region containing S.

Then

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}

where \nabla\times\mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle and d\mathbf{S} = \mathbf{n}\,dS.

What Stokes' theorem means

The curl at a point measures how much the field swirls around that point, like a tiny paddle wheel. Cover the surface with small patches and add up the swirl around each patch. On every shared edge, neighboring patches go in opposite directions and cancel, so only the swirl around the outer edge, the circulation along C, remains.

A striking consequence: the surface is not important, only its boundary. A hemisphere and a flat disk with the same circular edge have the same curl flux.

Why each hypothesis matters

Matching orientation. If C runs opposite to the right-hand rule, the line integral changes sign but the surface integral does not, so the equation fails by a factor of -1.

Smoothness on the surface. The field \mathbf{F} = \left\langle \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}, 0 \right\rangle has \nabla\times\mathbf{F} = \mathbf{0} wherever it is defined, but its circulation around the unit circle in the xy-plane is 2\pi. Every surface bounded by that circle must cross the z-axis, where \mathbf{F} is undefined, so Stokes' theorem does not apply.

A bounding curve. A closed surface such as a sphere has no boundary, so \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = 0 for it. That fits the divergence theorem, since \nabla\cdot(\nabla\times\mathbf{F}) = 0.

Worked examples

Example 1: a hemisphere versus its boundary. Let \mathbf{F} = \langle -y, x, z \rangle and let S be the upper half of the unit sphere with upward normal, so C is the unit circle in the xy-plane, counterclockwise from above.

Line integral. With \mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle:

\oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \left((-\sin t)(-\sin t) + (\cos t)(\cos t)\right) dt = \int_0^{2\pi} 1\,dt = 2\pi

Surface integral. \nabla\times\mathbf{F} = \langle 0 - 0,\; 0 - 0,\; 1 - (-1) \rangle = \langle 0, 0, 2 \rangle. Because only the boundary matters, replace the hemisphere by the flat unit disk with normal \mathbf{k}: the flux is 2 \cdot \pi = 2\pi. Both sides agree.

Example 2: a triangle in space. Evaluate \oint_C z\,dx + x\,dy + y\,dz, where C is the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1), traversed counterclockwise when viewed from above.

Step 1. For \mathbf{F} = \langle z, x, y \rangle, \nabla\times\mathbf{F} = \langle 1 - 0,\; 1 - 0,\; 1 - 0 \rangle = \langle 1, 1, 1 \rangle.

Step 2. Take S to be the flat triangle on the plane z = 1 - x - y with upward normal. For a graph z = g(x, y), d\mathbf{S} = \langle -g_x, -g_y, 1 \rangle\,dA = \langle 1, 1, 1 \rangle\,dA.

Step 3. (\nabla\times\mathbf{F})\cdot\langle 1, 1, 1 \rangle = 3, integrated over the projection, the triangle x, y \ge 0, x + y \le 1 of area \frac{1}{2}:

\oint_C \mathbf{F}\cdot d\mathbf{r} = 3\cdot\frac{1}{2} = \frac{3}{2}

One surface integral replaces three line integrals along the edges.

Example 3: conservative fields. If \mathbf{F} = \nabla f with f having continuous second partials, then \nabla\times\mathbf{F} = \mathbf{0} by Clairaut's theorem, so Stokes gives \oint_C \nabla f\cdot d\mathbf{r} = 0 for every closed curve bounding a surface. Gradient fields do no net work around closed loops.

Common mistakes

  1. Mismatched orientation. Decide the normal first, then orient C with the right-hand rule.
  2. Computing curl incorrectly. Write out the determinant carefully; the middle component is P_z - R_x, not R_x - P_z.
  3. Using the hard surface. If the curl is simple, a flat surface with the same boundary is usually much easier.

Stokes' theorem generalizes Green's theorem and sits beside the divergence theorem as one of the three great theorems of vector calculus. All three are higher-dimensional versions of the fundamental theorem of calculus. For the cross product behind curl, see vectors, dot and cross product, and for partial derivatives see partial derivatives.

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Stokes' theorem and the Divergence Theorem

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Frequently asked questions

What does Stokes' theorem say?

If S is an oriented, piecewise smooth surface with boundary curve C, oriented by the right-hand rule, and \mathbf{F} has continuous partial derivatives near S, then \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}.

How do you orient the curve in Stokes' theorem?

Point your right thumb along the chosen normal vector of S. Your fingers curl in the positive direction of C. Equivalently, walking along C with your head pointing along the normal, the surface stays on your left.

Why can you change the surface in Stokes' theorem?

The right side equals the line integral around C, which depends only on the curve. So any two surfaces with the same boundary and compatible orientation have the same curl flux, and you can pick the easiest one, often a flat disk.

How is Stokes' theorem related to Green's theorem?

If S is a flat region in the xy-plane with upward normal \mathbf{k}, then (\nabla\times\mathbf{F})\cdot\mathbf{k} = Q_x - P_y, and Stokes' theorem becomes Green's theorem.

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