Stokes' Theorem

In this lesson you’ll learn Stokes’ Theorem, which says that the total rotation (curl) inside a surface equals the total circulation around its boundary. This is the 3D generalization of Green’s Theorem and one of the most important results in vector calculus.

Green’s Theorem (Lesson 16) connected a line integral around a flat closed curve to a double integral over the enclosed region. Stokes’ Theorem does the same thing, but the surface doesn’t have to be flat. It can be any oriented surface in 3D space, as long as it has a boundary curve.

If S is an oriented surface with boundary curve C (oriented by the right-hand rule relative to the surface normal), and $\mathbf{F}$ has continuous partial derivatives, then

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

The left side is the flux of the curl of F through the surface S. The right side is the circulation of F around the boundary curve C. Stokes’ Theorem says they’re equal.

Just like Green’s Theorem, the power is in choosing the easier side. If the line integral around C is hard to compute, you can instead compute the surface integral of the curl. If the surface integral is hard, compute the line integral instead.

The boundary curve C must be oriented consistently with the surface normal. If you curl the fingers of your right hand in the direction you walk along C, your thumb should point in the direction of the surface normal n. This is the same right-hand rule from curl.

When the surface S is flat and lies in the xy-plane, Stokes’ Theorem reduces exactly to Green’s Theorem. The curl’s z-component $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is the integrand from Green’s Theorem, and the surface integral becomes a double integral over the flat region.

Example 1: Using Stokes’ to avoid a line integral

Evaluate $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = \langle -y,\; x,\; 0 \rangle$ and C is the unit circle in the xy-plane, traversed counterclockwise.

The setup is the same as Green’s Theorem: a flat disk D with boundary C. The curl of F points straight up through the disk. Stokes’ Theorem lets us integrate the curl over the disk instead of parametrizing the circle.

We already know from Lesson 14 that this line integral equals $2\pi$ (the rotational field does work = $2\pi$ around the full circle). Let’s verify using Stokes’ Theorem.

First, compute the curl:

$$\nabla \times \mathbf{F} = \langle 0,\; 0,\; 2 \rangle$$

(We computed this in Lesson 17.) The curl is a constant vector pointing straight up.

Choose S to be the flat unit disk in the xy-plane (the simplest surface with boundary C). The upward normal to this flat disk is $\mathbf{n} = \langle 0, 0, 1 \rangle,$ so $d\mathbf{S} = \langle 0, 0, 1 \rangle \, dA$

$$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_D \langle 0, 0, 2 \rangle \cdot \langle 0, 0, 1 \rangle \, dA = \iint_D 2 \, dA = 2 \cdot \pi(1)^2 = 2\pi$$

Both sides give $2\pi$. Stokes’ Theorem confirms what we computed directly.

Example 2: Choosing a different surface

Here’s the key insight: Stokes’ Theorem works for any surface with the same boundary. For the same boundary curve C (unit circle in the xy-plane), we could use the upper hemisphere as S instead of the flat disk. The answer would still be $2\pi$ because the theorem only cares about the boundary, not which surface you pick.

Both surfaces share the same boundary C, so Stokes’ Theorem gives the same answer for both. You always get to choose the easiest surface.

This means you always get to choose the easiest surface. For a flat boundary curve, the flat disk is usually simplest. But for a boundary curve in 3D, you might pick a surface that makes the curl computation cleaner.

Example 3: Stokes’ Theorem with a tilted surface

Let $\mathbf{F} = \langle z,\; x,\; y \rangle$ and let C be the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1), traversed counterclockwise when viewed from above.

Drag to rotate · Scroll/Pinch to zoom

The triangle lies in the plane x + y + z = 1. The orange arrow shows the outward normal. You can rotate to see how the three vertices sit on the coordinate axes. Instead of parametrizing three line segments, we integrate the curl over the triangular surface.

Computing the line integral directly would require parametrizing three line segments. Instead, use Stokes’ Theorem.

Compute the curl:

$$\nabla \times \mathbf{F} = \left\langle \frac{\partial y}{\partial y} - \frac{\partial x}{\partial z},\; \frac{\partial z}{\partial z} - \frac{\partial y}{\partial x},\; \frac{\partial x}{\partial x} - \frac{\partial z}{\partial y} \right\rangle = \langle 1, 1, 1 \rangle$$

The surface S is the triangular region in the plane x + y + z = 1. The outward normal to this plane (pointing upward) is $\mathbf{n} = \frac{1}{\sqrt{3}}\langle 1, 1, 1 \rangle$ and the surface element is $d\mathbf{S} = \langle 1, 1, 1 \rangle \, dA$ (where dA is the area element projected onto the xy-plane).

$$(\nabla \times \mathbf{F}) \cdot \langle 1, 1, 1 \rangle = \langle 1, 1, 1 \rangle \cdot \langle 1, 1, 1 \rangle = 3$$

The projection of the triangle onto the xy-plane is the region where $x \geq 0,\; y \geq 0,\; x + y \leq 1,$ which has area 1/2.

$$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_D 3 \, dA = 3 \cdot \frac{1}{2} = \frac{3}{2}$$

The circulation around the triangle is 3/2. Stokes’ Theorem turned three separate line integrals into one simple area calculation.

Stokes’ Theorem connects rotation inside a region to circulation around its boundary:

• Faraday’s Law in electromagnetism is literally Stokes’ Theorem applied to the electric field. The induced voltage around a loop equals the rate of change of magnetic flux through the loop.
• Fluid dynamics uses it to relate the vorticity (curl of velocity) inside a region to the circulation of the fluid around the boundary.
• Game engines use the principle behind Stokes’ Theorem when computing vortex effects: the swirl you see around the edge of a whirlpool is determined by the rotation inside it.
• Aerodynamics uses it to compute lift on an airfoil via the Kutta-Joukowski theorem, which relates circulation to lift force.