Math Core

Green's theorem relates circulation around a closed curve in the plane to the curl inside it. Stokes' theorem does the same thing for a surface floating in space: the circulation of $\mathbf{F}$ around the boundary curve equals the flux of $\operatorname{curl}\mathbf{F}$ through the surface. It lets you trade a hard line integral for an easier surface integral, or the reverse.

Orientation: the right-hand rule

Let $S$ be an oriented surface with unit normal $\mathbf{n}$ , bounded by a simple closed curve $C$ . The boundary gets the positive orientation that matches $\mathbf{n}$ : if you walk along $C$ with your head pointing in the direction of $\mathbf{n}$ , the surface is on your left. Equivalently, curl the fingers of your right hand in the direction of $C$ and your thumb points along $\mathbf{n}$ .

For an upward-oriented surface, this means $C$ runs counterclockwise when viewed from above.

The theorem

Stokes' theorem

Let $S$ be an oriented, piecewise-smooth surface bounded by a simple, closed, piecewise-smooth curve $C$ with positive orientation. If $\mathbf{F}$ has continuous partial derivatives on an open region containing $S$ , then

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}.

The idea behind it is the same cancellation as in Green's theorem. Chop $S$ into tiny patches. Around each patch, the circulation is about $(\operatorname{curl}\mathbf{F}\cdot\mathbf{n})\,\Delta S$ . Adding them, the circulations along shared interior edges cancel (each is traversed once in each direction), leaving only the outer boundary $C$ .

If $S$ is a flat region $D$ in the $xy$ -plane with $\mathbf{n} = \mathbf{k}$ , then $\operatorname{curl}\mathbf{F}\cdot\mathbf{k} = Q_x - P_y$ , and Stokes' theorem is exactly Green's theorem.

The surface doesn't matter, only its boundary

The right side depends on $S$ only through its boundary. So any two oriented surfaces with the same positively oriented boundary curve give the same value of $\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ . This is a powerful shortcut: replace a curved surface by the flat disk with the same edge.

Worked example: Verifying on a hemisphere

Let $\mathbf{F} = \langle -y,\ x,\ z\rangle$ and let $S$ be the upper hemisphere $x^2 + y^2 + z^2 = 4$ , $z \ge 0$ , oriented upward. Compute both sides of Stokes' theorem.

Line integral. The boundary is the circle $x^2 + y^2 = 4$ in the plane $z = 0$ , counterclockwise from above: $\mathbf{r}(t) = \langle 2\cos t,\ 2\sin t,\ 0\rangle$ . Then $\mathbf{F} = \langle -2\sin t,\ 2\cos t,\ 0\rangle$ and $\mathbf{r}' = \langle -2\sin t,\ 2\cos t,\ 0\rangle$ , so $\mathbf{F}\cdot\mathbf{r}' = 4$ and the integral is $4\cdot 2\pi = 8\pi$ .

Surface integral. $\operatorname{curl}\mathbf{F} = \langle 0 - 0,\ 0 - 0,\ 1 - (-1)\rangle = \langle 0, 0, 2\rangle$ . Since only the boundary matters, use the flat disk $x^2 + y^2 \le 4$ with $\mathbf{n} = \mathbf{k}$ : the flux is $2\cdot 4\pi = 8\pi$ . The two sides agree.

Worked example: A line integral around a triangle

Evaluate $\displaystyle\oint_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle z,\ x,\ y\rangle$ , where $C$ is the triangle with vertices $(1, 0, 0)$ , $(0, 1, 0)$ , $(0, 0, 1)$ , oriented counterclockwise when viewed from above.

Directly, this is three line integrals. Instead, $\operatorname{curl}\mathbf{F} = \langle 1 - 0,\ 1 - 0,\ 1 - 0\rangle = \langle 1, 1, 1\rangle$ . The triangle bounds the piece of the plane $z = 1 - x - y$ over the triangle $D$ with vertices $(0,0)$ , $(1,0)$ , $(0,1)$ . The upward normal is $\langle -g_x, -g_y, 1\rangle = \langle 1, 1, 1\rangle$ , so

\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D \langle 1, 1, 1\rangle\cdot\langle 1, 1, 1\rangle\,dA = 3\cdot\text{area}(D) = 3\cdot\frac{1}{2} = \frac{3}{2}.

Worked example: A curl flux through its boundary

Evaluate $\displaystyle\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ for $\mathbf{F} = \langle yz,\ -xz,\ e^{xy}\rangle$ , where $S$ is the part of the paraboloid $z = 5 - x^2 - y^2$ above the plane $z = 1$ , oriented upward.

Computing the curl and then a flux integral would be messy. The boundary is where $5 - x^2 - y^2 = 1$ : the circle $x^2 + y^2 = 4$ at height $z = 1$ , counterclockwise from above. Parametrize $\mathbf{r}(t) = \langle 2\cos t,\ 2\sin t,\ 1\rangle$ . On it, $\mathbf{F} = \langle 2\sin t,\ -2\cos t,\ e^{4\cos t\sin t}\rangle$ and $\mathbf{r}' = \langle -2\sin t,\ 2\cos t,\ 0\rangle$ , so

\mathbf{F}\cdot\mathbf{r}' = -4\sin^2 t - 4\cos^2 t = -4, \qquad \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = \int_0^{2\pi} -4\,dt = -8\pi.

The $e^{xy}$ term never mattered, because $\mathbf{r}'$ has no $z$ -component.

Consequences

If $\operatorname{curl}\mathbf{F} = \mathbf{0}$ , every closed-curve integral is $0$ . This is another view of why curl-free fields on $\mathbb{R}^3$ are conservative.
If $S$ is a closed surface (like a whole sphere), it has no boundary, so $\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = 0$ .
In a fluid, $\operatorname{curl}\mathbf{F}\cdot\mathbf{n}$ at a point measures the circulation per unit area around a tiny loop perpendicular to $\mathbf{n}$ . The paddle wheel spins fastest when its axle lines up with the curl.

Common mistake

Match the orientations. If the problem says the surface is oriented upward, the boundary must run counterclockwise when viewed from above (and vice versa). A mismatched pair gives the right number with the wrong sign.

Tip

When asked for $\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ over a complicated surface, first look at its boundary. Either compute the line integral around it, or swap in the simplest surface (often a flat disk) with the same boundary.

Practice

Practice 1

Let $\mathbf{F} = \langle -y,\ x,\ 0\rangle$ and let $S$ be the upper half of the unit sphere, oriented upward. Evaluate $\displaystyle\iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S}$ .