6.7 Stokes' Theorem

Big idea

Green’s theorem related a line integral around a closed curve in the plane to a double integral over the flat region inside it. Nothing in that argument really needed the region to be flat. If the closed curve wanders through space and you stretch any surface across it like a soap film, the same relationship holds: the circulation around the curve equals the flux of the curl through the surface. That statement is Stokes’ theorem, and Green’s theorem is the special case where the surface happens to lie in a plane.

The chapter’s organizing pattern is visible again. An integral of a derivative of the field over a region equals an integral of the field itself over the boundary of that region. For the fundamental theorem for line integrals the region was a curve and the boundary was two points. For Green’s theorem the region was a piece of plane and the boundary was a closed curve. Here the region is a curved surface sitting in space and the boundary is again a closed curve, but now the derivative is the full three dimensional curl and only the component along the surface normal contributes.

One consequence is startling the first time you see it. The left side of the equation mentions only the boundary curve, so the right side cannot depend on which surface you stretched across it. A flat disk, a bulging hemisphere and a long thin tube with the same rim all give the same flux of the curl. That freedom is the main practical tool: replace an awkward surface by an easier one with the same rim.

The second consequence is conceptual. If a field has zero curl on a region where every closed curve bounds a surface inside the region, then every circulation is zero and the field is conservative. That explains the simply connected hypothesis in the cross-partials test, and also why the punctured domain example failed: no surface inside that domain has a loop around the missing point as its boundary.

Decoder

For an oriented piecewise smooth surface with a simple closed boundary curve, the circulation of a field around that curve equals the flux of the curl of the field through the surface, provided the curve is oriented compatibly with the surface normal.

“Compatibly” is the word carrying the sign. The rule is the right hand rule: point the thumb of your right hand along the chosen unit normal, and the fingers curl in the required direction of travel around the boundary. An equivalent phrasing is that if you walk along the curve with your head pointing along the normal, the surface is on your left. Flip either the normal or the direction of travel and the equation acquires a minus sign; flip both and it is restored.

The hypotheses are not cosmetic. The surface must be orientable, so a Mobius band is excluded. The boundary must be a simple closed curve, traversed once. And the field needs continuous partial derivatives on an open region containing the surface, not merely on the curve, which is exactly the condition that fails when a singularity of the field sits somewhere inside the film you stretched.

The usual mistake is to compute the flux of $\mathbf{F}$ instead of the flux of $\operatorname{curl}\mathbf{F}$. The two sides of the equation involve different objects: the field itself along the curve, and its curl across the surface.

Definitions and results

Boundary of a surface. The boundary of an oriented surface $S$, written $\partial S$, is the curve or curves where the surface ends. A closed surface such as a sphere has empty boundary.

Induced orientation. Given a unit normal field $\mathbf{n}$ on $S$, the induced orientation on $\partial S$ is the one given by the right hand rule relative to $\mathbf{n}$.

Stokes’ theorem. Let $S$ be a piecewise smooth oriented surface with a piecewise smooth simple closed boundary curve $C$ carrying the induced orientation, and let $\mathbf{F}$ have continuous partial derivatives on an open region containing $S$. Then $$ \oint_{C} \mathbf{F}\cdot d\mathbf{r} = \iint_{S} \left(\operatorname{curl}\mathbf{F}\right)\cdot d\mathbf{S} $$

Green’s theorem as a special case. If $S$ lies in the $xy$-plane with upward normal $\mathbf{k}$ and $\mathbf{F} = \langle P, Q, 0\rangle$, then $(\operatorname{curl}\mathbf{F})\cdot\mathbf{k} = Q_x - P_y$ and the statement reduces exactly to the circulation form of Green’s theorem.

Surface independence. If two oriented surfaces share the same boundary curve with the same induced orientation, and the field has continuous partials on a region containing both, then the flux of the curl through them is equal. Choose whichever is easier.

Curl as circulation density. Shrinking a small disk with unit normal $\mathbf{n}$ around a point, the circulation around its rim divided by its area tends to $(\operatorname{curl}\mathbf{F})\cdot\mathbf{n}$. This is the definition of curl that does not mention coordinates.

Zero curl and conservative fields. If $\operatorname{curl}\mathbf{F} = \mathbf{0}$ on a simply connected open region, then every closed curve in the region bounds a surface in the region, so every circulation is zero and $\mathbf{F}$ is conservative there.

Closed surfaces. A closed surface has no boundary, so the flux of any curl through it is zero. That is the same fact as the identity saying the divergence of a curl vanishes, seen from the other side.

Worked examples

Both sides on a hemisphere

Let $\mathbf{F} = \langle -y, \; x, \; z\rangle$, and let $C$ be the unit circle in the plane $z=0$, counterclockwise seen from above.

The line integral first. With $\mathbf{r}(t) = \langle\cos t, \sin t, 0\rangle$ we get $\mathbf{r}'(t) = \langle-\sin t, \cos t, 0\rangle$ and $\mathbf{F}(\mathbf{r}(t)) = \langle -\sin t, \cos t, 0\rangle$, so the integrand is $\sin^2 t + \cos^2 t = 1$ and the circulation is $2\pi$.

Now the surface side, using the upper unit hemisphere $S$ with the outward normal. The curl is $$ \operatorname{curl}\mathbf{F} = \langle 0 - 0, \; 0 - 0, \; 1 - (-1)\rangle = \langle 0, 0, 2\rangle $$ On the unit sphere the outward unit normal at a point is the position vector, so $(\operatorname{curl}\mathbf{F})\cdot\mathbf{n} = 2z$. In spherical coordinates $z = \cos\phi$ and $dS = \sin\phi\,d\phi\,d\theta$, and the upper half is $0\le\phi\le\pi/2$: $$ \iint_S 2z\,dS = \int_0^{2\pi}\!\!\int_0^{\pi/2} 2\cos\phi\sin\phi\,d\phi\,d\theta = 2\pi\left[\sin^2\phi\right]_0^{\pi/2} = 2\pi $$ The two sides agree. Note that the flat disk $z=0$ inside the same circle, with upward normal, gives $\iint 2\,dA = 2\pi$ as well, which is surface independence in action: the bulge changed the surface but not the answer.

A triangle in a tilted plane

Let $\mathbf{F} = \langle z, \; x, \; y\rangle$, and let $C$ be the triangle with vertices $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, traversed in that order.

Take $S$ to be the flat triangle in the plane $x+y+z=1$. The vectors from the first vertex to the others are $\langle -1,1,0\rangle$ and $\langle -1,0,1\rangle$, and their cross product is $\langle 1,1,1\rangle$, so the normal compatible with the stated order is $\mathbf{n} = \langle1,1,1\rangle/\sqrt3$. The curl is $$ \operatorname{curl}\mathbf{F} = \langle \partial_y y - \partial_z x, \;\; \partial_z z - \partial_x y, \;\; \partial_x x - \partial_y z\rangle = \langle 1,1,1\rangle $$ so $(\operatorname{curl}\mathbf{F})\cdot\mathbf{n} = 3/\sqrt3 = \sqrt3$, a constant. The triangle is equilateral with side $\sqrt2$, so its area is $\frac{\sqrt3}{4}(\sqrt2)^2 = \frac{\sqrt3}{2}$, and the flux is $\sqrt3\cdot\frac{\sqrt3}{2} = \frac{3}{2}$.

Check by walking the boundary. On the first edge, $\mathbf{r}(t) = \langle 1-t, t, 0\rangle$ with $\mathbf{r}' = \langle -1,1,0\rangle$ and $\mathbf{F} = \langle 0, 1-t, t\rangle$, so the integrand is $1-t$ and the edge contributes $\frac12$. The other two edges are the same computation with the coordinates cycled, each contributing $\frac12$. The total is $\frac32$, matching.

Replacing a bad integral with a good one

Evaluate $\oint_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle -y^3, \; x^3, \; e^{z}\rangle$, where $C$ is the curve cut on the cylinder $x^2+y^2=1$ by the plane $z = y$, oriented counterclockwise seen from above.

Directly this is unpleasant: on the curve $z = \sin t$, and the third component contributes $e^{\sin t}\cos t$ against $dz$, on top of two sixth-degree trigonometric terms. Instead compute the curl, $$ \operatorname{curl}\mathbf{F} = \left\langle 0 - 0, \;\; 0 - 0, \;\; 3x^2 + 3y^2 \right\rangle $$ and take $S$ to be the flat elliptical region of the plane $z=y$ inside the cylinder. As a graph $z = g(x,y) = y$, the upward normal element is $\langle -g_x, -g_y, 1\rangle\,dA = \langle 0, -1, 1\rangle\,dA$, whose third component is positive, so it matches the stated orientation. The dot product keeps only the third curl component: $$ \left(\operatorname{curl}\mathbf{F}\right)\cdot\langle 0,-1,1\rangle = 3x^2+3y^2 $$ The shadow of $S$ on the $xy$-plane is the unit disk, so in polar coordinates $$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}\!\!\int_0^{1} 3r^2 \cdot r\,dr\,d\theta = 2\pi\cdot\frac34 = \frac{3\pi}{2} $$ The exponential term disappeared at the first step, because a function of $z$ alone in the third component contributes nothing to the curl. The tilt of the plane disappeared at the second, because the projection onto the $xy$-plane is a disk no matter how the plane leans.

Practice

These problems ask you to move in both directions: turn a circulation integral into a flux of the curl when the curve is awkward, and turn a flux of a curl into a line integral when the surface is awkward. Fix the orientation by the right hand rule before computing anything, and look for a simpler surface with the same boundary whenever the given one is inconvenient.

Quiz

Five items on applying Stokes’ theorem in both directions, orientation by the right hand rule, and choosing a convenient spanning surface.