5.4 Stokes' Theorem: Two Caps, One Boundary

The last section computed a circulation of $4.5 π$ around a circle and called it a hint. Here is the theorem it hints at. Stokes' Theorem says that the circulation of a field around a closed curve equals the flux of its curl through any surface that has the curve as boundary:

$\oint_{\partial S} 𝐅 \cdot d 𝐫 = \iint_{S} (\nabla \times 𝐅) \cdot d 𝐒,$

provided the curve's direction and the surface's normal are paired by the right-hand rule. The startling word is any. Hang two utterly different surfaces on the same rim and the theorem demands they collect identical curl flux — the cap is disposable, and only the boundary owns the answer. The figure stages exactly that confrontation.

A dark circle of radius 1.5 in the xy-plane, spanned by two translucent surfaces that share it as their rim: a flat yellow-toned disk and a dome-shaped purple-orange paraboloid rising above it. A field of arrows circulates counterclockwise around the z-axis, threading through both surfaces. — A dark circle of radius $1.5$ in the $x y$ -plane — the boundary — spanned by two translucent caps that share it as their rim: a flat yellow-toned disk and a purple-orange paraboloid dome, $z = 2.25 - r^{2}$ . Threading both, a field of arrows circulates counterclockwise around the $z$ -axis: $𝐅 = (- y, x, 0)$ , the non-conservative field from the previous section, its components independent of height.

Explore

Orbit slowly and confirm the two facts everything below rests on: both surfaces hang on the same dark rim, and the field looks identical at every height — no $z$ appears in its components.
Look straight down the $z$ -axis. The arrows circulate counterclockwise around the boundary. Curl your right hand's fingers that way: your thumb points $+ z$ . That pairing — counterclockwise rim, upward normals — fixes every sign below.
The dome clearly has more surface area than the disk, and both sit in the same swirling field. Predict, with one sentence of reasoning: does more curl flux pass through the dome?
Compute $∇ times 𝐅$ (the example below checks you). It comes out the same constant vertical vector everywhere — which makes the disk's flux a mental computation and sets up the trap in step 3.
Orbit to a low side view and look at the dome near its rim, where it rises steeply — nearly vertical. How much of a purely vertical curl vector passes through a nearly vertical patch of surface? So where on the dome is its flux actually collected?
Suppose the rim were traversed clockwise seen from above. What happens to the circulation's sign, and which way must both caps' normals point for the theorem to still balance?

The boundary owns the answer

Why should the cap not matter? Two caps on one rim, taken together (one normal flipped), enclose a solid region — and curl fields send zero net flux out of any closed surface. So whatever flux enters through one cap leaves through the other: hang any surface on the rim and the total is fixed by the rim alone. That is why the theorem writes $∂ S$ on one side: only boundary data survives. In practice you read the equation right-to-left — replace a hard surface with an easy one, or with the boundary curve itself.

Example 1 One rim, one number, three ways

For $𝐅 = (- y, x, 0)$ and the circle $x^{2} + y^{2} = 2.25$ in the $x y$ -plane, oriented counterclockwise from above: compute the curl, the flux through the flat disk, the circulation around the rim, and deduce the flux through the paraboloid dome $z = 2.25 - r^{2}$ .

The curl of $(P, Q, R) = (- y, x, 0)$ is

$\nabla \times 𝐅 = (R_{y} - Q_{z}, P_{z} - R_{x}, Q_{x} - P_{y}) = (0, 0, 1 - (- 1)) = (0, 0, 2),$

the same vertical vector at every point in space. The disk's upward unit normal is $𝐤$ , so its flux needs no integral: $2 \times area = 2 π (1.5)^{2} = 4.5 π$ . The rim circulation was the previous section's computation: $𝐅 (𝐫 (t)) \cdot 𝐫^{'} (t) = 2.25$ on $𝐫 open parenthesis t close parenthesis equals open parenthesis 1.5 cos t, 1.5 sin t, 0 close parenthesis$ , giving $2.25 \cdot 2 π = 4.5 π$ . Equality of the two is Stokes' Theorem, verified by hand. And the dome? It hangs on the same rim with the same orientation, so its curl flux is forced to be $4.5 π$ — not more, despite the extra area. Geometry agrees with the bookkeeping: near the rim the dome is nearly vertical, and a vertical patch collects nothing from a vertical curl vector; the dome gathers its $4.5 π$ near the top, where it flattens out — and its shadow on the floor is the very disk we already integrated.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.