It collapses to Green's theorem in the plane
Stokes is genuinely a generalisation: flatten the surface into the
xy-plane and it becomes Green's theorem.
Step 1 — take a planar field and a flat region. Let
\mathbf{F} = (P,\ Q,\ 0) with P, Q
functions of x, y only, and let S be a
flat region D of the xy-plane. Its
upward unit normal is \mathbf{n} = \mathbf{k} = (0, 0, 1) and
dS = dx\, dy.
Step 2 — compute the curl. With R = 0 and no
z-dependence, the \mathbf{i} and
\mathbf{j} components of
\nabla \times \mathbf{F} vanish, leaving only the
\mathbf{k} component:
\nabla \times \mathbf{F} = \left(0,\ 0,\ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right).
Step 3 — take the flux of that curl. Dotting into
\mathbf{n} = \mathbf{k} keeps only the
\mathbf{k} component:
\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\, dy.
Step 4 — the left side is a plane line integral. With
\mathbf{F} = (P, Q, 0) and
d\mathbf{r} = (dx, dy, 0),
\oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C P\, dx + Q\, dy.
Step 5 — equate. Stokes now reads
\oint_C P\, dx + Q\, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\, dy,
which is exactly Green's theorem. So the \mathbf{k}-component of the
curl is the "microscopic circulation" Green already measured; Stokes lifts it off the plane and
lets the surface bend into space.
Verifying both sides on an example
Take the swirl \mathbf{F} = (-y,\ x,\ 0) and let
S be the unit disk in the xy-plane,
x^2 + y^2 \le 1, oriented upward, so its boundary
C is the unit circle travelled counter-clockwise.
Left side — the circulation
Step 1 — parametrise the circle.
\mathbf{r}(t) = (\cos t,\ \sin t,\ 0),
0 \le t \le 2\pi, so
d\mathbf{r} = (-\sin t,\ \cos t,\ 0)\, dt.
Step 2 — evaluate \mathbf{F} on
C. With x = \cos t,
y = \sin t,
\mathbf{F} = (-\sin t,\ \cos t,\ 0).
Step 3 — dot and integrate.
\mathbf{F} \cdot d\mathbf{r} = (-\sin t)(-\sin t) + (\cos t)(\cos t)\, dt = (\sin^2 t + \cos^2 t)\, dt = dt,
\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} 1\, dt = 2\pi.
Right side — the flux of the curl
Step 4 — compute the curl. For
\mathbf{F} = (-y, x, 0) (as derived on the divergence-and-curl
page),
\nabla \times \mathbf{F} = (0,\ 0,\ 2).
Step 5 — take its flux through the disk. The upward normal is
\mathbf{k}, so
(\nabla \times \mathbf{F}) \cdot \mathbf{n} = 2, and
\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_D 2\, dA = 2 \cdot (\text{area of disk}) = 2 \cdot \pi(1)^2 = 2\pi.
Step 6 — compare. Both sides equal 2\pi. The total
circulation you would feel paddling once around the rim equals the total "spin" bottled up
across the surface — Stokes' theorem confirmed.
Let S be a piecewise-smooth oriented surface bounded by a
piecewise-smooth simple closed curve C = \partial S with matching
(right-hand-rule) orientation, and let \mathbf{F} have continuous
partial derivatives on an open set containing S. Then:
-
Circulation equals curl-flux:
\displaystyle \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}.
-
Green's theorem is the planar case: flat
S \subset xy-plane with
\mathbf{F} = (P, Q, 0) gives
\oint P\,dx + Q\,dy = \iint_D (Q_x - P_y)\, dA.
-
The surface is free: any two surfaces sharing the boundary
C give the same value — the integral depends only on
C.
-
Irrotational ⇒ path-independent: if
\nabla \times \mathbf{F} = \mathbf{0}, every closed-loop
circulation is 0, so \mathbf{F} is
conservative.
Stokes says only the boundary C governs the answer — a flat disk
and a tall bulging dome capping the same loop give identical curl-flux. Why? Take two
surfaces S_1, S_2 with the same boundary and glue them into one
closed surface \Sigma (flip one orientation so they
match along C). The difference of the two flux integrals is the
flux of \nabla \times \mathbf{F} out of the closed
\Sigma, which by
the divergence theorem
equals
\iiint_E \nabla \cdot (\nabla \times \mathbf{F})\, dV = \iiint_E 0\, dV = 0,
using the identity \nabla \cdot (\nabla \times \mathbf{F}) = 0.
The two fluxes are therefore equal. The very identity that looked like idle algebra on the
divergence-and-curl page is what makes "any capping surface" well defined here.
The two orientations must be compatible or the signs fight. The convention: if the surface
normal \mathbf{n} is your right thumb, your curling fingers give
the positive direction to walk the boundary C — equivalently,
walk C with the surface on your left and your head along
\mathbf{n}. Reverse either choice and both sides flip sign
together, so the equality survives; mismatch them by hand and you will be off by a sign.
This is the same boundary orientation Green's theorem uses (counter-clockwise around a
region viewed from above).