Stokes' theorem — statement, curl-circulation connection: Worked examples

This page is the drill floor for Stokes' theorem. The parent note built the statement and the "curl = circulation density" idea. Here we hit every kind of problem the theorem can throw at you — flat surfaces, curved surfaces, zero-curl traps, sign flips, degenerate loops, a physics word problem, and a sneaky exam twist.

Before we start, one reminder in plain words so no symbol is unearned:

Definition The three objects, in words

$F$ — a vector field: at every point of space it hands you an arrow $(P, Q, R)$ . Think of it as the velocity of a flowing fluid. See Line Integrals of Vector Fields.
$\oint_{C} F \cdot d r$ — the circulation: how much the flow $F$ pushes you along a closed loop $C$ as you walk all the way around. The little $d r$ is a tiny step-arrow along the curve; the dot product $F \cdot d r$ measures how much of $F$ points the same way as that step.
$\nabla \times F$ — the curl: a vector at each point whose length is how fast a tiny paddle-wheel would spin there, pointing along the paddle-wheel's axle. See Curl of a Vector Field.
$\iint_{S} (\nabla \times F) \cdot d S$ — the flux of the curl: add up the spin poking through a surface $S$ . The arrow $d S$ is a tiny patch of surface times its outward normal $\hat{n}$ . See Surface Integrals & Flux.

The scenario matrix

Every Stokes problem lives in one of these cells. The examples below are labelled with the cell(s) they cover, so by the end no cell is left empty.

#	Cell (scenario class)	What makes it tricky	Covered by
A	Flat surface, constant curl	easiest: flux = curl·n × area	Ex 1
B	Curved surface, same boundary swap	replace hard $S$ by a flat disk with same rim	Ex 2
C	Curl computation with all three components nonzero	must use the full determinant	Ex 3
D	Zero curl (conservative trap)	circulation must be $0$ — but only if no singularity inside	Ex 4
E	Sign / orientation flip	reversing traversal or normal flips the sign	Ex 5
F	Degenerate loop (limiting/zero area)	shrinking loop → curl at a point	Ex 6
G	Real-world word problem (physics)	Maxwell / work of a force field	Ex 7
H	Exam twist: singularity forces surface choice	field undefined at a point; must dodge it	Ex 8

Cell A — Flat surface, constant curl

Figure — Stokes' theorem — statement, curl-circulation connection

Figure s01, what to look at: the magenta arrows are the field $F = (- 3 y, 3 x, 0)$ — notice they wrap CCW, like a merry-go-round. The orange arrows on the edges show the walking direction of $C$ (also CCW). The violet dot with a ring ⊙ at the centre is the normal $\hat{n} = + \hat{k}$ pointing out of the page toward you — the right-hand rule (curl fingers along orange arrows, thumb out) confirms this pairing.

Step 1 — Check the domain conditions, then compute the curl. $F$ is a polynomial → smooth everywhere (condition 2 ✓); the triangle is a flat piecewise-smooth surface with boundary exactly $C$ (condition 1 ✓). Now: $\nabla \times F = (\partial_{y} 0 - \partial_{z} (3 x), \partial_{z} (- 3 y) - \partial_{x} 0, \partial_{x} (3 x) - \partial_{y} (- 3 y)) = (0, 0, 6) .$ Why this step? Stokes replaces the walk-around integral by the flux of the curl. So the very first move is always: turn $F$ into $\nabla \times F$ . Here only the $z$ -part survives: $Q_{x} - P_{y} = 3 - (- 3) = 6$ .

Step 2 — Pick the flat surface and its normal. The triangle lies in the $x y$ -plane, so take $S$ = the flat triangular region, with $\hat{n} = \hat{k} = (0, 0, 1)$ (upward, matching CCW by the right-hand rule — see Orientation & the Right-Hand Rule). Why this step? Because $C$ is already the rim of an obvious flat surface, we don't have to invent a curved one.

Step 3 — Dot and integrate. Recall $d S = \hat{n} d A$ , so: $(\nabla \times F) \cdot d S = (0, 0, 6) \cdot (0, 0, 1) d A = 6 d A (constant) .$ $\iint_{S} 6 d A = 6 \cdot Area (S) .$ The triangle has legs of length $2$ , so area $= \frac{1}{2} \cdot 2 \cdot 2 = 2$ . $\oint_{C} F \cdot d r = 6 \cdot 2 = 12.$ Why this step? A constant integrand means the integral is just the constant times the area — no calculus needed.

Verify: Direct line integral. On the bottom edge $y = 0$ , $F = (0, 3 x, 0)$ is perpendicular to $d r = (d x, 0, 0)$ → contributes $0$ . The hypotenuse and left edge also work out; the full CCW walk gives $12$ (checked in VERIFY). Units: circulation of a length-per-time field around a length loop — consistent. ✓

Cell B — Curved surface, boundary swap

Figure s02, what to look at: the violet bowl is the paraboloid cap $z = x^{2} + y^{2}$ ; the orange flat disk is the replacement surface $S^{'}$ at height $z = 4$ . Both are glued to the magenta rim $C$ (the circle drawn CCW). The two purple normal arrows show $\hat{n}$ pointing upward on each — because they share the same rim, Stokes gives the same flux through either. We integrate over the easy flat one.

Step 1 — Domain check, then curl. $F$ is smooth everywhere (condition 2 ✓); each candidate surface has boundary exactly $C$ (condition 1 ✓). $\nabla \times F = (\partial_{y} z^{3} - \partial_{z} x, \partial_{z} (- y) - \partial_{x} z^{3}, \partial_{x} x - \partial_{y} (- y)) = (0, 0, 2) .$ Why this step? Notice the $z^{3}$ contributes nothing to the curl (it only depends on $z$ , and its derivatives with respect to $x, y$ vanish). Already the scary part is gone.

Step 2 — Swap to the easiest surface with the same rim. The rim $C$ is the circle of radius $2$ at height $z = 4$ . The flat disk $S^{'}$ : ${x^{2} + y^{2} \leq 4, z = 4}$ has exactly that rim. By Green/Stokes, the flux of the curl is the same through any surface bounded by $C$ (parent Example 3). Why this step? We're allowed to slide the drum-skin as long as the rim stays put. Flat is easiest.

Step 3 — Flux through the disk. $\hat{n} = (0, 0, 1)$ , and $d S = \hat{n} d A$ , so $(\nabla \times F) \cdot d S = 2 d A$ . Disk area $= π (2)^{2} = 4 π$ . $\oint_{C} F \cdot d r = 2 \cdot 4 π = 8 π .$

Verify: Parametrize $C$ : $r = (2 cos t, 2 sin t, 4)$ , $d r = (- 2 sin t, 2 cos t, 0) d t$ . Then $F \cdot d r = (- 2 sin t) (- 2 sin t) + (2 cos t) (2 cos t) + 0 = 4 d t$ ; integrate $0$ to $2 π$ → $8 π$ . ✓ (The $z^{3}$ term rides along a curve of constant $z$ , so $d z = 0$ kills it.)

Cell C — All three curl components nonzero

Figure s03, what to look at: the three vertices $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ span the slanted violet triangle living on the plane $x + y + z = 1$ . The surface $S$ is this flat triangular patch; its purple normal arrow $\hat{n} = \frac{1}{3} (1, 1, 1)$ points away from the origin. The magenta arrows trace the boundary $C$ in the right-hand-rule direction matching that normal. Knowing this geometry is what lets us set up the flux integral in the first place.

Step 1 — Domain check, then full determinant curl. $F$ is a polynomial → smooth everywhere ✓; $S$ is the flat triangular patch on $x + y + z = 1$ with boundary $C$ ✓. $\nabla \times F = \hat{i} \partial_{x} y z \hat{j} \partial_{y} x z \hat{k} \partial_{z} x y = (\partial_{y} (x y) - \partial_{z} (x z), \partial_{z} (y z) - \partial_{x} (x y), \partial_{x} (x z) - \partial_{y} (y z)) .$ Compute each: $(x - x, y - y, z - z) = (0, 0, 0)$ . Why this step? We must use the whole determinant because all three components are present. The surprise: this field is curl-free — it is the gradient of $f = x y z$ .

Step 2 — Apply Stokes. $\oint_{C} F \cdot d r = \iint_{S} 0 \cdot d S = 0.$ Why this step? Zero curl over the whole slanted surface → zero flux → zero circulation. This is the conservative case: $F = \nabla (x y z)$ , and a closed loop in a conservative field does no net work.

Verify: Since $F = \nabla (x y z)$ and $C$ is closed, $\oint_{C} \nabla f \cdot d r = f (end) - f (start) = 0$ for a loop. Numerically we also confirm curl $= (0, 0, 0)$ in VERIFY. ✓

Cell D — Zero-curl trap (and when it's really zero)

Figure s04, what to look at: the magenta ellipse is $C$ ( $x^{2} /9 + y^{2} /4 = 1$ ), drawn with orange CCW arrows; the violet dot ring ⊙ at the centre marks $\hat{n} = + \hat{k}$ out of the page (right-hand rule: fingers along orange, thumb toward you). The magenta field arrows all point straight outward from the origin — perpendicular to the loop's motion at every point — so they never help you go around. That perpendicularity is why the circulation is zero.

Step 1 — Domain check, then curl. $F$ is smooth everywhere ✓; the flat elliptical disk has boundary exactly $C$ ✓. $\nabla \times (2 x, 2 y, 2 z) = (\partial_{y} 2 z - \partial_{z} 2 y, \partial_{z} 2 x - \partial_{x} 2 z, \partial_{x} 2 y - \partial_{y} 2 x) = (0, 0, 0) .$ Why this step? This is $\nabla (x^{2} + y^{2} + z^{2})$ , a pure gradient — no swirl anywhere. The paddle-wheel never spins.

Step 2 — Conclude. $\oint_{C} F \cdot d r = \iint_{S} 0 d S = 0.$ Why this step? The whole surface has zero curl and no singularity (the field is smooth everywhere), so Stokes gives a clean zero. Contrast this with Cell H, where a singularity ruins the shortcut.

Verify: Ellipse $r = (3 cos t, 2 sin t, 0)$ , $d r = (- 3 sin t, 2 cos t, 0) d t$ . $F \cdot d r = 2 (3 cos t) (- 3 sin t) + 2 (2 sin t) (2 cos t) = - 18 sin t cos t + 8 sin t cos t = - 10 sin t cos t$ ; integrate over $0$ to $2 π$ → $0$ . ✓

Cell E — Sign / orientation flip

Figure s05, what to look at: the magenta arrows show the Example-1 CCW walk with its out-of-page normal ⊙ (answer $+ 12$ ); the violet arrows show the reversed CW walk with its into-page normal ⊗ (answer $- 12$ ). Same triangle, same field — only the pairing of "walk direction ↔ normal" changed, and that flips the sign. Notice the magenta ⊙ (dot = coming toward you) versus the violet ⊗ (cross = going away).

Step 1 — Keep the same curl. $\nabla \times F = (0, 0, 6)$ (unchanged — the field didn't change).

Step 2 — Flip the normal to stay compatible with the new walk. Right-hand rule: thumb along traversal ⇒ for a clockwise walk (from above) the compatible normal points down, $\hat{n} = (0, 0, - 1)$ . Why this step? The circulation sign and the normal are locked together by the right-hand rule (domain condition 3). If you flip one, you must flip the other, or Stokes' two sides disagree.

Step 3 — Recompute the flux. $(0, 0, 6) \cdot (0, 0, - 1) d A = - 6 d A, \oint_{C} F \cdot d r = - 6 \cdot 2 = - 12.$ Why this step? Same size as Example 1, opposite sign — exactly the mistake the parent note warned about.

Verify: Reversing traversal direction negates every $d r$ , so the line integral negates: $+ 12 \to - 12$ . ✓

Cell F — Degenerate loop (limiting case → curl at a point)

Figure s06, what to look at: three nested magenta circles of shrinking radius $ε$ (large → medium → tiny) all centred at the origin, each with an orange CCW arrow. Beside each is its ratio (circulation ÷ area). The label on all three reads $2$ — the ratio never changes as the loop collapses to the central violet dot. That constant limiting value is the curl component at the point: this figure is the visual definition of "circulation per unit area."

Step 1 — Circulation of the small loop. By Stokes over the small disk $S_{ε}$ (area $π ε^{2}$ , normal $\hat{k}$ ), using $d S = \hat{n} d A$ : $\oint_{C_{ε}} F \cdot d r = \iint_{S_{ε}} (0, 0, 2) \cdot (0, 0, 1) d A = 2 \cdot π ε^{2} .$ Why this step? Even a degenerate (tiny) loop obeys Stokes; the curl is constant here so the flux is exactly $2 \times$ area.

Step 2 — Divide by area and take the limit. $\frac{1}{π ε ^{2}} \cdot 2 π ε^{2} = 2 \Rightarrow lim_{ε \to 0} (\dots) = 2.$ Why this step? This is why curl is called "circulation per unit area." Shrinking the loop and normalizing by area recovers exactly $(\nabla \times F) \cdot \hat{k} = 2$ at the origin — the local swirl. Degenerate case: as $ε \to 0$ the loop collapses to a point, and the ratio survives cleanly.

Verify: For any $ε$ , circulation/area $= 2 π ε^{2} / (π ε^{2}) = 2$ , independent of $ε$ , so the limit is $2$ . ✓

Cell G — Real-world word problem

Figure s07, what to look at: the magenta circle is the wire loop $C$ (radius $0.4$ m), with the orange CCW arrow giving its chosen walking direction. The right-hand rule then fixes the loop's own normal ⊙ pointing up ( $+ \hat{k}$ , violet). But the given curl $\nabla \times E = (0, 0, - 5)$ points down (violet downward arrow ⊗). Because the curl opposes the chosen normal, the dot product is negative — that is exactly why the EMF comes out negative (Lenz's law), not an arbitrary sign.

Step 1 — Recognize EMF as circulation. The electromotive force around a loop is precisely $\oint_{C} E \cdot d r$ — the work per unit charge to push around the loop. Stokes turns it into flux of $\nabla \times E$ . Why this step? This is literally Faraday's law written with Stokes' theorem; the "circulation = curl-flux" identity is the physics.

Step 2 — Flux of the curl through the flat disk. $\hat{n} = (0, 0, 1)$ (up, from the CCW walk), $d S = \hat{n} d A$ , area $= π R^{2} = π (0.4)^{2} = 0.16 π$ . $\oint_{C} E \cdot d r = (0, 0, - 5) \cdot (0, 0, 1) \cdot π R^{2} = - 5 \cdot 0.16 π = - 0.8 π \approx - 2.513 V .$ Why this step? Constant curl-flux = (curl $\cdot \hat{n}$ ) × area. The negative sign is the induced EMF opposing the change (Lenz's law) — the down-pointing curl against the up-pointing normal, exactly as the figure shows.

Verify: $- 5 \times π \times 0.16 = - 0.8 π \approx - 2.5133$ . Units: (V/m²)·(m²) = V ✓. Magnitude modest, as forecast.

Cell H — Exam twist: singularity forces a surface choice

Figure s08, what to look at: the magenta unit circle $C$ (orange CCW arrow) encloses the origin, marked with a violet ✕ "singularity" where the field blows up. The magenta field arrows swirl around this hole. Because the forbidden point sits inside any flat disk you'd bound by $C$ , domain condition 2 fails — you cannot lay a valid surface across it, so the Stokes shortcut is off-limits and we must integrate directly.

Step 1 — Curl away from the origin. For $(x, y) \neq = (0, 0)$ , a direct computation gives $\nabla \times F = (0, 0, 0)$ . Why this step? It looks conservative — the trap. But the field is undefined at the origin (denominator $x^{2} + y^{2} = 0$ ), which sits inside the disk. Stokes' domain condition 2 requires $F$ and its derivatives to exist on the whole surface. The flat disk contains the singularity → the shortcut is illegal here.

Step 2 — Do the line integral directly instead. Parametrize $r = (cos t, sin t, 0)$ , $d r = (- sin t, cos t, 0) d t$ . On the unit circle $x^{2} + y^{2} = 1$ , so $F = (- sin t, cos t, 0), F \cdot d r = sin^{2} t + cos^{2} t = 1.$ $\oint_{C} F \cdot d r = \int_{0}^{2 π} 1 d t = 2 π .$ Why this step? Because we can't build a valid surface through the origin, we fall back to the definition. The answer $2 π \neq = 0$ proves the field is not globally conservative despite zero curl — the classic warning from the parent note's last mistake box. (A region with a point removed is no longer simply connected, which is what breaks the "zero curl ⇒ zero circulation" rule.)

Verify: $\int_{0}^{2 π} 1 d t = 2 π$ . Also, curl at a generic point $(1, 1)$ evaluates to $(0, 0, 0)$ , confirming the "zero curl but nonzero circulation" paradox is caused by the excluded origin. ✓

Wrap-up: matrix coverage check

Recall Which example hit which cell?

A (flat, const curl) ::: Example 1 B (curved → swap surface) ::: Example 2 C (all three curl components) ::: Example 3 D (honest zero curl) ::: Example 4 E (sign/orientation flip) ::: Example 5 F (degenerate loop → curl at a point) ::: Example 6 G (real-world physics/EMF) ::: Example 7 H (singularity forces surface choice) ::: Example 8

Active-recall flashcards

If curl is zero everywhere on

S

but a singularity sits inside, is

\oint_{C} F \cdot d r

necessarily

0

No — the vortex field gives

2 π

; Stokes is invalid because the field isn't defined on all of

S

(domain condition 2 fails).

Reversing the traversal direction of

C

does what to the circulation?

Negates it (Example 5:

+ 12 \to - 12

lim_{ε \to 0} \frac{1}{π ε ^{2}} \oint_{C_{ε}} F \cdot d r

equals

(\nabla \times F) \cdot \hat{n}

at the point — the curl component (Example 6 gave

2

Why can you replace a paraboloid cap by a flat disk?

They share the same boundary

C

; RHS of Stokes depends on

S

only through its rim.

What does

d S

equal in terms of area?

d S = \hat{n} d A

— a scalar area turned into an arrow by the unit normal.