Unification — all three theorems as generalized Stokes: Worked examples

This page is the practice floor for the unification note. The parent told you why the four classical theorems are one identity $\int_{\partial M} ω = \int_{M} d ω$ . Here we run the identity by hand across every situation it can face — every dimension, every sign of orientation, every degenerate input — and we always compute both sides to watch them agree.

The scenario matrix

Before computing, let us name every kind of situation that this one identity has to handle. Each row is a "cell" the reader might land on; each worked example below is tagged with the cell it covers.

Cell	Dimension $k$	The twist being tested	Covered by
A	$k = 1$ (FTC)	orientation sign at endpoints, $f (b) - f (a)$	Example 1
B	$k = 2$ Green	curl term $Q_{x} - P_{y} \neq = 0$ (genuine circulation)	Example 2
C	$k = 2$ Green	degenerate: $Q_{x} - P_{y} = 0$ everywhere (conservative field) → answer $0$	Example 3
D	$k = 2$ Green as flux / 2D-divergence	expanding radial field, both readings	Example 4
E	$k = 3$ Divergence	full 3D flux out of a solid; boundary vs inside	Example 5
F	$k = 3$ Divergence	degenerate: $\nabla \cdot F = 0$ (incompressible) → net flux $0$	Example 6
G	any $k$	$d^{2} = 0$ trap: integrand is already an exact derivative of a derivative → $0$	Example 7
H	$k = 2$ Stokes (curl, surface in 3D)	non-flat surface, only boundary matters	Example 8
I	word problem	fluid leaving a tank (real-world divergence)	Example 9
J	exam twist	wrong orientation deliberately, watch the sign flip	Example 10

We will use these background rules the whole way (from Exterior Derivative and Differential Forms and the Wedge Product):

Cell A — Dimension 1: the Fundamental Theorem in Stokes clothing

Worked example Example 1 (Cell A)

Let $M = [1, 3]$ oriented left-to-right, and $ω = f$ the $0$ -form $f (x) = x^{2}$ . Verify $\int_{\partial M} ω = \int_{M} d ω$ .

Forecast: guess the single number both sides will give before reading on. (Hint: it's the change in $x^{2}$ across the interval.)

Step 1 — Find the boundary as signed points. The boundary of the oriented interval $[1, 3]$ is $\partial M = {3} - {1}$ : the endpoint you arrive at counts $+$ , the one you leave counts $-$ . Why this step? In dimension 1 the "boundary" is not an edge to walk — it is two points, and induced orientation is exactly the $\pm$ tags. That minus sign is the whole content of FTC.

Step 2 — Integrate the $0$ -form over those points (left side). Integrating a $0$ -form over a point just evaluates it: $\int_{\partial M} f = f (3) - f (1) = 9 - 1 = 8.$ Why this step? There is no length to integrate against at a point; the signed sum of values IS the integral.

Step 3 — Compute $d ω$ and integrate over the inside (right side). $d ω = f^{'} (x) d x = 2 x d x$ , so $\int_{M} d ω = \int_{1}^{3} 2 x d x = [x^{2}]_{1}^{3} = 9 - 1 = 8.$ Why this step? $d$ on a function is its ordinary derivative-times- $d x$ ; the right side is the familiar Riemann integral of $f^{'}$ .

Verify: left $= 8$ , right $= 8$ . ✅ This is literally $\int_{a}^{b} f^{'} = f (b) - f (a)$ — the Fundamental Theorem of Calculus is the $k = 1$ costume.

Cell B — Green with genuine circulation

Worked example Example 2 (Cell B)

Let $ω = - y d x + x d y$ over the unit disk $M$ (boundary = unit circle, counterclockwise). Verify Stokes.

Forecast: this field swirls counterclockwise. Do you expect the circulation to be positive, negative, or zero?

Step 1 — Read off $P$ and $Q$ . With $ω = P d x + Q d y$ we have $P = - y$ , $Q = x$ . Why this step? Green's Theorem needs the two component functions before we can build the curl term.

Step 2 — Build $d ω$ using antisymmetry. $d ω = (Q_{x} - P_{y}) d x \land d y = (1 - (- 1)) d x \land d y = 2 d x \land d y .$ Why this step? The four terms of $d P \land d x + d Q \land d y$ collapse to one because $d x \land d x = d y \land d y = 0$ and $d y \land d x = - d x \land d y$ . The surviving coefficient $Q_{x} - P_{y}$ is the 2D curl.

Step 3 — Integrate the inside (right side). $\int_{M} 2 d x d y = 2 \cdot (area of unit disk) = 2 π .$ Why this step? The integrand is the constant $2$ , so the double integral is just $2$ times the region's area; the unit disk has area $π$ . This is the right-side "add up the curl over every patch" reading.

Step 4 — Integrate around the boundary (left side). Parametrize the circle $x = cos t, y = sin t$ , $t \in [0, 2 π]$ , so $d x = - sin t d t$ , $d y = cos t d t$ : $\oint_{\partial M} (- y d x + x d y) = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = \int_{0}^{2 π} 1 d t = 2 π .$ Why this step? Walking the boundary directly is the "left side" reading; it must equal the inside sum.

Verify: both sides $= 2 π$ . ✅ Positive, as forecast — the field really does circulate counterclockwise. (This field is the classic "twice the area" detector.)

Cell C — Green, degenerate: a conservative field gives zero

Worked example Example 3 (Cell C)

Let $ω = 2 x y d x + x^{2} d y$ around any closed loop $\partial M$ (take the unit square for concreteness). Verify the answer is $0$ .

Forecast: notice $ω = d (x^{2} y)$ . What must the loop integral of an exact form be?

Step 1 — Compute the curl term. $P = 2 x y, Q = x^{2}$ , so $Q_{x} - P_{y} = 2 x - 2 x = 0$ . Why this step? When $Q_{x} - P_{y} \equiv 0$ the field is conservative (curl-free); this is the degenerate cell where the inside integrand vanishes identically.

Step 2 — Right side. $\int_{M} 0 d x d y = 0.$ Why this step? The inside integrand $Q_{x} - P_{y}$ is identically zero, so no matter which region $M$ the loop encloses, summing zero over every patch gives zero. The answer does not depend on the loop's shape.

Step 3 — Confirm with the exactness view. Since $ω = d (x^{2} y)$ is exact, applying $d$ again gives $d ω = d (d (x^{2} y)) = 0$ by $d^2=0$. So the right side is $\int_{M} d ω = \int_{M} d (d (x^{2} y)) = \int_{M} 0 = 0,$ which matches Step 2 and forces the boundary integral $\oint_{\partial M} ω = 0$ too. Why this step? It shows the zero is structural, not lucky: an exact form's exterior derivative vanishes ( $d^{2} = 0$ ), so its inside integral — and hence its loop integral — must be zero for every closed loop.

Verify: both readings give $0$ . ✅ Degenerate cell handled — a conservative field around any closed curve is always zero.

Cell D — 2D divergence: expanding radial field, both readings

Figure — Unification — all three theorems as generalized Stokes

Worked example Example 4 (Cell D)

Let $F = (x, y)$ on the unit disk. Compute the outward flux two ways. (This is the parent note's unification example — worked here in full.)

Forecast: the arrows all point outward and grow with radius. Positive or negative net flux?

Step 1 — Right side: divergence over the inside. $\nabla \cdot F = \partial_{x} x + \partial_{y} y = 1 + 1 = 2$ , so $\iint_{M} 2 d A = 2 \cdot π = 2 π .$ Why this step? In 2D the flux form's exterior derivative is $(\nabla \cdot F) d x \land d y$ ; look at the red outward arrows in the figure — the divergence measures how much they spread apart.

Step 2 — Left side: flux across the boundary circle. On the unit circle the outward unit normal is $\overset{n}{^} = (cos t, sin t)$ , and $F = (cos t, sin t)$ there, so $F \cdot \overset{n}{^} = cos^{2} t + sin^{2} t = 1$ . Arclength element is $d s = d t$ here and the total length is $2 π$ : $\oint_{\partial M} F \cdot \overset{n}{^} d s = \int_{0}^{2 π} 1 d t = 2 π .$ Why this step? The teal boundary arrows in the figure are the flux crossing the fence — this is the "walk the edge" reading, integrating flux density $F \cdot \overset{n}{^}$ against arclength $d s$ .

Verify: both $= 2 π$ . ✅ Positive, as forecast: outward-spreading field ⇒ positive net outflow.

Cell E — Full 3D Divergence theorem

Worked example Example 5 (Cell E)

Let $F = (x, y, z)$ and $M$ the solid unit ball. Verify $\iint_{\partial M} F \cdot d S = ∭_{M} \nabla \cdot F d V$ .

Forecast: the divergence is a constant. Multiply it by the ball's volume — what do you get?

Step 1 — Build the flux 2-form and its $d$ . $ω = x d y \land d z + y d z \land d x + z d x \land d y$ , and by the parent's rule $d ω = (\partial_{x} x + \partial_{y} y + \partial_{z} z) d x \land d y \land d z = 3 d V .$ Why this step? Each term of $d$ survives only when it hits the missing coordinate; the others wedge to zero. The surviving sum is the divergence — see Divergence Theorem.

Step 2 — Right side: integrate over the solid ball. Volume of the unit ball is $\frac{4}{3} π$ , so $∭_{M} 3 d V = 3 \cdot \frac{4}{3} π = 4 π .$ Why this step? The integrand is the constant divergence $3$ , so the triple integral is just $3$ times the ball's volume; this is the right-side "add up the source strength over every little chunk of the solid" reading.

Step 3 — Left side: flux out of the sphere. On the unit sphere the outward normal is $\overset{n}{^} = (x, y, z)$ (the position vector), so $F \cdot \overset{n}{^} = x^{2} + y^{2} + z^{2} = 1$ . Surface area of the unit sphere is $4 π$ , and $d S = \overset{n}{^} d S$ : $\iint_{\partial M} F \cdot d S = \int_{sphere} 1 d S = 4 π .$ Why this step? On the boundary the field is exactly the outward normal, so flux density is $1$ everywhere; total flux = surface area.

Verify: both $= 4 π$ . ✅ Constant divergence $\times$ volume matches the surface flux — the $k = 3$ costume works.

Cell F — 3D, degenerate: incompressible field, net flux zero

Worked example Example 6 (Cell F)

Let $F = (y, - x, 7)$ over any closed surface (take the unit ball). Verify net flux is $0$ from the divergence.

Forecast: the third component is a constant $7$ . Does a constant contribute to divergence?

Step 1 — Compute the divergence. $\nabla \cdot F = \partial_{x} (y) + \partial_{y} (- x) + \partial_{z} (7) = 0 + 0 + 0 = 0.$ Why this step? A field with zero divergence is incompressible — whatever flows into any region flows back out. This is the degenerate 3D cell.

Step 2 — Right side. $∭_{M} 0 d V = 0.$ Why this step? The inside integrand is the divergence, which is identically zero here, so summing it over every chunk of the solid gives zero regardless of the ball's size — the right side collapses to $0$ before we ever touch the surface.

Step 3 — Interpret the boundary side. By the Divergence Theorem, the total outward flux across the closed sphere must therefore also be $0$ : any patch's outflow is exactly cancelled by inflow elsewhere. Why this step? It shows zero divergence forces zero net flux regardless of the surface's shape — a strong structural fact, not a lucky cancellation.

Verify: flux $= 0$ . ✅ Degenerate 3D case handled. (Even the constant $7$ contributes nothing — its derivative is zero.)

Cell G — The $d^{2} = 0$ trap

Worked example Example 7 (Cell G)

Let $f (x, y) = x^{3} y - sin (x y)$ . Show that $\int_{M} d (df) = 0$ over any region $M$ , and that it equals $\oint_{\partial M} df$ .

Forecast: $df$ is already a derivative. What does the parent note say about applying $d$ twice?

Step 1 — First derivative. $df = f_{x} d x + f_{y} d y .$ Why this step? We need the $1$ -form $df$ before applying $d$ again.

Step 2 — Second derivative, watch the mixed partials. $d (df) = (f_{y x} - f_{x y}) d x \land d y .$ By equality of mixed partials $f_{x y} = f_{y x}$ , so $d (df) = 0$ identically. Why this step? This is $d^2=0$: the exterior derivative applied twice always vanishes. It is $\nabla \times \nabla f = 0$ in form language.

Step 3 — Boundary side. Stokes says $\oint_{\partial M} df = \int_{M} d (df) = 0$ . Directly: $df$ is exact, and $\int_{\partial M} df = f (end) - f (start) = 0$ around any closed loop. Why this step? Both the inside ( $d^{2} = 0$ ) and the boundary ( $\partial^{2} = 0$ , a closed loop has no endpoints) give zero — the two facts are dual through Stokes.

Verify: $d (df) = 0$ and its loop integral $= 0$ . ✅

Cell H — Stokes (curl) on a curved surface: only the boundary matters

Worked example Example 8 (Cell H)

Let $F = (- y, x, 0)$ . Take $M$ = the upper unit hemisphere ( $z \geq 0$ ), whose boundary $\partial M$ is the equator circle $x^{2} + y^{2} = 1$ in the plane $z = 0$ , oriented counterclockwise seen from above. Verify Stokes' Theorem (curl).

Forecast: the surface is curved and bulges up. But the boundary is just the flat equator. Which one is easier — and must they still agree?

Step 1 — Compute the curl. $\nabla \times F = (\partial_{y} 0 - \partial_{z} x, \partial_{z} (- y) - \partial_{x} 0, \partial_{x} x - \partial_{y} (- y)) = (0, 0, 2) .$ Why this step? Stokes' curl theorem integrates $(\nabla \times F) \cdot \overset{n}{^}$ over the surface; we need the curl vector first. It points straight up with magnitude $2$ .

Step 2 — Right side, using the flat-cap shortcut. Here is the payoff of Stokes: the surface integral of the curl depends only on the boundary, so we may replace the bulging hemisphere by the flat disk it bounds (same equator, from Orientation and Induced Boundary Orientation). On the flat disk $\overset{n}{^} = (0, 0, 1)$ , so $(\nabla \times F) \cdot \overset{n}{^} = 2$ and $d S = \overset{n}{^} d S$ : $\iint_{disk} (\nabla \times F) \cdot d S = \iint_{disk} 2 d S = 2 \cdot π = 2 π .$ Why this step? The plum boundary circle in the figure is the only thing the integral "sees"; the curved cap and the flat cap share it, so both give the same number.

Step 3 — Left side: circulation around the equator. With $x = cos t, y = sin t, z = 0$ : $F = (- sin t, cos t, 0)$ , $d r = (- sin t, cos t, 0) d t$ : $\oint_{\partial M} F \cdot d r = \int_{0}^{2 π} (sin^{2} t + cos^{2} t) d t = 2 π .$ Why this step? This is the "walk the boundary" left-side reading: we add up the field's push along the equator, and by Stokes it must equal the curl flux computed in Step 2.

Verify: both $= 2 π$ . ✅ The curved surface and the flat disk agree because Stokes only cares about the shared boundary.

Cell I — Word problem: fluid leaving a tank

Worked example Example 9 (Cell I)

Water in a cubical tank $[0, 1]^{3}$ (metres) has velocity field $v = (x, 2 y, 3 z)$ in metres per second. How many cubic metres per second leave the tank in total?

Forecast: the field speeds up in every direction, so water is being "created" (a source). Positive net outflow — but how much?

Step 1 — Recognise this as a flux question. Net outflow $= \iint_{\partial M} v \cdot d S$ , the flux of $v$ across the cube's surface, where $d S = \overset{n}{^} d S$ measures how much water crosses each area patch $d S$ . Why this step? "How much leaves per second across the boundary" is exactly the left side of the Divergence Theorem.

Step 2 — Convert to the easy inside integral. $\nabla \cdot v = 1 + 2 + 3 = 6 s^{- 1}$ (a constant), so $∭_{M} 6 d V = 6 \cdot (1)^{3} = 6.$ Why this step? Summing over six faces is tedious; the divergence theorem trades the surface for the volume, where the constant integrand makes it trivial — just $6$ times the cube's volume $1 m^{3}$ .

Verify — including units: $6 s^{- 1} \times 1 m^{3} = 6 m^{3} / s$ . Positive as forecast. ✅ The tank sheds $6$ cubic metres per second (a source field). The units work: $[s^{- 1}] \cdot [m^{3}] = [m^{3} / s]$ , a flow rate.

Cell J — Exam twist: wrong orientation flips the sign

Worked example Example 10 (Cell J)

Reconsider Example 2 ( $ω = - y d x + x d y$ , unit disk), but a student walks the boundary clockwise (the reversed orientation). What do they get, and why is it "wrong"?

Forecast: the field's true circulation is $+ 2 π$ . Reversing the walk should do what?

Step 1 — Reverse the parametrization. Clockwise: $x = cos t, y = - sin t$ , so $d x = - sin t d t$ , $d y = - cos t d t$ : $\oint_{cw} (- y d x + x d y) = \int_{0}^{2 π} (- sin^{2} t - cos^{2} t) d t = - 2 π .$ Why this step? Reversing the direction of travel negates every $d r$ , flipping the integral's sign.

Step 2 — Compare with the inside integral. The right side is fixed: $\int_{M} 2 d A = + 2 π$ (the region and its induced orientation didn't change). Why this step? Stokes only holds when $\partial M$ carries the induced orientation. The clockwise walk uses the opposite orientation, so $\int_{\partial M} ω = - 2 π \neq = + 2 π$ .

Step 3 — The fix. The induced boundary orientation for a region in the plane is counterclockwise ("region on your left"). Using it restores $+ 2 π = + 2 π$ . Why this step? This is the "ignore orientation" mistake from the parent note made concrete: a sign, not a magnitude, error — and it is total, not partial.

Verify: clockwise gives $- 2 π$ ; correct (ccw) gives $+ 2 π = \int_{M} d ω$ . ✅ Orientation is not decorative — it owns the sign.

Recall Which cell is which — quick self-test

Field $(x, y)$ on the disk, flux ::: Cell D, answer $2 π$ Field $(x, y, z)$ on the ball, flux ::: Cell E, answer $4 π$ $Q_{x} - P_{y} = 0$ around a closed loop ::: Cell C, answer $0$ $\nabla \cdot F = 0$ over a closed surface ::: Cell F, net flux $0$ Reversing boundary direction ::: Cell J, flips the sign of $\int_{\partial M} ω$ Integrating $d (df)$ over any region ::: Cell G, always $0$ by $d^{2} = 0$

Connections

Unification — all three theorems as generalized Stokes — the parent identity these examples exercise
Fundamental Theorem of Calculus — Cell A
Green's Theorem — Cells B, C, D
Stokes' Theorem (curl) — Cell H
Divergence Theorem — Cells E, F, I
Exterior Derivative and d squared equals zero — Cell G
Differential Forms and the Wedge Product — every $d ω$ computation
Orientation and Induced Boundary Orientation — Cell J