Chapter 13 · Line integrals and Green's theorem

The hook

The planimeter: area from the edge alone

A surveyor needs the area of an irregular field — a lake, a county, a leaf. There is a brass instrument, the planimeter, that gives it exactly: you trace its pointer once around the boundary, and a dial reads off the enclosed area. The arm never crosses the interior. It never knows what is inside. It only walks the fence-line — and somehow that is enough.

Traced the edge. Got the area. A one-dimensional walk around the rim recovers a two-dimensional quantity — the area inside. The boundary already encodes what the interior contains.

That is not a gimmick of brass and gears; it is a theorem. By the end of this chapter you will be able to derive the planimeter from

A = \tfrac12\oint_C (x\,dy - y\,dx),

an area computed from a pure boundary integral. And that single idea — the boundary knows what the inside contains — is the spine of everything here: line integrals, Green's theorem, and, in Beat 5, the moment the two halves of this entire course turn out to be one subject.

Why this matters

A line integral adds up a quantity along a path — most physically, the work a force field does on something moving through it. Green's theorem then says a loop integral around a closed curve equals an ordinary double integral over the region it encloses: it trades a walk around the edge for a sweep of the inside, in whichever direction is easier. That trade is one of the most useful moves in all of applied mathematics — and it is about to explain why Unit 1's contour integrals worked at all.

Work along a path

Line integrals: adding a field along a path

A vector field $\mathbf F = (P, Q) = P\,\mathbf i + Q\,\mathbf j$ assigns an arrow to every point — a force, a flow, a wind. The line integral of $\mathbf F$ along a curve $C$ totals the field's pull in the direction you travel:

Line integral = work done by

\mathbf F

along

C

\int_C \mathbf F\cdot d\mathbf r = \int_C P\,dx + Q\,dy, \qquad d\mathbf r = (dx, dy).

It is computed by one unbreakable ritual — the 1-D cousin of Chapter 10's sketch–shadow–slice:

The line-integral ritual

Parametrize the curve: $\mathbf r(t) = (x(t), y(t))$ , $t: a \to b$ .
Substitute everything into $t$ : $P$ , $Q$ , and $dx = x'(t)\,dt$ , $dy = y'(t)\,dt$ .
Integrate the single ordinary integral in $t$ from $a$ to $b$ .

Worked example · WE-1 · the ritual, once around a circle

Find the work of the rotational field $\mathbf F = (-y, x)$ once around the unit circle, counterclockwise. Parametrize: $\mathbf r(t) = (\cos t, \sin t)$ , $t: 0 \to 2\pi$ , so $dx = -\sin t\,dt$ , $dy = \cos t\,dt$ . Substitute: $P = -y = -\sin t$ , $Q = x = \cos t$ :

\int_C P\,dx + Q\,dy = \int_0^{2\pi}\big[(-\sin t)(-\sin t) + (\cos t)(\cos t)\big]\,dt = \int_0^{2\pi}(\sin^2 t + \cos^2 t)\,dt = \int_0^{2\pi} 1\,dt = 2\pi.

The field pushes you along at every step, so the work is positive all the way round. Hold onto the $2\pi$ — in Beat 3, Green's theorem gets this same answer in a single line.

Worked example · WE-2 · the thing school integrals never showed: path matters

Evaluate $\displaystyle\int_C x\,dy$ from $(0,0)$ to $(1,1)$ along two different paths.

Along the parabola

y = x^2

$\mathbf r(t) = (t, t^2)$ , $dy = 2t\,dt$ :

\int_0^1 t\cdot 2t\,dt = \int_0^1 2t^2\,dt = \frac{2}{3}.

Along the line

y = x

$\mathbf r(t) = (t, t)$ , $dy = dt$ :

\int_0^1 t\,dt = \frac{1}{2}.

Same endpoints, different values — $\tfrac23 \ne \tfrac12$ . A line integral depends on the path, not just where it starts and ends. (The special fields for which it doesn't — the conservative ones — are Beat 4, and they are exactly Chapter 7's path-independent integrals in disguise.)

The boundary equals the inside

Green's theorem: the edge equals the inside

Here is the planimeter's engine, stated in full. For a positively oriented (counterclockwise) simple closed curve $C$ bounding a region $R$ :

Green's theorem

\oint_C P\,dx + Q\,dy = \iint_R\!\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA.

The boundary loop integral on the left equals the double integral of one specific combination of derivatives — the field's "curl" — over the whole interior on the right.

Orientation is the mark-loser

Positive orientation means the region is on your left as you walk $C$ — counterclockwise for an ordinary curve. That direction is built into the theorem. Walk it clockwise and the whole integral changes sign. Every term is right and the answer is still wrong by a minus sign. The fix is a reflex: region on the left, or flip the sign.

The payoff promised in Beat 2: re-do WE-1 with Green's theorem. For $\mathbf F = (-y, x)$ , $\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = 1 - (-1) = 2$ , so

\oint_C (-y)\,dx + x\,dy = \iint_R 2\,dA = 2\cdot\text{Area}(R) = 2\pi.

One line, no parametrization. The boundary integral became "twice the area of the disk."

Worked example · WE-3 · Green both ways (the match is the theorem)

Evaluate $\displaystyle\oint_C (xy\,dx + x^2\,dy)$ over the triangle with vertices $(0,0)$ , $(1,0)$ , $(1,1)$ , counterclockwise. The point is to compute it both ways and watch them agree.

As a double integral (Green's)

$Q_x - P_y = 2x - x = x$ . The triangle is $0 \le x \le 1$ , $0 \le y \le x$ :

\iint_R x\,dA = \int_0^1\!\!\int_0^x x\,dy\,dx = \int_0^1 x^2\,dx = \frac{1}{3}.

As three edge integrals (direct)

Bottom $(0,0)\!\to\!(1,0)$ : $y=0,\,dy=0 \Rightarrow 0$ . Right $(1,0)\!\to\!(1,1)$ : $x=1,\,dx=0 \Rightarrow \int_0^1 1\,dy = 1$ . Diagonal back $(1,1)\!\to\!(0,0)$ : $\int (xy\,dx + x^2\,dy)$ with $x=y=1-t$ gives $-\tfrac23$ . Total $0 + 1 - \tfrac23 = \tfrac13$ .

Both give $\tfrac13$ . That agreement is Green's theorem — and it tells you which side to compute: pick whichever is easier. A messy loop integral becomes a clean area integral, or a hard area integral becomes an easy boundary walk.

End of sitting 1 — you can parametrize and evaluate a line integral, you know a line integral depends on its path, and you can run Green's theorem in both directions with the orientation sign under control. After the break: the area form (the planimeter, paid off), conservative fields — and then the reveal that the two units of this course were the same subject all along.

Two consequences

The area form, and fields that forget the path

Choose $P = -y$ , $Q = x$ in Green's theorem: $Q_x - P_y = 2$ , so $\oint_C (-y\,dx + x\,dy) = 2\,\text{Area}(R)$ . Halve it and you have the planimeter, exact:

Area from the boundary (the planimeter)

A = \tfrac12\oint_C (x\,dy - y\,dx) \qquad (\text{CCW}).

Worked example · WE-4 · the planimeter, paid off

Find the area of the ellipse $x = 2\cos t$ , $y = \sin t$ ( $t: 0 \to 2\pi$ ) from the boundary alone. With $dx = -2\sin t\,dt$ , $dy = \cos t\,dt$ ,

x\,dy - y\,dx = 2\cos t\cdot\cos t\,dt - \sin t\cdot(-2\sin t)\,dt = 2(\cos^2 t + \sin^2 t)\,dt = 2\,dt,

A = \tfrac12\int_0^{2\pi} 2\,dt = 2\pi = \pi a b\ \ (a=2,\,b=1).

The integrand collapsed to a constant, and the boundary trace returned $\pi ab$ — the ellipse-area formula, derived, not memorised.

The second consequence answers WE-2's loose end. Some fields give the same line integral on every path between two points; these are the conservative fields. On a simply-connected region the test is one equality of partials:

Conservative field

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \;\Longrightarrow\; \mathbf F = \nabla f, \quad \int_C \mathbf F\cdot d\mathbf r = f(\text{end}) - f(\text{start}), \quad \oint_C \mathbf F\cdot d\mathbf r = 0.

Worked example · WE-4b · a conservative field

Take $\mathbf F = (2xy,\ x^2)$ . Test: $P_y = 2x$ and $Q_x = 2x$ — equal, so it is conservative. A potential is $f = x^2 y$ (check: $f_x = 2xy$ , $f_y = x^2$ ). Therefore every closed loop integral is $0$ , and every path from one point to another gives the same value $f(\text{end}) - f(\text{start})$ .

You have seen this before

"Path-independent, and $\int_C = f(\text{end}) - f(\text{start})$ " should ring a bell. It is Chapter 7's path independence — $\int_C f\,dz = F(z_1) - F(z_0)$ for an analytic $f$ with antiderivative $F$ — wearing real-variable clothes. That is not a coincidence. The next beat shows exactly why.

The reveal · the course's payoff

The two units were one subject all along

Unit 1 was complex analysis; Unit 2 is vector calculus. They have looked like different courses. They are not. Watch a complex contour integral come apart into two real line integrals. Write $f = u + iv$ and $dz = dx + i\,dy$ ; multiply out:

\oint_C f\,dz = \oint_C (u + iv)(dx + i\,dy) = \underbrace{\oint_C u\,dx - v\,dy}_{\text{real part}} + \,i\underbrace{\oint_C v\,dx + u\,dy}_{\text{imaginary part}}.

Each of those is a real line integral of exactly the kind in Beat 2 — so apply Green's theorem to each. The real part has $P = u$ , $Q = -v$ ; the imaginary part has $P = v$ , $Q = u$ :

\text{real} \to \iint_R\big[(-v)_x - u_y\big]\,dA = \iint_R(-v_x - u_y)\,dA, \qquad \text{imag} \to \iint_R\big[u_x - v_y\big]\,dA.

Now the key move. If $f$ is analytic, $u$ and $v$ obey the Cauchy–Riemann equations from Chapter 3: $\;u_x = v_y\;$ and $\;u_y = -v_x$ . Substitute — and watch both interior integrands collapse to zero:

Cauchy–Riemann kills both integrands

Real part: $-v_x - u_y$ . The Cauchy–Riemann equation $u_y = -v_x$ means $-u_y = v_x$ , so $-v_x - u_y = -v_x + v_x = 0$ .
Imaginary part: $u_x - v_y$ . The Cauchy–Riemann equation $u_x = v_y$ makes this $v_y - v_y = 0$ .

Both double integrals are integrals of zero, hence zero. So for any analytic $f$ and any simple closed curve $C$ ,

\boxed{\ \oint_C f\,dz = 0.\ }

That is Cauchy's theorem (Cauchy–Goursat) — the cornerstone of Unit 1. And we just produced it out of Green's theorem plus the Cauchy–Riemann equations. Green's theorem $+$ the CR equations is exactly Cauchy–Goursat. The contour integrals of complex analysis were real line integrals the whole time; the analyticity that made them vanish was the Cauchy–Riemann equations forcing a curl of zero. Two units, one idea.

Worked example · WE-5 · where the singularity lives

Concretely, $\displaystyle\oint_{|z|=1} z\,dz = 0$ : $z$ is analytic everywhere, both real integrands vanish, done. But the famous $\displaystyle\oint_{|z|=1}\frac{1}{z}\,dz = 2\pi i$ is not zero. What broke?

$\frac1z$ blows up at the origin — and the unit circle encloses the origin. Green's theorem requires $\mathbf F$ to be smooth ( $C^1$ ) throughout $R$ ; at the puncture it isn't, so the hypothesis fails. The leftover $2\pi i$ is precisely the contribution of that one bad point.

The contour encloses the puncture. Green's theorem can't see across the hole — and the

2\pi i

it misses is the hole's whole story.

This is why Unit 1 needed residues. The enclosed singularity is exactly the $\oint (z - z_0)^n\,dz = 2\pi i$ (only when $n = -1$ ) you met in Chapter 7, and the residue theorem of Chapter 9 is the systematic bookkeeping for "how much do the enclosed bad points add back?" Green's theorem handles the smooth interior; residues handle the holes.

Interactive playground

The Green Machine

Pick a field and a closed curve, commit to an orientation, and predict the sign of the loop value before you reveal it. The Machine then shows the answer two ways at once — the direct loop integral and Green's double integral — side by side. They agree… except on the cells where the field has a singularity the curve encloses, where Green's theorem visibly fails and the Machine names exactly why. That failure is Beat 5, made interactive.

The Green Machine

Field:

Closed curve:

Orientation:

Predict the sign:

field & curve

Watch the vortex field $\dfrac{(-y, x)}{x^2+y^2}$ on a curve that encloses the origin: the two columns disagree. That gap is the singularity — the same $2\pi$ that becomes the $2\pi i$ of $\oint dz/z$ in Unit 1.

What to notice

Direct and Green agree — usually. For smooth fields the boundary walk and the area sweep give the same number; that is the theorem doing its job.
Clockwise flips the sign. The magnitude is unchanged; only the sign turns over. Region-on-the-left is the whole rule.
Enclose a singularity and Green's breaks. The vortex around the origin: Green's says $0$ , the true loop is $2\pi$ . The Machine flags it — that missing $2\pi$ is why residues exist.

Retrieval gauntlet

Commit to an answer before reading the feedback. These interleave the mechanics with the orientation trap and the reveal on purpose.

Quiz 1 · Green's theorem

$\displaystyle\oint_C (xy\,dx + x^2\,dy)$ over the triangle $(0,0), (1,0), (1,1)$ , counterclockwise, equals:

Quiz 2 · orientation — the classic lost mark

The same integral $\displaystyle\oint_C (xy\,dx + x^2\,dy)$ over that triangle, but taken clockwise, equals:

Predict the next step. Green's theorem is stated for a counterclockwise curve. Before revealing: when you instead walk the boundary clockwise, which way does the value go — and why?

Reversing the direction of travel reverses the sign of every $dx$ and $dy$ , so the whole line integral negates: $\oint_{\text{CW}} = -\oint_{\text{CCW}}$ . Green's theorem computes the counterclockwise value (region on the left); the clockwise loop is its negative. Here $\oint_{\text{CCW}} = \tfrac13$ , so $\oint_{\text{CW}} = -\tfrac13$ . The magnitude never changes — only the sign.

Quiz 3 · the area form

The area enclosed by the ellipse $x = 3\cos t$ , $y = 2\sin t$ , computed by $\tfrac12\oint_C (x\,dy - y\,dx)$ , is:

Quiz 4 · which fields are conservative? — decide, don't integrate

Which of these are conservative (path-independent)? (i) $(2xy,\ x^2)$ ; (ii) $(-y,\ x)$ ; (iii) $(y,\ x)$ .

Quiz 5 · the reveal, made concrete

The real vortex field $\left(\dfrac{-y}{x^2+y^2},\, \dfrac{x}{x^2+y^2}\right)$ integrated once counterclockwise around $|z| = 1$ gives:

Predict the next step (faded bridge). Green's theorem would turn this loop into

\iint_R(Q_x - P_y)\,dA

. Before revealing: compute

Q_x - P_y

for the vortex field away from the origin — and then explain why the loop is nonetheless

2\pi

, not

0

The curl is zero — almost everywhere. With $P = \dfrac{-y}{x^2+y^2}$ , $Q = \dfrac{x}{x^2+y^2}$ , a short quotient-rule computation gives

Q_x = \frac{y^2 - x^2}{(x^2+y^2)^2} = P_y, \qquad\text{so}\qquad Q_x - P_y = 0 \quad(\text{everywhere except the origin}).

So naïvely $\iint_R 0\,dA = 0$ . But that step is illegal: Green's theorem requires $P, Q$ to be $C^1$ on all of $R$ , and at the origin — which $C$ encloses — they are undefined. The theorem simply does not apply. Parametrize the actual loop instead ( $x = \cos t$ , $y = \sin t$ ) and it integrates to $\int_0^{2\pi} 1\,dt = 2\pi$ . The whole $2\pi$ is the contribution of the one excluded point — the residue.

Exam translator — what the paper actually asks

Exam shapes

In standard Indian engineering-math exams, and likely in MCC201A unless your class notes differ, this material wears five costumes — each one you have now seen worked:

"Evaluate $\int_C \mathbf F\cdot d\mathbf r$ " → parametrize → substitute → integrate (WE-1). First check whether $\mathbf F$ is conservative; if so it is path-independent.
"Verify Green's theorem for …" → compute both the line integral and the double integral and show they match (WE-3). The orientation sign is the classic lost mark.
"Using Green's theorem, evaluate $\oint$ …" → pick whichever side is easier; watch CCW vs CW.
"Find the area enclosed by …" → the area form $\tfrac12\oint(x\,dy - y\,dx)$ (WE-4).
"Explain Cauchy's theorem via Green's" → the Beat-5 derivation. This is examinable as a short theory question on the Unit-1 ↔ Unit-2 link.

Worked solutions are complete; nothing points off this page. Do them as listed — interleaved on purpose — not blocked by method. Two of them, PR5 and PR7, open with a faded bridge: part of the work is set out, the key step left for you to predict.

PR1 · a direct line integral

Shape · open path ⇒ parametrize → substitute → integrate

PR1. Evaluate $\displaystyle\int_C (x^2\,dx + xy\,dy)$ along $y = x^2$ from $(0,0)$ to $(1,1)$ .

$\mathbf r(t) = (t, t^2)$ , $dx = dt$ , $dy = 2t\,dt$ :

\int_0^1\big[t^2 + t\cdot t^2\cdot 2t\big]\,dt = \int_0^1 (t^2 + 2t^4)\,dt = \frac13 + \frac25 = \frac{11}{15}.

PR2 · Green's theorem, with the sign live

Shape · closed CCW curve ⇒ Green's; mind Q_x − P_y

PR2. Evaluate $\displaystyle\oint_C (y^2\,dx + xy\,dy)$ over the triangle $(0,0), (1,0), (0,1)$ , counterclockwise.

$Q_x - P_y = y - 2y = -y$ . The triangle is $0 \le x \le 1$ , $0 \le y \le 1-x$ :

\iint_R (-y)\,dA = -\int_0^1\!\!\int_0^{1-x} y\,dy\,dx = -\int_0^1 \frac{(1-x)^2}{2}\,dx = -\frac{1}{6}.

The negative answer is honest — the curl is negative over this region; nothing went wrong.

PR3 · the area form on a cusped curve

Shape · area enclosed ⇒ ½∮(x dy − y dx)

PR3. Find the area enclosed by the astroid $x = \cos^3 t$ , $y = \sin^3 t$ ( $t: 0 \to 2\pi$ ).

With $dx = -3\cos^2 t\sin t\,dt$ , $dy = 3\sin^2 t\cos t\,dt$ , the integrand $x\,dy - y\,dx = 3\cos^2 t\sin^2 t\,dt = \tfrac34\sin^2(2t)\,dt$ . So

A = \tfrac12\int_0^{2\pi}\tfrac34\sin^2(2t)\,dt = \tfrac38\cdot\pi = \frac{3\pi}{8}.

The area form handles the four cusps without any need to describe the interior region.

PR4 · decision-only · which tool, and why

Shape · no integration — name direct vs Green's (and the trap)

PR4. For each, say whether you would use a direct line integral or Green's theorem, and why. (a) $\oint$ of a simple field over a wiggly closed curve; (b) $\int$ along one straight segment; (c) $\oint$ of the vortex $(-y, x)/(x^2+y^2)$ around a curve enclosing the origin.

(a) Green's theorem. The curve is closed and wiggly; the area side is far simpler than parametrizing the boundary.

(b) Direct. A single open segment is not a closed curve, so Green's does not apply — parametrize and integrate.

(c) Neither naïvely. Green's theorem fails — the enclosed origin is a singularity, so its hypothesis breaks. The value is $2\pi$ by the Beat-5 result: the residue at the puncture, not anything Green's can compute directly.

PR5 · faded bridge · path independence

Shape · P_y = Q_x ⇒ conservative ⇒ value is f(end) − f(start)

PR5. For $\mathbf F = (2xy + y^2,\ 2xy + x^2)$ , evaluate $\int_C \mathbf F\cdot d\mathbf r$ from $(0,0)$ to $(1,1)$ .

Bridge: first test $P_y$ vs $Q_x$ . If they match, the field is conservative and you need a potential $f$ with $f_x = P$ , $f_y = Q$ — then the path doesn't matter. Predict the next step: is it conservative, and what is $f$ ?

Conservative? $P_y = 2x + 2y$ and $Q_x = 2x + 2y$ — equal, yes. Potential: $f = x^2 y + x y^2$ (check: $f_x = 2xy + y^2$ , $f_y = x^2 + 2xy$ ). So the integral is path-independent:

\int_C \mathbf F\cdot d\mathbf r = f(1,1) - f(0,0) = (1 + 1) - 0 = 2.

PR6 · Green's, polar on the area side

Shape · circular region ⇒ Green's, then polar for ∬

PR6. Evaluate $\displaystyle\oint_C (-y^3\,dx + x^3\,dy)$ over the unit circle, counterclockwise.

$Q_x - P_y = 3x^2 - (-3y^2) = 3(x^2 + y^2)$ . Over the unit disk this begs for polar, $x^2 + y^2 = r^2$ , $dA = r\,dr\,d\theta$ :

\iint_R 3(x^2+y^2)\,dA = \int_0^{2\pi}\!\!\int_0^1 3r^2\cdot r\,dr\,d\theta = 2\pi\cdot 3\cdot\tfrac14 = \frac{3\pi}{2}.

Green's turned a cubic boundary integral into a one-line polar area integral — Chapters 10 and 13 working together.

PR7 · faded bridge · the reveal, applied

Shape · ∮ f dz via two real line integrals + Green + CR

PR7. Using $\oint_C f\,dz = \oint_C(u\,dx - v\,dy) + i\oint_C(v\,dx + u\,dy)$ , show that $\displaystyle\oint_{|z|=1} z^2\,dz = 0$ .

Bridge: for $f = z^2$ , $u = x^2 - y^2$ and $v = 2xy$ . Predict the next step: what are the two Green integrands $-v_x - u_y$ and $u_x - v_y$ , and why do they vanish?

With $u = x^2 - y^2$ , $v = 2xy$ : $u_x = 2x$ , $u_y = -2y$ , $v_x = 2y$ , $v_y = 2x$ . The two Green integrands are

\text{real: } -v_x - u_y = -2y - (-2y) = 0, \qquad \text{imag: } u_x - v_y = 2x - 2x = 0.

Both are zero because the Cauchy–Riemann equations hold — $z^2$ is analytic. So both double integrals vanish and $\oint_{|z|=1} z^2\,dz = 0$ , exactly as Cauchy–Goursat promises.

Pre-flight checklist — run it before you integrate

\textbf{1.}\ \text{Line integral: parametrized the curve, substituted } dx, dy \text{ in } t?

\textbf{2.}\ \text{Conservative first? If } P_y = Q_x,\ \text{the integral is path-independent.}

\textbf{3.}\ \text{Green's: is } C \text{ counterclockwise (region on your left)? If clockwise, negate.}

\textbf{4.}\ \text{Green's: is } \mathbf F \text{ smooth on ALL of } R?\ \text{A singularity enclosed breaks it (use residues).}

\textbf{5.}\ \text{Area form: did you keep the } \tfrac12 \text{ in } A = \tfrac12\oint(x\,dy - y\,dx)?

Three things to carry out of this chapter

A line integral is work along a path; parametrize, substitute, integrate. Its value depends on the path — unless the field is conservative ( $P_y = Q_x$ ), when it depends only on the endpoints.
Green's theorem trades the boundary for the inside: $\oint_C P\,dx + Q\,dy = \iint_R (Q_x - P_y)\,dA$ , counterclockwise — clockwise negates. Compute whichever side is easier; the area form $\tfrac12\oint(x\,dy - y\,dx)$ is the special case that runs the planimeter.
Green's theorem $+$ the Cauchy–Riemann equations $=$ Cauchy's theorem. A complex contour integral is two real line integrals; analyticity makes both curls vanish, so $\oint f\,dz = 0$ — and an enclosed singularity is the one thing that breaks it, which is exactly what residues repair.

The anchor, one more time

Back to the surveyor. The planimeter recovers an area by walking only the fence-line because the boundary genuinely encodes the interior — that is Green's theorem in brass. The same idea, pushed into the complex plane, is why a contour integral of an analytic function is zero, and why the only thing that can make it nonzero is a hole the contour surrounds. Re-derive the Beat-5 result and re-do PR5 and PR7 from a blank page tomorrow — not re-read, re-do. The link between the two units is the single most important idea in this course.

What's next. Chapter 14 lifts line integrals off the plane: surfaces in 3-space, flux through them, and Stokes' theorem — Green's theorem's three-dimensional big sibling, where a boundary curve is traded for the surface it spans.

Same material, another voice

If a different explanation would help, these map onto this chapter — free, from MIT OpenCourseWare:

Read: MIT 18.02 (Auroux), Lecture Notes — the unit on double integrals and line integrals in the plane (Green's theorem).
Watch: MIT 18.02 Video Lectures (Auroux) — the Green's theorem lectures (work, line integrals, and Green's theorem in the plane).

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.