Three theorems, reviewed here as three separate facts. Green's theorem trades a flat region's inside for its boundary loop. The double integral of a curl-like quantity over the region equals a line integral of the field around the region's edge. Stokes' theorem makes the same trade, one dimension up. Now it is a capped surface's inside, traded for its boundary loop in 3D. The Divergence Theorem makes the same trade again, one dimension further. This time it is a solid's inside, traded for the surface that wraps around it. All three trade a "harder" integral for an "easier" one, across a boundary. Chapter 07 proves they are literally one theorem, stated once and then applied three separate ways.
Picture three different weather stations, with three different instruments. All three point to one underlying law. Whatever a field is doing throughout an entire region shows up, in total, on that region's edge. Here is the idea in one image. Count every gust inside a fenced field. You get the same number as counting the wind crossing the fence. That is true whether the "field" is a patch of ground, a dome-shaped tent, or a whole solid room.
Green's theorem is that fenced-field version. Add up the swirl everywhere inside a flat patch of ground. You get exactly the total push you would feel walking around its edge.
Stokes' theorem plays the same trick on a dome-shaped tent, instead of a flat patch. The swirl over the whole curved roof still matches the push around the rim, where the tent meets the ground.
The Divergence Theorem trades a room's whole inside for its walls, ceiling, and floor. Add up how much a field is "spreading out" everywhere inside the room. You get exactly how much of it is pushing through the walls.
Green's theorem is the statement that the double integral of a curl-like quantity over a flat 2D region equals a line integral of the field around that region's boundary loop. Stokes' theorem is the same statement one dimension up. Now it is a capped surface in 3D, instead of a flat region. The Divergence Theorem is the same statement again, over a solid instead of a surface. There, flux through the enclosing boundary surface sits on the right-hand side.
Notice the pattern each theorem shares, once you strip away its specific dimension. On one side sits an integral of some derivative of a field, over a region. On the other side sits an integral of the field itself, over that region's boundary. Those two sides are equal. Green's theorem integrates curl over an area and gets circulation on a curve. Stokes' theorem integrates curl over a surface and gets circulation on that surface's boundary curve. The Divergence Theorem integrates divergence over a solid and gets flux through that solid's boundary surface.
In every case, the boundary operation "differentiate, then integrate over the whole region" collapses to "just look at the edge." That is the entire reason a "harder" volume or area integral is worth trading for an "easier" boundary one. Chapter 07 makes "some derivative of a field" precise, as the exterior derivative d. It makes "region" and "boundary" precise too, as a chain and its boundary operator \partial.
Theorem
Green's Theorem
Recalled from Calculus III — Multivariable Calculus · Green's theorem — first taught there.
Let C be a positively oriented (counterclockwise), piecewise-smooth, simple closed curve in the plane, and let R be the region it encloses. If P and Q have continuous first partial derivatives on an open region containing R, then \oint_C P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA.
Equivalently, writing \mathbf{F} = P\,\mathbf{i} + Q\,\mathbf{j}, the circulation form is \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_R \operatorname{curl} \mathbf{F}\, dA, trading a circulation integral around the boundary for a curl integral over the whole enclosed region.
Sketch for a region R that is both x-simple and y-simple (the general case patches finitely many such regions together, with the shared interior edges canceling since each is traversed once in each direction). Write R = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}. Then \iint_R -\frac{\partial P}{\partial y}\, dA = -\int_a^b \big[P(x,g_2(x)) - P(x,g_1(x))\big]\, dx, by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly \oint_C P\,dx, splitting C into its bottom piece y=g_1(x) (traversed left to right) and top piece y=g_2(x) (traversed right to left, contributing the minus sign). The same argument with R described as y-simple gives \iint_R \frac{\partial Q}{\partial x}\, dA = \oint_C Q\,dy. Adding the two established identities gives the theorem.
Theorem
Stokes' Theorem
also known as: the curl theorem
Recalled from Calculus III — Multivariable Calculus · Stokes' theorem — first taught there.
Let S be an oriented, piecewise-smooth surface bounded by a simple, closed, piecewise-smooth curve C, with C's orientation matching S's unit normal \mathbf{n} by the right-hand rule. (Curl the fingers of the right hand along C's direction of travel, and the thumb points along \mathbf{n}.) If \mathbf{F} has continuous first partial derivatives on an open region containing S, then \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl} \mathbf{F}\cdot d\mathbf{S}, trading a curve's circulation for a surface's curl-flux.
Sketch for S a graph z=g(x,y) over a region R in the xy-plane, with C the boundary of R lifted onto the graph. Parametrize S by (x,y)\mapsto(x,y,g(x,y)), so d\mathbf{S} = (-g_x,-g_y,1)\,dA. Writing out \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} in terms of P, Q, R and their partials, then substituting z=g(x,y) throughout, the right side reduces — after applying the multivariable chain rule to track how P and Q change along the lifted curve — to exactly \oint_{\partial R} \big[P(x,y,g(x,y))+R(x,y,g(x,y))g_x\big]\,dx + \big[Q(x,y,g(x,y))+R(x,y,g(x,y))g_y\big]\,dy, which the circulation form of Green's Theorem converts into a double integral over R identical to the one obtained by substituting z=g(x,y) into the left-hand side's curl-flux integrand directly. A general piecewise-smooth S is patched together from finitely many such graphs, with shared interior edges canceling since each is traversed once in each direction.
Theorem
The Divergence Theorem
also known as: Gauss's Theorem
Recalled from Calculus III — Multivariable Calculus · The divergence theorem — first taught there.
Let E be a solid region in space bounded by a closed, piecewise-smooth surface S=\partial E, oriented outward. If \mathbf{F} has continuous first partial derivatives on an open region containing E, then \oiint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_E \operatorname{div} \mathbf{F}\, dV, trading a closed surface's flux for a solid's divergence integral.
Sketch for E a solid that is simultaneously x-simple, y-simple, and z-simple (a general E patches together finitely many such pieces, with shared interior faces canceling since each is traversed with opposite outward normals from the two adjoining pieces). It suffices to prove the three component identities separately and add them; take the z-component. Write E=\{(x,y,z): (x,y)\in R,\ h_1(x,y)\le z\le h_2(x,y)\}. Then \iiint_E \frac{\partial R}{\partial z}\, dV = \iint_R \big[R(x,y,h_2(x,y)) - R(x,y,h_1(x,y))\big]\, dA by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly the flux of R\,\mathbf{k} through S's top piece z=h_2(x,y) (outward normal has positive z-component) plus its bottom piece z=h_1(x,y) (outward normal has negative z-component, contributing the minus sign) — the side faces of E, if any, are vertical and contribute nothing to a \mathbf{k}-component flux. The same argument for the other two components, summed, gives the theorem.
Green's: interior curl equals boundary circulation
Drop the paddle wheel at several interior points, then the circulation circle around the whole region's boundary -- the boundary reading matches the sum of the interior readings.
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Divergence Theorem: interior divergence, boundary flux
Grow the sphere's radius slider -- the live divergence readout stays constant while the volume it's sampled over grows, matching total flux growing with enclosed volume.
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Worked example
Divergence Theorem, quick check
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1
Verify the Divergence Theorem's two sides agree in direction for \mathbf{F}=\langle x,y,z\rangle on the unit ball.
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2
\operatorname{div}\mathbf{F} = 1+1+1 = 3, constant everywhere. The volume integral is 3\cdot\text{(ball volume)}=3\cdot\frac{4}{3}\pi = 4\pi.
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3
On the unit sphere, \mathbf{F}\cdot\mathbf{n} = \mathbf{F}\cdot\mathbf{F} = x^2+y^2+z^2 = 1 everywhere. The outward normal has length 1, and \mathbf{F} points straight out from the center with length 1 on the unit sphere. So the flux is 1\cdot\text{(sphere area)} = 4\pi.
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4
\boxed{4\pi = 4\pi} — both sides agree, confirming the theorem on a case simple enough to check by hand.
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Where this goes next
Chapter 7 of this course
The generalized Stokes theorem, \int_{\partial S}\omega = \int_S d\omega, specializes to Green's when S is a 2D region, to classical Stokes' when S is a capped surface, and to the Divergence Theorem when S is a solid — one line of chapter 07 recovers all three statements above.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
Using Green's Theorem, find \oint_C \mathbf{F}\cdot d\mathbf{r} for \mathbf{F}(x,y)=(y^2,\,x^2) around the square with vertices (0,0),(2,0),(2,2),(0,2) (counterclockwise).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Q_x-P_y = 2x-2y. Integrating over the square: \iint_0^2\int_0^2 (2x-2y)\,dx\,dy. By symmetry (swap x\leftrightarrow y flips the sign of the integrand but the square is symmetric under that swap), this integral is 0.
Correct answer
Also accepted:
Why
Set up the double integral of 2x-2y over the square and look for symmetry before grinding through the algebra.
From Calculus III — Multivariable Calculus
Using Green's Theorem, find the circulation of \mathbf{F}(x,y)=(-y,x) around a square of side length 4 (counterclockwise), centered anywhere.
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Why
Q_x-P_y=2 everywhere. The square's area is 4^2=16. Circulation =2\cdot16=32 — the same regardless of where the square is centered, since curl is constant.
Correct answer
Also accepted:
Why
The curl is constant, so circulation is just curl times enclosed area, regardless of the region's exact position or even its exact shape.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
Using Stokes' Theorem with a flat square of side 4 (area 16) as S, find \oint_C \mathbf{F}\cdot d\mathbf{r} for \mathbf{F}(x,y,z)=(3z-y,\,x+2z,\,0), C the square's boundary, oriented to match \mathbf{n}=\mathbf{k}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\operatorname{curl}\mathbf{F}'s \mathbf{k}-component is Q_x-P_y=1-(-1)=2. Dotted with \mathbf{n}=\mathbf{k}, the integrand is constant 2. Circulation =2\cdot16=32.
Correct answer
Also accepted:
Why
Only the \mathbf{k}-component of curl survives on a flat, \mathbf{k}-oriented surface -- find it and multiply by area.
From Calculus III — Multivariable Calculus
Green's Theorem (tangential form) is Stokes' Theorem's special case when...
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Any component pattern of \mathbf{F} still needs S to be flat for the reduction to Green's Theorem.
Green's Theorem applies to any positively oriented simple closed curve, not just circles.
With curl identically zero, both sides of Stokes' Theorem are trivially 0 -- that's not the reduction to Green's Theorem itself.
Why
With S a flat region in the xy-plane, d\mathbf{S}=\mathbf{k}\,dA and \operatorname{curl}\mathbf{F}\cdot\mathbf{k} collapses to Q_x-P_y, giving exactly Green's Theorem's tangential form.
Think about what happens to d\mathbf{S} and \operatorname{curl}\mathbf{F}\cdot\mathbf{n} when S is forced flat in the xy-plane.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
The Divergence Theorem is the 3D analogue of which earlier result?
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The tangential form's 3D analogue is Stokes' Theorem, which pairs with curl, not the Divergence Theorem.
That theorem is a 0-dimensional-boundary statement, further from this one in the family.
The conservative test is a different kind of statement -- an equivalence, not a boundary-to-interior integral trade.
Why
Both trade a boundary flux integral for an interior divergence integral -- the normal form of Green's Theorem in 2D, the Divergence Theorem in 3D.
Look for the earlier theorem that also trades a boundary flux integral for an interior divergence integral, just one dimension down.
From Calculus III — Multivariable Calculus
Using the Divergence Theorem, find \oiint_S \mathbf{F}\cdot d\mathbf{S} for \mathbf{F}(x,y,z)=(2x,\,2y,\,2z), S the sphere of radius 3.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\operatorname{div}\mathbf{F}=2+2+2=6. Volume of the ball of radius 3 is \frac{4}{3}\pi(27)=36\pi. Flux =6\cdot36\pi=216\pi\approx678.6.
Correct answer
Also accepted:
Why
Compute the constant divergence, then multiply by the ball's volume \frac{4}{3}\pi r^3.