6.6 Curl, divergence, Stokes' theorem and Gauss's theorem

Curl measures how much a field spins and divergence how much it spreads; Stokes' theorem says the circulation around the edge of a surface is the flux of the curl through it, the divergence theorem says the flux out of a closed surface is the integral of the divergence inside, and both are the fundamental theorem of calculus one dimension up.

In 6.1 the fundamental theorem of calculus became a statement about a curve: the integral of $\nabla f$ along a path is the change of $f$ between its ends. Green’s theorem (6.4) was the next step: the circulation around a loop is the spin inside it. This chapter takes both steps into space. A loop that is the edge of a curved surface, and a closed surface that is the skin of a solid, each get a theorem. The first is Stokes’ theorem and the second is the divergence theorem, also called Gauss’s theorem. They are what Maxwell’s equations are written with (6.8), and they are the end of Unit 6 in MATH 376.

Before you start

6.4: Green’s theorem. 6.5: flux, and $n\,dS = (r_u\times r_v)\,du\,dv$ with a sign check. 5.5: mixed partials of a $C^2$ function are equal. 6.2: the line integral of a gradient around a closed path is $0$ .

Quick checkGreen's theorem for a positively oriented Jordan curve

C

bounding

R

says

\oint_C P\,dx + Q\,dy

equals which integral over

R

Quick checkIn a flux integral through a surface parametrized by

r(u, v)

, which expression stands for

n\,dS

when

r_u\times r_v

points the way

n

does?

Quick checkFor a

C^2

function

f

, what is

\partial_y(f_z) - \partial_z(f_y)

The picture: a wheel in the flow, and a skin round a solid

Start with the rotation field and press Play: the wheel turns at the rate the curl predicts. Switch to the shear field $(y, 0, 0)$ . Every arrow points the same way and nothing circles the origin, but the wheel turns, because the flow is faster at the top of the wheel than at the bottom. Then choose the vortex: the flow circles the origin and the wheel does not turn. So curl does not mean ‘the field goes round something’. It means ‘a small wheel at this point would spin’. Last, choose the layers field and tilt the axis: the spin is the dot product of the curl with the axis.

The torus with $F = (x, y, z)$ : the arrows leave the whole surface and the flux is $6\pi^2Ra^2$ , which is three times the volume of the solid. That is the divergence theorem for this field, because $\operatorname{div}F = 3$ .

The idea, in words

Look at a tiny cube in a flowing fluid. Count what leaves through the six faces minus what enters through the opposite faces. On the faces perpendicular to the $x$ axis the net outflow is $\big(P(x + \Delta x) - P(x)\big)\Delta y\,\Delta z\approx P_x\,\Delta x\,\Delta y\,\Delta z$ . Do the same for the other two directions and the net outflow per unit volume is $P_x + Q_y + R_z$ , the divergence. Now take a tiny circle in a plane and add up the part of the field along it. On a small square in the $xy$ -plane the circulation is $\big(Q(x + \Delta x) - Q(x)\big)\Delta y - \big(P(y + \Delta y) - P(y)\big)\Delta x\approx(Q_x - P_y)\,\Delta x\,\Delta y$ , the curl in the $z$ direction. Big surfaces and big solids are built from tiny squares and cubes. On each piece the circulation (or the flux) equals the local density times the size. When you add the pieces, the edges (or faces) inside cancel in pairs, and only the outer boundary survives. That is the entire content of both theorems.

Why· Why the inner edges cancel

Two neighbouring squares share an edge. One walks it left to right, the other right to left, so the line integrals along that edge are negatives of each other. Two neighbouring cubes share a face. One has the outward normal pointing into the other, so the flux through the shared face counts $+$ for one cube and $-$ for the other. Everything that is not on the outer boundary cancels. This is the same cancellation as in the telescoping sum $(f(x_1) - f(x_0)) + (f(x_2) - f(x_1)) + \dots = f(x_n) - f(x_0)$ , which is why these theorems are the fundamental theorem of calculus in higher dimensions.

Where this came from· Stokes' theorem: a letter, an exam question, a footnote

In a letter of July 1850 William Thomson, later Lord Kelvin, told George Stokes about the relation between the line integral around a loop and the surface integral over a surface it bounds. Stokes put it as question 8 in the Smith’s Prize examination at Cambridge in February 1854, a competitive examination for the best mathematics students of the university. James Clerk Maxwell sat that examination and was placed joint first. When Maxwell printed the theorem in his Treatise on Electricity and Magnetism (1873) he added a footnote saying that Stokes had used it as that examination question, and that footnote is the usual explanation of why it carries Stokes’ name, although the idea came to Stokes from Thomson. Neither Thomson nor Stokes published a proof; the first published proof is attributed to Hermann Hankel in 1861, in a work on the general theory of fluid motion. What the theorem says is what 6.8 uses it for: the circulation of a field around a loop equals the flux of its curl through any surface the loop bounds.

Where this came from· The divergence theorem: Lagrange, Gauss, Ostrogradsky

Lagrange had a special case in 1762, in his work on fluid mechanics. Around 1813 Gauss met surface integrals while studying the attraction of an ellipsoid, and proved special cases of the theorem, adding more in 1833 and 1839. The first proof of the general theorem is usually credited to the Russian mathematician Mikhail Ostrogradsky, who presented it to the Paris Academy of Sciences in 1826 as part of his work on the flow of heat (it was printed a few years later). Green found the planar analogue in 1828. Today it is called the divergence theorem, or the Gauss theorem, or the Gauss and Ostrogradsky theorem, depending on the book. The shape of the problem is the one in the heat chapter 6.8: how much heat leaves a body is the total of the heat produced inside it.

Curl and divergence

Name the components of a vector field $F = (P, Q, R)$ , not $F_x, F_y, F_z$ : the subscript $x$ looks like a partial derivative and invites errors. Write $P_y$ for $\partial P/\partial y$ and so on. The symbol $\nabla = (\partial_x, \partial_y, \partial_z)$ is a vector whose entries are derivative operators; it has no meaning alone, but it combines with $F$ the way a vector would.

Definition T

Curl

The curl of a differentiable vector field $F = (P, Q, R)$ is the vector field

\operatorname{curl}F = \nabla\times F = \det\begin{bmatrix}e_1 & e_2 & e_3\\ \partial_x & \partial_y & \partial_z\\ P & Q & R\end{bmatrix} = (R_y - Q_z,\ \ P_z - R_x,\ \ Q_x - P_y).

It is the cross product formula of 6.5 with $\partial_x, \partial_y, \partial_z$ in the place of $a_1, a_2, a_3$ and $P, Q, R$ in the place of $b_1, b_2, b_3$ . So the middle component has the flipped sign: $P_z - R_x$ , not $R_x - P_z$ . The third component $Q_x - P_y$ is the integrand of Green’s theorem.

In one lineFor $F = (P, Q, R)$ , $\operatorname{curl}F = \nabla\times F = (R_y - Q_z,\ P_z - R_x,\ Q_x - P_y)$ : a vector field that records how much, and about which axis, $F$ spins at each point.

Why it existsIt is the spin density of the field: the circulation per unit area around a tiny loop, so it is what Stokes' theorem integrates.

Definition T

Divergence

The divergence of $F = (P, Q, R)$ is the scalar field $\operatorname{div}F = \nabla\cdot F = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z}$ . It is positive where the field spreads out (a source) and negative where it gathers (a sink).

In one lineFor $F = (P, Q, R)$ , $\operatorname{div}F = \nabla\cdot F = P_x + Q_y + R_z$ : a scalar field that records how much $F$ spreads out from each point.

Why it existsIt is the net outflow per unit volume, so it is what the divergence theorem integrates.

Definition T

Vorticity

For a velocity field of a fluid the curl is called the vorticity. A rigid rotation with angular velocity $\omega$ has vorticity $2\omega$ .

In one lineThe vorticity of a velocity field $v$ is $\operatorname{curl}v$ .

Why it existsIt is the name fluid dynamics gives to the curl, and it is zero exactly when the flow has no local spin.

Definition T

Irrotational field

A field with $\operatorname{curl}F = 0$ everywhere is irrotational. By the identity below every gradient field is irrotational; on a simply connected domain such as all of $\mathbb R^3$ every irrotational field is a gradient (Apostol, label A), and then the fluid flow is called a potential flow.

In one lineA field with curl $0$ everywhere on its domain. A flow with this property is called a potential flow.

Why it existsOn a simply connected domain these are exactly the gradient fields, so the name tells you a potential exists.

Definition T

Solenoidal field

A field with $\operatorname{div}F = 0$ everywhere is solenoidal (or divergence-free). A fluid whose velocity field is solenoidal is incompressible, and the magnetic field $B$ is solenoidal (6.8).

In one lineA field with divergence $0$ everywhere on its domain; a flow with this property is incompressible.

Why it existsIt means no net source or sink anywhere: whatever flows into a region flows out, the property of an incompressible fluid and of the magnetic field.

Definition A

Laplacian

The Laplacian of a $C^2$ scalar field $f$ is $\nabla^2f = \operatorname{div}(\nabla f) = f_{xx} + f_{yy} + f_{zz}$ . A function with $\nabla^2f = 0$ is called harmonic.

In one lineFor a scalar field $f$ the Laplacian is $\nabla^2 f = \operatorname{div}\nabla f = f_{xx} + f_{yy} + f_{zz}$ .

Why it existsIt is the divergence of the gradient, and the operator in the heat equation and in Poisson's equation (6.8).

Worked example MATH 376 worksheet 13 problems 1, 2 and 7

Compute the divergence of

F = (e^x\sin y,\ e^x\cos y,\ z^2)

, and the curl of

F = (x^2, y^2, z^2)

and of

F = (-y, x, z)

Solution.

Name the components of the first field. $P = e^x\sin y$ , $Q = e^x\cos y$ , $R = z^2$ . Names first, then derivatives; $F_x$ would clash with a partial derivative.
Differentiate. $P_x = e^x\sin y$ , $Q_y = -e^x\sin y$ , $R_z = 2z$ . The derivative of $\cos y$ in $y$ is $-\sin y$ .
Add. $\operatorname{div}F = e^x\sin y - e^x\sin y + 2z = 2z$ .
Curl of $(x^2, y^2, z^2)$ . $R_y - Q_z = 0 - 0$ , $P_z - R_x = 0 - 0$ , $Q_x - P_y = 0 - 0$ . Each component depends on one variable only, so every cross derivative vanishes. So the curl is $0$ , and indeed $F = \nabla\frac{x^3 + y^3 + z^3}3$ .
Curl of $(-y, x, z)$ . $R_y - Q_z = 0 - 0 = 0$ , $P_z - R_x = 0 - 0 = 0$ , $Q_x - P_y = 1 - (-1) = 2$ . $\partial_xx = 1$ and $\partial_y(-y) = -1$ .

Answer. $\operatorname{div}F = 2z$ ; the curl of $(x^2,y^2,z^2)$ is $0$ ; the curl of $(-y, x, z)$ is $(0, 0, 2)$ .

Worked example MATH 376 worksheet 28 problem 8; worksheet 14 problems 1 and 2

Find the curl and divergence of

F = (yz, xz, xy)

, the vorticity of the velocity field

v = (yz, xz, xy + z^2)

, and the divergence of the heat flow

q = -\nabla T

for

T = x^2 + y^2 + z^2

Solution.

Curl of $F$ . With $P = yz$ , $Q = xz$ , $R = xy$ : $R_y - Q_z = x - x = 0$ , $P_z - R_x = y - y = 0$ , $Q_x - P_y = z - z = 0$ . So $\operatorname{curl}F = 0$ .
Divergence of $F$ . $P_x + Q_y + R_z = 0 + 0 + 0 = 0$ . $yz$ has no $x$ , $xz$ no $y$ , $xy$ no $z$ . (This $F$ is $\nabla(xyz)$ , and $xyz$ is harmonic.)
Vorticity of $v$ . The same three differences for $R = xy + z^2$ : $R_y - Q_z = x - x = 0$ , $P_z - R_x = y - y = 0$ , $Q_x - P_y = z - z = 0$ . The vorticity is $0$ : a potential flow, with $v = \nabla(xyz + z^3/3)$ .
Heat flow. $\nabla T = (2x, 2y, 2z)$ , so $q = (-2x, -2y, -2z)$ and $\operatorname{div}q = -2 - 2 - 2 = -6$ . Negative divergence means the flow gathers: heat flows in toward each point, a sink.

Answer. curl $0$ and div $0$ ; vorticity $0$ ; $\operatorname{div}q = -6$ .

Your answerLet

F = (x^2y,\ yz^2,\ xz)

. What is

\operatorname{div}F

at the point

(1, 2, 3)

Quick checkWhat is the second component of

\operatorname{curl}(P, Q, R)

Identities

Theorem T MATH 376 lecture of 2 March, Lemma 5.13; homework 5 problem 5; Apostol 12.12 and 12.14

Identities for curl and divergence

Let $F, G$ be $C^1$ vector fields, $f$ a $C^1$ function, and $a, b$ numbers. Then $\operatorname{div}(aF + bG) = a\operatorname{div}F + b\operatorname{div}G$ and the same for curl, and the product rules hold. If $F$ is $C^2$ then $\operatorname{div}\operatorname{curl}F = 0$ , and if $f$ is $C^2$ then $\operatorname{curl}\nabla f = 0$ .

In one lineCurl and divergence are linear; $\operatorname{div}(\operatorname{curl}F) = 0$ for $C^2$ fields; $\operatorname{curl}(\nabla f) = 0$ for $C^2$ functions; $\operatorname{div}(fF) = f\operatorname{div}F + \nabla f\cdot F$ and $\operatorname{curl}(fF) = f\operatorname{curl}F + \nabla f\times F$ .

Why it is trueThe two zero identities explain why potentials have no spin and why a magnetic field has no source, and the product rules are used to compute with fields.

The hypothesis $C^2$ matters: the two zero identities use the equality of mixed partials (5.5), which needs continuous second partials. Without it the cancellation can fail at a point.

ProofMATH 376 homework 5 problem 5; lecture Lemma 5.13

The idea: Expand, group the terms by function, and see that every second derivative appears twice with opposite signs, as a pair of mixed partials.

\operatorname{curl}F = (R_y - Q_z,\ P_z - R_x,\ Q_x - P_y)

Definition of the curl of

F = (P, Q, R)

\operatorname{div}\operatorname{curl}F = \partial_x(R_y - Q_z) + \partial_y(P_z - R_x) + \partial_z(Q_x - P_y)

Definition of the divergence, applied to the three components of the curl.

= R_{yx} - Q_{zx} + P_{zy} - R_{xy} + Q_{xz} - P_{yz}

Differentiate each term;

R_{yx}

means

\partial_x(\partial_yR)

= (R_{yx} - R_{xy}) + (P_{zy} - P_{yz}) + (Q_{xz} - Q_{zx})

Regroup so that each bracket has the same function.

= 0 + 0 + 0 = 0

Equality of mixed partials: each bracket is two mixed partials of one

C^2

function.This is the step that uses

C^2

\operatorname{curl}\nabla f = (f_{zy} - f_{yz},\ f_{xz} - f_{zx},\ f_{yx} - f_{xy})

For

\operatorname{curl}\nabla f

, take

P = f_x

Q = f_y

R = f_z

in the curl formula.

= (0, 0, 0)

Equality of mixed partials again.The two zero identities are the same fact. In 6.7 they will both be 'd squared is zero'.

\operatorname{div}(fF) = f(P_x + Q_y + R_z) + (f_xP + f_yQ + f_zR) = f\operatorname{div}F + \nabla f\cdot F

Product rule for the divergence:

\partial_x(fP) = fP_x + f_xP

, and similarly for

y

and

z

The bracket

f_yR - f_zQ

is the first component of

\nabla f\times F = (f_yR - f_zQ,\ f_zP - f_xR,\ f_xQ - f_yP)

, and the other two components are the same with the letters cycled. So

\operatorname{curl}(fF) = f\operatorname{curl}F + \nabla f\times F

\blacksquare

The same product rule in the first component of

\operatorname{curl}(fF)

\partial_y(fR) - \partial_z(fQ) = f(R_y - Q_z) + (f_yR - f_zQ)

.Linearity of div and curl is the same kind of remark: each partial derivative of

aP_1 + bP_2

a

times one plus

b

times the other, so each component of div and curl is linear. Problem 'Prove that div and curl are linear' at the end writes it out.

Worked example new (identities, MATH 376 lecture Lemma 5.13)

Check the product rule

\operatorname{div}(fF) = f\operatorname{div}F + \nabla f\cdot F

for

f = xyz

and

F = (x, yz, z^2)

Solution.

Left side. $fF = (x^2yz,\ xy^2z^2,\ xyz^3)$ , and $\partial_x(x^2yz) = 2xyz$ , $\partial_y(xy^2z^2) = 2xyz^2$ , $\partial_z(xyz^3) = 3xyz^2$ . So $\operatorname{div}(fF) = 2xyz + 5xyz^2$ .
Right side, first term. $\operatorname{div}F = 1 + z + 2z = 1 + 3z$ , so $f\operatorname{div}F = xyz + 3xyz^2$ .
Right side, second term. $\nabla f = (yz, xz, xy)$ , so $\nabla f\cdot F = xyz + xyz^2 + xyz^2 = xyz + 2xyz^2$ .
Compare. $xyz + 3xyz^2 + xyz + 2xyz^2 = 2xyz + 5xyz^2$ .

Answer. Both sides are $2xyz + 5xyz^2$ .

Worked example MATH 376 homework 5 problem 4 (the cross product part) and new (vector potentials; compare Apostol 12.15 exercise 5 and 12.16)

Is there a field

A

with

\operatorname{curl}A = (x, y, -2z)

? What about

(x, y, z)

? Is there a vector

v

with

(1, 2, 1)\times v = (3, 1, 5)

Solution.

Test the divergence. $\operatorname{div}(x, y, -2z) = 1 + 1 - 2 = 0$ , but $\operatorname{div}(x, y, z) = 3\ne0$ . If $G = \operatorname{curl}A$ then $\operatorname{div}G = \operatorname{div}\operatorname{curl}A = 0$ . So $(x, y, z)$ is not a curl.
Find $A$ for the first. Try $A = (P, Q, 0)$ . We need $-Q_z = x$ , $P_z = y$ , $Q_x - P_y = -2z$ . So $Q = -xz$ , $P = yz$ , and then $Q_x - P_y = -z - z = -2z$ . Choose the third component of $A$ to be $0$ and integrate in $z$ .
Check. $A = (yz, -xz, 0)$ has $R_y - Q_z = 0 + x = x$ , $P_z - R_x = y - 0 = y$ , $Q_x - P_y = -z - z = -2z$ .
The cross product question. $(1, 2, 1)\times v$ is orthogonal to $(1, 2, 1)$ for every $v$ , but $(1,2,1)\cdot(3,1,5) = 3 + 2 + 5 = 10\ne0$ . So no such $v$ exists. The same shape as the field question: a curl has divergence zero, a cross product is orthogonal to its factors.

Answer. $(x, y, -2z)$ is the curl of $(yz, -xz, 0)$ ; $(x, y, z)$ is not a curl; no such $v$ .

Beyond the course: for understanding· The converse: when is a field with curl zero a gradient?

The converses of the two zero identities are true on a nice domain (label A/B). On a star-shaped domain such as all of $\mathbb R^3$ or a ball: $\operatorname{curl}F = 0$ implies $F$ is a gradient, and $\operatorname{div}G = 0$ implies $G$ is a curl. The domain matters. The field $F = \dfrac{(-y, x, 0)}{x^2 + y^2}$ on $\mathbb R^3$ minus the $z$ -axis has $\operatorname{curl}F = 0$ , but its circulation around the unit circle in $z = 0$ is $\int_0^{2\pi}(\sin^2t + \cos^2t)\,dt = 2\pi\ne0$ , so it is not a gradient (this is homework 5 problem 4.1: the answer to ‘curl zero implies conservative?’ is false). The field $G = (x, y, z)/\rho^3$ on $\mathbb R^3$ minus the origin, with $\rho = \sqrt{x^2 + y^2 + z^2}$ , has $\operatorname{div}G = 0$ , but its flux out of the unit sphere is $4\pi$ (shown below), whereas the flux of a curl through a closed surface is always $0$ (also below). So $G$ is not a curl, although its divergence is zero. Holes in the domain are what stops the converses, and counting them is what 6.7 and a course in topology do.

Stokes’ theorem

We need to say which way round the edge of the surface goes, given which side of the surface is called positive.

Definition T

Induced orientation

Let $S$ be an oriented surface with boundary curve $\partial S$ . The induced orientation of $\partial S$ is the one for which, if you walk along $\partial S$ with your head pointing along $n$ , the surface $S$ is on your left. Equivalently: right-hand fingers along $\partial S$ , thumb along $n$ . For a flat region in the $xy$ -plane with $n = (0, 0, 1)$ this is the counterclockwise direction seen from above, the positive orientation of Green’s theorem.

In one lineIf $S$ is an oriented surface with unit normal $n$ , its boundary curve $\partial S$ is given the direction for which, walking along it with your head pointing in the direction of $n$ , the surface is on your left. This is the right-hand rule: curl the fingers of the right hand along $\partial S$ and the thumb points along $n$ .

Why it existsThe line integral on the left of Stokes' theorem changes sign with the direction of travel and the surface integral on the right changes sign with $n$ ; this rule makes the two signs agree.

Theorem T MATH 376 lecture of 2 March, Theorem 5.14; Apostol 12.11

Stokes' theorem

Let $r: T\to S$ be a smooth one-to-one parametrization of a surface $S$ , where $T$ is bounded by a piecewise smooth Jordan curve $\Gamma$ . Let $C = \partial S = r(\Gamma)$ be the boundary curve, oriented as the image of $\Gamma$ traversed counterclockwise, and $n$ the unit normal with which it is compatible (the induced orientation). Let $F$ be a vector field of class $C^1$ on an open set containing $S$ . Then

\oint_C F\cdot dr = \iint_S(\operatorname{curl}F)\cdot n\,dS = \iint_T(\operatorname{curl}F)(r(u, v))\cdot(r_u\times r_v)\,du\,dv.

The lecture states it with $C^2$ ; the textbook needs only $C^1$ . The last equality holds when $r_u\times r_v$ points the way $n$ does, by the flux theorem.

In one lineIf $S$ is a piecewise smooth oriented surface whose boundary curve $\partial S$ carries the induced orientation and $F$ is $C^1$ on an open set containing $S$ , then $\oint_{\partial S}F\cdot dr = \iint_S(\operatorname{curl}F)\cdot n\,dS$ .

Why it is trueIt turns a line integral around a loop into a surface integral over any surface the loop bounds (or the reverse), whichever is easier, and it is Green's theorem for surfaces in space.

What fails without each hypothesis, one by one.

Hypothesis	Why it is needed	What goes wrong without it
$S$ is orientable with a continuous $n$	The right side needs one unit normal at each point	The Moebius strip has no such $n$
$\partial S$ is a piecewise smooth, closed curve	The left side is a line integral over a closed curve	An open curve bounds no surface in this sense
$\partial S$ is oriented compatibly with $n$	It fixes the sign of the left side	The two sides differ by a minus sign
$F$ is $C^1$ at every point of $S$ , the inside included	The proof differentiates $F$ across the whole surface	The example below

For the last row, take $F = \dfrac{(-y, x, 0)}{x^2 + y^2}$ , which is not defined on the $z$ -axis, and $\operatorname{curl}F = 0$ where it is defined. Its circulation around the unit circle in $z = 0$ is $2\pi$ . The upper unit hemisphere has this circle as its boundary, and $\iint\operatorname{curl}F\cdot n\,dS = 0$ . The two sides disagree, because the hemisphere meets the $z$ -axis at $(0, 0, 1)$ where $F$ is not defined. This is the same failure as for Green’s theorem on the punctured plane in 6.4.

Why it is true: the proof for a graph

ProofMATH 376 lecture of 2 March (the affine case and a tiling argument); the same idea is Apostol 12.11

The idea: Write the line integral over the boundary of the surface as a line integral in the xy-plane, using that z depends on x and y along the curve. Apply Green's theorem there. Differentiate with the chain rule, and watch the terms with second derivatives of g cancel. What is left is the dot product of the curl with the normal (-g_x, -g_y, 1).

The boundary

C

S

is the image of

\partial D

under

(x, y)\mapsto(x, y, g(x, y))

, with

\partial D

run counterclockwise.

Setup:

S

is the graph

z = g(x, y)

over a region

D

bounded by a piecewise smooth Jordan curve, with

g

of class

C^2

and the upward normal;

F = (P, Q, R)

C^1

, and

P, Q, R

below are evaluated at

(x, y, g(x, y))

.Orientation check, which is where signs go wrong. The upward normal and the right-hand rule make

C

run counterclockwise as seen from above. The map

(x, y)\mapsto(x, y, g)

does not change the

x

and

y

coordinates, so

\partial D

must be run counterclockwise too, which is the positive orientation Green's theorem needs.

\displaystyle\iint_S(\operatorname{curl}F)\cdot n\,dS = \iint_D\Big[-(R_y - Q_z)g_x - (P_z - R_x)g_y + (Q_x - P_y)\Big]dx\,dy

Flux formula of 6.5:

n\,dS = (-g_x, -g_y, 1)\,dx\,dy

for the upward normal of a graph.

\displaystyle\oint_CF\cdot dr = \oint_{\partial D}(P + Rg_x)\,dx + (Q + Rg_y)\,dy

Along

C

we have

z = g(x, y)

, hence

dz = g_x\,dx + g_y\,dy

(chain rule), so

P\,dx + Q\,dy + R\,dz

becomes a form in

dx

and

dy

alone.

\displaystyle= \iint_D\big[\partial_x(Q + Rg_y) - \partial_y(P + Rg_x)\big]\,dx\,dy

Green's theorem (6.4) on

D

, applied to the

C^1

pair

p = P + Rg_x

and

q = Q + Rg_y

(this is where

g\in C^2

is used).

\partial_x(Q + Rg_y) = Q_x + Q_zg_x + (R_x + R_zg_x)g_y + Rg_{yx}

and

\partial_y(P + Rg_x) = P_y + P_zg_y + (R_y + R_zg_y)g_x + Rg_{xy}

Chain rule for functions of

(x, y, g(x, y))

: for instance

\partial_x[Q(x, y, g)] = Q_x + Q_zg_x

The difference is

(Q_x - P_y) + (Q_z - R_y)g_x + (R_x - P_z)g_y

Subtract. The terms

Rg_{yx}

and

Rg_{xy}

cancel because mixed partials of

g

are equal, and the terms

R_zg_xg_y

cancel.This cancellation is the same cancellation as in 'curl of a gradient is zero'.

= -(R_y - Q_z)g_x - (P_z - R_x)g_y + (Q_x - P_y)

, so the two sides are equal.

\blacksquare

Compare with the second line: the integrands are the same.

Proof workshopRebuild the proof of Stokes' theorem for a graphMATH 376 lecture of 2 March; Apostol 12.11

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Why· The picture behind the proof, and the general surface

Project the surface straight down onto the $xy$ -plane. The boundary curve of $S$ projects to the boundary of $D$ , and Green’s theorem lives on $D$ . The surface’s slope enters through $g_x$ and $g_y$ : they are exactly the components of the normal that tilt the flat spin $Q_x - P_y$ into the spin about the tilted normal. A general surface is cut into pieces, each a graph over some plane. On each piece the theorem holds by the argument above, and the inner edges cancel in pairs because neighbouring pieces traverse them in opposite directions (the ‘why the inner edges cancel’ box). The lecture’s own proof does the same with tiny affine pieces: for a field that is a linear function and a surface that is a linear graph both sides are constants times the area, and then the general case is a limit of sums.

Worked example MATH 376 lecture of 2 March, Example 5.14 (the affine case of the lecture's proof)

Check Stokes' theorem for the field

F = (a_1x + a_2y + a_3z,\ 0,\ 0)

on the linear graph

S = \{(u, v, b_1u + b_2v) : (u, v)\in T\}

, where

T

is a rectangle.

Solution.

Curl. $P = a_1x + a_2y + a_3z$ , $Q = R = 0$ , so $\operatorname{curl}F = (R_y - Q_z,\ P_z - R_x,\ Q_x - P_y) = (0,\ a_3,\ -a_2)$ . Only the derivatives of $P$ in $z$ and $y$ survive.
Surface side. The graph normal is $N = (-b_1, -b_2, 1)$ , pointing up, so $(\operatorname{curl}F)\cdot N = -a_3b_2 - a_2$ , a constant. The surface integral is $-(a_2 + a_3b_2)\operatorname{area}(T)$ .
Line side, set up. Only $P\,dx$ is non-zero, and on $S$ we have $x = u$ , $y = v$ , $z = b_1u + b_2v$ , so $P = (a_1 + a_3b_1)u + (a_2 + a_3b_2)v$ . The boundary of $S$ is the image of the boundary of $T$ run counterclockwise, and $dx = du$ along it.
Line side, evaluate. $\oint_{\partial T}u\,du = 0$ because $u\,du$ is the exact differential of $u^2/2$ . And $\oint_{\partial T}v\,du = -\operatorname{area}(T)$ by Green’s theorem with $P = v$ , $Q = 0$ , where $Q_u - P_v = -1$ . So the line integral is $-(a_2 + a_3b_2)\operatorname{area}(T)$ .

Answer. Both sides equal $-(a_2 + a_3b_2)\operatorname{area}(T)$ . The lecture’s proof does this computation for affine fields on linear pieces, which is all you need on a tiny patch where $F$ and the surface look affine, then adds the patches: the inner edges cancel in pairs and the outer edge is left.

How to use Stokes’ theorem

The exam standard for ‘use Stokes to compute’ is to name every object. Quoting the theorem and writing the determinant for the curl is only the first two lines.

Why· Full-credit template for 'use Stokes' theorem to compute'

Name the surface $S$ : give an equation and the parameter region. Choose the simplest surface with boundary $C$ (usually the flat disc or polygon).
Name the normal $n$ : give its components.
Name the boundary and its direction: check the right-hand rule. Walking along $C$ with your head along $n$ , $S$ must be on your left.
Name the field $F = (P, Q, R)$ with the component formulas.
State the hypotheses and check them: $S$ orientable and piecewise smooth, $C$ its boundary, $F$ of class $C^1$ on an open set containing all of $S$ , not only $C$ .
Compute the curl of the full field, component by component, writing each partial derivative.
Dot $\operatorname{curl}F$ with $n$ .
Integrate over $S$ with the right $dS$ or $n\,dS$ , showing antiderivatives.
Conclude in a sentence.

Worked example MATH 376 Midterm 2 problem 3, 5 points; see [A.3](/vol2/a3-midterm-2/)

Use Stokes' theorem to compute

\oint_CF\cdot d\alpha

for

F(x, y, z) = (y, -x, z^2)

and

C

the circle

x^2 + y^2 = 4

in the plane

z = 0

, counterclockwise when viewed from above.

Solution.

Surface. $S = \{(x, y, 0): x^2 + y^2\le4\}$ , the flat disc of radius $2$ . It is the simplest surface with boundary $C$ .
Normal. $n = (0, 0, 1)$ . Counterclockwise from above and the right-hand rule: thumb up.
Boundary. $\partial S = C$ , counterclockwise from above, matching the problem.
Field. $P = y$ , $Q = -x$ , $R = z^2$ ; all are polynomials, so $F$ is $C^1$ on $\mathbb R^3$ and Stokes’ hypotheses hold on any surface.
Curl. $R_y - Q_z = \partial_y(z^2) - \partial_z(-x) = 0$ ; $P_z - R_x = \partial_z(y) - \partial_x(z^2) = 0$ ; $Q_x - P_y = -1 - 1 = -2$ . So $\operatorname{curl}F = (0, 0, -2)$ .
Dot. $(\operatorname{curl}F)\cdot n = -2$ .
Integrate. $\oint_CF\cdot d\alpha = \iint_S(-2)\,dS = -2\operatorname{area}(S) = -2\cdot4\pi = -8\pi$ .
Check by the line integral. $\alpha(t) = (2\cos t, 2\sin t, 0)$ , $\alpha' = (-2\sin t, 2\cos t, 0)$ , $F(\alpha) = (2\sin t, -2\cos t, 0)$ , $F\cdot\alpha' = -4\sin^2t - 4\cos^2t = -4$ , and $\int_0^{2\pi}(-4)\,dt = -8\pi$ . Both ways agree.

Answer. $-8\pi$ .

Worked example MATH 376 practice Midterm 2 problem 3; see [A.5](/vol2/a5-practice-exams/). Worksheet 14 problem 5 is the same with vertices $(3, 0, 0)$, $(0, 3, 0)$, $(0, 0, 3)$ and gives $\frac{27}2$; homework 6 problem 1 is the field $(yz, xz, xy)$ and gives $0$.

Use Stokes' theorem for

\oint_C(z, x, y)\cdot d\alpha

, where

C

is the boundary of the triangle with vertices

(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

, oriented counterclockwise as viewed from the direction of

(1, 1, 1)

Solution.

Surface. $S$ is the triangle in the plane $x + y + z = 1$ , a graph $z = 1 - x - y$ over $D = \{x, y\ge0,\ x + y\le1\}$ . Each vertex satisfies $x + y + z = 1$ .
Normal. $N = (-g_x, -g_y, 1) = (1, 1, 1)$ , pointing toward the viewer at $(1, 1, 1)$ . Right-hand rule: counterclockwise as seen from $(1, 1, 1)$ matches a normal that points toward that viewer.
Curl. $P = z$ , $Q = x$ , $R = y$ : $R_y - Q_z = 1$ , $P_z - R_x = 1$ , $Q_x - P_y = 1$ , so $\operatorname{curl}F = (1, 1, 1)$ .
Dot and integrate. $n\,dS = N\,dA$ , so $(\operatorname{curl}F)\cdot N = 3$ and $\iint_D3\,dA = 3\cdot\frac12 = \frac32$ .
Check along the three edges. $(1,0,0)\to(0,1,0)$ : $\alpha = (1 - t, t, 0)$ , $F = (0, 1 - t, t)$ , $F\cdot\alpha' = 1 - t$ , integral $\frac12$ . The other two edges give $\frac12$ each by symmetry. The sum is $\frac32$ .

Answer. $\dfrac32$ . With vertices $(3, 0, 0)$ , $(0, 3, 0)$ , $(0, 0, 3)$ the base $D$ has area $\frac92$ , so the answer is $3\cdot\frac92 = \frac{27}2$ . With $F = (yz, xz, xy)$ the curl is $0$ (every component is $z - z$ or similar), so the circulation is $0$ .

Worked example MATH 376 worksheet 13 problem 8 and worksheet 27 problem 10

Use Stokes' theorem for

\oint_C F\cdot d\alpha

with

F = (-y, x, z)

and

C

the unit circle

x^2 + y^2 = 1

at height

z = 0

and at height

z = 3

, counterclockwise from above.

Solution.

Curl. $\operatorname{curl}F = (0, 0, 2)$ from the example above. It does not depend on $z$ .
Surface. The flat unit disc in the plane of the circle, with $n = (0, 0, 1)$ .
Dot and integrate. $(\operatorname{curl}F)\cdot n = 2$ , and $\iint_S2\,dS = 2\pi$ .
Check. $\alpha = (\cos t, \sin t, 3)$ , $\alpha' = (-\sin t, \cos t, 0)$ , $F\cdot\alpha' = \sin^2t + \cos^2t = 1$ , integral $2\pi$ .

Answer. $2\pi$ at both heights. The third component of $F$ is orthogonal to $\alpha'$ on a horizontal circle, so it does not matter.

Worked example new (choosing the surface; two surfaces with the same boundary)

Let

F = (y, z, x)

and

C

the unit circle in the plane

z = 0

, counterclockwise from above. Compute

\oint_CF\cdot dr

using (a) the flat disc and (b) the paraboloid

z = 1 - x^2 - y^2

z\ge0

, as the surface.

Solution.

Curl. $P = y$ , $Q = z$ , $R = x$ : $R_y - Q_z = 0 - 1$ , $P_z - R_x = 0 - 1$ , $Q_x - P_y = 0 - 1$ . So $\operatorname{curl}F = (-1, -1, -1)$ .
(a) The disc. $n = (0, 0, 1)$ , $(\operatorname{curl}F)\cdot n = -1$ , and the integral is $-\pi$ .
(b) The paraboloid. It is the graph of $g = 1 - x^2 - y^2$ over the unit disc with the upward normal $N = (2x, 2y, 1)$ , and its boundary is the same circle. Same boundary, compatible orientation: upward normal, counterclockwise from above.
(b) Dot. $(\operatorname{curl}F)\cdot N = -2x - 2y - 1$ .
(b) Integrate. $\iint_D(-2x - 2y - 1)\,dA = -\pi$ . By symmetry $\iint x\,dA = \iint y\,dA = 0$ over the disc, and $\iint1\,dA = \pi$ .
Direct check. On $C$ , $z = 0$ , so $F\cdot\alpha' = y\,x'$ and $\oint y\,dx = \int_0^{2\pi}\sin t(-\sin t)\,dt = -\pi$ .

Answer. $-\pi$ by all three. Two surfaces with the same boundary give the same flux of the curl, which is why you may replace a curved surface by a flat one.

Worked example MATH 376 lecture of 2 March (Stokes for the hemisphere)

Let

F = (-y, x, 0)

. Compute the circulation around the unit circle in the plane

z = 0

, and the flux of

\operatorname{curl}F

through the upper unit hemisphere with outward normal.

Solution.

Circulation. $\alpha = (\cos t, \sin t, 0)$ , $F\cdot\alpha' = \sin^2t + \cos^2t = 1$ , so $\oint = 2\pi$ .
Curl. $\operatorname{curl}F = (0, 0, 2)$ .
Hemisphere. The outward $N = \sin\varphi\,(\sin\varphi\cos\theta,\ \sin\varphi\sin\theta,\ \cos\varphi)$ from 6.5, with $0\le\varphi\le\pi/2$ . So $(\operatorname{curl}F)\cdot N = 2\sin\varphi\cos\varphi$ .
Integrate. $\int_0^{2\pi}\!\int_0^{\pi/2}2\sin\varphi\cos\varphi\,d\varphi\,d\theta = 2\pi\cdot1 = 2\pi$ .

Answer. $2\pi$ both ways. The boundary of the upper hemisphere with outward normal is the equator traversed counterclockwise from above, as the right-hand rule says.

Worked example MATH 376 worksheet 14 problem 6

Ampere's law: a long straight wire has cross-section the disc of radius

2

and current density

J = (0, 0, 20r)

, where

r

is the distance from the axis. Find the circulation of

B

around the edge of the cross-section, given

\nabla\times B = \mu_0J

Solution.

Surface. $S$ is the cross-sectional disc of radius $2$ , with $n = (0, 0, 1)$ along the current; its boundary $C$ is the edge, with the induced direction.
Stokes, then Ampere. $\oint_CB\cdot dr = \iint_S(\nabla\times B)\cdot n\,dS = \mu_0\iint_SJ\cdot n\,dS$ .
Dot. $J\cdot n = 20r$ .
Integrate in polar coordinates. $\int_0^{2\pi}\!\int_0^2\mu_0\cdot20r\cdot r\,dr\,d\theta = 2\pi\mu_0\cdot20\cdot\frac83 = \dfrac{320\pi\mu_0}3$ . The extra $r$ is the Jacobian $dS = r\,dr\,d\theta$ of polar coordinates.

Answer. $\dfrac{320\pi\mu_0}3$ . The circulation of the magnetic field around the wire equals $\mu_0$ times the current through it.

Your answerUse Stokes' theorem to find the circulation of

F = (-y, x, z^3)

counterclockwise around the ellipse

\frac{x^2}4 + y^2 = 1

in the plane

z = 0

Quick checkWhich orientation of the boundary circle of the upper hemisphere (outward normal) makes Stokes' theorem true?

The mistake of taking the curl of a restricted field

Worked example MATH 376 Final problem 3, 5 points; see [A.4](/vol2/a4-final/) and [A.1](/vol2/a1-what-you-lost-points-on/)

Find

\oint_CF\cdot d\alpha

for

F = (yz\cos(xyz),\ xz\cos(xyz),\ xy\cos(xyz))

and

C

the unit circle in

z = 0

, counterclockwise.

Solution.

Method 1: a gradient. $\nabla\sin(xyz) = (yz\cos(xyz),\ xz\cos(xyz),\ xy\cos(xyz)) = F$ . So $F$ is a gradient field on $\mathbb R^3$ and the integral around the closed curve $C$ is $0$ (6.2).
Method 2: Stokes with the full curl. Write $w = xyz$ . Then $R_y = x\cos w - x^2yz\sin w$ and $Q_z = x\cos w - x^2yz\sin w$ , so the first component is $0$ . In the same way $P_z - R_x = 0$ and $Q_x - P_y = 0$ . So $\operatorname{curl}F = 0$ and the flux through the unit disc is $0$ .
Method 3: directly. On $C$ we have $z = 0$ and $\alpha' = (\alpha_1', \alpha_2', 0)$ . Substituting after the dot product is safe: $F\cdot\alpha' = 0\cdot\alpha_1' + 0\cdot\alpha_2' + xy\cdot0 = 0$ .

Answer. $0$ by any of the three.

Watch out· Restrict AFTER you differentiate

On the 376 final, a student answered $0$ , which was correct, but the method was not: they wrote $F(x, y, 0) = (0, 0, xy)$ and took the curl of that, getting $(x, -y, 0)$ . The curl contains derivatives in the $z$ direction ( $Q_z$ and $P_z$ ), and a $z$ -derivative needs the values of the field at nearby heights. Setting $z = 0$ first throws them away. The true curl is $0$ everywhere, while the restricted curl at $(1, 1, 0)$ is $(1, -1, 0)$ . Their answer survived only because the disc in $z = 0$ has $n = (0, 0, 1)$ , which uses only $Q_x - P_y$ , and $x$ - and $y$ -derivatives are tangent to the plane and survive the restriction. It fails as soon as the normal has an $x$ or $y$ component: for $F = (0, z, 0)$ the true curl is $(-1, 0, 0)$ , and the flux through the unit disc in the plane $x = 0$ with $n = (1, 0, 0)$ is $-\pi$ , while restricting to $z = 0$ would give $F = 0$ and flux $0$ . Restrict to a curve or surface only after all derivatives are taken; restricting inside a line integral is safe because no derivative is taken there.

Deducing one theorem from another

Worked example MATH 376 Midterm 2 problem 4(b), 3 points (scored 0.5); see [A.3](/vol2/a3-midterm-2/) and [A.1](/vol2/a1-what-you-lost-points-on/)

Deduce Green's theorem from Stokes' theorem.

Solution.

Hypotheses of Green. $D\subset\mathbb R^2$ is bounded by a positively oriented piecewise smooth Jordan curve $C$ , parametrized by $\alpha(t) = (x(t), y(t))$ , and $(P, Q)$ is $C^1$ on an open set containing $D$ . What we are given.
Surface. $S = D\times\{0\}$ , the copy of $D$ in the plane $z = 0$ , flat and so piecewise smooth, with boundary $C\times\{0\}$ .
Normal. $n = (0, 0, 1)$ . Counterclockwise from above and the right-hand rule: the orientations are compatible.
Field. $F(x, y, z) = (P(x, y),\ Q(x, y),\ 0)$ . Lift the planar field into space with a zero third component. It is $C^1$ on an open set containing $S$ .
The line integral. On $C\times\{0\}$ , $d\alpha = (dx, dy, 0)$ , so $F\cdot d\alpha = P\,dx + Q\,dy$ and $\oint_{C\times\{0\}}F\cdot d\alpha = \oint_CP\,dx + Q\,dy$ . The third component drops out.
The curl. $R = 0$ and $P, Q$ do not depend on $z$ , so $R_y - Q_z = 0$ and $P_z - R_x = 0$ . Hence $\operatorname{curl}F = (0, 0, Q_x - P_y)$ and $(\operatorname{curl}F)\cdot n = Q_x - P_y$ .
The surface element. $S$ is parametrized by $r(x, y) = (x, y, 0)$ with $r_x\times r_y = (0, 0, 1)$ of length $1$ , so $dS = dx\,dy$ .
Apply Stokes. $\oint_CP\,dx + Q\,dy = \iint_S(\operatorname{curl}F)\cdot n\,dS = \iint_D(Q_x - P_y)\,dx\,dy$ . All hypotheses were checked above. This is Green’s theorem.

Answer. Stokes’ theorem on the flat surface $D\times\{0\}$ with $n = (0, 0, 1)$ for the field $(P, Q, 0)$ is Green’s theorem. ‘Deduce’ means specialize: say what $S$ , $n$ and $F$ are, and compute the curl.

Watch out· Midterm 2 problem 4(b) lost 2.5 of 3 points: quoting is not deducing

On the 376 midterm, a student wrote $\oint_CF\cdot d\alpha = \iint_S\operatorname{curl}F\cdot dS$ and the determinant for the curl, and then stopped. That is the statement of the theorem, not an argument: the rubric item applied was ‘stated Green’s theorem or Stokes’ theorem, otherwise no helpful steps’. The question said to deduce one theorem from another, and the work is in the specializing: which surface, which normal and why that orientation, which field in which dimension, what the curl of that field is, and what $dS$ is. Also, they used $F_x, F_y, F_z$ for components; writing $P, Q, R$ (or $F_1, F_2, F_3$ ) avoids confusion with partial derivatives.

Worked example MATH 376 practice Midterm 2 problem 4(b); see [A.5](/vol2/a5-practice-exams/)

Let

a, b\in\mathbb R^2

and

P

the parallelogram

\{sa + tb : s, t\in[0, 1]\}

. Deduce from Stokes' theorem that its signed area is

\det(a, b)

Solution.

Setup. Write $a = (a_1, a_2)$ , $b = (b_1, b_2)$ , $D = a_1b_2 - a_2b_1$ . Embed $P$ in the plane $z = 0$ with $n = (0, 0, 1)$ .
Choose a field with curl $(0, 0, 1)$ . $F = (-\frac y2, \frac x2, 0)$ has $\operatorname{curl}F = (0, 0, \frac12 - (-\frac12)) = (0, 0, 1)$ . Then $(\operatorname{curl}F)\cdot n = 1$ and the flux is the area.
Stokes. $\operatorname{area}(P) = \iint_P1\,dS = \oint_{\partial P}F\cdot dr$ , with $\partial P$ traversed $0\to a\to a + b\to b\to0$ , which is counterclockwise when $D>0$ .
Side 1 ( $r = ta$ ): $F\cdot r' = -\frac{ta_2}2a_1 + \frac{ta_1}2a_2 = 0$ . Side 4 ( $r = (1 - t)b$ , $r' = -b$ ) is $0$ in the same way, because $F$ is perpendicular to the position vector.
Side 2 ( $r = a + tb$ , $r' = b$ ): $F\cdot r' = \frac{-(a_2 + tb_2)b_1 + (a_1 + tb_1)b_2}2 = \frac{a_1b_2 - a_2b_1}2 = \frac D2$ . The terms with $t$ cancel. Side 3 ( $r = (1 - t)a + b$ , $r' = -a$ ) also gives $\frac D2$ .
Add. $0 + \frac D2 + \frac D2 + 0 = D$ .

Answer. The signed area is $\det(a, b)$ : positive if the traversal is counterclockwise, negative if clockwise, and $0$ if $a$ and $b$ are parallel.

What curl means, now proved

Proposition B Apostol 12.12 (curl and rotation); the paddle-wheel picture

Curl is circulation per unit area

Let $F$ be $C^1$ near $p$ and $n$ a unit vector. Let $D_\varepsilon$ be the flat disc of radius $\varepsilon$ with centre $p$ perpendicular to $n$ , and $C_\varepsilon = \partial D_\varepsilon$ with the induced orientation. Then $\displaystyle\lim_{\varepsilon\to0}\frac1{\pi\varepsilon^2}\oint_{C_\varepsilon}F\cdot dr = (\operatorname{curl}F)(p)\cdot n$ .

In one lineFor a $C^1$ field $F$ , a point $p$ and a unit vector $n$ , the circulation of $F$ around the circle of radius $\varepsilon$ about $p$ in the plane perpendicular to $n$ (counterclockwise from the tip of $n$ ), divided by its area $\pi\varepsilon^2$ , tends to $(\operatorname{curl}F)(p)\cdot n$ as $\varepsilon\to0$ .

Why it is trueIt is the definition-free meaning of curl, and the reason the paddle wheel turns at half of the curl along its axis.

ProofApostol 12.12

The idea: Stokes turns the circulation into a flux of the curl, and a continuous function over a small disc is close to its value at the centre.

\oint_{C_\varepsilon}F\cdot dr = \iint_{D_\varepsilon}(\operatorname{curl}F)\cdot n\,dS

Stokes' theorem on the flat oriented disc

D_\varepsilon

with normal

n

Then

\Big\lvert\iint_{D_\varepsilon}\big[(\operatorname{curl}F)\cdot n - (\operatorname{curl}F)(p)\cdot n\big]dS\Big\rvert\le\eta\,\pi\varepsilon^2

The curl is continuous because

F

C^1

: given

\eta>0

there is

\varepsilon_0

with

\lvert(\operatorname{curl}F)(q)\cdot n - (\operatorname{curl}F)(p)\cdot n\rvert<\eta

for

q\in D_\varepsilon

\varepsilon<\varepsilon_0

\Big\lvert\dfrac1{\pi\varepsilon^2}\oint_{C_\varepsilon}F\cdot dr - (\operatorname{curl}F)(p)\cdot n\Big\rvert\le\eta

for all small

\varepsilon

\blacksquare

The integral of the constant

(\operatorname{curl}F)(p)\cdot n

over the disc is that constant times the area

\pi\varepsilon^2

This is why the wheel in the picture at the top spins at half of $\operatorname{curl}F\cdot n$ (the rim moves at the average of the field on the rim, so the angular speed is the circulation divided by the rim length $2\pi\varepsilon$ and by the radius $\varepsilon$ , giving circulation per area over $2$ ), and why the curl is largest along the axis in which the circulation density is largest. The curl vector points along the axis of fastest rotation, and its length is that rotation rate (twice the angular speed).

The divergence theorem

Definition T

Solid region

A solid region $E\subset\mathbb R^3$ is a bounded set with non-empty interior whose boundary $\partial E$ is a piecewise smooth surface. A ball, a cube, the region under a paraboloid and a torus (the doughnut, not only the skin) are solid regions.

In one lineA bounded region $E$ of space with non-empty interior whose boundary $\partial E$ is a piecewise smooth surface.

Why it existsIt is the shape the divergence theorem integrates over: a body with a skin.

Definition T

Closed surface

A surface is closed if it has no boundary curve: it encloses a solid. A sphere, a torus and the surface of a cube are closed; a hemisphere, a bowl and a cube with its top removed are not. (In 4.1 ‘closed set’ means that the set contains its boundary points, which is a different use of the word.)

In one lineA surface without a boundary curve of its own, such as the boundary of a solid region: a sphere, the surface of a cube or a torus. An open surface such as a bowl or a cube without a lid has a boundary curve.

Why it existsStokes' theorem is about surfaces with an edge and the divergence theorem about surfaces without one, and mixing them up is a common error (practice Final 7(d)).

Theorem T MATH 376 lecture of 4 March, Theorem 5.16; Midterm 2 problem 4(a); Apostol 12.19

The divergence theorem

Let $E\subset\mathbb R^3$ be a solid region whose boundary $S = \partial E$ is a closed, orientable, piecewise smooth surface, and let $n$ be the outward unit normal. Let $F$ be $C^1$ on an open set containing $E$ . Then $\iint_SF\cdot n\,dS = \iiint_E\operatorname{div}F\,dV$ . (The lecture states it with $C^2$ .)

In one lineIf $E$ is a solid region with boundary a closed piecewise smooth surface $S = \partial E$ with outward unit normal $n$ , and $F$ is $C^1$ on an open set containing $E$ , then $\iint_SF\cdot n\,dS = \iiint_E\operatorname{div}F\,dV$ .

Why it is trueIt trades a flux through a closed surface for a volume integral of the divergence (or the reverse), and it says that what a field produces inside a body is what leaves through its skin.

Midterm 2 problem 4(a) asked only for this statement, and the marks were for the hypotheses: a solid region, a closed piecewise smooth boundary, the outward normal, and a $C^1$ field on a neighbourhood of all of $E$ . What fails without each one:

Hypothesis	What goes wrong without it
$S$ is closed and bounds a solid	An open surface (a cube without a lid) has no solid inside it. Take $F = (0, 0, 1)$ and the upper unit hemisphere: $\operatorname{div}F = 0$ , but the flux is $\pi$ . Close it first.
outward normal	With the inward normal, the left side changes sign.
$F$ is $C^1$ on all of $E$ , the inside included	$F = r/\rho^3$ has divergence $0$ away from the origin and flux $4\pi$ out of the unit sphere, because it is not defined at the origin.

Why it is true: the proof for a standard region

ProofApostol 12.19; MATH 376 lecture of 4 March (the affine case on a box)

The idea: Do the third component of the field alone, in the z direction: the fundamental theorem of calculus in z turns the volume integral into a difference of values of R on the top and bottom, and that difference is exactly the flux of (0, 0, R) through the top and bottom. The first two components are the same with the axes renamed.

A standard region is of this form in all three directions, and a box is the simplest case (

g_1

and

g_2

constant).

Setup:

E = \{(x, y, z): (x, y)\in D,\ g_1(x, y)\le z\le g_2(x, y)\}

for a region

D

and continuous, piecewise

C^1

functions

g_1\le g_2

, and the field

(0, 0, R)

with

R

of class

C^1

\displaystyle\iiint_ER_z\,dV = \iint_D\Big(\int_{g_1}^{g_2}R_z\,dz\Big)dA = \iint_D\big[R(x, y, g_2) - R(x, y, g_1)\big]dA

Fubini (6.3) with

z

innermost, then the fundamental theorem of calculus in

z

The flux of

(0, 0, R)

through the top is

\iint_DR(x, y, g_2)\,dA

The top is the graph of

g_2

with the upward (outward) normal,

n\,dS = (-g_{2x}, -g_{2y}, 1)\,dA

(6.5); the dot product with

(0, 0, R)

keeps only the

1

The flux through the bottom is

-\iint_DR(x, y, g_1)\,dA

The bottom is the graph of

g_1

and outward means downward, so

n\,dS = -(-g_{1x}, -g_{1y}, 1)\,dA

The flux through the side wall is

0

The side wall is vertical, so its outward normal is horizontal and

(0, 0, R)\cdot n = 0

\iint_S(0, 0, R)\cdot n\,dS = \iiint_ER_z\,dV

Add the three fluxes and compare with the second line.If the region is not of this form, cut it into pieces that are; the shared faces carry opposite normals and cancel.

\iint_SF\cdot n\,dS = \iiint_E(P_x + Q_y + R_z)\,dV

\blacksquare

Repeat with

x

innermost for the field

(P, 0, 0)

and with

y

innermost for

(0, Q, 0)

, which a standard region allows, and add the three results.

Proof workshopGive the reason for each line: the divergence theorem for the z componentApostol 12.19

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\displaystyle\iiint_ER_z\,dV = \iint_D\Big(\int_{g_1(x,y)}^{g_2(x,y)}R_z\,dz\Big)dA$ .
$= \iint_D\big[R(x, y, g_2) - R(x, y, g_1)\big]dA$ .
The flux of $(0, 0, R)$ through the top surface $z = g_2$ is $\iint_DR(x, y, g_2)\,dA$ .
The flux through the bottom surface $z = g_1$ is $-\iint_DR(x, y, g_1)\,dA$ .
The flux through the vertical side wall is $0$ .
So $\iint_S(0, 0, R)\cdot n\,dS = \iiint_ER_z\,dV$ .

Worked example MATH 376 lecture of 4 March, Example 5.15

Check the divergence theorem on the box

V = [a_1, b_1]\times[a_2, b_2]\times[a_3, b_3]

for

F = (\alpha x + \beta y + \gamma z,\ 0,\ 0)

Solution.

Volume side. $\operatorname{div}F = \alpha$ , a constant, so $\iiint_V\alpha\,dV = \alpha\operatorname{Vol}(V)$ .
Four faces contribute nothing. The faces with normals $\pm e_2$ and $\pm e_3$ have $F\cdot n = 0$ , because $F$ points along $e_1$ , which is orthogonal to those normals.
The two remaining faces. The face $x = b_1$ has $n = e_1$ and the face $x = a_1$ has $n = -e_1$ . Parametrize both by $(y, z)$ in the rectangle $W = [a_2, b_2]\times[a_3, b_3]$ . The flux is $\iint_W(\alpha b_1 + \beta y + \gamma z)\,dy\,dz - \iint_W(\alpha a_1 + \beta y + \gamma z)\,dy\,dz$ .
Cancel. The terms $\beta y + \gamma z$ are identical in the two integrals and cancel, leaving $\alpha(b_1 - a_1)\operatorname{area}(W) = \alpha\operatorname{Vol}(V)$ . Width in $x$ times the area of the face is the volume of the box.

Answer. Both sides are $\alpha\operatorname{Vol}(V)$ . This is the box case of the proof above with the faces written out: the top and bottom of the proof are the faces $x = b_1$ and $x = a_1$ here, and the shared faces of neighbouring small boxes cancel in the general case.

Why· The picture: the torus and the divergence

In the picture at the top of the chapter the torus has flux $6\pi^2Ra^2$ for $F = (x, y, z)$ . Divergence $3$ times volume $2\pi^2Ra^2$ . For the field $(y, -x, z)$ the torus flux is $2\pi^2Ra^2$ , because its divergence is $0 + 0 + 1 = 1$ . For $(0, 0, 1)$ it is $0$ : the field enters through the bottom of the torus and leaves through the top in equal measure. Try all four fields on the torus in the picture and compare with $\operatorname{div}F\times\text{volume}$ .

Worked example MATH 376 Midterm 2 problem 4(a), 2 points; see [A.3](/vol2/a3-midterm-2/)

State the divergence theorem, with all hypotheses.

Solution.

Region. Let $E\subset\mathbb R^3$ be a solid region whose boundary $S = \partial E$ is a piecewise smooth closed surface. Hypothesis on the region: ‘solid’ means bounded with an interior, ‘closed’ means no edge.
Orientation. Let $n$ be the outward-pointing unit normal field on $S$ .
Field. Let $F$ be a $C^1$ vector field on an open set containing $E$ . It must be defined and continuously differentiable on all of $E$ , boundary included.
Conclusion. Then $\iint_SF\cdot n\,dS = \iiint_E\operatorname{div}F\,dV$ , where $\operatorname{div}F = \partial_xF_1 + \partial_yF_2 + \partial_zF_3$ .

Answer. The four lines in order: region, orientation, field, conclusion. The same pattern for Stokes’ theorem (practice Midterm 2 problem 4(a)): oriented surface, boundary orientation by the right-hand rule, $C^1$ field, conclusion.

Worked example MATH 376 Final problem 2, 5 points (scored 4); see [A.4](/vol2/a4-final/)

Compute the flux of

F = \big(\frac x3 + \sin(yz),\ \frac y3 + e^{xz},\ \frac z3 + \cos(xy)\big)

through the surface

\max(\lvert x\rvert, \lvert y\rvert, \lvert z\rvert) = 1

with the outward normal.

Solution.

Read the surface. $\max(\lvert x\rvert, \lvert y\rvert, \lvert z\rvert) = 1$ means $\lvert x\rvert, \lvert y\rvert, \lvert z\rvert\le1$ with at least one equal to $1$ . The maximum is $1$ exactly when no coordinate exceeds $1$ in absolute value and one reaches it.
Identify it. It is the surface of the solid cube $E = [-1, 1]^3$ , side $2$ , six faces. Sketch it before computing.
Divergence. $\partial_x(\frac x3 + \sin(yz)) = \frac13$ , $\partial_y(\frac y3 + e^{xz}) = \frac13$ , $\partial_z(\frac z3 + \cos(xy)) = \frac13$ . So $\operatorname{div}F = 1$ . $\sin(yz)$ has no $x$ , $e^{xz}$ has no $y$ , $\cos(xy)$ has no $z$ .
Hypotheses. $E$ is a solid, $S = \partial E$ is closed and piecewise smooth (six squares), $n$ is outward, and $F$ is $C^1$ on $\mathbb R^3$ .
Apply. $\iint_SF\cdot n\,dS = \iiint_E1\,dV = \operatorname{Vol}(E) = 2\cdot2\cdot2 = 8$ .
Cross-check by faces. On $x = 1$ with $n = (1, 0, 0)$ , $F\cdot n = \frac13 + \sin(yz)$ . On $x = -1$ with $n = (-1, 0, 0)$ , $F\cdot n = \frac13 - \sin(yz)$ . The $\sin(yz)$ terms cancel when the two faces are added, leaving $2\cdot\frac13\cdot4 = \frac83$ . The other two pairs of faces also give $\frac83$ , with $e^{xz}$ and $\cos(xy)$ cancelling. Total $8$ .

Answer. $8$ . For $\max(\lvert x\rvert, \lvert y\rvert, \lvert z\rvert) = 2$ the answer is $4^3 = 64$ . The trap about this problem is in the Traps section below.

Worked example MATH 376 worksheet 17 problem 4 and worksheet 27 problem 11

Find the flux of

F = (2x + y,\ 3y + z,\ 4z + x)

out of a closed surface enclosing a solid of volume

10

, and the flux of

F = (x, y, z)

out of the ball of radius

R

both ways.

Solution.

First field. $\operatorname{div}F = 2 + 3 + 4 = 9$ , constant. $\partial_x(2x + y) = 2$ , $\partial_y(3y + z) = 3$ , $\partial_z(4z + x) = 4$ .
Apply. The flux is $9\cdot10 = 90$ .
Ball, by the theorem. $\operatorname{div}(x, y, z) = 3$ , so the flux is $3\cdot\frac43\pi R^3 = 4\pi R^3$ .
Ball, directly. On the sphere $n = (x, y, z)/R$ , so $F\cdot n = R$ , constant, and the flux is $R\cdot4\pi R^2 = 4\pi R^3$ . Constant $F\cdot n$ times the area.

Answer. $90$ and $4\pi R^3$ . If the divergence is a constant $c$ , the flux out of any closed surface is $c$ times the enclosed volume.

Worked example MATH 376 homework 6 problems 2 and 3

Use the divergence theorem to find the flux of

F = (x^3, y^3, z^3)

out of the sphere

x^2 + y^2 + z^2 = a^2

, and of

F = (x, y, z)

out of the boundary of

V = \{0\le z\le4 - x^2 - y^2\}

Solution.

First field, divergence. $\operatorname{div}F = 3x^2 + 3y^2 + 3z^2 = 3\rho^2$ with $\rho^2 = x^2 + y^2 + z^2$ .
Spherical coordinates. $dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$ over the ball: $\int_0^{2\pi}\!d\theta\int_0^\pi\sin\varphi\,d\varphi\int_0^a3\rho^2\cdot\rho^2\,d\rho$ . The integrand is a product of a function of $\rho$ , a function of $\varphi$ and a constant in $\theta$ , with constant limits, so it splits.
Evaluate. $2\pi\cdot2\cdot\frac{3a^5}5 = \frac{12\pi a^5}5$ .
Second field. $\operatorname{div}F = 3$ , and $V$ is bounded above by the paraboloid and below by the disc $x^2 + y^2\le4$ in $z = 0$ , so $\partial V$ is closed. In cylindrical coordinates $dV = r\,dz\,dr\,d\theta$ : $3\int_0^{2\pi}\!\int_0^2\!\int_0^{4 - r^2}r\,dz\,dr\,d\theta = 3\cdot2\pi\int_0^2(4r - r^3)\,dr = 3\cdot2\pi\cdot4 = 24\pi$ .

Answer. $\dfrac{12\pi a^5}5$ and $24\pi$ (three times the volume $8\pi$ ).

Your answerUse the divergence theorem to find the outward flux of

F = (xy,\ yz,\ zx)

through the surface of the cube

[0, 2]^3

Your answerFind the outward flux of

F = (x, 2y, 3z)

through the sphere of radius

3

centred at the origin.

When the surface is not closed

If $S$ is not closed, add flat pieces $S'$ so that $S\cup S'$ bounds a solid, apply the theorem, and subtract the flux through $S'$ : flux through $S$ = flux through $S\cup S'$ minus flux through $S'$ (with the outward normal on all of $S\cup S'$ ).

Worked example MATH 376 worksheet 14 problem 7

Find the flux of

v = (x^2, y^2, z^2)

out of the open box

0\le x, y, z\le1

with no lid (the lid is the face

z = 1

Solution.

Close it. The full cube $[0, 1]^3$ has a closed boundary. $\operatorname{div}v = 2x + 2y + 2z$ , and $\iiint_{[0,1]^3}(2x + 2y + 2z)\,dV = 3\cdot2\cdot\frac12 = 3$ . Each term integrates to $2\cdot\frac12\cdot1\cdot1 = 1$ .
The lid. On $z = 1$ the outward normal is $(0, 0, 1)$ and $v\cdot n = z^2 = 1$ , so the flux through the lid is $1\cdot\text{area} = 1$ .
Subtract. The flux through the open box is $3 - 1 = 2$ .

Answer. $2$ . Gauss does not apply to the open box directly. This is the closing technique.

Worked example new (compare 6.5, where the same flux is computed directly)

Find the upward flux of

F = (x, y, z)

through the cap

z = 4 - x^2 - y^2

z\ge0

, by closing it with a disc.

Solution.

Close. The cap and the disc $x^2 + y^2\le4$ in the plane $z = 0$ bound the solid $V$ under the paraboloid, with outward normal up on the cap and down on the disc.
Volume integral. $\operatorname{div}F = 3$ and $\operatorname{Vol}(V) = \int_0^{2\pi}\!\int_0^2(4 - r^2)r\,dr\,d\theta = 2\pi\cdot4 = 8\pi$ . So the flux out of the closed surface is $24\pi$ .
Disc. On the bottom disc $n = (0, 0, -1)$ and $F\cdot n = -z = 0$ . $z = 0$ there. So the disc contributes $0$ .
Subtract. The cap’s flux is $24\pi - 0 = 24\pi$ .

Answer. $24\pi$ . In 6.5 the direct computation with $N = (2s^2\cos\theta, 2s^2\sin\theta, s)$ gives $2\pi\int_0^2(s^3 + 4s)\,ds = 2\pi(4 + 8) = 24\pi$ , the same.

Worked example MATH 376 worksheet 14 problem 8

A sphere of radius

2

and the field

v = r/\lVert r\rVert

(so

v = n

on the sphere). Find the outward flux and the average divergence inside.

Solution.

Flux. On the sphere $v\cdot n = n\cdot n = 1$ , so the flux is the area $4\pi\cdot2^2 = 16\pi$ .
Average divergence. The volume of the ball is $\frac43\pi\cdot8 = \frac{32\pi}3$ . If the divergence theorem applied, the average divergence would be $\frac{16\pi}{32\pi/3} = \frac32$ .
Check the singularity. $\operatorname{div}(r/\rho) = 2/\rho$ , which is not defined at the origin but is integrable: $\int_0^2\frac2\rho\,4\pi\rho^2\,d\rho = 8\pi\cdot\frac{4}{2} = 16\pi$ . The field is not $C^1$ at the origin, so the theorem as stated does not apply; the direct volume integral over the shells is what supports the value.

Answer. Flux $16\pi$ and average divergence $\frac32$ .

Singular fields and the excised ball

Worked example MATH 376 lecture of 4 March (the counterexample); new (the excision)

Let

G = (x, y, z)/\rho^3

with

\rho = \sqrt{x^2 + y^2 + z^2}

. Show

\operatorname{div}G = 0

away from the origin, and find the outward flux of

G

through the unit sphere and through the surface of the cube

[-1, 1]^3

Solution.

Divergence. $\partial_x(x\rho^{-3}) = \rho^{-3} - 3x^2\rho^{-5}$ , and the same for $y$ and $z$ . Adding: $3\rho^{-3} - 3\rho^2\rho^{-5} = 0$ .
Unit sphere. There $n = (x, y, z)$ and $G\cdot n = \rho^2/\rho^3 = 1$ , so the flux is $4\pi$ . The divergence theorem would predict $0$ , and it does not apply since $G$ is undefined at the origin, inside the ball.
The cube. Let $B_\varepsilon$ be the ball of radius $\varepsilon<1$ about the origin and $E$ the region between the sphere of radius $\varepsilon$ and the cube. On $E$ the field is $C^1$ with $\operatorname{div}G = 0$ , so the flux out of $\partial E$ is $0$ . The theorem applies to the region with the hole.
Split the boundary. $\partial E$ is the cube (outward normal) and the small sphere (with normal pointing into the hole, the outward normal of $E$ ). So (flux out of the cube) $-$ (flux out of the small sphere) $= 0$ .
Small sphere. The flux out of the sphere of radius $\varepsilon$ is $\frac1{\varepsilon^2}\cdot4\pi\varepsilon^2 = 4\pi$ .

Answer. The flux through the cube is $4\pi$ , and the same for every closed surface that encloses the origin. Moving the origin outside the surface makes the flux $0$ . This is Gauss’s law for a point charge in 6.8.

Worked example new (symmetry; the torus of 6.5)

Find the outward flux of

F = (x + yz,\ y + zx,\ z + xy)

through the surface of the cube

[-1, 1]^3

, and the flux of

F = (x, y, z)

out of the torus with radii

R > a

Solution.

Divergence. $\partial_x(x + yz) = 1$ , $\partial_y(y + zx) = 1$ , $\partial_z(z + xy) = 1$ . So $\operatorname{div}F = 3$ , and the flux is $3\cdot8 = 24$ . The terms $yz$ , $zx$ , $xy$ do not contain the variable being differentiated, so they drop out; face by face they cancel in pairs, as in the exam problem.
Torus. The solid torus has volume $2\pi^2Ra^2$ (cross-section area $\pi a^2$ times the length $2\pi R$ of the core circle, by Pappus; or $\int\!\!\int\!\!\int r\,dr\,dz\,d\theta$ ) and $\operatorname{div}F = 3$ , so the flux is $6\pi^2Ra^2$ .
Compare with 6.5. The direct computation in 6.5 with $r_u\times r_v$ gave $\iint F\cdot N = 6\pi^2Ra^2$ . The two methods agree.

Answer. $24$ and $6\pi^2Ra^2$ .

Worked example MATH 376 lecture of 6 March, Example 5.19

Use the divergence theorem to show that the volume of the ball of radius

R

\frac43\pi R^3

Solution.

Choose a field with divergence $1$ . $F = \frac13(x, y, z)$ , so $\operatorname{div}F = \frac13 + \frac13 + \frac13 = 1$ and $\iiint_E1\,dV = \operatorname{Vol}(E)$ . Any field with divergence $1$ would do; this one is symmetric.
Flux through the sphere. On the sphere of radius $R$ , $n = (x, y, z)/R$ so $F\cdot n = \frac13R$ , constant. The area is $4\pi R^2$ .
Apply. $\operatorname{Vol}(E) = \frac13R\cdot4\pi R^2 = \frac43\pi R^3$ . The area formula of 6.5 is used, so this is a check of that formula against the volume formula.

Answer. $\frac43\pi R^3$ . The same idea gives the volume of any solid as $\frac13\iint r\cdot n\,dS$ .

What divergence means, now proved

Proposition B Apostol 12.20

Divergence is flux per unit volume

Let $F$ be $C^1$ near $p$ , $B_\varepsilon$ the ball of radius $\varepsilon$ about $p$ , $S_\varepsilon = \partial B_\varepsilon$ with the outward normal. Then $\displaystyle\lim_{\varepsilon\to0}\frac1{\operatorname{Vol}(B_\varepsilon)}\iint_{S_\varepsilon}F\cdot n\,dS = (\operatorname{div}F)(p)$ .

In one lineFor a $C^1$ field $F$ and a point $p$ , the outward flux of $F$ through the sphere $S_\varepsilon$ of radius $\varepsilon$ about $p$ , divided by the volume of the ball, tends to $(\operatorname{div}F)(p)$ as $\varepsilon\to0$ .

Why it is trueIt is the definition-free meaning of divergence: net outflow per unit volume, the picture of the 'idea in words' made exact.

The proof is the same as for the curl: the divergence theorem turns the flux into $\iiint_{B_\varepsilon}\operatorname{div}F\,dV$ , and since $\operatorname{div}F$ is continuous it differs from $(\operatorname{div}F)(p)$ by less than any given $\eta$ on small balls, so the average over $B_\varepsilon$ is within $\eta$ of $(\operatorname{div}F)(p)$ .

Corollary A Apostol 12.15 exercises 12 and 14 (the tangent and normal forms of Green's theorem)

Green's theorem in normal form

With $r(t) = (x(t), y(t))$ counterclockwise, the outward unit normal is $n = (y', -x')/\lVert r'\rVert$ and $ds = \lVert r'\rVert\,dt$ . So $\oint_CF\cdot n\,ds = \oint_C(P\,y' - Q\,x')\,dt = \oint_C(-Q)\,dx + P\,dy$ , and Green’s theorem for the pair $(-Q, P)$ gives $\iint_D(P_x - (-Q)_y)\,dA = \iint_D(P_x + Q_y)\,dA$ . For $F = (x, y)$ and the circle of radius $R$ : $F\cdot n = R$ , so the left side is $R\cdot2\pi R = 2\pi R^2$ , and the right side is $2\cdot\pi R^2$ .

In one lineFor a positively oriented Jordan curve $C$ bounding $D$ and a $C^1$ planar field $F = (P, Q)$ , $\oint_CF\cdot n\,ds = \iint_D(P_x + Q_y)\,dA$ , where $n$ is the outward unit normal and $ds$ is arc length.

Why it is trueIt is the divergence theorem in two dimensions, and it shows that Green's theorem has two forms: one with curl and one with divergence.

Quick checkA solid region has a closed boundary and the divergence of

F

-3

everywhere inside. Which is possible for the outward flux?

Quick checkStokes' theorem relates the flux of a vector field through a closed surface to the divergence of the field in the volume enclosed. True or false?

Quick checkWhich condition is the one that fails for

F = r/\rho^3

on the unit ball when the divergence theorem is applied?

Quick checkYou must compute the flux of a field with constant divergence

5

out of a cube of side

2

with a face missing. What do you do first?

Which theorem, which side

Theorem	Left side	Right side	Orientation rule
Fundamental theorem for gradients (6.2)	$\int_Cd\varphi$ along a path	$\varphi(B) - \varphi(A)$	direction $A\to B$
Green (plane)	$\oint_CP\,dx + Q\,dy$	$\iint_D(Q_x - P_y)\,dA$	$C$ counterclockwise
Stokes (space)	$\oint_{\partial S}F\cdot dr$	$\iint_S(\operatorname{curl}F)\cdot n\,dS$	right-hand rule between $\partial S$ and $n$
Divergence (space)	$\iint_{\partial E}F\cdot n\,dS$	$\iiint_E\operatorname{div}F\,dV$	$n$ outward

Stokes with a flat $S$ in $z = 0$ and $n = (0, 0, 1)$ is Green. Choosing the easy side:

You are asked for	Ask	Then
a line integral around a closed curve in space	Does $C$ bound an easy surface (a disc, a polygon)? Is $\operatorname{curl}F$ simpler than $F$ ?	Stokes with the simplest surface. If $\operatorname{curl}F = 0$ on a simply connected domain, the answer is $0$ .
the flux of $F$ through a closed surface	Is $\operatorname{div}F$ constant or simple? Is the solid easy to integrate over?	Divergence theorem. If $\operatorname{div}F = 0$ and $F$ is $C^1$ inside, the answer is $0$ .
the flux of $F$ through an open surface	Can I close it with flat pieces?	Close, apply the theorem, subtract. Or compute directly with $r_u\times r_v$ (6.5).
the flux of a curl through a surface	Is it a curl? Where does the surface end?	Stokes: it equals the circulation round the edge, and the answer depends only on the edge. The flux of a curl through a closed surface is $0$ .
a field with a singular point	Is the singular point inside or outside?	Outside: use the theorems. Inside: cut out a small ball (or disc) around it and use the theorem on the rest.
symmetric regions and odd integrands	Does the integrand change sign under $x\mapsto-x$ ?	The integral over a symmetric region is $0$ .

Quick checkThe flux of

\operatorname{curl}A

through the surface of any solid is

Where it is used

Used inside maths· Inside maths: harmonic functions and the fundamental theorem of calculus

If $\nabla^2f = 0$ (a harmonic function) on a solid $E$ , then the flux of $\nabla f$ out of $\partial E$ is $\iiint_E\operatorname{div}\nabla f\,dV = \iiint_E\nabla^2f\,dV = 0$ . This is the first step in the proof that a harmonic function has no interior maximum and satisfies the mean value property: its average over a sphere equals its value at the centre. In one variable the divergence theorem on an interval $[a, b]$ with ‘outward normal’ $+1$ at $b$ and $-1$ at $a$ reads $f(b) - f(a) = \int_a^bf'(x)\,dx$ , the fundamental theorem. In two variables it is Green’s theorem in normal form. All of them say the integral of a derivative over a region is the integral of the function over its boundary, which is the single sentence of 6.7.

Used in the real world· Used in the real world: the circulation of a hurricane

Meteorologists model a hurricane’s winds by a Rankine vortex: inside the eyewall radius $R$ the air rotates rigidly with angular velocity $\Omega$ , and outside it the speed falls like $1/r$ . Take $R = 30$ km and the strongest wind $50$ m/s at the eyewall. Then $\Omega = 50/30\,000 = 1.67\times10^{-3}$ s $^{-1}$ , so the vorticity inside the eyewall is $2\Omega = 3.3\times10^{-3}$ s $^{-1}$ (twice the angular velocity, from the example). Outside, $v = \Omega R^2/r$ and the curl is $0$ (the field is the vortex of the picture). Stokes’ theorem gives the circulation around any loop enclosing the eyewall as $\oint v\cdot dr = \text{vorticity}\times\text{area} = 2\Omega\cdot\pi R^2 = 2\pi Rv_{\max} = 2\pi\cdot30\,000\cdot50 = 9.4\times10^6$ m $^2$ /s, and it is the same for every larger loop, because the extra area carries no vorticity. A forecaster who measures the winds around a loop of 100 km radius and finds the circulation $9.4\times10^6$ m $^2$ /s has measured the total spin of the core without going into it.

The divergence theorem is the audit of conservation. If the velocity of an incompressible liquid in a pipe network is $v$ , then $\operatorname{div}v = 0$ , so the flux out of any closed surface is $0$ : what enters a junction leaves it. A leak detector that adds the measured flows $\iint v\cdot n\,dS$ over the faces of a control volume and finds $0.025$ m $^3$ /s of net outflow from a cube of side $0.5$ m, volume $0.125$ m $^3$ , has measured an average divergence of $0.2$ s $^{-1}$ : some of the liquid is appearing there, by a leak from outside or a source, at $0.2$ per second per unit volume.

Your answerA Rankine vortex has eyewall radius

30

km and maximum wind

50

m/s at the eyewall. What is the circulation (in m

^2

/s) of the wind around a circle of radius

100

km about the centre? Give the exact value in terms of

\pi

Python labThe divergence theorem on a lopsided surface, and when it fails

import numpy as np

a, b, c = 1.0, 1.3, 0.8                      # ellipsoid x^2/a^2 + y^2/b^2 + z^2/c^2 = 1
def F(x, y, z):    return (x * y + z**2, y * z + x**2, z * x + y**2 + 0.5 * z**3)
def divF(x, y, z): return y + z + x + 1.5 * z**2       # by hand: y + z + (x + 1.5 z^2)

n = 400
ph = (np.arange(n) + 0.5) * np.pi / n;  th = (np.arange(2 * n) + 0.5) * np.pi / n
PH, TH = np.meshgrid(ph, th, indexing="ij"); w = (np.pi / n) ** 2
x, y, z = a * np.sin(PH) * np.cos(TH), b * np.sin(PH) * np.sin(TH), c * np.cos(PH)
N = (b * c * np.sin(PH)**2 * np.cos(TH), a * c * np.sin(PH)**2 * np.sin(TH), a * b * np.sin(PH) * np.cos(PH))  # r_phi x r_theta, outward
flux = sum(f * m for f, m in zip(F(x, y, z), N)).sum() * w

m = 200                                      # inside: x = a t sin(ph) cos(th) ..., Jacobian a b c t^2 sin(ph)
t = (np.arange(m) + 0.5) / m
T, P2, T2 = np.meshgrid(t, ph[::2], th[::2], indexing="ij")
X, Y, Z = a * T * np.sin(P2) * np.cos(T2), b * T * np.sin(P2) * np.sin(T2), c * T * np.cos(P2)
vol = (divF(X, Y, Z) * a * b * c * T**2 * np.sin(P2)).sum() * (1 / m) * (2 * np.pi / n) ** 2
print("flux out of the ellipsoid :", round(flux, 5))
print("integral of div F inside  :", round(vol, 5))

def point_source(x0):                        # F = (r - x0)/|r - x0|^3, flux through the unit sphere (n = position)
    X0, Y0, Z0 = x0; k = 400
    p = (np.arange(k) + 0.5) * np.pi / k; q = (np.arange(2 * k) + 0.5) * np.pi / k
    A, B = np.meshgrid(p, q, indexing="ij")
    nx, ny, nz = np.sin(A) * np.cos(B), np.sin(A) * np.sin(B), np.cos(A)
    dx, dy, dz = nx - X0, ny - Y0, nz - Z0; d3 = (dx**2 + dy**2 + dz**2) ** 1.5
    return ((dx * nx + dy * ny + dz * nz) / d3 * np.sin(A)).sum() * (np.pi / k) ** 2
for x0 in [(0, 0, 0), (0.5, 0.2, 0), (2, 0, 0)]:
    print("source at", x0, ":", round(point_source(x0), 4))
print("4 pi =", round(4 * np.pi, 4))

Try this. Run it. The flux of the field out of the ellipsoid (computed from a grid of 320000 small patches) and the integral of the divergence over its inside (computed with 200 by 200 by 400 small boxes) agree to about four digits, 0.8364. Neither is a number you could guess, and they match, so the divergence theorem is doing real work here. The second part uses the field of a point source, F = (x, y, z) divided by its distance cubed, with the source at x0 and the unit sphere around the origin: a source at the centre gives 4*pi = 12.566. Move the source to (0.5, 0.2, 0), still inside, and the flux is the same 12.566. Move it outside, to (2, 0, 0), and the flux is 0. The divergence is 0 everywhere except at the source, so the flux only counts the sources inside. Try the source at (0.99, 0, 0), barely inside, and at (1.01, 0, 0), barely outside, and see how sharply the flux changes. Change a, b, c, or the field F and its divergence (compute the divergence by hand first).

Traps

Watch out· Name the theorem you are using, and the right one

Stokes’ theorem is about a surface with an edge: circulation round the edge equals flux of the curl. The divergence theorem is about a closed surface: flux through the skin equals the integral of the divergence inside. A statement such as ‘Stokes’ theorem relates the flux through a closed surface to the divergence in the volume enclosed’ (practice Final problem 7(d)) describes the divergence theorem and is false for Stokes. Write which theorem and why its hypotheses hold before you use it.

Watch out· Read the set literally and sketch it

On the 376 final, a student read the set $\max(\lvert x\rvert, \lvert y\rvert, \lvert z\rvert) = 1$ of problem 2 as a unit cube in the first octant, because the absolute values were dropped, and their answer was $1$ instead of $8$ (4 of 5 points). The set is the surface of the cube $[-1, 1]^3$ , of side $2$ . Their own computation was consistent for the cube they drew, so the slip was in reading the problem, and the regrade request was declined for that reason. Sketch the set before choosing a method, and say in words what it is. On that problem the structure of the field was also a hint: the terms $x/3$ , $y/3$ , $z/3$ give divergence $1$ and the other terms drop out, which points straight at the divergence theorem.

Watch out· Hypotheses are part of the answer

Stokes needs $F$ to be $C^1$ on an open set containing the whole surface $S$ , the inside included, not only the curve. The divergence theorem needs $F$ to be $C^1$ on all of the solid. A field with a singularity inside (the angle field for Stokes, $r/\rho^3$ for Gauss) can have curl or divergence $0$ and still have a non-zero circulation or flux. The marks in a ‘state the theorem’ question are for the hypotheses: solid region, closed piecewise smooth boundary, outward normal, $C^1$ on a neighbourhood.

Watch out· Check orientation: the right-hand rule, and outward

The boundary direction in Stokes and the normal must be linked by the right-hand rule, and the divergence theorem needs the outward normal. If you reverse one and not the other the answer has the wrong sign. For ‘counterclockwise from above’ use the upward normal; for ‘counterclockwise as viewed from the direction of $(1, 1, 1)$ ’ use the normal pointing toward that viewer. Reversing both the surface orientation and the curve keeps the equation true.

Watch out· Compute the whole curl, with the flipped middle sign

The middle component of the curl is $P_z - R_x$ . A wrong sign there is invisible when the normal is $(0, 0, 1)$ , because only the third component is used, and it ruins a problem with a tilted normal. Write all three differences with their partial derivatives, and test the formula on a field you know ( $F = (z, 0, 0)$ should have curl $(0, 1, 0)$ ).

Problems

checknew (MATH 376 worksheet 13 style)

Compute the curl and the divergence of $F = (x^2y,\ y^2z,\ z^2x)$ .

Solution

$P = x^2y$ , $Q = y^2z$ , $R = z^2x$ . Curl: $R_y - Q_z = 0 - y^2 = -y^2$ ; $P_z - R_x = 0 - z^2 = -z^2$ ; $Q_x - P_y = 0 - x^2 = -x^2$ . So $\operatorname{curl}F = (-y^2, -z^2, -x^2)$ . Divergence: $P_x + Q_y + R_z = 2xy + 2yz + 2zx$ .

checknew (identities)

For $f = x^2 + y^2 + z^2$ compute $\operatorname{curl}(\nabla f)$ and $\operatorname{div}(\nabla f)$ .

Solution

$\nabla f = (2x, 2y, 2z)$ . The curl is $(0 - 0,\ 0 - 0,\ 0 - 0) = 0$ , as it must be for any gradient. The divergence is $2 + 2 + 2 = 6 = \nabla^2f$ .

checkMATH 376 practice Final problem 7(c)

True or false: if $f, g$ are $C^2$ vector fields on $\mathbb R^3$ , then $\operatorname{div}(f + g) = \operatorname{div}f + \operatorname{div}g$ .

Solution

True. Write $f = (f_1, f_2, f_3)$ , $g = (g_1, g_2, g_3)$ . $\operatorname{div}(f + g) = \partial_x(f_1 + g_1) + \partial_y(f_2 + g_2) + \partial_z(f_3 + g_3)$ , and a partial derivative of a sum is the sum of the partial derivatives, so this is $\operatorname{div}f + \operatorname{div}g$ .

checkMATH 376 homework 6 problem 4 (statement 1)

True or false: reversing the orientation of $S$ in Stokes’ theorem keeps the equation true.

Solution

True. Reversing the orientation replaces $n$ by $-n$ , so the surface integral changes sign. The right-hand rule then requires the opposite direction on $\partial S$ , so the line integral also changes sign. Both sides flip.

checkMATH 376 homework 6 problem 4 (statements 3 and 4)

True or false: (a) if $F = \nabla f$ then $\oint_CF\cdot d\alpha = 0$ for every closed curve $C$ . (b) The divergence theorem says that for a closed piecewise smooth surface $S$ bounding a solid $E$ , with outward normal $n$ and $F$ of class $C^1$ on a neighbourhood of $E$ , $\iint_SF\cdot n\,dS = \iiint_E\operatorname{div}F\,dV$ .

Solution

(a) True. By the fundamental theorem for line integrals (6.2) $\oint_C\nabla f\cdot d\alpha = f(\text{end}) - f(\text{start}) = 0$ on a closed curve. It also follows from Stokes, since $\operatorname{curl}\nabla f = 0$ whenever $C$ bounds a surface. (b) True, this is the theorem as stated above, with all four hypotheses (the exact wording in the original homework is not in the sources, so compare it with your copy).

useMATH 376 homework 6 problem 4 (statements 2 and 5)

True or false: (a) the integral $\iint_S(\operatorname{curl}F)\cdot n\,dS$ depends only on the area of $S$ . (b) The flux of $F$ through $S$ depends only on the area of $S$ .

Solution

Both false. (a) Take $F = (0, x, 0)$ , so $\operatorname{curl}F = (0, 0, 1)$ . The flat unit disc in the $xy$ -plane with $n = (0, 0, 1)$ gives $(\operatorname{curl}F)\cdot n = 1$ and the integral $\pi$ . The flat unit disc in the $yz$ -plane with $n = (1, 0, 0)$ has the same area and $(\operatorname{curl}F)\cdot n = 0$ , so the integral $0$ . (b) Take $F = (0, 0, 1)$ . A disc of area $A$ in the $xy$ -plane with $n = (0, 0, 1)$ has flux $A$ , and the same disc in the $yz$ -plane with $n = (1, 0, 0)$ has flux $0$ . Direction matters.

useMATH 376 worksheet 14 problem 5 style

Use Stokes’ theorem to compute the work of $F = (x^2, 2x, z)$ around the boundary $C$ of the square $[0, 1]^2$ in the plane $z = 0$ , counterclockwise from above. Check by a line integral.

Solution

$P = x^2$ , $Q = 2x$ , $R = z$ . $\operatorname{curl}F = (0 - 0,\ 0 - 0,\ 2 - 0) = (0, 0, 2)$ . With $S$ the flat square, $n = (0, 0, 1)$ : the circulation is $2\cdot\operatorname{area} = 2$ . Check: bottom ( $y = 0$ , $r = (t, 0)$ ) $F\cdot r' = t^2$ , integral $\frac13$ ; right ( $r = (1, t)$ ) $F = (1, 2, 0)$ , $r' = (0, 1, 0)$ , integral $2$ ; top ( $r = (1 - t, 1)$ , $r' = (-1, 0, 0)$ ) $F\cdot r' = -(1 - t)^2$ , integral $-\frac13$ ; left ( $r = (0, 1 - t)$ ) $F = (0, 0, 0)$ , integral $0$ . Total $\frac13 + 2 - \frac13 + 0 = 2$ .

usenew (divergence theorem with a paraboloid, compare homework 6 problem 3)

Find the outward flux of $F = (x^3, y^3, z^3)$ through the surface of the solid cylinder $x^2 + y^2\le1$ , $0\le z\le2$ (both discs included).

Solution

$\operatorname{div}F = 3x^2 + 3y^2 + 3z^2$ . In cylindrical coordinates $\iiint(3r^2 + 3z^2)\,r\,dz\,dr\,d\theta = 2\pi\int_0^1\!\int_0^2(3r^3 + 3z^2r)\,dz\,dr = 2\pi\int_0^1(6r^3 + 8r)\,dr = 2\pi\big(\frac64 + 4\big) = 11\pi$ .

usenew (closing a surface)

Find the upward flux of $F = (x, y, 1)$ through the upper unit hemisphere $x^2 + y^2 + z^2 = 1$ , $z\ge0$ , by closing it with the unit disc in $z = 0$ .

Solution

$\operatorname{div}F = 2$ , the solid half-ball has volume $\frac{2\pi}3$ , so the flux out of the closed surface is $\frac{4\pi}3$ . The bottom disc has $n = (0, 0, -1)$ and $F\cdot n = -1$ , so its flux is $-\pi$ . The hemisphere’s flux is $\frac{4\pi}3 - (-\pi) = \frac{7\pi}3$ .

usenew (a field and its curl: Stokes on a curved surface)

Let $F = (z, x, y)$ and let $S$ be the part of the paraboloid $z = 4 - x^2 - y^2$ with $z\ge0$ , upward normal. Find the flux of $\operatorname{curl}F$ through $S$ by Stokes, and check it directly.

Solution

$\operatorname{curl}F = (1, 1, 1)$ . The boundary is the circle $x^2 + y^2 = 4$ , $z = 0$ , counterclockwise from above. Use the flat disc: $n = (0, 0, 1)$ , $(\operatorname{curl}F)\cdot n = 1$ , so the flux is $4\pi$ . Directly on the paraboloid: $N = (2x, 2y, 1)$ , $(\operatorname{curl}F)\cdot N = 2x + 2y + 1$ , and the integral over $x^2 + y^2\le4$ is $0 + 0 + 4\pi = 4\pi$ . Along the circle: $\alpha = (2\cos t, 2\sin t, 0)$ , $F = (0, 2\cos t, 2\sin t)$ , $\alpha' = (-2\sin t, 2\cos t, 0)$ , $F\cdot\alpha' = 4\cos^2t$ , integral $4\pi$ .

proveMATH 376 homework 5 problem 5

Prove that $\operatorname{div}$ and $\operatorname{curl}$ are linear, and that $\operatorname{div}\operatorname{curl}F = 0$ for $C^2$ fields $F$ .

Solution

Let $F = (P_1, Q_1, R_1)$ , $G = (P_2, Q_2, R_2)$ and $a, b$ real. $\operatorname{div}(aF + bG) = \partial_x(aP_1 + bP_2) + \partial_y(aQ_1 + bQ_2) + \partial_z(aR_1 + bR_2)$ , and each partial derivative is linear, so this equals $a\operatorname{div}F + b\operatorname{div}G$ . The first component of $\operatorname{curl}(aF + bG)$ is $\partial_y(aR_1 + bR_2) - \partial_z(aQ_1 + bQ_2) = a(R_{1,y} - Q_{1,z}) + b(R_{2,y} - Q_{2,z})$ , and the other two components are the same with the letters cycled. For the last claim see the proof in the section on identities: expand and apply the equality of mixed partials (this uses $C^2$ ).

provenew (circulation is path independent: Stokes shows the surface does not matter)

Let $S_1$ and $S_2$ be two oriented surfaces with the same boundary curve $C$ , with orientations inducing the same direction on $C$ . Show that $\iint_{S_1}(\operatorname{curl}F)\cdot n\,dS = \iint_{S_2}(\operatorname{curl}F)\cdot n\,dS$ for $C^1$ fields $F$ defined on both. Deduce that the flux of a curl through a closed surface is $0$ .

Solution

By Stokes, both integrals equal $\oint_CF\cdot dr$ . For a closed surface $S$ , cut it along a closed curve $C$ into two surfaces $S_1$ and $S_2$ whose outward orientations induce opposite directions on $C$ . The flux through $S$ is the flux through $S_1$ plus that through $S_2$ , which equals the circulation along $C$ in one direction plus the circulation in the other, so $0$ . (Or: it is $\iiint\operatorname{div}\operatorname{curl}F\,dV = 0$ by the divergence theorem.)

provenew (flux of a gradient)

If $\nabla^2f = 0$ on a solid $E$ with closed boundary $S$ (and $f$ is $C^2$ ), show that $\iint_S\nabla f\cdot n\,dS = 0$ . Is the converse true?

Solution

By the divergence theorem the flux is $\iiint_E\operatorname{div}\nabla f\,dV = \iiint_E\nabla^2f\,dV = 0$ . The converse is false for a single solid: the flux can be $0$ while $\nabla^2f$ is positive in one part of $E$ and negative in another (the integral of $\nabla^2f$ over $E$ is $0$ while $\nabla^2f$ is not).

provenew (area from Green in normal form)

Use Green’s theorem in normal form with $F = (x, y)$ to show that the area of a region $D$ bounded by a positively oriented curve $C$ is $\frac12\oint_CF\cdot n\,ds$ . Check for the ellipse $x = a\cos t$ , $y = b\sin t$ .

Solution

With $F = (x, y)$ , $P_x + Q_y = 2$ , so $\oint_CF\cdot n\,ds = 2\operatorname{area}(D)$ . For the ellipse, $n\,ds = (y', -x')\,dt = (b\cos t, a\sin t)\,dt$ and $F\cdot n\,ds = (ab\cos^2t + ab\sin^2t)\,dt = ab\,dt$ , so $\frac12\oint = \frac12ab\cdot2\pi = \pi ab$ .

stretchnew (Gauss's law for several sources, compare 6.8)

Let $G_1 = (r - p_1)/\lvert r - p_1\rvert^3$ and $G_2 = (r - p_2)/\lvert r - p_2\rvert^3$ be two point sources at $p_1$ and $p_2$ , and $S$ a closed surface through neither. Show that the outward flux of $G_1 + 2G_2$ through $S$ is $4\pi$ times (the number of sources inside weighted by $1$ and $2$ ).

Solution

Each $G_i$ has divergence $0$ away from $p_i$ and is $C^1$ there. For a source outside $S$ the divergence theorem applies on the solid inside and the flux is $0$ . For a source inside, remove a small ball around it; the theorem on the rest gives that the flux through $S$ equals the flux through the small sphere, which is $4\pi$ as computed in the text. Flux is linear in the field, so the flux of $G_1 + 2G_2$ is $4\pi(\epsilon_1 + 2\epsilon_2)$ where $\epsilon_i = 1$ if $p_i$ is inside and $0$ if outside.

stretchnew (vorticity, compare worksheet 14 problem 1)

A velocity field is $v = (-\omega y,\ \omega x,\ 0)$ with $\omega$ a constant. Find the circulation of $v$ around the circle of radius $\rho$ in the $xy$ -plane, directly and by Stokes, and compare with the vorticity.

Solution

Directly: $r = (\rho\cos t, \rho\sin t, 0)$ , $v\cdot r' = \omega\rho^2$ , circulation $2\pi\omega\rho^2$ . By Stokes: $\operatorname{curl}v = (0, 0, 2\omega)$ , flux through the disc $2\omega\cdot\pi\rho^2$ . The vorticity $2\omega$ is twice the angular velocity, and circulation divided by area is $2\omega$ .

Summary

Curl $= (R_y - Q_z,\ P_z - R_x,\ Q_x - P_y)$ , the determinant with $\nabla$ in the second row, with the flipped middle sign. It is circulation per unit area about the axis, and a paddle wheel spins at half of $\operatorname{curl}F\cdot n$ . Divergence $= P_x + Q_y + R_z$ is flux per unit volume.
$\operatorname{curl}\nabla f = 0$ and $\operatorname{div}\operatorname{curl}F = 0$ for $C^2$ data, both because mixed partials are equal. The converses hold on simply connected or star-shaped domains and fail on domains with holes: the angle field and $r/\rho^3$ .
Stokes: $\oint_{\partial S}F\cdot dr = \iint_S\operatorname{curl}F\cdot n\,dS$ , with $\partial S$ and $n$ linked by the right-hand rule and $F$ of class $C^1$ on all of $S$ . Proof for a graph: Green plus the chain rule, with the second derivatives of $g$ cancelling.
To use Stokes, name the surface, the normal, the boundary orientation, the field, check the hypotheses, compute the full curl, dot, integrate. Never take the curl of a field restricted to the curve or the surface. ‘Deduce’ means specialize.
Divergence theorem: $\iint_{\partial E}F\cdot n\,dS = \iiint_E\operatorname{div}F\,dV$ , closed surface, outward normal, $F$ of class $C^1$ on all of $E$ . Proof: Fubini and the fundamental theorem of calculus in each direction. Read the set literally (a max-norm sphere is a cube).
Close an open surface with flat pieces and subtract their flux. Cut a small ball around a singular point. If the divergence is a constant $c$ , the flux is $c$ times the volume.
Green, Stokes and Gauss are one idea, the fundamental theorem of calculus one dimension higher: the integral of a derivative over a region is the integral of the function over its boundary, because inner boundaries cancel. 6.7 states it for all dimensions.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

6.6 Curl, divergence, Stokes' theorem and Gauss's theorem

The picture: a wheel in the flow, and a skin round a solid

The idea, in words

Curl and divergence

Curl

Divergence

Vorticity

Irrotational field

Solenoidal field

Laplacian

Identities

Identities for curl and divergence

Stokes’ theorem

Induced orientation

Stokes' theorem

Why it is true: the proof for a graph

How to use Stokes’ theorem

The mistake of taking the curl of a restricted field

Deducing one theorem from another

What curl means, now proved

Curl is circulation per unit area

The divergence theorem

Solid region

Closed surface

The divergence theorem

Why it is true: the proof for a standard region

When the surface is not closed

Singular fields and the excised ball

What divergence means, now proved

Divergence is flux per unit volume

Green's theorem in normal form

Which theorem, which side

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere