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VECTOR CALCULUS

17

17.9The Divergence Theorem

In this section, we will learn about:

The Divergence Theorem for simple solid regions,

and its applications in electric fields and fluid flow.

VECTOR CALCULUS

2

INTRODUCTION

In Section 17.5, we rewrote Green’s Theorem in a vector version as:

where C is the positively oriented boundary curve of the plane region D.

div ( , )C

D

ds x y dA⋅ =∫ ∫∫F n F

3

INTRODUCTION

If we were seeking to extend this theorem to vector fields on , we might make the guess that

where S is the boundary surface of the solid region E.

3

div ( , , )S E

dS x y z dV⋅ =∫∫ ∫∫∫F n F

Equation 1

4

DIVERGENCE THEOREM

It turns out that Equation 1 is true, under appropriate hypotheses, and is called the Divergence Theorem.

5

DIVERGENCE THEOREM

Notice its similarity to Green’s Theorem and Stokes’ Theorem in that:

It relates the integral of a derivative of a function (div F in this case) over a region to the integral of the original function F over the boundary of the region.

6

DIVERGENCE THEOREM

At this stage, you may wish to review the various types of regions over which we were able to evaluate triple integrals in Section 16.6

7

SIMPLE SOLID REGION

We state and prove the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3.

We call such regions simple solid regions.

For instance, regions bounded by ellipsoids or rectangular boxes are simple solid regions.

8

SIMPLE SOLID REGIONS

The boundary of E is a closed surface.

We use the convention, introduced in Section 17.7, that the positive orientation is outward.

That is, the unit normal vector n is directed outward from E.

9

THE DIVERGENCE THEOREM

Let: E be a simple solid region and let S be

the boundary surface of E, given with positive (outward) orientation. F be a vector field whose component functions

have continuous partial derivatives on an open region that contains E.

Then, div

S E

d dV⋅ =∫∫ ∫∫∫F S F

10


Thus, the Divergence Theorem states that:

Under the given conditions, the flux of Facross the boundary surface of E is equal to the triple integral of the divergence of Fover E.

11

GAUSS’S THEOREM

The Divergence Theorem is sometimes called Gauss’s Theorem after the great German mathematician Karl Friedrich Gauss (1777–1855).

He discovered this theorem during his investigation of electrostatics.

12

OSTROGRADSKY’S THEOREM

In Eastern Europe, it is known as Ostrogradsky’s Theorem after the Russian mathematician Mikhail Ostrogradsky (1801–1862).

He published this result in 1826.

13


Let F = P i + Q j + R k

Then,

Hence,

div P Q Rx y z

∂ ∂ ∂= + +∂ ∂ ∂

F

Proof

divE

E E E

dV

P Q RdV dV dVx y z

∂ ∂ ∂= + +

∂ ∂ ∂

∫∫∫

∫∫∫ ∫∫∫ ∫∫∫

F

14


If n is the unit outward normal of S, then the surface integral on the left side of the Divergence Theorem is:

( )S S

S

S S S

d d

P Q R dS

P dS Q dS R dS

⋅ = ⋅

= + + ⋅

= ⋅ + ⋅ + ⋅

∫∫ ∫∫

∫∫

∫∫ ∫∫ ∫∫

F S F n S

i j k n

i n j n k n

Proof

15


So, to prove the theorem, it suffices to prove these equations:

S E

S E

S E

PP dS dVxQQ dS dVyRR dS dVz

∂⋅ =

∂

∂⋅ =

∂

∂⋅ =

∂

∫∫ ∫∫∫

∫∫ ∫∫∫

∫∫ ∫∫∫

i n

j n

k n

Proof—Eqns. 2-4

16


To prove Equation 4, we use the fact that E is a type 1 region:

where D is the projection of Eonto the xy-plane.

( ) ( ) ( ) ( ){ }1 2, , , , , ,

E

x y z x y D u x y z u x y

=

∈ ≤ ≤

Proof

17


By Equation 6 in Section 15.6, we have:

( )( )

( )2

1

,

,, ,

u x y

u x yE D

R RdV x y z dz dAz z

∂ ∂ = ∂ ∂ ∫∫∫ ∫∫ ∫

Proof

18


Thus, by the Fundamental Theorem of Calculus,

( )( ) ( )( )2 1, , , , , ,E

D

R dVz

R x y u x y R x y u x y dA

∂∂

= −

∫∫∫

∫∫

Proof—Equation 5

19


The boundary surface S consists of three pieces: Bottom surface S1

Top surface S2

Possibly a vertical surface S3, which lies above the boundary curve of D(It might happen that S3 doesn’t appear, as in the case of a sphere.)

Proof

Fig. 17.9.1, p. 113620


Notice that, on S3, we have k ∙n = 0, because k is vertical and n is horizontal.

Thus,

Proof

3

3

0 0

S

S

R dS

dS

⋅

= =

∫∫

∫∫

k n

Fig. 17.9.1, p. 113621


Thus, regardless of whether there is a vertical surface, we can write:

1 2S S S

R dS R dS R dS⋅ = ⋅ + ⋅∫∫ ∫∫ ∫∫k n k n k n

Proof—Equation 6

22


The equation of S2 is z = u2(x, y), (x, y) D, and the outward normal n points upward.

So, from Equation 10 in Section 17.7 (with F replaced by R k), we have:

( )( )2

2, , ,S

D

R dS

R x y u x y dA

⋅ =∫∫

∫∫

k n

Proof

∈

Fig. 17.9.1, p. 113623


On S1, we have z = u1(x, y).

However, here, n points downward.

So, we multiply by –1:

Proof

( )( )1

1, , ,S

D

R dS

R x y u x y dA

⋅ =

−

∫∫

∫∫

k n

Fig. 17.9.1, p. 113624


Therefore, Equation 6 gives:

( )( ) ( )( )2 1, , , , , ,S

D

R dS

R x y u x y R x y u x y dA

⋅

= −

∫∫

∫∫

k n

Proof

25


Comparison with Equation 5 shows that:

Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2 or type 3 region, respectively.

S E

RR dS dVz

∂⋅ =

∂∫∫ ∫∫∫k n

Proof

26


Notice that the method of proof of the Divergence Theorem is very similar to that of Green’s Theorem.

27

DIVERGENCE

Find the flux of the vector field F(x, y, z) = z i + y j + x k

over the unit sphere x2 + y2 + z2 = 1

First, we compute the divergence of F:

Example 1

( ) ( ) ( )div 1z y xx y z∂ ∂ ∂

= + + =∂ ∂ ∂

F

28

DIVERGENCE

The unit sphere S is the boundary of the unit ball B given by: x2 + y2 + z2 ≤ 1

So, the Divergence Theorem gives the flux as:

( ) ( )343

div 1

413

S B B

F dS dV dV

V B ππ

⋅ = =

= = =

∫∫ ∫∫∫ ∫∫∫F

Example 1

29

DIVERGENCE

Evaluate where: F(x, y, z) = xy i + (y2 + exz2) j + sin(xy) k

S is the surface of the region E bounded by the parabolic cylinder z = 1 – x2

and the planesz = 0, y = 0, y + z = 2

S

d⋅∫∫F SExample 2

Fig. 17.9.2, p. 113730

DIVERGENCE

It would be extremely difficult to evaluate the given surface integral directly.

We would have to evaluate four surface integrals corresponding to the four pieces of S.

Also, the divergence of F is much less complicated than F itself:

Example 2

( ) ( ) ( )22div sin

2 3

xzxy y e xyx y z

y y y

∂ ∂ ∂= + + +∂ ∂ ∂

= + =

F

31

DIVERGENCE

So, we use the Divergence Theorem to transform the given surface integral into a triple integral.

The easiest way to evaluate the triple integral is to express E as a type 3 region:

( ){ }2, , 1 1,0 1 ,0 2

E

x y z x z x y z

=

− ≤ ≤ ≤ ≤ − ≤ ≤ −

Example 2

32

DIVERGENCE

Then, we have:

21 1 2

1 0 0

div

3

3

S

E

E

x z

d

dV

y dV

y dy dz dx− −

−

⋅

=

=

=

∫∫

∫∫∫

∫∫∫

∫ ∫ ∫

F S

F

Example 2

33

DIVERGENCE

( )

( )

( )

( )

2

2

21 1

1 0

131

10

1 3212 1

1 6 4 2 184350

23

2

232 3

1 8

3 3 7

x

x

zdz dx

zdx

x dx

x x x dx

−

−

−

−

−

−=

−= −

= − + −

= − + + − =

∫ ∫

∫

∫

∫

Example 2

34

UNIONS OF SIMPLE SOLID REGIONS

The Divergence Theorem can also be proved for regions that are finite unions of simple solid regions.

The procedure is similar to the one we used in Section 17.4 to extend Green’s Theorem.

35


For example, let’s consider the region Ethat lies between the closed surfaces S1

and S2, where S1 lies inside S2.

Let n1 and n2 be outward normals of S1 and S2.

36


Then, the boundary surface of E is: S = S1 S2

Its normal n is given by:

n = –n1 on S1

n = n2 on S2

Fig. 17.9.3, p. 113837

UNIONS OF SIMPLE SOLID RGNS.

Applying the Divergence Theorem to S, we get:

( )1 2

1 2

1 2

divE S

S

S S

S S

dV d

dS

dS dS

d d

= ⋅

= ⋅

= ⋅ − + ⋅

= − ⋅ + ⋅

∫∫∫ ∫∫

∫∫

∫∫ ∫∫

∫∫ ∫∫

F F S

F n

F n F n

F S F S

Equation 7

38

APPLICATIONS—ELECTRIC FIELD

Let’s apply this to the electric field (Example 5 in Section 16.1):

where S1 is a small sphere with radius a and center the origin.

( ) 3Qε

=E x xx

39


You can verify that div E = 0 (Exercise 23).

Thus, Equation 7 gives:

2 1

1

2

divS S E

S

S

d d dV

d

dS

⋅ = ⋅ +

= ⋅

= ⋅

∫∫ ∫∫ ∫∫∫

∫∫

∫∫

E S E S E

E S

E n40


The point of this calculation is that we can compute the surface integral over S1 because S1 is a sphere.

41


The normal vector at x is x/|x|.

Therefore.

since the equation of S1 is |x| = a.

2 23 4Q Q Q Q

aε ε ε ε

⋅ = ⋅ = ⋅ = =

xE n x x xxx x x

42


Thus, we have:

( )2 1 1

2

12

22 4

4

S S S

Qd dS dSa

Q A SaQ a

aQ

ε

ε

ε π

πε

⋅ = ⋅ =

=

=

=

∫∫ ∫∫ ∫∫E S E n

43


This shows that the electric flux of E is 4πεQthrough any closed surface S2 that containsthe origin.

This is a special case of Gauss’s Law (Equation 11 in Section 17.7) for a single charge.

The relationship between ε and ε0 is ε = 1/4πε0.

44

Divergence Theorem Examples

Gauss' divergence theorem relates triple integrals and surface integrals.

GAUSS' DIVERGENCE THEOREM

Let be a vector field. Let be a closed surface, and let be the region inside of . Then:F W We

(( ((( a bW

F A F† . œ .Ze

div

EXAMPLE 1 Evaluate , where is the sphere .(( a bW

# # #$B #C † . W B C D œ *i j A

SOLUTION We could parameterize the surface and evaluate the surface integral, but it is much faster touse the divergence theorem. Since:

diva b a b a b a b$B #C œ $B #C ! œ &` ` `

`B `C `Di j

the divergence theorem gives:

(( (((a b a bW

$B #C † . œ & .Z œ & ‚ œ ")!i j Ae

the volume of the sphere 1 è

EXAMPLE 2 Evaluate , where is the boundary of the cube defined by(( ˆ ‰W

# $C D C BD † . Wi j k A

" Ÿ B Ÿ " " Ÿ C Ÿ " ! Ÿ D Ÿ #, , and .

SOLUTION Since:

divˆ ‰ ˆ ‰ ˆ ‰ a bC D C BD# $i j k œ C D C BD œ $C B` ` `

`B `C `D# $ #


(( (((ˆ ‰ ˆ ‰

( ( ( ˆ ‰

(

W

# $ #

! " "

# " "#

"

"#

C D C BD † . œ $C B .Z

œ $C B .B .C .D

œ # 'C .C œ )

i j k Ae

è

EXAMPLE 3 Let be the region in bounded by the paraboloid and the plane ,e ‘$ # #D œ B C D œ "

and let be the boundary of the region . Evaluate .W C B D † .e (( ˆ ‰W

#i j k A

SOLUTION Here is a sketch of the region in question:

r

z (1, 1)

z = r2

z = 1

Since:

divˆ ‰ ˆ ‰a b a bC B D œ C B D œ #D` ` `

`B `C `Di j k# #


(( (((W

#D † . œ #D .Zk Ae

It is easiest to set up the triple integral in cylindrical coordinates:

((( ( ( (

( ’ “

( ˆ ‰

Œ

e

1

#D .Z œ #D < .D .< .

œ # D < .<

œ # < < .<

œ # œ" " #

# ' $

! ! <

# " "

!

"#

Dœ<

"

!

"&

#

#

)

1

1

11

è

In general, you should probably use the divergence theorem whenever you wish to evaluate a vectorsurface integral over a closed surface.

The divergence theorem can also be used to evaluate triple integrals by turning them into surfaceintegrals. This depends on finding a vector field whose divergence is equal to the given function.

EXAMPLE 4 Find a vector field whose divergence is the given function .F 0 Ba b (a) (b) (c)0 B œ " 0 B œ B C 0 B œ B Da b a b a b È# # #

SOLUTION The formula for the divergence is:

diva bF Fœ f † œ `J `J

`B `C `D

`JB DC

We get to choose , , and , so there are several possible vector fields with a given divergence.J J JB C D

(This is similar to the freedom enjoyed when finding a vector field with a given rotation.)

(a) works, as does , , and so forth.F i F j F kœ B œ C œ D

(b) Three possible solutions are , , and .F i F j F kœ B C œ B C œ B CD" "

$ #$ # # #

(c) It is difficult to integrate with respect to or , but we can integrate with respect to ÈB D B D C# #

to get .F jœ C B DÈ # # è

EXAMPLE 5 Let be the region defined by . Use the divergence theorem toe B C D Ÿ "# # #

evaluate .(((e

D .Z#

SOLUTION Let be the unit sphere . By the divergence theorem:W B C D œ "# # #

((( ((e

D .Z œ † .#

W

F A

where is any vector field whose divergence is . One possible choice is :F F kD œ D"

$# $

((( ((e

D .Z œ D † ."

$# $

W

k A

All that remains is to compute the surface integral .((W

$"

$D † .k A

We have parameterized the sphere many times by now:

r

z

> œ B œ ? >

< œ ? C œ ? >

D œ ? D œ ?

! Ÿ > Ÿ # Ÿ ? Ÿ# #

)

11 1

cos cos sin

sin sin

cos cos

so

and

This gives:

. œ .> .? œ .> .? ? > ? > !

? > ? > ?

œ ? >ß

A

i j k i j kâ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â â

ˆ ‰

`B `D`> `> `>

`C

`B `D`? `? `?

`C

#

cos sin cos cos

sin cos sin sin cos

cos cos cos sin cos sin#? >ß ? ? .> .?

so:

(( ( ( a b

(

” •

W ! Î#

$# Î#

$

Î#

Î#%

&

Î#

Î#

" "

$ $D † . œ ? ? ? .? .>

œ ? ? .?#

$

œ ?# "

$ &

œ%

"&

k A1 1

1

1

1

1

1

sin cos sin

sin cos

sin

1

1

1è

Of course, in the last example it would have been faster to simply compute the triple integral. In reality,the divergence theorem is only used to compute triple integrals that would otherwise be difficult to setup:

EXAMPLE 6 Let be the surface obtained by rotating the curveW

< œ ?

D œ #? Ÿ ? Ÿ# #

cos

sin

1 1

around the -axis:D

r

z

Use the divergence theorem to find the volume of the region inside of .W

SOLUTION We wish to evaluate the integral , where is the region inside of . By the(((e

.Z We

divergence theorem:

((( ((e

.Z œ † .W

F A

where is any vector field whose divergence is . Because of the cylindrical symmetry, and areF i j" B C

poor choices for . We therefore let :F F kœ D

((( ((e

.Z œ D † .W

k A

All that remains is to evaluate the surface integral .((W

D † .k A

We were essentially given the parameterization of the surface:

r

z

> œ B œ ? >

< œ ? C œ ? >

D œ #? D œ #?

! Ÿ > Ÿ # Ÿ ? Ÿ# #

)

11 1

cos cos sin

sin sin

cos cos

so

and

Thus:

. œ .> .? œ .> .? ? > ? > !

? > ? > # #?

œ # ? #

A

i j k i j kâ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â ââ â â â

a b

`B `D`> `> `>

`C

`B `D`? `? `?

`C

cos sin cos cos

sin cos sin sin cos

cos cos ? >ß # ? #? >ß ? ? .> .?cos cos cos sin cos sin

so:

(( ( ( a b

(

” •a b

W ! Î#

# Î#

Î#

Î#

Î#

Î#

#

D † . œ #? ? ? .? .>

œ # #? ? ? .?

œ # %? %?"

"'

œ#

k A1 1

1

1

1

1

1

sin cos sin

sin cos sin

sin

1

1

1

(according to my calculator)

è

4/28/2004, DIVERGENCE THEOREM Math21a, O. Knill

HOMEWORK. Section 13.8: 6,10,14,30

DIV. The divergence of a vector field F is div(P, Q, R) = ! · F = Px + Qy + Rz. It is a scalar field.The flux integral of a vector field F through a surface S = r(R) was defined as

! !S

F · dS =! !

RF (r(u, v)) ·

ru " rv dudv.Recall also that the integral of a scalar function f on a region R is

! ! !R

f dV =! ! !

Gf(x, y, z) dxdydz.

GAUSS THEOREM or DIVERGENCE THEOREM. Let G be a region in space bounded by a surface S andlet F be a vector field. Then

" " "G

div(F ) dV =

" "S

F · dS .

The orientation of S is such that the normal vector ru " rv points outside of G.

EXAMPLE. Let F (x, y, z) = (x, y, z) and let S be sphere. The divergence of F is constant 3 and! ! !G

div(F ) dV = 3 · 4!/3 = 4!. The flux through the boundary is! !

Sr · ru " rv dudv =! !

S|r(u, v)|2 sin(v) dudv =

! !

0

! 2!

0sin(v) dudv = 4! also.

CONTINUITY EQUATION. If " is the density of a fluid and let v be the velocity field of the fluid, thenby conservation of mass, the flux

! !S

v · dS of the fluid through a closed surface S bounding a region G isd/dt

! ! !G

"dV , the change of mass inside G. But this flux is by Gauss theorem equal to! ! !

Gdiv(v) dV .

Therefore,! ! !

G" # div(v) dV = 0. Taking a very small ball G around a point (x, y, z), where

! ! !G

f dV $f(x, y, z) gives " # div(v). This is called the continuity equation.

EXAMPLE. What is the flux of the vector field F (x, y, z) = (2x, 3z2 + y, sin(x)) through the box G = [0, 3] "[0, 2]" [#1, 1]?Answer: Use the divergence theorem: div(F ) = 2 and so

! ! !G

div(F ) dV = 2! ! !

GdV = 2Vol(G) = 24.

Note: Often, it is easier to evaluate a three dimensional integral than a flux integral because the later needs aparameterization of the boundary, which requires the calculation of ru " rv etc.

PROOF OF THE DIVERGENCE THEOREM. Consider a small box[x, x + dx] " [y, y + dy] " [z, z + dz]. Call the sides orthogonal tothe x axis x-boundaries etc. The flux of F = (P, Q, R) through thex-boundaries is [F (x + dx, y, z) · (1, 0, 0) + F (x, y, z) · (#1, 0, 0)]dydz =P (x + dx, y, z) # P (x, y, z) = Pxdxdydz. Similarly, the flux throughthe y-boundaries is Pydydxdz and the flux through the z-boundaryis Pzdzdxdy. The total flux through the boundary of the box is(Px + Py + Pz)dxdydz = div(F )dxdydz.

For a general body, approximate it with a union of small little cubes.The sum of the fluxes over all the little cubes is sum of the fluxes throughthe sides which do not touch an other box (fluxes through touching sidescancel). The sum of all the infinitesimal fluxes of the cubes is the fluxthrough the boundary of the union. The sum of all the div(F )dxdydz isa Riemann sum approximation for the integral

! ! !G

div(F )dxdydz. Inthe limit, where dx, dy, dz goes to zero, we obtain Gauss theorem.

VOLUME CALCULATION. Similarly as the planimeter allowed to calculate the area of a region by passingalong the boundary, the volume of a region can be determined as a flux integral. Take for example the vectorfield F (x, y, z) = (x, 0, 0) which has divergence 1. The flux of this vector field through the boundary of a regionis the volume of the region.

! !"G

(x, 0, 0) · dS = Vol(G).

GRAVITY INSIDE THE EARTH. How much do we weight deepin earth at radius r from the center of the earth? (Relevant inthe movie ”The core”) The law of gravity can be formulated asdiv(F ) = 4!", where " is the mass density. We assume that theearth is a ball of radius R. By rotational symmetry, the gravita-tional force is normal to the surface: F (x) = F (r)x/||x||. The fluxof F through a ball of radius r is

! !Sr

F (x) · dS = 4!r2F (r). By

the divergence theorem, this is 4!Mr = 4!! ! !

Br

"(x) dV ,where Mr is the mass of the material inside Sr. We have(4!)2"r3/3 = 4!r2F (r) for r < R and (4!)2"R3/3 = 4!r2F (r) forr % R. Inside the earth, the gravitational force F (r) = 4!"r/3.Outside the earth, it satisfies F (r) = M/r2 with M = 4!R3"/3.

0.5 1 1.5 2

0.2

0.4

0.6

0.8

1

WHAT IS THE BOUNDARY OF A BOUNDARY? The fundamental theorem for lineintegral, Green’s theorem,Stokes theorem and Gauss theorem are all of the form

!A

dF =!

"AF , where dF is a derivative of F and #A is

a boundary of A. They all generalize the fundamental theorem of calculus. There is some similarity in how dand # behave:

f scalar field ddf = curl grad(f) = 0F vector field ddF = div curl(F ) = 0

S surface in space #S is union of closed curves ##S = &G region in space #G is a closed surface ##G = &

The question when div(F ) = 0 implies F = curl(G) or whether curl(F ) = 0 implies G = grad(G) is interesting.We look at it Friday.

STOKES AND GAUSS. Stokes theorem was found by Ampere in 1825. George GabrielStokes: (1819-1903) was probably inspired by work of Green and rediscovers the identity around1840. Gauss theorem was discovered 1764 by Joseph Louis Lagrange. Carl FriedrichGauss, who formulates also Greens theorem, rediscovers the divergence theorem in 1813. Greenalso rediscovers the divergence theorem in 1825 not knowing of the work of Gauss and Lagrange.

Carl Friedrich Gauss George Gabriel Stokes Joseph Louis Lagrange Andre Marie Ampere

GREEN IDENTITIES. If G is a region in space bounded by a surface S and f, g are scalar functions, then with!f = !2f = fxx + fyy + fzz, one has as a direct consequence of Gauss theorem the first and second Greenidentities (see homework)! ! !

G(f!g + !f ·!g) dV =

! !S

f!g · dS! ! !

G(f!g # g!f) dV =

! !S(f!g # g!f) · dS

These identities are useful in electrostatics. Example: if g = f and !f = 0 and either f = 0 on the boundaryS or !f is orthogonal to S, then Green’s first identity gives

! ! !G|!f |2 dV = 0 which means f = 0. This can

be used to prove uniqueness for the Poisson equation !h = 4!" when applying the identity to the di"erencef = h1 # h2 of two solutions with either Dirichlet boundary conditions (h = 0 on S) or von Neumannboundary conditions (!h orthogonal to S).

GAUSS THEOREM IN HIGHER DIMENSIONS. If G is a n-dimensional ”hyperspace” bounded by a (n # 1)dimensional ”hypersurface” S, then

!G

div(F )dV =!

SF · dS.

DIV. In dimension d, the divergence is defined div(F ) = ! · F =#

i $Fi/$xi. The proof of the n-dimensionaldivergence theorem is done as in three dimensions.

By the way: Gauss theorem in two dimensions is just a version of Green’s theorem. Replacing F = (P, Q)with G = (#Q, P ) gives curl(F ) = div(G) and the flux of G through a curve is the lineintegral of F along thecurve. Green’s theorem for F is identical to the 2D-divergence theorem for G.

4/30/2004, DIVERGENCE THEOREM II Math21a, O. Knill

Problems: 1,2,3,4,5 in Homework (see homework handout).

ZWICKY’S VIEW. Fritz Zwicky was an astronomer who also devised the morphological method for creativ-ity. The method is one of many tools to solve problems which require new ideas. Zwicky was highly creativehimself. The idea of the supernovae, dark matter and many pulsar discoveries can be attributed to him. In1948 for example, Zwicky suggested to use extraterrestrial sources to reconstruct the universe. This shouldbegin with changing other planets, moons and asteroids by making them inhabitable and to change their orbitsaround the sun in order to adjust their temperature. He even suggested to alter the fusion of the sun to makedisplace our own space solar system towards an other planetary system. Zwickies morphological method isto organize ideas in boxes like a spreadsheet. For example, if we wanted to make order in the zoo of integraltheorems we have seen now, we would one coordinate to display the dimension, in which we work and the secondcoordinate the maximal dimension along which we integrate in the theorem.

1 2 3

1 Fundamental theorem of calculus - -2 Fundamental theorem of line integrals Greens theorem -3 Fundamental theorem of line integrals Stokes theorem Gauss theorem

Fundamental thm of line inte-grals. We integrate over 1 and0 dimensional objects. The 0 di-mensional object has no bound-ary.

Green and Stokes theorems. Weintegrate over 2 and 1 dimen-sional objects. The 1 dimen-sional object is a closed curveand has no boundary.

Divergence theorem. We inte-grate over 3 and 2 dimensionalobjects. The 2 dimensional ob-ject is a closed surface and hasno boundary.

One could build now for each of the di!erential operators grad, curl and div such a matrix and see whether thereis a theorem. Many combinations do not make sense like integrating the curl over a three dimensional object.But there is a theorem, which seem have escaped: if we integrate in two dimensions the div(F ) = Mx +Ny overa region R, then there this can be written as an integral over the boundary. There is indeed such a theorem,but it is just a version of Greens theorem in disguise (turn the vector field by 90 degrees and replace the line bya 2D version of the flux integral). It can be seen as a special case of the divergence theorem in three dimensionsand it does not make sense to put it on the footing of the other theorems. Anyway, the morphological methodwas used in management like in Ciba, it is today just one of a variety of methods to come up with new ideas.

IDENTITIES. While direct computations can verify the identities to the left, they become evident with Nabla

calculus from formulas for vectors like !v ! !v = !0,!v · !v ! !w = 0 or u ! (v ! w) = v(u · w) " (u · v)w.

div(curl(F )) = 0.curl(grad(F )) = !0curl(curl(F )) = grad(div(F ) ""(F )).

# ·#! F = 0.#!#F = !0.#!#! F = #(# · F ) " (# ·#)F .

QUIZZ. Is there a vector field G such that F = (x + y, z, y2) = curl(G)? Answer: no because div(F ) = 1 isincompatible with divcurl(G)) = 0.

BOXED OVERVIEW. All integral theorems are of the form!

RF ! =

!!R

F , where F ! is a ”derivative” and "Ris a ”boundary”. There are 2 such theorems in dimensions 2, three theorems in dimensions 3, four in dimension4 etc.

dim dim(R) theorem1D 1 Fund. thm of calculus

2D 1 Fund. thm of line integrals2D 2 Green’s theorem

dim dim(R) theorem3D 1 Fundam. thm of line integrals3D 2 Stokes theorem3D 3 Divergence theorem

1 $% 1 f ! derivative1 $% 2 #f gradient2 $% 1 #! F curl

1 $% 3 #f gradient3 $% 3 #! F curl3 $% 1 # · F divergence

MAXWELL EQUATIONS. c is the speed of light.

div(B) = 0 No monopoles there are no magnetic monopoles.curl(E) = " 1

cBt Faraday’s law change of magnetic flux induces voltage

curl(B) = 1

cEt + 4"c j Ampere’s law current or change of E produces magnetic field

div(E) = 4#$ Gauss law electric charges are sources for electric field

MAGNETOSTATICS: curl(B) = 0 so that B = grad(f). But since also div(B) = 0, we have

"f = div(grad(f)) = 0

ELECTROSTATICS. curl(E) = 0 so that E = grad(f). But since also div(E) = 0, we have

"f = div(grad(f)) = 0

Static electric and magnetic fields have a harmonic potential.

FLUID DYNAMICS. v velocity, $ density of fluid.

Continuity equation $ + div($v) = 0 no fluid get lostIncompressibility div(v) = 0 incompressible fluidsIrrotational curl(v) = 0 no vorticityPotential fluids v = grad(f),"(f) = 0 incompressible, irrotational fluids

Incompressible, irrotational fluid velocity fields have a potential which is harmonic

HARMONIC FUNCTIONS. A function f(x, y, z) is called harmonic, if "f(x, y, z) = 0. Harmonic functionsplay an important role in physics. For example time independent solutions of the Schrodinger equation ihft ="f or the heat equation ft = "f are harmonic. In fluid dynamics, incompressible and irrotational fluidswith velocity distribution v satisfy v = grad(f) and since div(v) = 0, also "f = 0. Harmonic functions haveproperties which can be seen nicely using the divergence theorem.

MAXIMUM PRINCIPLE. A harmonic func-tion in D can not have a local maximum insideD.

MEAN VALUE PROPERTY. The average of a har-monic function on a sphere of radius r around a pointP is equal to the value of the function at the point P .

You deal with such problems in the homework. They are direct consequences of the divergence theorem.

THE PIGEON PROBLEM. A farmer drives a closed van throughthe countryside. Inside the van are dozens of pigeons. The vanweights 200 pounds, the pigeon, an other 100 pounds. The driverhas to pass a bridge, which can only sustain 250 pounds. Thedriver has an idea: if all the pigeon would fly while crossing thebridge, a passage would be possible. Question: does the bridgehold? What happens if the cage is open?

The flight of the pigeon produces a wind distribution in the cage. The question is whether the flying pigeonwill still produces a weight on the van.

Green’s Theorem Revisited

Green’s Theorem:∮C

M(x , y) dx + N(x , y) dy =

∫∫R

(∂N∂x− ∂M∂y

)dA

R

T

C

n

x

y

J. Robert Buchanan The Divergence Theorem

Green’s Theorem Vector Form (1 of 3)

Simple closed curve C is described by the vector-valuedfunction

r(t) = 〈x(t), y(t)〉 for a ≤ t ≤ b.

The unit tangent vector and unit (outward) normal vector to Care respectively

T(t) =1

‖r′(t)‖〈x ′(t), y ′(t)〉 and n(t) =

1‖r′(t)‖

〈y ′(t),−x ′(t)〉.



If the vector field F(x , y) = M(x , y)i + N(x , y)j, then along thesimple closed curve C:

F · n = 〈M(x(t), y(t)),N(x(t), y(t))〉 · 1‖r′(t)‖

〈y ′(t),−x ′(t)〉

=(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

) 1‖r′(t)‖

.

Now consider the line integral∮C

F · n ds.

Note: this is a line integral with respect to arc length.



∮C

F · n ds =

∫ b

a(F · n)(t) ‖r′(t)‖dt

=

∫ b

a

(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

) ‖r′(t)‖‖r′(t)‖

dt

=

∫ b

a

(M(x(t), y(t))y ′(t)− N(x(t), y(t))x ′(t)

)dt

=

∮C

M(x , y) dy − N(x , y) dx

=

∫∫R

(∂M∂x

+∂N∂y

)dA (by Green’s Theorem)

=

∫∫R∇ · F dA


Summary and Objective

Green’s Theorem in vector form states∮C

F · n ds =

∫∫R∇ · F(x , y) dA.

A double integral of the divergence of a two-dimensional vectorfield over a region R equals a line integral around the closedboundary C of R.

The Divergence Theorem (also called Gauss’s Theorem) willextend this result to three-dimensional vector fields.


Divergence Theorem

Remark: the Divergence Theorem equates surface integralsand volume integrals.

Theorem (Divergence Theorem)

Let Q ⊂ R3 be a region bounded by a closed surface ∂Q andlet n be the unit outward normal to ∂Q. If F is a vector functionthat has continuous first partial derivatives in Q, then∫∫

∂QF · n dS =

∫∫∫Q∇ · F dV .


Proof (1 of 7)

Suppose F(x , y , z) = M(x , y , z)i + N(x , y , z)j + P(x , y , z)k,then the Divergence Theorem can be stated as∫∫

∂QF · n dS

=

∫∫∂Q

M(x , y , z)i · n dS +

∫∫∂Q

N(x , y , z)j · n dS

+

∫∫∂Q

P(x , y , z)k · n dS

=

∫∫∫Q

∂M∂x

dV +

∫∫∫Q

∂N∂y

dV +

∫∫∫Q

∂P∂z

dV

=

∫∫∫Q∇ · F(x , y , z) dV .


John_2

Typewritten Text

must prove this middle equality

Proof (2 of 7)

Thus the theorem will be proved if we can show that∫∫∂Q

M(x , y , z)i · n dS =

∫∫∫Q

∂M∂x

dV∫∫∂Q

N(x , y , z)j · n dS =

∫∫∫Q

∂N∂y

dV∫∫∂Q

P(x , y , z)k · n dS =

∫∫∫Q

∂P∂z

dV .

All of the proofs are similar so we will focus only on the third.


Proof (3 of 7)

Suppose region Q can be described as

Q = {(x , y , z) |g(x , y) ≤ z ≤ h(x , y), for (x , y) ∈ R}

where R is a region in the xy -plane.

Think of Q as being bounded by three surfaces S1 (top), S2(bottom), and S3 (side).


Proof (4 of 7)

S1: z=hHx,yL

S2: z=gHx,yL S3

x

y

z

On surface S3 the the unit outward normal is parallel to thexy -plane and thus∫∫

∂QP(x , y , z) k · n︸︷︷︸

=0

dS =

∫∫∂Q

0 dS = 0.


Proof (5 of 7)

Now we calculate the surface integral over S1.

S1 = {(x , y , z) | z − h(x , y) = 0, for (x , y) ∈ R}

Unit outward normal:

n =∇(z − h(x , y))

‖∇(z − h(x , y))‖=

−hx(x , y)i− hy (x , y)j + k√[−hx(x , y)]2 + [−hy (x , y)]2 + 1

andk · n =

1√[hx(x , y)]2 + [hy (x , y)]2 + 1


Proof (6 of 7)

∫∫S1

P(x , y , z)k · n dS =

∫∫S1

P(x , y , z)√[hx(x , y)]2 + [hy (x , y)]2 + 1

dS

=

∫∫R

P(x , y ,h(x , y)) dA

In a similar way we can show the surface integral over S2 is∫∫S2

P(x , y , z)k · n dS = −∫∫

RP(x , y ,g(x , y)) dA.


Proof (7 of 7)

Finally,∫∫∂Q


=

∫∫S1

P(x , y , z)k · n dS +

∫∫S2


+

∫∫S3


=

∫∫R

P(x , y ,h(x , y)) dA−∫∫

RP(x , y ,g(x , y)) dA

=

∫∫R[P(x , y ,h(x , y))− P(x , y ,g(x , y))] dA

=

∫∫R

P(x , y , z)

∣∣∣∣z=h(x ,y)

z=g(x ,y)

dA

=

∫∫R

∫ h(x ,y)

g(x ,y)

∂P∂z

dz dA =

∫∫∫Q

∂P∂z

dV .


vector calculus - interactive-mathvision.com · vector calculus. 2. introduction in section 17.5,...

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