VECTOR CALCULUS 17. 2 17.9 The Divergence Theorem In this section, we will learn about: The Divergence Theorem for simple solid regions, and its applications

VECTOR CALCULUSVECTOR CALCULUS

17

2

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

3

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

4

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

5

DIVERGENCE THEOREM

It turns out that Equation 1 is true,

under appropriate hypotheses, and

is called the Divergence Theorem.

6

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.

7

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

8

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.

9

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.

10

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

11


Thus, the Divergence Theorem states

that:

Under the given conditions, the flux of F across the boundary surface of E is equal to the triple integral of the divergence of F over E.

12

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.

13

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.

14


Let F = P i + Q j + R k

Then,

Hence,

divP Q R

x y z

F

Proof

divE

E E E

dV

P Q RdV dV dVx y z

F

15


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

16


So, to prove the theorem, it suffices to prove

these equations:

S E

S E

S E

PP dS dV

x

QQ dS dV

y

RR dS dV

z

i n

j n

k n

Proof—Eqns. 2-4

17


To prove Equation 4, we use the fact that

E is a type 1 region:

where D is the projection of E

onto the xy-plane.

1 2, , , , , ,

E

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

Proof

18


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

19


Thus, by the Fundamental Theorem of

Calculus,

2 1, , , , , ,

E

D

RdVz

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

Proof—Equation 5

20


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. 1136

21


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. 1136

22


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

23


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. 1136

24


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. 1136

25


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

26


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 dV

z

k n

Proof

27


Notice that the method of proof of

the Divergence Theorem is very similar

to that of Green’s Theorem.

28

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

29

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

41

3

S B B

F dS dV dV

V B

F

Example 1

30

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

dF S

Example 2

Fig. 17.9.2, p. 1137

31

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

32

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

33

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

34

DIVERGENCE

2

2

21 1

1 0

131

1

0

1 3212 1

1 6 4 2 184350

23

2

23

2 3

1 8

3 3 7

x

x

zdz dx

zdx

x dx

x x x dx

Example 2

35

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.

36


For example, let’s consider the region E

that lies between the closed surfaces S1

and S2, where S1 lies inside S2.

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

37


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. 1138

38

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

39

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.

3

QE x xx

40


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 n

41


The point of this calculation is that

we can compute the surface integral

over S1 because S1 is a sphere.

42


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

Therefore.

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

2 23 4

Q Q Q Q

a

xE n x x x

xx x x

43


Thus, we have:

2 1 1

2

12

224

4

S S S

Qd dS dS

a

QA S

aQ

aaQ

E S E n

44


This shows that the electric flux of E is 4πεQ

through any closed surface S2 that contains

the 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.

45

APPLICATIONS—FLUID FLOW

Another application of the Divergence

Theorem occurs in fluid flow.

Let v(x, y, z) be the velocity field of a fluid with constant density ρ.

Then, F = ρv is the rate of flow per unit area.

46


Suppose:

P0(x0, y0, z0) is a point in the fluid.

Ba is a ball with center P0 and very small

radius a.

Then, div F(P) ≈ div F(P0) for all points in Ba since div F is continuous.

47


We approximate the flux over the boundary

sphere Sa as follows:

0

0

div

div

div

a a

a

S B

B

a

d dV

P dV

P V B

F S F

F

F

48


This approximation becomes better as

a → 0 and suggests that:

00

1div lim

a

aa S

P dV B

F F S

Equation 8

49

SOURCE AND SINK

Equation 8 says that div F(P0) is the net rate

of outward flux per unit volume at P0.

(This is the reason for the name divergence.)

If div F(P) > 0, the net flow is outward near P and P is called a source.

If div F(P) < 0, the net flow is inward near P and P is called a sink.

50

SOURCE

For this vector field, it appears that the vectors

that end near P1 are shorter than the vectors

that start near P1.

Thus, the net flow is outward near P1.

So, div F(P1) > 0 and P1 is a source.

Fig. 17.9.4, p. 1139

51

SINK

Near P2, the incoming arrows are longer

than the outgoing arrows.

The net flow is inward.

So, div F(P2) < 0 and P2 is a sink.

Fig. 17.9.4, p. 1139

52

SOURCE AND SINK

We can use the formula for F to confirm

this impression.

Since F = x2 i + y2 j, we have div F = 2x + 2y, which is positive when y > –x.

So, the points above the line y = –x are sources and those below are sinks.

Documents

VECTOR CALCULUS 17. 2 17.9 The Divergence Theorem In this section, we will learn about: The Divergence Theorem for simple solid regions, and its applications