Transcript divergence theorem

17
VECTOR CALCULUS
VECTOR CALCULUS
17.9
The 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.
2
INTRODUCTION
In Section 17.5, we rewrote Green’s Theorem
in a vector version as:

C
F  n ds   div F( x, y) dA
D
where C is the positively oriented
boundary curve of the plane region D.
3
Equation 1
INTRODUCTION
If we were seeking to extend this theorem to
vector fields on
3
, we might make the guess
that
 F  n dS   div F( x, y, z) dV
S
E
where S is the boundary surface
of the solid region E.
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,
 F  dS   div FdV
S
E
10
THE DIVERGENCE THEOREM
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.
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
THE DIVERGENCE THEOREM
Proof
Let F = P i + Q j + R k
P Q R
 Then, div F 


x y z
 Hence,
 div F dV
E
P
Q
R
 
dV  
dV  
dV
x
y
z
E
E
E
14
THE DIVERGENCE THEOREM
Proof
If n is the unit outward normal of S,
then the surface integral on the left side
of the Divergence Theorem is:
 F  dS   F  n dS
S
S
   P i  Q j  R k   n dS
S
  P i  n dS   Q j  n dS   R k  n dS
S
S
S
15
THE DIVERGENCE THEOREM
Proof—Eqns. 2-4
So, to prove the theorem, it suffices to prove
these equations:
P
dV
S P i  n dS  
x
E
Q
dV
S Q j  n dS  
y
E
R
dV
S R k  n dS  
z
E
16
THE DIVERGENCE THEOREM
Proof
To prove Equation 4, we use the fact that
E is a type 1 region:
E
 x, y, z   x, y   D, u  x, y   z  u
1
2
 x, y 
where D is the projection of E
onto the xy-plane.
17
THE DIVERGENCE THEOREM
Proof
By Equation 6 in Section 15.6,
we have:
R
 u2  x , y  R

dV    
x, y, z  dz  dA


u1  x , y  z

z


E
D
18
THE DIVERGENCE THEOREM
Proof—Equation 5
Thus, by the Fundamental Theorem of
Calculus,
R
dV

z
E
   R  x, y, u2  x, y    R  x, y, u1  x, y    dA
D
19
THE DIVERGENCE THEOREM
Proof
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.)
20
Fig. 17.9.1, p. 1136
THE DIVERGENCE THEOREM
Proof
Notice that, on S3, we have k ∙n = 0,
because k is vertical and n is horizontal.
 Thus,
 R k  n dS
S3
  0 dS  0
S3
21
Fig. 17.9.1, p. 1136
THE DIVERGENCE THEOREM
Proof—Equation 6
Thus, regardless of whether there is
a vertical surface, we can write:
 R k  n dS   R k  n dS   R k  n dS
S
S1
S2
22
THE DIVERGENCE THEOREM
Proof
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:
 R k  n dS 
S2
 R  x, y, u  x, y   dA
2
D
23
Fig. 17.9.1, p. 1136
THE DIVERGENCE THEOREM
Proof
On S1, we have z = u1(x, y).
However, here, n points downward.
 So, we multiply
by –1:
 R k  n dS 
S1
  R  x, y, u1  x, y   dA
D
24
Fig. 17.9.1, p. 1136
THE DIVERGENCE THEOREM
Proof
Therefore, Equation 6 gives:
 R k  n dS
S
   R  x, y, u2  x, y    R  x, y, u1  x, y    dA
D
25
THE DIVERGENCE THEOREM
Proof
Comparison with Equation 5 shows
that:
R
dV
S R k  n dS  
z
E
 Equations 2 and 3 are proved in a similar manner
using the expressions for E as a type 2 or type 3
region, respectively.
26
THE DIVERGENCE THEOREM
Notice that the method of proof of
the Divergence Theorem is very similar
to that of Green’s Theorem.
27
DIVERGENCE
Example 1
Find the flux of the vector field
F(x, y, z) = z i + y j + x k
over the unit sphere
x2 + y 2 + z2 = 1
 First, we compute the divergence of F:



div F   z    y    x   1
x
y
z
28
Example 1
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:
 F  dS   div F dV   1dV
S
B
B
4
 V  B    1 
3
4
3
3
29
Example 2
DIVERGENCE
Evaluate
where:
F

d
S

S
2
 F(x, y, z) = xy i + (y2 + exz ) j + sin(xy) k
 S is the surface of
the region E bounded
by the parabolic
cylinder z = 1 – x2
and the planes
z = 0, y = 0, y + z = 2
30
Fig. 17.9.2, p. 1137
Example 2
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:



 2 xz 2

div F   xy  
y e
  sin xy 
x
y
z
 y  2 y  3y
31
DIVERGENCE
Example 2
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:
E
2
x
,
y
,
z

1

x

1,
0

z

1

x
, 0  y  2  z



32
Example 2
DIVERGENCE
Then, we have:
 F  dS
S
  div F dV
E
  3 y dV
E
 3
1
1 x 2
 
1 0
2 z
0
y dy dz dx
33
Example 2
DIVERGENCE
 3
1

1 x
1 0
2
2  z
2
2
dz dx
1 x 2
3

2  z 

3 1
  

2 1 
3 
0
dx
3
2

    x  1  8 dx
1 

1
2
1
    x  3x  3x  7  dx  184
35
1
6
4
2
0
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
UNIONS OF SIMPLE SOLID REGIONS
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.
36
UNIONS OF SIMPLE SOLID REGIONS
Then, the boundary surface of E is:
S = S1
S2
Its normal n is given
by:
n = –n1 on S1
n = n2 on S2
37
Fig. 17.9.3, p. 1138
UNIONS OF SIMPLE SOLID RGNS. Equation 7
Applying the Divergence Theorem to S,
we get:
 div F dV   F  dS
E
S
  F  n dS
S
  F   n1  dS   F  n 2 dS
S1
S2
   F  dS   F  dS
S1
S2
38
APPLICATIONS—ELECTRIC FIELD
Let’s apply this to the electric field
(Example 5 in Section 16.1):
E x 
Q
x
3
x
where S1 is a small sphere with
radius a and center the origin.
39
APPLICATIONS—ELECTRIC FIELD
You can verify that div E = 0 (Exercise 23).
Thus, Equation 7 gives:
 E  dS   E  dS   div E dV
S2
S1
E
  E  dS
S1
  E  n dS
S2
40
APPLICATIONS—ELECTRIC FIELD
The point of this calculation is that
we can compute the surface integral
over S1 because S1 is a sphere.
41
APPLICATIONS—ELECTRIC FIELD
The normal vector at x is x/|x|.
Therefore.
 x  Q
Q Q
E  n  3 x     4 x  x  2  2
a
x
x
x x
Q
since the equation of S1 is |x| = a.
42
APPLICATIONS—ELECTRIC FIELD
Thus, we have:
 E  dS   E  n dS 
S2
S1

Q
a
2
Q
2
 dS
S1
A  S1 
a
Q
2
 2 4 a
a
 4 Q
43
APPLICATIONS—ELECTRIC FIELD
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.
44
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.
45
APPLICATIONS—FLUID FLOW
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.
46
APPLICATIONS—FLUID FLOW
We approximate the flux over the boundary
sphere Sa as follows:
 F  dS   div F dV
Sa
Ba
  div F  P0  dV
Ba
 div F  P0  V  Ba 
47
APPLICATIONS—FLUID FLOW
Equation 8
This approximation becomes better as
a → 0 and suggests that:
1
div F  P0   lim
F  dS

a 0 V  B 
a Sa
48
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.
49
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.
50
Fig. 17.9.4, p. 1139
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.
51
Fig. 17.9.4, p. 1139
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.
52