15
Vector Analysis
Copyright © Cengage Learning. All rights reserved.
15.7
Divergence Theorem
Copyright © Cengage Learning. All rights reserved.
Objectives
 Understand and use the Divergence
Theorem.
 Use the Divergence Theorem to calculate
flux.
3
Divergence Theorem
4
Divergence Theorem
5
6
Example 1 – Using the Divergence Theorem
Let Q be the solid region bounded by the coordinate planes
and the plane 2x + 2y + z = 6, and let F = xi + y2j + zk.
Find
where S is the surface of Q.
Solution:
From Figure 15.56,
you can see that Q is bounded
by four subsurfaces.
Figure 15.56
7
Example 1
F = xi + y2j + zk
cont’d
So, you would need four surface integrals to evaluate
However, by the Divergence Theorem, you need only one
triple integral. Because
2x + 2y + z = 6
xy-trace: z=0=> x+y=3
you have
8
Example 1 – Solution
cont’d
9
10
Compare this one-line calculation to long and hard spherical coordinates
surface integral!!!
11
My solution is
much better
S1 : z  g ( x, y )  0  g x  g y  0 
 
F  dS  2 z , x, y 2    0,0,1   y 2
 
2
F
  dS    y dA
S1
R
S 2 : z  g ( x, y )  4  x 2  y 2  g x  2 x, g y  2 y
 
F  dS  2 z , x, y 2    2 x,2 y,1  4 xz  2 xy  y 2
 
2
F
  dS   4 xz  2 xy  y dA 
S2
R
 
 
 
F

d
S

F

d
S

F


  dS   4 xz  2 xydA 
S
S1
 4 x(4  x
2
S2
R
 y )  2 xydA  [ 0 all odd in symmetric R]
2
R
2
2π
0
0
  dr r  dθ 4r cos (θ )( 4  r 2 )  2r 2 cos (θ ) sin (θ )  0
2
 cos( )d  sin(  )
2
0
0
2
2
0
1
0 sin( 2 )d   2 cos(2 ) 0  0

OR  div ( F )dV   0dV  0 - - same in just one line.
Q
Q
12
13
14

div ( F )  2 x  x  0  3 x
 

F
d
S

div
(
F
) dV 


S
Q
2
2
6  r cos( )
0
0
0
 d  dr r 3r cos( )  dz 
2
2
0
0
3  d  dr r 2 cos( )( 6  r cos( )) 
2
2
r3
r4
3  d 6 cos( )  cos 2 ( ) 
3
4
0
0
2
3  d 16 cos( )  4
0
2
3  d 16 cos( )  4
0
1  cos( 2 )

2
1  cos( 2 )

2
2
sin( 2 ) 

 48 sin(  )  6 
  12
2 0

15
16
Divergence Theorem
Even though the Divergence Theorem was stated for a
simple solid region Q bounded by a closed surface, the
theorem is also valid for regions that are the finite unions of
simple solid regions.
For example, let Q be the
solid bounded by the closed
surfaces S1 and S2, as shown
in Figure 15.59.
To apply the Divergence
Theorem to this solid,
let S = S1 U S2.
Figure 15.59
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Divergence Theorem
The normal vector N to S is given by
−N1 on S1 and by N2 on S2.
So, you can write
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Now, let S1 be a small sphere around central electric field F and S2 be any surface,
which completely encloses small sphere S1


kq
r
 x, y, z  inverse square field
F  kq  3  2
( x  y 2  z 2 )3 / 2
|| r ||
2y
3
1
2x
3
1

y


x

 ( x 2  y 2  z 2 )3 / 2

2 ( x 2  y 2  z 2 )5 / 2
2 ( x 2  y 2  z 2 )5 / 2 ( x 2  y 2  z 2 )3 / 2
div ( F )  kq
2z
3
1

z

2
2
/
3
2
2
2
(x  y  z )
2 ( x  y 2  z 2 )5 / 2


 


  F  N1dS   F  N 2 dS   F  NdS   div ( F )dV  0 
S1
S2
S


0



Q


 F  N 2 dS   F  N1dS  4kq  [general case q   qi - superposit ion of charges inside surface S2 ]
S2
S1
Where we used this derivation of flux integral of
electric field on sphere S1 (from 15.6)


 
 
r
r
S FdS S F  NdS  S kq || r ||3  || r || dS 

kq
kq
1
|| r ||2
4a 2  4kq

dS

dS
kq

dS
kq


2
2 
S || r ||4 S || r ||2
a
a S
This shows that the flux of electric field is 4*pi*k*q through ANY closed surface that 19
contains the origin.
Flux and the Divergence Theorem
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Flux and the Divergence Theorem
To help understand the Divergence Theorem, consider the
two sides of the equation
You know that the flux integral on the left determines the
total fluid flow across the surface S per unit of time.
This can be approximated by summing the fluid flow across
small patches of the surface.
The triple integral on the right measures this same fluid
flow across S, but from a very different
perspective—namely, by calculating the flow of fluid into (or
out of) small cubes of volume ∆Vi .
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Flux and the Divergence Theorem
The flux of the ith cube is approximately
Flux of ith cube ≈ div F(xi, yi, zi) ∆Vi
for some point (xi, yi, zi) in the ith cube.
Note that for a cube in the interior of Q, the gain (or loss) of
fluid through any one of its six sides is offset by a
corresponding loss (or gain) through one of the sides of an
adjacent cube.
After summing over all the cubes in Q, the only fluid flow
that is not canceled by adjoining cubes is that on the
outside edges of the cubes on the boundary.
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Flux and the Divergence Theorem
So, the sum
approximates the total flux into (or out of) Q, and therefore
through the surface S.
To see what is meant by the
divergence of F at a point,
consider ∆Vα to be the volume
of a small sphere Sα of
radius α and center (x0, y0, z0),
contained in region Q,
as shown in Figure 15.60.
Figure 15.60
23
Flux and the Divergence Theorem
Applying the Divergence Theorem to Sα produces
where Qα is the interior of Sα. Consequently, you have
and, by taking the limit as α → 0, you obtain the divergence
of F at the point (x0, y0, z0).
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Flux and the Divergence Theorem
The point (x0, y0, z0) in a vector field is classified as a
source, a sink, or incompressible, as follows.
Figure 15.61(a)
Figure 15.61(b)
Figure 15.61(c)
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Example 4 – Calculating Flux by the Divergence Theorem
Let Q be the region bounded by the sphere x2 + y2 + z2 = 4.
Find the outward flux of the vector field
F(x, y, z) = 2x3i + 2y3j + 2z3k
through the sphere.
Solution:
By the Divergence Theorem, you have
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Example 4
cont’d
27
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