Transcript Chap16_Sec9

16
VECTOR CALCULUS
VECTOR CALCULUS
16.9
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.
INTRODUCTION
In Section 16.5, 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.
Equation 1
INTRODUCTION
If we were seeking to extend this theorem to
3
vector fields on ° , 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.
DIVERGENCE THEOREM
It turns out that Equation 1 is true,
under appropriate hypotheses, and
is called the Divergence Theorem.
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.
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 15.6
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.
SIMPLE SOLID REGIONS
The boundary of E is a closed surface.
We use the convention, introduced in
Section 16.7, that the positive orientation
is outward.
 That is, the unit normal vector n is directed
outward from E.
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

d
S

div
F
dV


S
E
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.
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.
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.
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
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
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
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
where D is the projection of E
onto the xy-plane.
2
 x, y 
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
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
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.)
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
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
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 16.7
(with F replaced by
R k), we have:
 R k  n dS 
S2
 R  x, y, u  x, y   dA
2
D
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
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
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.
THE DIVERGENCE THEOREM
Notice that the method of proof of
the Divergence Theorem is very similar
to that of Green’s Theorem.
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
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
Example 2
DIVERGENCE
Evaluate
where:
 F  dS
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
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
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



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
Example 2
DIVERGENCE
 3
1

1 x
2
2  z
1 0
2
3 1  2  z
  
2 1 
3
3
2
dz dx
1 x 2


 0
dx
  12   x  1  8 dx
1 

1
3
2
    x  3 x  3 x  7  dx  184
35
1
0
6
4
2
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 16.4 to extend Green’s Theorem.
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.
UNIONS OF SIMPLE SOLID REGIONS
Then, the boundary surface of E is:
S = S1
Its normal n is given
by:
n = –n1 on S1
n = n2 on S2
S2
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
APPLICATIONS—ELECTRIC FIELD
Let’s apply this to the electric field
(Example 5 in Section 16.1):
E  x 
where S1 is a small sphere with
radius a and center the origin.
Q
x
3
x
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  dS
S1
  E  n dS
S2
E
APPLICATIONS—ELECTRIC FIELD
The point of this calculation is that
we can compute the surface integral
over S1 because S1 is a sphere.
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.
APPLICATIONS—ELECTRIC FIELD
Thus, we have:
Q
 E  dS   E  n dS  a  dS
2
S2
S1
S1

Q
2
A  S1 
a
Q
2
 2 4 a
a
 4 Q
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 16.7) for a single charge.
 The relationship between ε and ε0 is ε = 1/4πε0.
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.
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.
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 
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
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.
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.
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.
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.