Lecture notes on MAT122_Lec12_curl and divergence.ppt

Curl and Divergence

1. Lecture 12
Curl and Divergence
VECTOR CALCULUS
In this lecture, we will learn about:
The operations of curl and divergence
and how they can be used to obtain
vector forms of Green’s Theorem.

2. VECTOR CALCULUS
Here, we define two operations
that:
Can be performed on vector fields.
Play a basic role in the applications of vector
calculus to fluid flow, electricity, and magnetism.

3. VECTOR CALCULUS
Each operation resembles differentiation.
However, one produces a vector field
whereas the other produces a scalar field.

4. CURL
Suppose:
F = P i + Q j + R k is a vector field on .
The partial derivatives of P, Q, and R all exist
 °
3

5. CURL
Then, the curl of F is the vector field on
defined by:
curl
R Q P R Q P
y z z x x y

          
        
         
F
i j k
Equation 1
 °
3

6. CURL
As a memory aid, let’s rewrite Equation 1
using operator notation.
We introduce the vector differential operator
(“del”) as: 
x y z
  
  
  
i j k

7. CURL
It has meaning when it operates on a scalar
function to produce the gradient of f :
f f f
f
x y z
f f f
x y z
  
   
  
  
  
  
i j k
i j k

8. CURL
If we think of as a vector with
components ∂/∂x, ∂/∂y, and ∂/∂z, we can
also consider the formal cross product of
with the vector field F as follows.



9. CURL
curl
x y z
P Q R
R Q P R Q P
y z z x x y

  

  
          
         
         

F
i j k
i j k
F

10. CURL
Thus, the easiest way to remember
Definition 1 is by means of the symbolic
expression
curl F F
Equation 2

11. CURL
If F(x, y, z) = xz i + xyz j – y
2
k
find curl F.
Using Equation 2, we have the following
result.
Example 1

12. CURL Example 1

 
  

2
2
2
curl
2 0 0
2
x y z
xz xyz y
y xyz
y z
y xz xyz xz
x z x y
y xy x yz
y x x yz
  
 
  

  
  
 
  
      
    
  
      
      
   
i j k
F F
i
j k
i j k
i j k

13. CURL
Most computer algebra systems (CAS)
have commands that compute the curl and
divergence of vector fields.
If you have access to a CAS, use these commands
to check the answers to the examples and exercises
in this section.

14. CURL
Recall that the gradient of a function f of
three variables is a vector field on .
So, we can compute its curl.
The following theorem says that the curl
of a gradient vector field is 0.
3


15. GRADIENT VECTOR FIELDS
If f is a function of three variables that has
continuous second-order partial derivatives,
then
curlf 0
Theorem 3

16. GRADIENT VECTOR FIELDS
By Clairaut’s Theorem,
 
2 2 2 2
2 2
curl
0 0 0
f f
x y z
f f f
x y z
f f f f
y z z y z x x z
f f
x y y x
  
  
  
  
  
      
   
   
          
  
 
 
    
   
i j k
i j
k
i j k 0
Proof

17. GRADIENT VECTOR FIELDS
Notice the similarity to what we know
earlier:
a x a = 0 for every three-dimensional (3-D)
vector a.

18. CONSERVATIVE VECTOR FIELDS
A conservative vector field is one for which

So, Theorem 3 can be rephrased as:
If F is conservative, then curl F = 0.
This gives us a way of verifying that
a vector field is not conservative.
fF

19. CONSERVATIVE VECTOR FIELDS
Show that the vector field
F(x, y, z) = xz i + xyz j – y
2
k
is not conservative.
In Example 1, we showed that:
curl F = –y(2 + x) i + x j + yz
k
This shows that curl F ≠ 0.
So, by Theorem 3, F is not conservative.
Example 2

20. CONSERVATIVE VECTOR FIELDS
The converse of Theorem 3 is not true in
general.
The following theorem, though, says that
it is true if F is defined everywhere.
More generally, it is true if the domain is
simply-connected—that is, “has no hole.”

21. CONSERVATIVE VECTOR FIELDS
If F is a vector field defined on all of
whose component functions have continuous
partial derivatives and curl F = 0, then
F is a conservative vector field.
 °
3
Theorem 4

22. CONSERVATIVE VECTOR FIELDS
a.Show that
F(x, y, z) = y
2
z
3
i + 2xyz
3
j + 3xy
2
z
2
k
is a conservative vector field.
b.Find a function f such that .
Example 3
fF

23. CONSERVATIVE VECTOR FIELDS
As curl F = 0 and the domain of F is ,
F is a conservative vector field by Theorem 4.
Example 3 a
  
 
2 3 3 2 2
2 2 2 2 2 2
3 3
curl
2 3
6 6 3 3
2 2
x y z
y z xyz xy z
xyz xyz y z y z
yz yz
  
 
  
   
 

i j k
F F
i j
k
0
3


24. CONSERVATIVE VECTOR FIELDS
The technique for finding f was given in one of
our previous lectures
We have:
f
x(x, y, z) = y
2
z
3
f
y(x, y, z) = 2xyz
3
f
z(x, y, z) = 3xy
2
z
2
E. g. 3 b—Eqns. 5-7

25. CONSERVATIVE VECTOR FIELDS
Integrating Equation 5 with respect to x,
we obtain:
f(x, y, z) = xy
2
z
3
+ g(y, z)
E. g. 3 b—Eqn. 8

26. CONSERVATIVE VECTOR FIELDS
Differentiating Equation 8 with respect to y,
we get:
f
y(x, y, z) = 2xyz
3
+ g
y(y, z)
So, comparison with Equation 6 gives:
g
y(y, z)
= 0
Thus, g(y, z) = h(z) and
f
z
(x, y, z) = 3xy
2
z
2
+
h’(z)
Example 3 b

27. CONSERVATIVE VECTOR FIELDS
Then, Equation 7 gives:
h’(z) = 0
Therefore,
f(x, y, z) = xy
2
z
3
+ K
Example 3 b

28. CURL
The reason for the name curl is that
the curl vector is associated with rotations.

29. CURL
Particles near (x, y, z) in the fluid tend
to rotate about the axis that points in
the direction of curl F(x, y, z).
The length of
this curl vector is
a measure of
how quickly
the particles move
around the axis.

30. F = 0 (IRROTATIONAL CURL)
If curl F = 0 at a point P, the fluid is free
from rotations at P.
F is called irrotational at P.
That is, there is no whirlpool or eddy at P.

31. F = 0 & F ≠ 0
If curl F = 0, a tiny paddle wheel moves with
the fluid but doesn’t rotate about its axis.
If curl F ≠ 0, the paddle wheel rotates about
its axis.
We give a more detailed explanation in Section 12.8
as a consequence of Stokes’ Theorem.

32. DIVERGENCE
If F = P i + Q j + R k is a vector field on
and ∂P/∂x, ∂Q/∂y, and ∂R/∂z exist,
the divergence of F is the function of three
variables defined by:
Equation 9
div
P Q R
x y z
  
  
  
F
 °
3

33. CURL F VS. DIV F
Observe that:
Curl F is a vector field.
Div F is a scalar field.

34. DIVERGENCE
In terms of the gradient operator
the divergence of F can be written
symbolically as the dot product of and F:
x y z
      
      
      
i j k
div F F
Equation 10

35. DIVERGENCE
If F(x, y, z) = xz i + xyz j – y
2
k
find div F.
By the definition of divergence (Equation 9 or 10)
we have:

2
div
xz xyz y
x y z
z xz

  
   
  
 
F F
Example 4

36. DIVERGENCE
If F is a vector field on , then curl F is
also a vector field on .
As such, we can compute its divergence.
The next theorem shows that the result is 0.
3

3


37. DIVERGENCE
If F = P i + Q j + R k is a vector field on
and P, Q, and R have continuous second-
order partial derivatives, then
div curl F = 0
Theorem 11
 °
3

38. DIVERGENCE
By the definitions of divergence and curl,
The terms cancel in pairs by Clairaut’s Theorem.
Proof

2 2 2 2 2 2
div curl
0
R Q P R Q P
x y z y z x z x y
R Q P R Q P
x y x z y z y x z x z y

             
          
            
     
      
           
F
F

39. DIVERGENCE
Note the analogy with the scalar triple
product:
a
.
(a x b) = 0

40. DIVERGENCE
Show that the vector field
F(x, y, z) = xz i + xyz j – y
2
k
can’t be written as the curl of another vector
field, that is, F ≠ curl G
In Example 4, we showed that
div F = z + xz
and therefore div F ≠ 0.
Example 5

41. DIVERGENCE
If it were true that F = curl G, then Theorem 11
would give:
div F = div curl G = 0
This contradicts div F ≠ 0.
Thus, F is not the curl of another vector field.
Example 5

42. DIVERGENCE
Again, the reason for the name divergence
can be understood in the context of fluid flow.
If F(x, y, z) is the velocity of a fluid (or gas),
div F(x, y, z) represents the net rate of change
(with respect to time) of the mass of fluid (or gas)
flowing from the point (x, y, z) per unit volume.

43. INCOMPRESSIBLE DIVERGENCE
In other words, div F(x, y, z) measures
the tendency of the fluid to diverge from
the point (x, y, z).
If div F = 0, F is said to be incompressible.

44. GRADIENT VECTOR FIELDS
Another differential operator occurs when
we compute the divergence of a gradient
vector field .
If f is a function of three variables,
we have:
f

2 2 2
2 2 2
divf f
f f f
x y z
 
  
  
  

45. LAPLACE OPERATOR
This expression occurs so often that
we abbreviate it as .
The operator is called
the Laplace operator due to its relation to
Laplace’s equation
2
f
2
 
2 2 2
2
2 2 2
0
f f f
f
x y z
  
    
  

46. LAPLACE OPERATOR
We can also apply the Laplace operator
to a vector field
F = P i + Q j + R k
in terms of its components:
2 2 2 2
P Q R   F i j k

47. VECTOR FORMS OF GREEN’S THEOREM
The curl and divergence operators
allow us to rewrite Green’s Theorem
in versions that will be useful in our
later work.

48. VECTOR FORMS OF GREEN’S THEOREM
We suppose that the plane region D, its
boundary curve C, and the functions P and Q
satisfy the hypotheses of Green’s Theorem.

49. VECTOR FORMS OF GREEN’S THEOREM
Then, we consider the vector field
F

=

P i + Q j
Its line integral is:

Fdr
C—
PdxQdy
C—

50. VECTOR FORMS OF GREEN’S THEOREM
Regarding F as a vector field on with
third component 0, we have:
 °
3

curl
, , 0
x y z
P x y Q x y
Q P
x y
  

  
  
 
 
  
i j k
F
k

51. VECTOR FORMS OF GREEN’S THEOREM
Therefore,
 curl
Q P
x y
Q P
x y
  
   
 
  
 
 
 
F k k k

52. VECTOR FORMS OF GREEN’S TH.
Hence, we can now rewrite the equation
in Green’s Theorem in the vector form

Fdr
C—
curl FkdA
D

Equation 12

53. VECTOR FORMS OF GREEN’S TH.
Equation 12 expresses the line integral of
the tangential component of F along C as
the double integral of the vertical component
of curl F over the region D enclosed by C.
We now derive a similar formula involving
the normal component of F.

54. VECTOR FORMS OF GREEN’S TH.
If C is given by the vector equation
r(t) = x(t) i + y(t) j a ≤ t ≤ b
then the unit tangent vector (Section 13.2)
is:





' '
' '
x t y t
t
t t
 T i j
r r

55. VECTOR FORMS OF GREEN’S TH.
You can verify that the outward unit normal
vector to C is given by:





' '
' '
y t x t
t
t t
 n i j
r r

56. VECTOR FORMS OF GREEN’S TH.
Then, from Equation 3 in Section 12.2,
by Green’s Theorem, we have:

Fnds
C—
Fntr'tdt
a
b


Pxt,yt y't
r't

Qxt,yt x't
r't






a
b

r'tdt

57. VECTOR FORMS OF GREEN’S TH.
   , ' , '
b
a
C
D
P x t y t y t dt Q x t y t x t dt
Pdy Qdx
P Q
dA
x y
 
 
  
 
 
  




58. VECTOR FORMS OF GREEN’S TH.
However, the integrand in that double
integral is just the divergence of F.
So, we have a second vector form
of Green’s Theorem—as follows.

59. VECTOR FORMS OF GREEN’S TH.
This version says that the line integral of
the normal component of F along C is equal
to the double integral of the divergence of F
over the region D enclosed by C.

Fnds
C—
div Fx,ydA
D

Equation 13

60. Acknowledgement
•David Calvis
•Bob Greer
•Fenny Lee