16
VECTOR CALCULUS
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.
VECTOR CALCULUS
Each operation resembles differentiation.
However, one produces a vector field
whereas the other produces a scalar field.
VECTOR CALCULUS
16.5
Curl and Divergence
In this section, we will learn about:
The operations of curl and divergence
and how they can be used to obtain
vector forms of Green’s Theorem.
CURL
Suppose:
 F = P i + Q j + R k is a vector field on
°
3
.
 The partial derivatives of P, Q, and R all exist
CURL
Equation 1
Then, the curl of F is the vector field on °
defined by:
3
curl F 
 R Q   P R   Q P 
 y  z  i   z  x  j   x  y  k
 

 

CURL
As a memory aid, let’s rewrite Equation 1
using operator notation.
 We introduce the vector differential operator
 (“del”) as:



  i  j k
x
y
z
CURL
It has meaning when it operates on a scalar
function to produce the gradient of f :
f
f
f
f  i  j  k
x
y
z
f
f
f
 i  j k
x
y
z
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.
CURL
F
i


x
P
j
k

y
Q

z
R
 R Q   P R   Q P 


i


k
 j


 y z   z x   x y 
 curl F
CURL
Equation 2
Thus, the easiest way to remember
Definition 1 is by means of the symbolic
expression
curl F    F
CURL
If
Example 1
F(x, y, z) = xz i + xyz j – y2 k
find curl F.
 Using Equation 2, we have the following
result.
CURL
Example 1
i

curl F    F 
x
xz
k
j


z
y
xyz  y 2



2
    y    xyz   i
z

 y







2
    y    xz   j    xyz    xz   k
y
z
  x
 x

  2 y  xy  i   0  x  j   yz  0  k
  y  2  x  i  x j  yz k
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.
CURL
Recall that the gradient of a function f of
three variables is a vector field on ° 3 .
So, we can compute its curl.
 The following theorem says that the curl
of a gradient vector field is 0.
GRADIENT VECTOR FIELDS
Theorem 3
If f is a function of three variables that has
continuous second-order partial derivatives,
then
curl f   0
GRADIENT VECTOR FIELDS
Proof
By Clairaut’s Theorem,
i

curl  f      f  
x
f
x
j

y
f
y
k

z
f
z
2 f   2 f
2 f 
 2 f


i

j


 y z z y   z x x z 
2 f 
 2 f


k

 x y y x 
 0i  0 j 0k  0
GRADIENT VECTOR FIELDS
Notice the similarity to what we know
from Section 12.4:
a x a = 0 for every three-dimensional (3-D)
vector a.
CONSERVATIVE VECTOR FIELDS
A conservative vector field is one for which
F  f
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.
CONSERVATIVE VECTOR FIELDS
Example 2
Show that the vector field
F(x, y, z) = xz i + xyz j – y2 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.
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.”
CONSERVATIVE VECTOR FIELDS
Theorem 4 is the 3-D version of
Theorem 6 in Section 16.3
 Its proof requires Stokes’ Theorem and
is sketched at the end of Section 16.8
CONSERVATIVE VECTOR FIELDS
Theorem 4
If F is a vector field defined on all of °
3
whose component functions have continuous
partial derivatives and curl F = 0, then
F is a conservative vector field.
CONSERVATIVE VECTOR FIELDS
Example 3
a. Show that
F(x, y, z) = y2z3 i + 2xyz3 j + 3xy2z2 k
is a conservative vector field.
b. Find a function f such that F  f .
CONSERVATIVE VECTOR FIELDS
i

curl F    F 
x
2 3
y z
j
Example 3 a
k


y
z
3
2 2
2 xyz 3xy z
  6 xyz 2  6 xyz 2  i   3 y 2 z 2  3 y 2 z 2  j
  2 yz 3  2 yz 3  k
0
3
°
 As curl F = 0 and the domain of F is ,
F is a conservative vector field by Theorem 4.
CONSERVATIVE VECTOR FIELDS
E. g. 3 b—Eqns. 5-7
The technique for finding f was given in
Section 16.3
We have:
fx(x, y, z) = y2z3
fy(x, y, z) = 2xyz3
fz(x, y, z) = 3xy2z2
CONSERVATIVE VECTOR FIELDS
E. g. 3 b—Eqn. 8
Integrating Equation 5 with respect to x,
we obtain:
f(x, y, z) = xy2z3 + g(y, z)
CONSERVATIVE VECTOR FIELDS
Example 3 b
Differentiating Equation 8 with respect to y,
we get:
fy(x, y, z) = 2xyz3 + gy(y, z)
 So, comparison with Equation 6 gives:
gy(y, z) = 0
 Thus, g(y, z) = h(z) and
fz(x, y, z) = 3xy2z2 + h’(z)
CONSERVATIVE VECTOR FIELDS
Example 3 b
Then, Equation 7 gives:
h’(z) = 0
 Therefore,
f(x, y, z) = xy2z3 + K
CURL
The reason for the name curl is that
the curl vector is associated with rotations.
 One connection is explained in Exercise 37.
 Another occurs when F represents the velocity
field in fluid flow (Example 3 in Section 16.1).
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.
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.
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 16.8
as a consequence of Stokes’ Theorem.
DIVERGENCE
Equation 9
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:
P Q R
div F 


x y z
3
CURL F VS. DIV F
Observe that:
 Curl F is a vector field.
 Div F is a scalar field.
DIVERGENCE
Equation 10
In terms of the gradient operator
      

   i    j  k
 x   y   z 
the divergence of F can be written
symbolically as the dot product of  and F:
div F    F
DIVERGENCE
If
Example 4
F(x, y, z) = xz i + xyz j – y2 k
find div F.
 By the definition of divergence (Equation 9 or 10)
we have:
div F    F



2
  xz    xyz     y 
x
y
z
 z  xz
DIVERGENCE
If F is a vector field on ° 3, then curl F is
also a vector field on ° 3 .
As such, we can compute its divergence.
 The next theorem shows that the result is 0.
DIVERGENCE
Theorem 11
If F = P i + Q j + R k is a vector field on °
3
and P, Q, and R have continuous secondorder partial derivatives, then
div curl F = 0
DIVERGENCE
Proof
By the definitions of divergence and curl,
div curl F
     F 
  R Q    P R    Q P 


 

 
 


x  y z  y  z x  z  x y 
 2 R  2Q  2 P  2 R  2Q  2 P
0






x y x z y z y x z x z y
 The terms cancel in pairs by Clairaut’s Theorem.
DIVERGENCE
Note the analogy with the scalar triple
product:
a . (a x b) = 0
DIVERGENCE
Example 5
Show that the vector field
F(x, y, z) = xz i + xyz j – y2 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.
DIVERGENCE
Example 5
 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.
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.
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.
GRADIENT VECTOR FIELDS
Another differential operator occurs when
we compute the divergence of a gradient
vector field f .
 If f is a function of three variables,
we have:
div  f      f 
2 f 2 f 2 f
 2  2  2
x
y
z
LAPLACE OPERATOR
This expression occurs so often that
we abbreviate it as 2 f .
The operator 2    is called
the Laplace operator due to its relation to
Laplace’s equation
2
2
2

f

f

f
2
 f  2  2  2 0
x
y
z
LAPLACE OPERATOR
We can also apply the Laplace operator
to a vector field
F=Pi+Qj+Rk
in terms of its components:
2F  2 P i  2Q j  2 R k
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.
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.
VECTOR FORMS OF GREEN’S THEOREM
Then, we consider the vector field
F=P i + Q j
 Its line integral is:
—
 F  dr  —
 P dx  Q dy
C
C
VECTOR FORMS OF GREEN’S THEOREM
3
Regarding F as a vector field on ° with
third component 0, we have:
i
j


curl F 
x
y
P  x, y  Q  x, y 
 Q P 


k

 x y 
k

z
0
VECTOR FORMS OF GREEN’S THEOREM
Therefore,
 Q P 
 curl F   k     k  k
 x y 
Q P


x y
VECTOR FORMS OF GREEN’S TH. Equation 12
Hence, we can now rewrite the equation
in Green’s Theorem in the vector form
F

dr

—

C
curl
F

k
dA



D
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.
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  
i
j
r ' t 
r ' t 
VECTOR FORMS OF GREEN’S TH.
You can verify that the outward unit normal
vector to C is given by:
y ' t 
x ' t 
n t  
i
j
r ' t 
r ' t 
VECTOR FORMS OF GREEN’S TH.
Then, from Equation 3 in Section 16.2,
by Green’s Theorem, we have:
—
 F  n ds
  F  n t  r ' t  dt
 P x t , y t  y ' t  Q x t , y t x ' t 
 

 r ' t  dt
r ' t 
r ' t 


C
b
a
b
a
VECTOR FORMS OF GREEN’S TH.
b
  P  x  t  , y  t   y '  t  dt  Q  x  t  , y  t   x '  t  dt
a
  P dy  Q dx
C
 P Q 
  

dA

x y 
D 
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.
VECTOR FORMS OF GREEN’S TH. Equation 13
F

n
ds

div
F
x,
y
dA


—


C
D
 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.