7. Vector Integration
7.1. Line Integrals
Consider a vector field F  F1 ( x, y, z )i  F2 ( x, y, z ) j  F3 ( x, y, z )k and a smooth curve C. Let A and
B be two points on the curve C.
B
Pi 1
 r 1  r i 1  r i
Pi
r i 1
A
ri
O
Divide AB into n small segments. Let the ith segment be from Pi (Position vector ri ) to Pi 1 (position
vector ri 1  ri   ri ). At Pi let the vector field be Fi . The line integral of the second kind of F
along C is defined to be

C
N
F  dr   F  dr  lim  Fi  ri
N 
AB
i 1
Example
If E is an electric field and a unit charge moves from A to B along C,
Ei  ri  Ei cos ri
represents the work done by the
field in moving the charge from
Pi to Pi 1 (approximately).
Ei
Hence
Pi 1


E  dr
 ri
Pi
AB
represents the total work done in moving the charge from A to B i.e. The potential difference between
A and B.
The curve C is called the path of the integral. If the two end-points coincide (i.e. A = B) we say we
have a closed path and usually write
 F  dr
C
A
B
1
By convention we go anticlockwise around a closed path.
Usually a curve is given in parametric form
x  xt ,
y  y t ,
and we may write
z  z t 
t1  t  t 2
r t   xt i  y t  j  z t k
e.g. t may be arc length.
Example
x(t )  a cos t , y (t )  z sin( t ), z  1, 0  t  2 describes the circle x 2  y 2  a 2 lying in the plane
z  1 in a positive sense (anticlockwise). Note that t is not arc length here.
We may then evaluate the line integral as

C
dr
dt
dt
dz 
dy
 dx
 F3 dt
 F2
   F1
C
dt 
dt
 dt
F  dr   F 
C
Example
Determine the work done by the force F  x 2  y i  y 2  z  j  z 2  x k in moving a particle from
A(0,0,0) to B(1,1,1) along (i) the straight line x = y = z, (ii) the curve y  x 2 , z  x 3
Sol:
(i) Choose the parametric representation
x  t,

C1
y  t,
t
  3t
1
F  dr  
2
t 0
1
t 0
2
0  t 1
 
 t   dt


 
 t  1  t 2  t  1  t 2  t  1 dt
1
2
0  t 1

(ii) Choose x  t , y  t 2 , z  t 3
 F  dr   t
  3t
1
1

 
 3t dt
 
 t 2  1  t 4  t 3 2t  t 6  t 3t 2 dt
8
 2t 5  2t 4
t 0


2
t 0
3
29
60
Note that the value of the integral in this case depends on the path taken (but not the parametric
representation chosen).
If we had integrated from B to A the results would have been 

F  dr    F  dr along the same path
AB
BA
2
1
29
and 
2
60
respectively, i.e.
z
B(1,1,1)
C1
C2
A(0,0,0)
y
x
Hence, if C is the close path from A to B along C1 and then back to A again along C2 , the result is
 F  dr  
C
C1
F  dr   F  dr
C2
1  29 
    
2  60 
1

60
And this result is independent of the starting (=finishing) point.
An alternative notation for line integrals is

C
F  dr   F1i  F2 j  F3 k   dxi  dyj  dzk 
C
  F1 dx  F2 dy  F3 dz
C
Example
Evaluate

C
y 2 dx  2 xydy from the origin (0,0,0) to the point (4,2,0) along the paths:
(i) a straight line
(ii) the parabola y 2  x
(iii)
part of the x-axis and then the line x = 4
This is equivalent to evaluating

C
F  dr where F  y 2 i  2 xyj
Sol:
1
x
2
C1 : y 
(i)
y
dy 1

dx 2

C1
2
y 2 dx  2 xydy
C1
2
4 1 
1 1
   x  dx  2 x x  dx
x 0 2


2 2
4 3

x 2 dx  16
x 0 4
4
3
x
(ii) C 2 : y 2  x
2y


y
dy
1
dx
y 2 dx  2 xyfy
C2
2
C2
1
xdx  2 x dx
x 0
2
4
4
  2 xdx  16
4, 2 


(iii)
0, 0 
4
x 0
x
4
x 0
y
y 2 dx  2xydy
2
0dx  2 x  0  0  
2
y 0
y 2  0  2  4 ydy
2
  8 ydy  16
y 0
C3
x
4
Is this just a coincidence? Unlikely!
If a field F is such that

F  dr is independent of the path from A to B, then F is said to be a
AB
conservative field.
Consider
 F  dr
around any
C
B
closed curve in a conservative
field. Then

C
Q
F  dr   F  dr
APBQA

APB
F  dr   F  dr
BQA
A
  F  dr   F  dr
APNB
 F  dr  0
C
AQB
in a conservative field
C
If F   , where  is some scalar field, then
4
P
 

   dx
dy
dz 
F  dr   
i
j
k    i 
j  k dt
AB x
y
z   dt
dt
dt 

  dx  dy  dt 
dt
  


AB x dt
y dt z dz 


AB
d
dt
AB dt


AB
F  dr   ( B)   ( A)
independen t of the part from A to B
Hence if F   , F is conservative.
Example
Show that F  y 2 i  2 xyj is a conservative field and hence evaluate

C
F  dr along any path C
from (0,0) to (4,2).
Sol:
F is a conservative if F  
i.e. (i) y 2 

x
and (ii) 2 xy 

y
Integrating (i)
  y 2 x  f ( y) , where f ( y ) – an arbitrary function

 2 yx  f ' ( y )
y
Matching with (ii) shows that f ' ( y )  0,
f ( y)  c
  y2x  c
Hence

C
F  dr   (4,2)   (0,0)
 16
Note that
 F  dr
is the circulation of F around C and so an alternative definition of curl F is
C
 1

 1

 1

curlF  lim   F  dr i  lim 
F  dr  j  lim   F  dr k

A1 0 A C1
A3 0 A
 1
 A2 0 A2 C2

 3 C3

where C1 , C2 and C 3 are curves parallel to the yz-plane, xz-plane and xy-plane respectively around
a point ( x, y , z ) and A1 , A2 and A3 are the areas enclosed by these three curves respectively. But
 F  dr = 0 if F is conservative, and hence curl F = 0 if F is conservative.
C
In fact it is also true that if curl F = 0 everywhere, then F is conservative (and F may be written
as   ) and this is sometimes taken as an alternative definition of a conservative field.
5
This gives us a very convenient test for deciding whether F is conservative or not. We see if curl F is
zero or not.
Hence the following are all equivalent:
(i) F is conservative.
(ii)
(iii)

F  dr is independent of the path.
AB
 F  dr = 0 around any closed path (circulation is zero).
C
F =   for some scalar field  .
(iv)
(v)

F  dr   ( B)   ( A) .
AB
(vi) curl F = 0 (F is irrotational).
Example
The electric field E and potential  due to a point charge at the origin are E 

Q
r
and
40 r 3
Q 1
where E   . Hence E is a conservative field. This is consistent with the work done
4 0 r
in moving a unit charge from A to B being the potential difference  ( A)   ( B) between the points
and independent of the path taken between them.
Example
Show that F  yzi  ( xz  2 y ) j  xyk is a conservative field and find a scalar function f ( x, y , z )
such that F  f .
Proof:
i

curl F 
x
yz
j

y
xz  2 y
k

z
xy
 ix  x   j  y  y   k z  z 
0
hence F is conservative and a scalar function f exists such that F  f .
i.e. (i)
f
 yz
x
(ii)
f
 xz  2 y
y
(iii)
f
 xy
z
Integrate (i) f  xyz  g ( y , z ) . Match with (ii)
f
g
 xz 
 xz  2 y
y
y
Hence
6
g
 2y
y
g  y 2  h z 
and
f  xyz  y 2  h z 
Match with (iii)
f
 xy  h  z   xy
z
Hence
h  z   0
h z   c
c- constant
and
f  xyz  y 2  c
Example (optional)
A positively charged ring of radius R with a charge of q Coulombs, which is evenly distributed, lies in
the xy-plane. Compute the electric field intensity at a point P on the axis of the ring, which is suppose
on z-axis.
Sol:
To calculate the electric field intensity at a point P from a continuous distribution, we calculate the
electric field intensity due to one arbitrary element of charge dq, at the point P and then integrate this
over the entire region, the positively charged ring of radius R, under consideration.
The electric field intensity, due to one arbitrary element of charge dq 
density of charge),
q
q
ds , (
is the line
2R
2R
is given by:
q
ds
r
dE  2R 2 , where r is the vector from dq to P and r  r . Note that the x-components and the
40 r r
y-components of the electric field intensity due to charge elements dq on opposite sides of the ring
will cancel because they are of equal magnitude, but opposite in direction. Hence the net electric field
q
ds
r
2

R
intensity will point along the z-axis and will be due to the z component of dE 
, which is
2
40 r r
q
ds
r
of value dE  k  2R 2  k .
40 r r
7
Therefore,
E  k   dE  k 
C
q
ds
r  k , where C is the path of the ring of radius R.

8 R 0 C r 3
2
Let the center of the ring of radius R is at (0,0,0) and a point on the charge element dq is (x, y, 0)
and the point P on the axis of the ring is (0, 0, z).
Thus,
r 2  R 2  z 2 & r  (0  x)i  (0  y ) j  ( z  0)k .
r k  z
E  k   dE  k 
C
qz

8 2 R 0 R 2  z

 ds 
3
2 2 C
2Rqz

8 2 R 0 R 2  z
qz


3
2 2

40 R 2  z 2

3
2
Or using polar coordinate, we have
x  R cos ,. y  R sin  , ds 
dx 2  (dy) 2
 R 2 sin 2  (d ) 2  R 2 cos 2  (d ) 2
 Rd 
E  k   dE  k 
C

qz

8 2 R 0 R 2  z
 ds 

3
2 2 C
2
qz

8 2 R 0 R 2  z

3
2 2
 Rd 
0
2Rqz

8 2 R 0 R 2  z 2

3
2
qz

40 R 2  z 2

3
2
It follows that E  ( E  k )k 
qz

40 R  z
2

3
2 2
k
qz
(Note: if z  R , then R 2  z 2  z 2 & E  k 

40 R  z
E  ( E  k )k 
q
4 0 z 2
2

3
2 2

qz
40 z
3

q
40 z 2
. Thus
k)
Example (optional)
Positive charges is distributed uniformly over the entire xy-plane, with a charge per unit area, or
surface density of charge, . Find the electric field intensity at a point P.
Sol:
Let z-axis pass through the point P(0,0,z).
Then by the previous example, the projection of the electric field intensity at the point P on k due to
the strip of the ring of radius r is:
8

C
2
z

40 r  z
2

3
2 2
ds 

0
rz

40 r  z
2

3
2 2
d
Therefore,
2
rz
Ek  

2
2
0 0
 4 0 r  z


z r 2  z 2 

1
4 0


1
2

0




2rz
dr 
d

3


0
2
4 0 r 2  z 2


1

z 2

r  z2  2
2 0

0

z
2 0

dr 
3
2
z  d r 2  z 2 
3
4 0 0 2
2 2
r
z


1
1 
z







2
2
z 2  2 0 z 2 0
  z
2
E  ( E  k )k 

k
2 0
Another type of line integral (the line integral of the first kind) is
I   f  x, y, z ds
C
where f is a scalar function and s is arc length along C. If C has the parametric representation
( x(t ), y(t ), z(t )), t1  t  t 2 , we have that
ds 
dx 2  dy 2  dz 2
and hence
I 
t2
t t1
f  xt , y t , z t 
2
2
2
 dx   dy   dz 
        dt
 dt   dt   dt 
Example
Find the mass of an ideal wire with density  ( x, y, z )  kz if it has the shape of the helix,
x  3 cos t , y  3 sin t , z  4t
Sol:
9
x  3 cos t ,
Mass   ds
y  3 sin t ,
0t 
z  4t
C

 k  4t 9 sin 2 t  9 cos 2 t  16dt
t 0

 20k  tdt
t 0
 10k 2
Example (optional)
Let the line density of the curve C: x 2  y 2  ax, a  0 be f ( x, y )  x 2  y 2 . Find the mass of the
curve C: x 2  y 2  ax, a  0 .
Sol:
x 2  y 2  ax  2 x  2 y
dy
dy a  2 x
a

dx
dx
2y
2
 a  2x 
 dx   dy 
 dx 
ds       dx  1  
 dx   dx 
 2y 
2

2
4 y 2  a 2  4ax  4 x 2
dx
4y2
4ax  a 2  4ax
a
dx 
dx
2
2y
4y
2
a
a2

, a  0 , we have
For the upper part of the circle x  y  ax   x    y 2 
2
4

2
ds 
2
a
a
dx 
dx,0  x  a
2y
2 ax  x 2
a
mass of the line   x 2  y 2 ds  2  ax
C
2 ax  x 2
dx  2 a
a
x
dx

2 0 ax  x 2
ax a
2
0  2a
1
ax
0
2
(Note: the mass of the upper part and the lower part of the circle
a
 a a
1
0
a
a
d (a  x)  a a
2
a
a2

x 2  y 2  ax   x    y 2 
, a  0 are equal.)
2
4

(Note:
ax
a
a
2 ax  x
2
is not defined at x=0 and x= a so the integral

0
in the sense of improper integral.)
10
ax
a
2 ax  x
2
dx we get is
Example (optional)
x 2  y 2  z 2  a 2
Let the line density of the curve C: 
be f ( x, y, z )  x 2 . Find the mass of the
x

y

z

0

curve C:
x 2  y 2  z 2  a 2

 x yz 0
Sol:
We observe that x 2  y 2  z 2  a 2 is a sphere of radius a with center at (0,0,0) and x  y  z  0 is
plane passing through (0,0,0). So their intersection C is a circle with center at (0,0,0) and radius a.
From the property of symmetry, we know that
x
2
ds   y 2 ds   z 2 ds . Thus
C
C
2
 x ds 
1
1 2
a 2 2a 2a 3
2
2
2
x

y

z
ds

a
ds


.
3 C
3 C
3
3
C
C


7.2. Surface integrals
Consider a vector field F and a surface S
in the field.
divide the surface into
N small area elements
 S i (approximately flat).
ni
Fi

 Si
Define  Si as the vector
S
whose magnitude is  S i
and whose direction is normal
to the element of surface  S i
(usually outwards if S is closed).
Then  Si   Si ni where ni is a unit vector normal to the region  S i .
Let Fi be the vector field evaluated at some point on  S i . Then
Fi   Si  Fi   Si ni  Fi  Si cos 
represents the flux of F through the element of surface  S i . We define the total flux of F through
S to be the surface integral of the second kind
11
 Fi  d S  lim
N
N 
S
F  S
i 1
i
i
Example
If v is the velocity field of a fluid, 
(i)
 v  d S represents the total volume of fluid crossing S in a
S
unit time. Replacing v by  v (  = density) gives mass flux.
If J is an electric current vector, 
(ii)

S
J  d S represents the rate at which electric charge crosses
S.
  E  d S represents the electric flux through S.
(iii)
If E is an electric field vector,
(iv)
If q is a heat conduction vector, 
S
 qdS
S
represents the rate at which heat flows through S.
Example
  F dS
Evaluate
S
where F  x 2 i  xyj  z 2 k and S is the surface of the cube bounded by x = 0,
x = 2, y = 0, y = 2, z = 0 and z = 2.
Sol:
We integrate over each face in turn and sum the results.
For the face ABCD:
dS  dydz , the outward normal n  i , x  2 .
 F  d S 
ABCD

2
z
2 G
ABCD

2
x 2 dydz  
z 0 y 0
2

2
z 0 y 0
4dydz  16
2
y
For the face BEFC:
dS  dxdz , the outward normal n  j , y  2 .
 F  d S   F  jdS
BEFC

BEFC

2
x 0 z 0
xydxdz  
Similarly,
2
2
 2 xdxdz  8
x 0 z 0
 F  d S  16,  F  d S   F  d S   F  d S  0
CFGD
OEFG
2
O
x
2
C
D
 F  idS
F
OADG
OABE
12
A
E
B
y
y
Hence
  F  d S  40
S
If part of S is not parallel to one of the coordinate planes, we proceed as follows:
If the surface S is written as  ( x, y, z )  constant then    grad is normal to the surface and
hence the unit normals are
i
n



 j
k
x
y
z
2
        
  

  

 x   y   z 
2
2
If the surface is given in the form z   ( x, y ) , i.e. z   ( x, y )  0 , the unit normals are
n
  X i  Y j  k
 X 2  Y 2  1
There are two normals to the open surface at each point which are of the same magnitude but are of
opposite directions. The ways to decide the required normal are explained through examples:
Example
Given  ( x, y, z )  x 2  y 2  z 2 ,
Consider the surface S:  ( x, y, z )  x 2  y 2  z 2  a 2 , z  0 (the upper sphere)
At the point (x, y, z), there are two unit vectors (normals) which are perpendicular to the surface S:
 ( x, y, z )  x 2  y 2  z 2  a 2 , where z  0 . They are the vectors


X
Y
Z
 
i
j
k
 2 2 2
 X2  Y2   Z2
 X2  Y2   Z2
 X2  Y2   Z2 
Y
Z
 X
 2 xi  2 yj  2 zk 
y
z 
x

  i  j  k 
2a
a
a 
a
 (cos  )i  (cos  ) j  (cos  )k 



  X i  Y j   Z k 
where  is the angle between

and the positive direction of x-axis (or the angle between 

and the positive direction of x-axis ) ,  is the angle between
y-axis and  is the angle between

and the positive direction of


and the positive direction of z-axis.

13
So cosine of the angle between the normals and the positive direction of x-axis is
 cos   
X
  Y2   Z2
2
X
So cosine of the angle between the normals and the positive direction of y-axis is
 cos   
Y
  Y2   Z2
2
X
So cosine of the angle between the normals and the positive direction of z-axis is
 cos   
Z
 X2   Y2   Z2
We observe that the outward normal at the point (x, y, z ) to the upper sphere i.e
x 2  y 2  z 2  a 2

z0

makes an acute angle with the positive direction of z-axis, so cosine of that acute angle  0 ,
y
z 
x
therefore the normal we need is  i  j  k 
a
a 
a
As for the inward normal at the point (x, y, z ) to the upper sphere i.e
x 2  y 2  z 2  a 2
makes an

z0

obtuse angle with the positive direction of z-axis, so cosine of that angle  0 , therefore the normal
y
z 
x
we need is   i  j  k 
a
a 
a
Similarly, the outward normal at the point (x, y, z ) to the lower sphere i.e
makes an obtuse angle
with the positive direction of z-axis, so cosine of that obtuse angle  0 ,
therefore the normal at the point (x, y, z ) to the lower sphere i.e
y
z 
x
 i  j  k  (note: z
a
a 
a
x 2  y 2  z 2  a 2

z0

x 2  y 2  z 2  a 2
we need is

z0

is a negative #)
As for the inward normal at the point (x, y, z ) to the lower sphere i.e
x 2  y 2  z 2  a 2
makes an

z

0

acute angle with the positive direction of z-direction, so cosine of that acute angle  0 , therefore
14
the normal at the point (x, y, z ) to the lower sphere i.e
y
z 
x
  i  j  k  ( note: z
a
a 
a
x 2  y 2  z 2  a 2
we need is

z

0

is a negative #)
We call the normal to a surface S: which makes an acute angle with the positive direction of z-axis an
upper normal.
Also we call the normal to a surface S: which makes an obtuse angle with the positive direction of
z-axis a lower normal.
If the surface S given is an explicit function z   ( x, y ) , then the normals to surface S at the point
( x, y, z )  ( x, y,  ( x, y )) is:
( x, y.z )  z   ( x, y)  0 
  X i  Y j   Z k 
 X  Y   Z
2
2
2

   X i  Y j  k 
 X 2  Y 2  1
Once the normal to the open surface has been decided, we have

F  d S   F  ndS   F 
S
S
S
 
dS

This is now in the form  g ( x, y , z )dS , where g ( x, y , z ) is a scalar field, which is known as the
S
surface integral of the first kind, and can also arise in its own right.
Example
(i) If g ( x, y, z )  1 , then

S
dS  surface area of S
(ii) if g ( x, y, z )  electrostatic charge density

S
gdS = total charge on S
To evaluate this integral suppose we project S onto one of the coordinate planes where the surface
integral becomes a double integral.
15
z
Suppose the surface is given implicitly by
 ( x, y, z )  0 and we decide to project S
k

n
S
onto the xy-plane where
its projection is the region S xy
y
Let  be the angle between
the normal n to  S and
k, i.e. cos   n  k .
xy
x
Now  x y   S cos    S n  k , therefore
S 
 x y
nk
and in the limit, we have
1
 g ( x, y, z)dS   g ( x, y, z) n  k dxdy 
S
S
  g x, y, z ( x, y ) 
 x2   y2   z2
z
 xy
If the surface is given explicitly by z   ( x, y ) , then dS 
and
 g ( x, y, z )dS   g ( x, y, z )
S
dxdy
 X2  Y2  1
1
dxdy  1   X2  Y2 dxdy
1   X2   Y2 dxdy
xy
Example
Compute
 ydS where S is the part of
z  z ( x, y)  x  y 2 whose projection onto xy-plane is the
S
region  xy : 0  x  1,0  y  2 .
Sol:
The projection of S onto xy-plane is the region  xy : 0  x  1,0  y  2 .
So dS  1  z X2  zY2 dxdy  1  1  (2 y ) 2 dxdy  2  4 y 2 dxdy .
Therefore
16
 ydS 
S
1 2


2
  y 2  4 y 2 dy dx
y
2

4
y
dxdy





 xy
00

21
   
2  4y2 d 2  4y2
8
00

1
1

1 2
dx     2  4 y 2

83
0




3
2 2
0
dx


3
3
3
3

 1
1 2
1 2
52 2 13 2
2 


  18  2 dx  18  2 2  
27(2 2 )  2 2 

12 
12 
12
3
0

 12
1
Example
Compute
 ( x  y  z)dS over the surface S:
x2  y2  z 2  a2 , z  0 .
S
Sol:
Since we are going to project the open surface S: x 2  y 2  z 2  a 2 , z  0 onto xy-plane, we will use
the formula
 X2   Y2   Z2
dxdy
S g ( x, y, z )dS   g ( x, y, z )
Z
xy
x 2  y 2  z 2  a 2 , z  0 onto xy-plane is  xy , which is x 2  y 2  a 2 .
The projection of
Let  ( x, y, z )  x 2  y 2  z 2  a 2 , then the surface S is just the
 ( x, y, z )  0
equation : 
, i.e. x 2  y 2  z 2  a, z  0 and this equation defines z as an implicit function
z0

of x, y.
Also,



 2 x.
 2 y,
 2z
x
y
z
And
dS 

 X2  Y2   Z2
4x 2  4 y 2  4z 2
4a 2
2a
dxdy 
dxdy 
dxdy 
dxdy
Z
2z
2z
2z
a
dxdy 
z
a
a  x2  y2
2
dxdy
Or you can figure out in the following way, of course, the following method requires more being able
in mathematics:
2
 z   z 
 x  y
dS  1       dxdy  1       dxdy 
 x   y 
z  z
2
2
2
So
17
a
a  x2  y2
2
dxdy
 ( x  y  z )dS   x  y 
S
a2  x2  y2

a
a2  x2  y2
xy

 a x
    x  y  a2  x2  y2
a 
 a 2  x 2
2
a
2

dxdy
a  a2  x2

dy dx    
a 2  x 2  y 2 
a 
 a 2  x 2
a

 ady dx

a2  x2  y2
a ( x  y )dy

1
1
 2 2


 
2
2
d y2
a  a2  x2
 a  x axdy  2 d x  y


 
ax
2
  
 ady dx     

 a  dy dx
2
2
2
a2  x2  y2
a2  x2  y2
a  a 2  x 2
a  a 2  x 2  a  x  y

 



 


a
 ax  2a


a
 
a
a 2  x 2 dx  4a  a 2  x 2 dx  4a
a
0
Or
 ( x  y  z )dS   x  y 
S
a2  x2  y2
xy

a 2

    r cos   r sin   a 2  r 2
o
0 

a
  2ardr  ar 2

a
0


a 2
4
 a 3
a
a  x2  y2
2
dxdy
a

 ra sin 

ra cos 
rd dr   

 ra 
a2  r2
a2  r2
a2  r 2
0


a
2
0
dr
 a 3
0
Now we are going to make a few concluding remarks on the ways of computing the surface integral
of second kind, that is, compute  F ( x, y, z )  d S , where F ( x, y, z ) is a vector field defined on S.
S
Suppose the surface is given implicitly by  ( x, y, z )  0 and we are going to project S onto xy-plane
with the projection, the region  xy . Assume we have chosen the normal n to S at the point (x, y, z) by
considering the cosine of the included angle between the normal and the positive direction of z-axis.
Then

F  d S   F  ndS   F 
S
S
  F    
 xy
S
1
z
 
  
dS   F 
dxdy

  Z
 xy
dxdy
Whether the positive sign or the negative sign is chosen is depending on the problem raised.
In addition, if the surface is given explicitly by z   ( x, y )  0 and still we are going to project S onto
xy-plane with the projection, the region  xy , then
18
 F  d S   F  ndS   F 
S
S
   X i  Y j  k 
 X 2  Y 2  1
S
dS 
   X i  Y j  k   X  Y  1
F
dxdy   F     X i  Y j  k dxdy

1
 2   2 1
 xy
 xy
2
X
2
Y
Example
Evaluate
  F dS
where S is the closed surface of the region bounded by the cylinder
S
x 2  z 2  4 and the planes x = 0, y = 0, z = 0 and y = 4, and F  zi  zj  yk .
z
2 C
B
4
A y
0
2
x E
D
Sol:
Consider first the integral over the curved surface BCED which is the level surface   x 2  z 2  4 .
The curved surface is projected on xy-plane with the projection  xy which is the rectangle OADE
bounded by y  0, y  4, x  0, x  2 .
  x 2  z 2 ,   2 xi  2 zk ,  z  2 z

BCED

2

F  dS  
4
x 0 y 0

AOED
( zi  zj  yk )  ( 2 xi  2 zk )
1
dxdy
2z
( x  y )dydx
 8
The other surfaces are all planes and give the results
  F  dS  8,
OABC

OADE
F  dS  16,

OEC
F  dS  
8
F  dS
 sum
with
ABD
3
8
3

  F  dS  0
S
Example
Calculate
 F  d S
where F ( x, y, z )  2i  2 j  2k , the open surface S is the part of the plane
S
x + z =1 which is enclosed by the cylinder x 2  y 2  1 and the normal to S is pointing to xy-plane.
Sol:
19
Let  ( x, y, z )  x  z , then
 






i
j
k  i  k and
 1,
 0,
 1.
x
y
z
x
y
z
Since the normal to S makes an obtuse angle with the positive direction of

or the direction    i  k .
n
 
z axis, we choose
And we are going to project S onto the xy- plane and the projection, say,  xy .
The following formula will be used:

F dS 
S
  F 
 xy
 
z
dxdy   - F   dxdy (note that  z  1)
 xy
  - F  dxdy   (2i  2 j  2k )  (i  k )dxdy 
 xy
 xy
1 2
 4dxdy   
0 0
x  y 1
2
2
4rddr  4
Example
 F ( x, y, z)  d S , where
Find
F ( x, y, z )  x 2 yi  3xy 2 j  4 y 3 k and S is the part of
S
z  z ( x, y)  x 2  y 2  9 whose projection onto xy-plane is the region  xy :
0  x  2,0  y  1 . The normal to S at (x, y, z) we choose is the lower normal, that is the normal
which makes an obtuse angle with the positive direction of z-axis.
Sol:
The direction of the normal is the same as the direction of
 (2 xi  2 yj  k )
since the normal to S at (x, y, z) makes an obtuse angle with the positive direction of z-axis.
 x
2

 xy

 xy
S
 

yi  3xy2 j  4 y 3 k  d S   x 2 yi  3xy2 j  4 y 3 k  2 xi  2 yj  k dxdy
1

2 x y  6 xy  4 y dxdy     2 x 3 y  6 xy3  4 y 3 dy  dx
0 0


3
3
3

2


2
 x 4 3x 2

3
 3 2 3 4


4 1
   x y  xy  y  dx    x 3  x  1dx   
 x  02  4  3  2  1
2
2
4
0

 4

0
0
2
20
Example (optional)
 F  d S
Find
where F ( x, y, z)  ( y  z)i  z  x j  ( x  y)k , S is the close surface which is
S
bounded by x 2  y 2  z 2 & z  h where h  0 .
Sol:
For a close surface if the direction of the normal is not specifically mentioned, then it is understood
that the normal is assumed to be outward.
x 2  y 2  z 2
S consists of two parts, say, S1 : x 2  y 2  z 2 , where 0  z  h & S 2 : 
zh

 F  d S   F  d S   F  d S
S
For
S1
S2
 F  d S , there are two directions which are perpendicular to
S1 , they are  gradient, i.e.
S1
  :
 

 
 ( x, y, z )  z 2  x 2  y 2     i 
j
k    2 xi  2 yj  2 zk 
y
z 
 x
Since the question requires the direction which makes an obtuse angle with the positive direction of
 

 
z-axis,     i 
j
k    2 xi  2 yj  2 zk   2 xi  2 yj  2 zk is the direction we
y
z 
 x
want.
We project S1 : x 2  y 2  z 2 , where 0  z  h onto xy-plane, we have  xy which is
 xy : x 2  y 2  h .
Then
 F  d S   F 
S1
xy
 
Z
dxdy 
( y  z )i  z  x  j  ( x  y )k  


xy

(2 xi  2 yj  2 zk )
dxdy
2z

 x( y  x 2  y 2 ) y x 2  y 2  x

  

 ( x  y ) dxdy


x2  y2
x2  y2
 xy 

2 h
  r cos  (r sin   r ) r sin  r  r cos  
 
   

 r cos   r sin  rdr d
r
r
 
0 0 
21
2
h

    r cos  sin   r cos   r sin   r cos  sin   r cos   r sin  rdr d
0 0

2
2
h

2
 2

2
2
     2r cos   2r sin  dr d    r 3 cos   r 3 sin   0h d
3
3

0 0
0 

2


 3 h  cos  sin  d  3 h  sin   cos 

2
2
3
3
2
0
0
0
For
 F  d S ,
S 2 is the part of z  h (note : z  h is a constant function of x, y) whose
S2
projection of onto
xy-plane is  xy : x 2  y 2  h .
 


i
j  1k    k
y
 x

 ( x, y )  h  the normal to S 2 is   
Since the outward normal to S 2 makes 0 radian with the positive direction of z-axis so k is chosen.
xy-plane, is  xy : x 2  y 2  h .
The projection of S 2 onto
So
 F  d S   F  kdxdy   x  y dxdy
S2
xy
2
xy
2


1
    r cos   r sin  rdr d   cos   sin   r 3 0h d
3
0 0
0

2

h
 3 h cos  sin  d  0
1
3
0
Finally we have
 F  d S   F  d S   F  d S  0
S
S1
S2
7.3 Divergence
The net outward flux of a vector v through a closed surface S is
  v  ds
S
If we take a point P and surround it by a surface S enclosing a volume V we have
Average flux per unit volume =
22
1
V
  v  dS
S
Now let V  0 so that the volume shrinks to the point P. The limiting value of the average flux per
unit volume is then the divergence of v at P. This gives an alternative definition of div v
1
div v  lim 
V 0 V


  v  dS 
S
7.4 The Divergence Theorem (Gauss’s Theorem)
For any closed surface S, enclosing a region V in a vector field F,
 divFdxdydz   F  d S
V
i.e.
S
the total divergence of F from a volume V = total outward flux of F through S.
Proof:
Divide the volume V into N
Vi
S i
V
S
Small blocks Vi with surface S i
For each small volume, We have approximately div F 
or
div FVi  

Si
1
Vi
 
Si
F dS
F dS
i.e. div FVi = total outward flux of F across S i .
Summing over all blocks and letting N   as Vi  0 we have
LHS  
  div Fdxdydz
V
For the RHS we observe that flux out of a face of one block equals the flux into the face of an
adjacent block, unless that face is part of the surface S. Hence for all internal faces the flux cancels,
leaving RHS  
 F dS
S
and the theorem is proved.
Example
Verify the divergence theorem for F = zi + yj + zk and the region bounded by z  4  x 2  y 2 and
the plane z = 0.
Sol:
div F = 3
23
   divFdV
 

V
2
 4 x 2
x  2 y   4  x 2
4 x 2  y 2
x 0
z
3dzdydx
4 x  y
2

2

2

4 x 2
x  2 y   4  x 2
3z 
S2
4
2
dydx
0
 34  r rdrd
2
2
r 0  0
Now consider
y
2
2

r4 
 2  3  2r 2  
4 0

 24
2
S1
x
  F dS
S
For the circular “base” we have
  xi  yj  zk   kdxdy
    zdxdy
S1
S1
 0 S i n cze = 0 oSn1
For the curved surface   z  x 2  y 2  4
  2xi  2 yj  k

S2
F  dS    2
x  y 2 4
 2
x  y 2 4
 2
x  y 2 4

r 0
xi  yj  zk   2 xi  2 yj  k 1 dxdy
2 x
2 x
1

2
 2 y 2  z dxdy
2
 2 y 2  4  x 2  y 2 dxdy

 4  r rdrd
2
2
z  1
2
0
2
 2 r4 
 2  2r  
4 0

 24
Hence
  F dS    
S
V
Example
Calculate
 F  d S

div FdV

where F ( x, y, z)  xyi  y 2  e xz j  sin( xy)k , S is the close surface which
S
is bounded by z  1  x 2 , z  0, y  0 & y  z  2 .
24
2
Sol:
It is difficult to compute
 F  d S
directly, by Divergence theorem
we compute
S


   xy  y 2  e xz
 sin xy 

divFdxdydz



dxdydz   3 ydxdydz instead, where V is the


 x


y

z
V
V 
V

2
solid enclosed by S.
The projection of V onto xz-plane is  xz which is bounded by z  0, z  1  x 2
So
 2 z

divFdxdydz   3 ydxdydz     3 ydy dxdz

V
V
 xz  0

1  1 x
 1 x  2 z

 
3
2
      3 ydy  dz dx     2  z  dz dx




2
1  0  0
1  0
 

2
1
2
1
1



2
3
1
1
3
   2  z  10 x dx    x 2  1  8 dx
2
2 1
1
1



1


1
x 6  3 x 4  3 x 2  7 dx    x 6  3 x 4  3 x 2  7 dx
2 1
0
 x 7 3x 5

 

 x 3  7 x  10
5
 7

1 3
184
   1 7 
7 5
35
If you prefer to project V on xy-plane,
 divFdxdydz should be as follows:
V
z  1  x2
We observe that the surface z  1  x and the plane y  z  2 intersect at the line 
and the
y z  2
2
projection of the line on xy-plane is y  1  x 2  2  y  x 2  1
. The projection of V onto xy-plane is  xy which consists of two parts, say,  xy   xy   xy .
 xy
is bounded by x=-1, x=1, y=0, y  x 2  1 and  xy is bounded by x= -1 , x=1, y  x 2  1 , y=2. And
over  xy the upper surface of the solid is
upper surface of the solid is
z  1  x 2 and the lower surface is z=0. Over  xy
y  z  2 and the lower surface is z=0. Thus
25
the
1 x

2  y

3
ydxdydz

3
y
dz
dxdy

3
y
dz




  0 
  0 dxdy
V


 xy 

 xy
2
1  2
1 x 2  1 x 2
 
 3 ydz dy  dx  

 0  0   1   2
1 
 
 
1 x
1
 2 y
 
 3 ydz dy  dx
 
 
 0
 
1
1 x

 2

2
    3 y 1  x dy  dx     3 y 2  y dy  dx
1 x 2
 0

1 
1 

2
1






2
3

   1  x 2 (1  x 2 ) dx   3 y 2  y 3

1  2
1
1

1

1



1
3
2 2
2
2
1

x
(
1

x
)
dx

1 2
1 4  3 1  x
1


1

2
1 x 2
dx
  1  x  dx
2
2 3

 

  3 1  2 x  x (1  x )dx  2  4  3 1  2 x 2  x 4  1  3x 2  3x 4  x 6 dx
2
4
2
0
1
0


1


1


  3 1  x 2  x 4  x 6 dx  2  2  3x 2  x 6 dx   7  3x 2  3x 4  x 6 dx
0
0
0
3
1 
3 1 184

  7 x  x 3  x 5  x 7  10  7  1   
5
7 
5 7 35

Example – Gauss’s Law
Gauss’s law for an electric field states that the total displacement flux (of  0 E ) through a closed
surface equals the total charge within the volume enclosed by the surface
i.e.
 
S
0
E dS  

V
 dV
where E is the electric intensity and  is the charge density.
Applying the divergence theorem to the LHS gives
 
S
0
E dS  
   div EdV
V
0
i.e.

V
0
divE  dV  
   dV
V
Since the region V is arbitrary, it must be true that
 0 div E  
i.e. div E   /  0 – one of Mxwell’s equations of electromagnetism.
Since E   where  is the electrostatic potential,  div grad   /  0 or
2
2
2
which is Poisson’s equation.


x 2 y 2 z 2
In a region of no charge (  0) , the electrostatic potential satisfies Laplace’s equation  2   0 .
2    /  0
2 
These are important equations in electromagnetic field theory.
26
7.5 Stockes’s Theorem
For any open surface S, having a closed curve C as its edge, in a vector field F
  curl F  d S   F  dr
S
C
i.e. the flux of curl F through an open surface S = the circulation around the boundary curve C.
Proof:
Divide the surface S into N small patches S i with edge C i and normal ni
For each small area S i
curl Fi
we have approximately
S i
1
F

dr
ni
curl Fi  ni 
S i Ci
S
or curl Fi  ni S i   F  dr
Ci
Summing over all patches and letting
C i
C
N   as S i  0
we have
LHS  
 curl F  dS
S
For the RHS we observe that the flow along an edge of C i is equal and opposite to the flow along
that edge of an adjacent element, unless that edge is part of the boundary C. Hence for all internal
edges the flow cancels leaving RHS   F  dr and the theorem is proved.
C
Example
Verify Stokes’s theorem for F  2 xyi  x 2 j  xk where S is the hemisphere x 2  y 2  z 2  1, y  0
i
j


 F 
x y
2 xy x 2
k

z
x
 j
Sol:
We project S onto the xz-plane where S xz is the disc x 2  z 2  1 .
  x 2  y 2  z 2  1,   2 xi  2 yj  2zk ,  y  2 y
 curlF  dS  
x 2  z 2 1
S
  2
x  x 2 1
j  2 xi  2 yj  2 zk  
dxdz

27
1
dxdz
2y
C is the circle x 2  z 2  1 is the xz-plane. Let the parametric equations of C
z  cos t , x  sin t , y  0, 0  t  2

C
F  dr  
2
t 0
2 xyi  x
2
be

dy
dz 
 dx
j  xk   i 
j  k dt
dt
dt 
 dt
2
  sin 2 tdt
t 0

Hence the theorem is verified.
Note the positive direction of C taken in the zx-plane, relative to the normal of S,
C
n
Example (optional)
Compute
 F  dr where
F  ( y  z )i  ( z  x) j  ( x  y )k and C is the ellipse
C

x2  y2  a2

x z
 a  h  1, where a, h  0
The direction of the path C is anticlockwise when looking from the positive direction of x-axis.
Sol:
For Stoke’s Theorem, the index finger of the right hand represents the direction of the line integral
along the curve, the middle finger of the right hand points to the interior of the open surface which is
enclosed by the curve, that is, the boundary of the open surface and the thumb finger of the right hand
represents the normal to the surface which is closed the boundary. Notice the three vectors form a so
called positive triple.
Remark:
The index finger of the right hand represents the positive direction of x-axis, the middle finger of the
right hand represents the positive direction of y-axis and the thumb finger of the right hand represents
the positive direction of z-axis. The three vectors form a so called positive triple.
By Stoke’s Theorem
 F  dr   curlF  d S where S is the open surface with C as its boundary.
C
 i


curlF  det 
 x
yz

j

y
zx
S
k 

 
 2i  2 j  2k
z 
x  y 
The normal to S :
28
1 
1


1
1
ah i  k 
i
k
i k
x z
h 
z  a
h   a
 ( x, y, z )    n   x
a h
1
1
1
1
a2  h2


a2 h2
a2 h2


h
a
 
i
k 
2
2
a2  h2 
 a h
Since the direction of the line integral along the path C is anticlockwise when looking from the
positive direction of x-axis, according to Stoke’s Theorem, we have to choose the upper normal


h
a
n  
i
k 
2
2
a2  h2 
 a h
So
 F  dr   curlF  d S    2i  2 j  2k   d S
C
S
S


h
a
   2i  2 j  2k   ndS    2i  2 j  2k   
i
k dS
2
2
2
2
a h 
 a h
S
S

 1
h
a
   2i  2 j  2k   
i
k 
dxdy
2
2
2
2
a  h  z
 a h
 xy

h
a
 2  

2
2
a2  h2
 xy  a  h
 a2  h2

dxdy

a

(h  a ) 2
h

 2    1dxdy  2
a   2a ( h  a )
a
a


 xy
Note:
 g ( x, y, z)dS 
S
Or
 F     

1
xy
 g ( x, y, z( x, y))
 xy
 X2  Y2  Z2
dxdy
Z
dxdy
z
Example – Green’s Theorem in the Plane
If the region S lies completely in the xy-plane, and
F  F1i  F2 j ,
29
 F F 
curlF   2  1 k
y 
 x
dS  kdxdy
y
S
 F2 F1 
S curlF  dS  S  x  y dxdy
C
x
and

C
F  dr   F1i  F2 j   dxi  dyj 
C
  F1 dx  F2 dy
C
giving
 F2 F1 
dxdy   F1 dx  F2 dy

S
C
x
y 
 
which is known as Green’s Theorem in the Plane.
Example
Faraday’s Law for a time-varying electric field states that the tangential component of E, the electric
intensity, measured around a closed curve C equals the negative rate of change with respect to time of
the total magnetic flux through any surface bounded by C
i.e.

C
E  dr  

B  dS
t S
where B is the magnetic flux density.
Sol:
Applying Stockes’s theorem to the LHS gives

C
E  dr   curlE  dS
S
Hence,
B
 curlE  dS    t  dS
S
S
and since the surface S is arbitrary.
curlE  
This is another of Maxwell’s equations.
30
B
t