ENGI 5432
2.4
2.4 Surface Integrals - Projection Method
Page 2.19
Surface Integrals - Projection Method
Surfaces in 3
In 3 a surface can be represented by a vector parametric equation
r  x  u, v  ˆi  y  u, v  ˆj  z  u, v  kˆ
where u, v are parameters.
Example 2.4.1
The unit sphere, centre O, can be represented by
r  ,    sin cos , sin sin  , cos
0     and 0    2


declination
azimuth
If every vertical line (parallel to the z-axis) in 3 meets the surface no more than once,
then the surface can also be parameterized as
r  x, y   x, y , f  x, y 
or as z  f  x, y 
Example 2.4.2
z 
4  x2  y 2 ,
 x, y  | x
2
 y 2  4
is a hemisphere, centre O.
A simple surface does not cross itself.
If the following condition is true:
{ r  u1, v1   r  u2 , v2    u1, v1    u2 , v2  for all pairs of points in the domain}
then the surface is simple.
The converse of this statement is not true.
This condition is sufficient, but it is not necessary for a surface to be simple.
The condition may fail on a simple surface at coordinate singularities. For example, one
of the angular parameters of the polar coordinate systems is undefined everywhere on the
z-axis, so that spherical polar (2, 0, 0) and (2, 0, ) both represent the same Cartesian
point (0, 0, 2). Yet a sphere remains simple at its z-intercepts.
ENGI 5432
2.4 Surface Integrals - Projection Method
Page 2.20
Tangent and Normal Vectors to Surfaces
A surface S is represented by r(u, v). Examine the neighbourhood of a point Po at
r(uo, vo). Hold parameter v constant at vo (its value at Po) and allow the other parameter
u to vary. This generates a slice through the two-dimensional surface, namely a onedimensional curve Cu containing Po and represented by a vector parametric equation
r  r  u, vo  with only one freely-varying parameter (u).
Cu : r  u, vo 
Cv : r  uo , v 
If, instead, u is held constant at uo and v is allowed to vary, we obtain a different slice
containing Po, the curve Cv : r  uo , v  .
On each curve a unique tangent vector can be defined.
At all points along Cu, a tangent vector is defined by Tu 

 r  u , vo   .
u
[Note that this is not necessarily a unit tangent vector.]

At Po the tangent vector becomes Tu P 
 r  uo , vo   .
o
u

 r  uo , vo   .
Po
v
If the two tangent vectors are not parallel and neither of these tangent vectors is the zero
vector, then they define the orientation of tangent plane to the surface at Po.
Similarly, along the other curve Cv, the tangent vector at Po is Tv

ENGI 5432
2.4 Surface Integrals - Projection Method
N  Tu  Tv 
A normal vector to the tangent plane is

 ˆi

x
 det 
 u
 x

 v
where
  x, y 
  u, v 
ˆj
y
u
y
v

kˆ 

z 
u 
z 

v  uo ,vo 

x
 u
is the Jacobian det 
x
 v

Page 2.21
r r

u v uo ,vo 
  y , z    z , x    x, y 
,
,
 u, v   u, v   u, v 
,
uo ,vo 
y
u 
.
y
 v 
Cartesian parameters
With u = x, v = y, z = f (x, y) , the components of the normal vector
N  N1 ˆi  N2 ˆj  N3 kˆ are:
 y, z 
N1 

  x, y 
0
1
f
x
f
y
 z, x 
N2 

  x, y 
f
 
x
N3 
  x, y 

  x, y 

a normal vector to the surface z = f (x, y) at (xo, yo, zo) is
1 0
 1
0 1
N 

f
f
,
, 1
x  y
 xo , yo 
f
x
f
y
1
 
0
f
y
ENGI 5432
2.4 Surface Integrals - Projection Method
Page 2.22
If the normal vector N is continuous and non-zero over all of the surface S, then the
surface is said to be smooth.
Example 2.4.3
A sphere is smooth.
A cube is piecewise smooth (six smooth faces)
A cone is not smooth ( N is undefined at the apex)
Surface Integrals (Projection Method)
This method is suitable mostly for surfaces which can be expressed easily in the
Cartesian form z = f (x, y).
The plane region D is the projection of the surface S : f (r) = c onto a plane (usually the
xy-plane) in a 1:1 manner.
The plane containing D has a constant unit normal n̂ .
N is any non-zero normal vector to the surface S.
A  S cos 
but
cos  
N nˆ
N
ENGI 5432

2.4 Surface Integrals - Projection Method

dS 

g  r  dS 
Page 2.23
 N nˆ dA
N
D
S
and
S

D
g r 
N
N nˆ
dA
For z = f (x, y) and D = a region of the xy-plane,
z
z
N   ,
, 1 and nˆ  kˆ
x  y

N nˆ  1 and
S g  r  dS

 g r 
D
2
2
 z   z 
  
  1 dA
 x    y 
which is the projection method of integration of g(x, y, z) over the surface z = f (x, y) .
Advantage:
Region D can be geometrically simple (often a rectangle in
2
).
Disadvantage: Finding a suitable D (and/or a suitable 1:1 projection) can be difficult.
You may have to split the surface into pieces (such as splitting a sphere into two
hemispheres) in order to obtain separate 1:1 projections. The projection fails if part of
the surface is vertical (such as a vertical cylinder onto the xy plane).
ENGI 5432
2.4 Surface Integrals - Projection Method
Page 2.24
Example 2.4.4
Evaluate
 z dS , where the surface S is the section of the cone
z2 = x2 + y2 in the first
S
octant, between z = 2 and z = 4.
z2 = x2 + y2
 2z

z
 2x  0
x
z
x


x
z
x
x  y2
2
By symmetry,
z
y


y
z
2
dS 
2
 z   z 
      1 dA 
 x   y 
 x2   y 2 
 2    2   1 dA 
z  z 
y
x2  y2
 z2 
 2   1 dA 
z 
Use the polar form for dA :
2  r  4 , 0    2
dA  r dr d ,
r 

x2  y 2  z

S
z dS 
 /2
4
0  2 r
2 r dr d
2 dA
ENGI 5432
2.4 Surface Integrals - Projection Method
Example 2.4.4 (continued)

 z dS
S


 /2
2
0
1 d 
4
2 r
 /2
2
dr 


2  

0
 2 56

 64 8 
2   0    

2
3
2
 3 3 


S
z dS 
28 2
3
4
 r3 
 
 3 2
Page 2.25
ENGI 5432
2.5
2.5 Surface Integrals - Surface Method
Page 2.26
Surface Integrals - Surface Method
When a surface S is defined in a vector parametric form r = r(u, v), one can lay a
coordinate grid (u, v) down on the surface S.
r r

A normal vector everywhere on S is N 
.
u v
dS  dS  N du dv
 g  r  dS

S

g r 
r r

du dv
u  v
S
Advantage:
 only one integral to evaluate
Disadvantage:
 it is often difficult to find optimal parameters (u, v).
The total flux of a vector field F through a surface S is
 

S
F dS 

S
ˆ dS 
FN

F
r r
 du dv
u  v
S
(which involves the scalar triple product F
r r
 ).
u  v
ENGI 5432
2.5 Surface Integrals - Surface Method
Page 2.27
Example 2.5.1: (same as Example 2.4.4, but using the surface method).
Evaluate
 z dS , where the surface S is the section of the cone
z2 = x2 + y2 in the first
S
octant, between z = 2 and z = 4.
Choose a convenient parametric net:
u  r 
x2  y 2  z
and
v  
then
r  r cos , r sin  , r
 2  r  4,

ˆj
sin 
r cos 
r
 r sin  , r cos  , 0

kˆ
1   r cos  ,  r sin  , r
0
 N  N  r cos2   sin 2   1  r 2


S
z dS 

S
z N dr d 
 /2
4
0  2 r

2 r dr d
 z dS
S


r
 cos  , sin  , 1
r
and
ˆi
 N   cos 
r sin 
0    2
(as before)
28 2
3
ENGI 5432
2.5 Surface Integrals - Surface Method
Page 2.28
Just as we used line integrals to find the mass and centre of mass of [one dimensional]
wires, so we can use surface integrals to find the mass and centre of mass of [two
dimensional] sheets.
Example 2.5.2
Find the centre of mass of the part of the unit sphere (of constant surface density) that lies
in the first octant.
Cartesian equation of the sphere:
x 2  y 2  z 2  1;
x  0, y  0, z  0
The radius of the sphere is r = 1.
For the parametric net, use the two angular
coordinates of the spherical polar coordinate
system  r ,  ,   .
x  sin  cos 
0    2
0    2
y  sin  sin 
z  cos 

and
r
 cos  cos  , cos  sin  ,  sin 

r
  sin  sin  , sin  cos  , 0

ˆi
ˆj
 N   cos  cos  cos  sin 
 sin  sin  sin  cos 
kˆ
 sin 
0

  sin 2  cos  , sin 2  sin  , sin  cos  cos 2   sin 2 
  sin  sin  cos  , sin  sin  , cos   sin  r
The outward normal is clearly N   sin  r
 N  N  sin 
r  sin 

ENGI 5432
2.5 Surface Integrals - Surface Method
Example 2.5.2
(continued)
  dS   
Mass: m 
S
 
 
 /2
 /2
0
0
 

 /2
N d d
S
sin  d d
sin  d 
0

Page 2.29
   0  1 2  0

 /2
0
 /2
 /2
d     cos  0    0


 m 
2
OR
Note that the mass of a complete spherical shell of radius r and constant density  is
4 


.
4 r 2  . Therefore the mass of one eighth of a shell of radius 1 is
8
2
By symmetry, the three Cartesian coordinates of the centre of mass are all equal:
xyz.
Taking moments about the xy plane:
M 

z  dS  
S
 

 /2 1
0
2
 /2
 /2
0
0
 
sin 2 d 

 /2
0
 cos   sin  d d
 /2
 /2
 cos 2 
d    
   0

4 0

1 
1
1 1 

      0  
 m
2 2
2
4 4  2

M
1

m
2
Therefore the centre of mass is at

z 
 x, y, z 

 12 , 12 , 12

ENGI 5432
2.5 Surface Integrals - Surface Method
Page 2.30
Example 2.5.3
Find the flux of the field F  x, y,  z across that part of x + 2y + z = 8 that lies in
the first octant.
The Cartesian coordinates x, y will
serve as parameters for the surface:
r  x, y, 8  x  2 y
r
 1, 0,  1
x
r
and
 0, 1,  2
y

ˆi
 N   1
0
ˆj
0
1
kˆ
1   1, 2, 1
2
Choose N to point “outwards”.
N  1, 2,1
Range of parameter values:
In the xy plane:
 0  x  8 and 0  y  4
But the area is a triangle, not a rectangle, so these inequalities do not provide the correct
limits for the inner integral.
ENGI 5432
2.5 Surface Integrals - Surface Method
Page 2.31
Example 2.5.3 (continued)
 F dS   F
Net flux =  
N
S
dS 
S
  F Nˆ   N dA   F N dA
S
S
(where dA = dx dy)
F N  x, y,   8  x  2 y  1, 2, 1  x  2 y  8  x  2 y  2  x  2 y  4 
  
 2
 2
 2
 4



4
8  2 y 

0
4


2
2

4
0
4
8 2 y
0 0
2  x  2 y  4  dx dy
8 2 y
 x2

dy
  2 xy  4 x 
2
 x0


 2  8  2 y  y  4  8  2 y     0  0  0   dy




 32  16 y  2 y 2  16 y  4 y 2  32  8 y  dy
0
4
4y  y 
2
0
4

y3 

64 

dy  4  2 y 2    4   32     0  0  
3 0
3 



Therefore the net flux is
 
128
3
The iteration could be taken in the other order:
 
 2

4 x /2
8
0 0

8
0

2  x  2 y  4  dy dx
4 x /2
 x y  xy 2  4 y 
dx
y0

8
 4 x  12 x  dx
2
0


128
3
ENGI 5432
2.5 Surface Integrals - Surface Method
Page 2.32
Example 2.5.4
Find the total flux  of the vector field F  z kˆ through the simple closed surface S
x2 y 2 z 2


 1
a 2 b2 c 2
Use the parametric grid  ,   , such that the displacement vector to any point on the
ellipsoid is
r  a sin  cos  , b sin  sin  , c cos
This grid is a generalisation of the spherical polar coordinate grid and covers the entire
surface of the ellipsoid for 0     , 0    2 .
One can verify that x  a sin  cos  , y  b sin  sin  , z  c cos  does lie on the ellipsoid
x2 y 2 z 2


 1
a 2 b2 c 2
for all values of  ,   :
x2 y 2 z 2
a 2 sin 2  cos 2  b2 sin 2  sin 2  c 2 cos2 





a 2 b2 c 2
a2
b2
c2
 sin 2  cos 2   sin 2  sin 2   cos 2   sin 2   cos 2   sin 2    cos 2 
 sin 2   cos2   1
 and 
The tangent vectors along the coordinate curves  = constant and  = constant are
dr
 a cos  cos  , b cos  sin  ,  c sin  and
d
dr
 a sin  sin  , b sin  cos  , 0 .
d
The normal vector at every point on the ellipsoid follows:
ˆi
ˆj
kˆ
dr dr
N 

 a cos  cos  b cos  sin  c sin 
d d
a sin  sin  b sin  cos 
0
 bc sin 2  cos  , ac sin 2  sin  , ab sin  cos   cos 2   sin 2  
(and this vector points away from the origin).
ENGI 5432
2.5 Surface Integrals - Surface Method
Page 2.33
Example 2.5.4 (continued)
On the ellipsoid, F  z kˆ  c cos  kˆ
 F N  c cos   ab sin  cos    abc sin  cos 2 
The total flux of F through the surface S is therefore
 
S
F dS 
2

0 0
F N d d  abc
2

 0  0 sin  cos  d d
2
Let u = cos , then du = –sin  d and  = 0  u = +1,  =   u = –1

2
S F dS  abc 0
1
2  u 3 


1 d 
u du  abc   

0
1
 3  1
 1 1
 abc  2  0     
 3 3

1
2

 
4 abc
3
ENGI 5432
2.5 Surface Integrals - Surface Method
Page 2.34
For vector fields F(r),
Line integral:
 F dr
C
Surface integral:
r r
 F r  dS   F r  N dS   F N du dv    F u  v du dv
S
S
S
S
On a closed surface, take the sign such that N points outward.
Some Common Parametric Nets
1)
The circular plate
 x  xo 
2
  y  yo   a 2 in the plane z = zo.
2
Let the parameters be r,  where 0 < r  a , 0   < 2
x = xo + r cos  , y = yo + r sin  , z = zo
ˆi
ˆj
kˆ
 r r 
N        cos 
sin 
0   r kˆ

r




r sin  r cos  0
2)
The circular cylinder
 x  xo 
2
  y  yo   a 2 with zo  z  z1 .
2
Let the parameters be z,  where zo  z  z1 , 0   < 2
x = a cos  , y = a sin  , z = z
ˆi
ˆj
kˆ
 r r 
N      
0
0
1    a cos  ˆi  a sin  ˆj

z




a sin  a cos  0


Outward normal: N  a cos  ˆi  a sin  ˆj
3)
The frustrum of the circular cone w  wo  a
 u  uo 
2
  v  vo 
2
where
w1  w  w2 and wo  w1 . Let the parameters here be r,  where
w1  wo
w  wo
 r  2
,
0    2
a
a
x  u  uo  r cos , y  v  vo  r sin  , z  w  wo  a r
ˆi
ˆj
kˆ
 r r 
N        cos 
sin 
a
  r  
r sin  r cos  0
   a r cos   ˆi   a r sin   ˆj  r kˆ 
Outward normal: N  ar cos  ˆi  ar sin  ˆj  r kˆ
ENGI 5432
4)
2.5 Surface Integrals - Surface Method
The portion of the elliptic paraboloid
2
2
z  z o  a 2  x  xo   b 2  y  y o 
with
Page 2.35
zo  z1  z  z2
Let the parameters here be r,  where
z1  zo
z2  zo
 r 
, 0    2
2
2
2
2
2
2
a cos   b sin 
a cos   b2 sin 2 
x = xo + r cos  ,
y = yo + r sin  ,
z = zo + r2 (a2 cos2 + b2 sin2)
ˆi
ˆj
kˆ
 r r 
N        cos 
sin  2r  a 2 cos 2   b 2 sin 2  

r




r sin  r cos  2r 2  b 2  a 2  sin  cos 
   2a 2 r 2 cos   ˆi   2b 2 r 2 sin   ˆj  r kˆ 
Outward normal: N   2a 2 r 2 cos   ˆi   2b 2 r 2 sin   ˆj  r kˆ
5)
The surface of the sphere x  x 2   y  y 2  z  z 2  a 2 .
Let the parameters here be ,  where 0     , 0    2
x = xo + a sin  cos  , y = yo + a sin  sin  , z = zo + a cos 
ˆi
ˆj
kˆ
 r r 
N        a cos  cos  a cos  sin  a sin 
   
a sin  sin  a sin  cos 
0
  a 2 sin   sin  cos   ˆi   sin  sin   ˆj   cos   kˆ 
Outward normal: N  a 2 sin   sin  cos   ˆi   sin  sin   ˆj   cos   kˆ 
6)
The part of the plane Ax  x   B y  y   Cz  z   0 in the first octant with
A, B, C > 0 and Axo+ Byo+ Czo > 0.
Let the parameters be x, y where
Ax  By  Cz  By
Ax  By  Cz
0 x 
;
0 y 
A
B
ˆi
ˆj
kˆ
 r r 
Bˆ
Aˆ
ˆ
N  

   1 0  AC    i  j  k
C
C

x y
B
0 1 
C
ENGI 5432
2.6
2.6 Gauss, Stokes; Potential Functions
Page 2.36
Theorems of Gauss and Stokes; Potential Functions
Gauss’ Divergence Theorem
Let S be a piecewise-smooth closed surface enclosing a volume V in 3 and let F be a
vector field. Then
the net flux of F out of V is
 F dS   F
N
S
dS .
S
But the divergence of F is a flux density, or an “outflow per unit volume” at a point.
Integrating div F over the entire enclosed volume must match the net flux out through
the boundary S of the volume V. Gauss’ divergence theorem then follows:
 F dS    F dV
S
V
Example 2.6.1 (Example 2.5.4 repeated)
Find the total flux  of the vector field F  z kˆ through the simple closed surface S
x2 y 2 z 2


 1
a 2 b2 c 2
div F dV
S F dS  
V
Use Gauss’ Divergence Theorem:
F is differentiable everywhere in
 
div F   z kˆ 

  
, ,
x y z
div F dV   1 dV

V
V
3
, so Gauss’ divergence theorem is valid.
0, 0, z  0  0  1  1
 V 
4 abc
– the volume of the ellipsoid !
3
Therefore
 
S
F dS 
4 abc
3
In fact, the flux of F  z kˆ through any simple closed surface is just the volume enclosed
by that surface.
ENGI 5432
Example 2.6.2
2.6 Gauss, Stokes; Potential Functions
Page 2.37
Archimedes’ Principle
Gauss’ divergence theorem may be used to derive Archimedes’ principle for the buoyant
force on a body totally immersed in a fluid of constant density  (independent of depth).
Examine an elementary section of the surface S of the immersed body, at a depth z < 0
below the surface of the fluid:
The pressure at any depth z is the weight of fluid per unit area from the column of fluid
above that area. Therefore
pressure = p =   g z
 g is the weight of the column
– z is the height of the column (note z < 0).
The normal vector N to S is directed outward, but the hydrostatic force on the surface
(due to the pressure p) acts inward. The element of hydrostatic force on S is
 pressure    area    direction 
 
ˆ     g z S  N
ˆ
    g z   S  N
The element of buoyant force on S is the component of the hydrostatic force in the
direction of k (vertically upwards):
   g z S Nˆ  kˆ
ˆ dS .
Define F   g z kˆ and dS  N
Summing over all such elements S, the total buoyant force on the immersed object is
  g z kˆ Nˆ dS   F dS    F dV
S
S
V
(by the Gauss Divergence Theorem)
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Page 2.38
Example 2.6.2 Archimedes’ Principle (continued)

    g z kˆ  dV  
V

  g dV
V
V
  
, ,
x y z
0, 0,  g z dV


  g   0 
 provided
z


= weight of fluid displaced
Therefore the total buoyant force on an object fully immersed in a fluid equals the weight
of the fluid displaced by the immersed object (Archimedes’ principle).
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Page 2.39
Gauss’ Law
A point charge q at the origin O generates an electric field
q
q
E 
r 
rˆ
3
4 r
4 r 2
If S is a smooth simple closed surface not enclosing the charge, then the total flux
through S is
 E dS    E dV
S
(Gauss’ divergence theorem)
V
1 
But Example 1.4.1 showed that   3 r   0 r  0 .
r 
Therefore
 E dS  0 .
S
There is no net outflow of electric flux through any closed surface not enclosing the
source of the electrostatic field.
If S does enclose the charge, then one cannot use Gauss’ divergence theorem, because
 E is undefined at the origin.
Remedy:
Construct a surface S1 identical to S except for a small hole cut where a narrow tube T
connects it to another surface S2, a sphere of radius a centre O and entirely inside S. Let
S *  S1 T S 2 (which is a simple closed surface), then O is outside S* !
Applying Gauss’ divergence theorem to S*,
 E dS    E dV
S
*

V
 0
*
 E dS   E dS   E dS  0
S1
T
S2
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Gauss’ Law (continued)
As the tube T approaches zero thickness,
 E dS  0
and therefore
T
 E dS    E dS
S1
S2
But S 2 is a sphere, centre O, radius a.
Using parameters  ,   on the sphere,
r  a sin  cos  , sin  sin  , cos
Finding
r r
as before leads to N   a sin  r .
,
 
But the “outward normal” to S 2 actually points towards O.
 N   a sin  r on the sphere S 2
and E 
q
4 a 3
r everywhere on S 2 .
Also r  a rˆ  r r  a 2
 EN 
q
4 a
3
r  a sin  r  
q sin 
q sin 
rr 
2
4 a
4
Recall that dS  N dS  N N d d  N d d
q 2 
E dS 
E N d d 
sin  d d
0
0
4

S
S

2



 
2
q
q
q

2
 1  1 2  0   
  cos   0    0 
4
4

 E dS    E dS   
q
S1
S2
Page 2.40
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Page 2.41
Gauss’ Law (continued)
But, as a  0

S2  O  , S1  S
The surface S 1 looks more and more like the surface S as the tube T collapses to a line
and the sphere S 2 collapses into a point at the origin. Gauss’ law then follows.
Gauss’ law for the net flux through any smooth simple closed surface S, in the presence
of a point charge q at the origin, then follows:

S
q

E dS   
 0
if S encloses O
otherwise
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Page 2.42
Example 2.6.3 Poisson’s Equation
The exact location of the enclosed charge is immaterial, provided it is somewhere inside
the volume V enclosed by the surface S. The charge therefore does not need to be a
concentrated point charge, but can be spread out within the enclosed volume V. Let the
charge density be  (x, y, z), then the total charge enclosed by S is
q 
  dV
V
Gauss’ law 

E dS 
S
q

Apply Gauss’ divergence theorem to the left hand side, substitute for q on the right hand
side and assume that the permittivity  is constant throughout the volume:


  E dV    dV
V


V
V


  E   dV  0


V
This identity will hold for all volumes V only if the integrand is zero everywhere.
Poisson’s equation then follows:
E 
E   V
and  V   2V



 2V  


This reduces to Laplace’s equation 2V  0 when   0 .
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Page 2.43
Stokes’ Theorem
Let F be a vector field acting parallel to the xy-plane.
components by F  f1 ˆi  f 2 ˆj  f1 , f 2 , 0 . Then
ˆi
ˆj
Represent its Cartesian
kˆ



 f2
 f1
    F  kˆ 

x  y z
x
y
f1
f2
0
Green’s theorem can then be expressed in the form
F dr 
  F kˆ dA
F 


C
D
Now let us twist the simple closed curve C and its enclosed surface out of the xy-plane, so
that the normal vector k is replaced by a more general normal vector N.
If the surface S (that is bounded in 3 by the simple closed curve C) can be represented by
z = f (x, y), then a normal vector at any point on S is
N 

z
z
,  ,1
x y
C is oriented coherently with respect to S if, as one travels along C with N pointing from
one’s feet to one’s head, S is always on one’s left side. The resulting generalization of
Green’s theorem is Stokes’ theorem:
 F dr     F N dS    curl F  dS
C
S
S
This can be extended further, to a non-flat surface S with a non-constant normal vector N.
Example 2.6.4
Find the circulation of F  xyz, xz, e xy around C : the unit square in the xz-plane.
Because of the right-hand rule, the positive orientation
around the square is OGHJ (the y axis is directed into
the page).
In the xz plane y = 0  F  0, xz,1
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Example 2.6.4 (continued)
Computing the line integral around the four sides of the square:
OG : r  0, 0, t
 0  t  1
and F  0, 0, 1
 F


F dr 
OG

1
0
dr
 0, 0, 1
dt

dr
 0, 0, 1 0, 0, 1  1
dt
1 dt  t  0  1  0  1
1
In a similar way (Problem Set 6 Question 6), it can be shown that
 F dr
 0,
GH

 F dr
 F dr
  1 and
HJ
 F dr
 0
JO
 1 0 1 0  0
C
OR use Stokes’ theorem:
On D   F 
ˆi
ˆj

x
0

y
xz
kˆ

  x, 0, z
z
1
dA  ˆj dA
   F dA   x, 0, z

   F dA  0
D
0,1, 0 dA  0 dA

 F dr
 0
C
Note that this vector field F is not conservative, because   F  0 .
Page 2.44
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Page 2.45
Domain
A region  of 3 is a domain if and only if
1)
For all points Po in , there exists a sphere, centre Po, all of whose interior points
are inside ; and
2)
For all points Po and P1 in , there exists a piecewise smooth curve C, entirely in
, from Po to P1.
A domain is simply connected if it “has no holes”.
Example 2.6.5
Are these regions simply-connected domains?
The interior of a sphere.
YES
The interior of a torus.
NO
The first octant.
YES
On a simply-connected domain the following statements are either all true or all false:




F is conservative.
F  
F  0
 F dr    Pend     Pstart  - independent of the path between the two points.
C

 F dr
 0 C  
C
Example 2.6.6
Find a potential function  (x, y, z) for the vector field F  2 x, 2 y, 2 z .
First, check that a potential function exists at all:
ˆi
ˆj
kˆ
curl F    F 

x
2x

y
2y

z
2z
Therefore F is conservative on

3
.
0, 0, 0  0
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Example 2.6.6
 F   

 2x
x

Page 2.46
(continued)
  
,
,
x y z
   x 2  g  y, z 

g
 0
 2y
y
y
 g  y, z   y 2  h  z 
   x2  y 2  h  z 


dh
 00
 2z
z
dz
 h  z   z2  c
   x2  y 2  z 2  c
We have a free choice for the value of the arbitrary constant c . Choose c = 0, then
  x, y, z   x 2  y 2  z 2  r 2
ENGI 5432
2.6 Gauss, Stokes; Potential Functions
Page 2.47
Maxwell’s Equations (not examinable in this course)
We have seen how Gauss’ and Stokes’ theorems have led to Poisson’s equation, relating
the electric intensity vector E to the electric charge density  :

E 

Where the permittivity is constant, the corresponding equation for the electrical flux
density D is one of Maxwell’s equations:  D   .
Another of Maxwell’s equations follows from the absence of isolated magnetic charges
(no magnetic monopoles):  H  0   B  0 , where H is the magnetic
intensity and B is the magnetic flux density.
Faraday’s law, connecting electric intensity with the rate of change of magnetic flux

E dr  
B dS .
density, is
Applying Stokes’ theorem to the left side

t
C
S


produces
E  
Ampère’s circuital law, I 
 H dl , leads to
B
t
  H  J  J d , where
C
the current density is J   E  V v ,  is the conductivity, V is the volume charge
density; and the displacement charge density is J d 
D
t
The fourth Maxwell equation is
H  J 
D
t
The four Maxwell’s equations together allow the derivation of the equations of
propagating electromagnetic waves.
END OF CHAPTER 2