VC.10
Surface Integrals and the Divergence
Theorem (Gauss’ Theorem)
(Day 1)
VC.06: Measuring the Flow of a Vector Field
ACROSS a Closed Curve
Let C be a closed curve with a counterclockwise parameterization. Then the net
flow of the vector field ACROSS the closed curve is measured by:
Ñ
 Fie ld( x , y )  o u te ru n itn o rm al d s
C
b

 Fie ld( x( t ), y( t ))  ( y '( t ),  x '( t ))d t
a

Ñ
 C  n( x , y )d x  m ( x , y )d y
Let region R be the interior of C. If the vector field has no singularities in R, then
we can use Gauss-Green:
 m
n 
  

 dx dy
x
y 
R 

 d iv Fie ld d x
R
dy
Le t d iv Fie ld( x , y ) 
m
x

n
y
.
VC.06: The Flow of A Vector Field ACROSS a
Closed Curve:
Let C be a closed curve parameterized counterclockwise. Let
Field(x,y) be a vector field with no singularities on the interior
region R of C. Then:
Ñ
 Fie ld( x , y )  o u te ru n itn o rm al d s
C

 d iv Fie ld d x
dy
R
This measures the net flow of the vector field ACROSS the closed
curve.
W e d e fin e th e d iv e rg e n ce o f th e v e cto r fi e ld as:
d iv Fie ld( x , y ) 
m
x

n
y
 D[m[ x , y ], x ]  D[n[ x , y ], y ]
VC.06: Measuring the Flow of a Vector Field
ALONG a Closed Curve
Let C be a closed curve with a counterclockwise parameterization. Then the net
flow of the vector field ALONG the closed curve is measured by:
Ñ
 Fie ld( x , y )  u n ittan d s
C
b

 Fie ld( x( t ), y( t ))  ( x '( t ), y '( t ))d t
a

Ñ
 C m ( x , y )d x  n( x , y )d y
Let region R be the interior of C. If the vector field has no singularities in R, then
we can use Gauss-Green:
 n m 
  

 dx dy
x
y 
R 

 ro tFie ld d x
R
dy
Le t ro tFie ld( x , y ) 
n
x

m
y
.
VC.06: The Flow of A Vector Field ALONG a
Closed Curve:
Let C be a closed curve parameterized counterclockwise. Let
Field(x,y) be a vector field with no singularities on the interior
region R of C. Then:
Ñ
 Fie ld( x , y )  u n ittan d s
C

 ro tFie ld d x
dy
R
This measures the net flow of the vector field ALONG the closed
curve.
W e d efin e th e ro tatio n o f th e vecto r field as:
ro tField( x , y ) 
n
x

m
y
 D[n[ x , y ], x ]  D[m[ x , y ], y ]
VC.10: The Flow of A Vector Field ACROSS a
Closed Surface:
Constructing a three-dimensional analog of using
the Gauss-Green Theorem to compute the flow
of a vector field across a closed curve is not
difficult. This is because the notion of
divergence extends to three dimensions pretty
naturally.
We will save the three-dimensional analog of flow
ALONG for next chapter…
VC.10: The Flow of A Vector Field ACROSS a
Closed Surface:
Let R be a solid in three dimensions with boundary surface (skin)
C with no singularities on the interior region R of C. Then the
net flow of the vector field Field(x,y,z) ACROSS the closed
surface is measured by:
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C

 d iv Fie ld d x
dy dz
R
L e t Fie ld( x , y , z )   m ( x , y , z ), n( x , y , z ), p( x , y , z )  .
W e d e fin e th e d iv e rg e n ce o f th e v e cto r fi e ld as:
d iv Fie ld( x , y , z ) 
m
x

n
y

p
z
 D[m[ x , y , z ], x ]  D[n[ x , y , z ], y ]  D[p[ x , y , z ], z ]
More Traditional Notation: The Divergence
Theorem (Gauss’ Theorem)
Let V be a solid in three dimensions with boundary surface (skin)
S with no singularities on the interior region V of S. Then the
net flow of the vector field F(x,y,z) ACROSS the closed surface
is measured by:
Ò
  F  n  d S
S

    F  d V
V
L e t F( x , y , z )  m ( x , y , z )  n( x , y , z )  p( x , y , z ).
 

 
Let   
,
,
 b e k n o w n as " d e l" , o r th e d iffe re n t ial o p e rato r.

x

y

z


N o te d iv Fie ld( x , y , z )    F 
m
x
Fin ally , le t n = o u te ru n itn o rm al.

n
y

p
z
.
Example 1: Avoiding Computation Altogether
Le t Fie ld( x , y )   7 x  2 , y  6  an d le t C b e a clo s e d cu rv e g iv e n b y

2
C (t)  ( x( t ), y( t ))  sin ( t ), co s( t )  sin ( t )

fo r 

 t 
3
.
4
4
Is th e flo w o f th e v e cto r fie ld acro ss t h e cu rv e fro m in sid e to o u tsid e o r
o u tsid e to in sid e ?
d iv Fie ld( x , y ) 
Ñ
 C  n( x , y )d x  m ( x , y )d y
m
x


n
y
 71  8
 d iv Fie ld( x , y ) d x
R
dy 
 8 d x
dy
R
Sin ce d ivField (x,y) is A LW A YS p ositive f or all (x,y) an d th ere are n o
sin g u larities for an y (x,y), th is in teg ral is p ositive for an y closed cu rve.
Th at is, for A N Y closed cu rve, th e n et f low of th e vector field across
th e cu rve is from in sid e to ou tsid e.
Example 2: Avoiding Computation Altogether
L e t Fie ld( x , y , z ) 
x  y
2
3
 co s(z ) ,  y  x z ,  3 z  8 x  3e
y
.
L e t C b e th e a b o u n d in g su rface o f a so lid re g io n .
Is th e flo w o f th e v e cto r fie ld acro ss t h e cu rv e fro m
in sid e to o u tsid e o r o u tsid e to in s id e ?
d iv Fie ld( x , y , z ) 
m
x

n
y

p
z
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C
2
2
 1  3y  3  3y  4

 d iv Fie ld d x
dy dz
R
Sin ce d ivField (x,y,z) is A LW A Y S n eg ative at all (x,y,z) (sin ks) an d th ere are
n o sin g u larities, th is in teg ral is n eg ative for an y closed su rface.
Th at is, for A N Y closed su rface, th e n et flow of th e vector field across
th e su rface is from ou tsid e to in sid e.
Example 3: Avoiding Computation Altogether

5
3
L e t Fie ld( x , y , z )  sin (y )  z , z x co s( x ), sin (5 x y )  3 x e
xy  sin ( x )
.
L e t C b e th e b o u n d in g su rface o f a so lid re g io n .
Is th e flo w o f th e v e cto r fie ld acro ss t h e cu rv e fro m
in sid e to o u tsid e o r o u tsid e to i n sid e ?
d iv Fie ld( x , y , z ) 
m
x

n
y

p
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C
z

 0
 d iv Fie ld d x
dy dz
R
Sin ce d ivField (x,y,z) is A LW A Y S 0 for all (x,y,z) an d th ere are n o
sin g u larities for an y (x,y,z), th is in te g ral is 0 for an y closed su rface.
Th at is, for A N Y closed su rface, th e n et flow of th e vector field across
th e su rface is 0.
C
Example 4: Find the Net Flow of a Vector
Field ACROSS a Closed Curve
L e t Fie ld( x , y ) 
x
2
2
 2 xy ,  y  x

a n d le t C b e th e r e c ta n g le b o u n d e d b y
x   2 , x  5 , y   1 , a n d y  4 . M e a su re th e flo w o f th e v e c to r fie ld a c ro ss
th e c u rv e .
d iv Fie ld( x , y ) 
m
x

 n( x , y )d x  m ( x , y )d y 
n
 2x  4 y
y
 d iv Fie ld( x , y ) d x
dy
R
4

5
  2x  4 y  dx
dy
1 2
  105
N eg ative. Th e net flow of th e vector fie ld across our closed curve
is from outsid e to insid e.
Example 5: Find the Net Flow of a Vector
Field ACROSS a Closed Surface

2
L e t Fie ld( x , y , z )  2 x y ,  y , 5 z  4 x z

a n d le t C b e th e re c ta n g u la r p rism
b o u n d e d b y  1  x  4 ,  2  y  3 , a n d 0  z  5 . M e a su re th e flo w o f
th e v e c to r fie ld a c ro ss th e c lo se d su rfa c e .
d iv Fie ld( x , y ) 
m
x

n
y

p
z
 5  4x
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A

C
 d iv Fie ld( x , y , z ) d x
R
5 3

dy dz
4
   5  4 x  dx
dy dz
0 2 1
 1375
Positive. T h e n et flow of th e vector fie ld across ou r closed
su rface is from in sid e to ou tsid e.
Summary: The Divergence Locates Sources
and Sinks
Let R be a solid in three dimensions with boundary surface (skin) C with no
singularities on the interior region R of C. Then:
If d iv Fie ld (x,y,z)> 0 fo r all p o in ts in C , th e n all th e se p o in ts
are so u rce s an d th e n e t flo w o f th e v e cto r fie ld acro ss C
is fro m in sid e to o u tsid e .
If d iv Fie ld (x,y,z)< 0 fo r all p o in ts in C , th e n all o f th e se
p o in ts are sin k s an d th e n e t flo w o f th e v e cto r fie ld
acro ss C is fro m o u tsid e to in sid e .
If d ivField( x , y , z )  0 for all p oin ts in C, th en th e n et flow of
th e vector field across C is 0.
The Flow of a Vector Field Across an Open
Surface
The Divergence Theorem is great for a closed surface, but it is not useful at all
when your surface does not fully enclose a solid region. In this situation, we
will need to compute a surface integral. For a parameterized surface, this is
pretty straightforward:
 Fi e ld( x , y , z )  o u te rn o rm al d A
C

t2
 
t1
s2
s1
Fie ld( x( s , t ), y( s , t ), z( s , t ))  n o rm al( s , t ) d s d t
W h at is th e n o rm al ve cto r??
Normal Vectors: Curves Versus Surfaces
2-D im e n sio n s:
3 -D im e n sio n s:
 Fie ld( x , y )  o u te ru n itn o rm a l d s
 Fi e ld( x , y , z )  o u te rn o rm al d A
C
b

 Fie ld( x ( t ), y ( t ))  o u te rn o rm a l d t
C

t2
 
t1
s2
s1
Fie ld( x( s , t ), y( s , t ), z( s , t ))  n o rm al( s , t ) d s d t
a
b

 Fie ld( x ( t ), y ( t ))  ( y '( t ),  x '( t ))d t
a
In 2 d im en sion s, ou tern orm al  ( y '( t),  x '( t)).
Th is is m ore su b tle in 3 d im en sion s...
Normal Vectors: Curves Versus Surfaces
t2
 Fi e ld( x , y , z )  o u te rn o rm al d A

Fo r a cu rv e C p aram e te rize d b y
 x( s , t ), y( s , t ), z( s , t )  , yo u
C
 
s2
t1
s1
Fie ld( x( s , t ), y( s , t ), z( s , t ))  n o rm al( s , t ) d s d t
can fin d tw o lin e arly-
in d e p e n d e n t tan g e n t v e cto rs to th e cu rv e u sin g p artial d e riv ativ e s :
 x y z 
 x y z 
,
,
,
,

 an d 


s

s

s

t

t
t 



U se th ese tw o vectors tan g en t to th e cu rve to
g en erate you r n orm al vector:
 x y z   x y z 
n o rm al( s , t )  
,
,
,
,


 s s s   t t t 
 x y z 
,
,



s

s

s


 x y z 
,
,



t

t

t


Example 6: Using a Substitute Surface
When the Divergence is 0

Le t Fie ld( x , y , z )  z  y , z  x , x
2
.
Le t C b e th e b o u n d in g su rface o f th e so lid re g io n p ictu re d b e lo w , w h e re C is
th e u n io n o f th e p o in ty cap , C 1 , an d th e e llip tical b ase C 2 . Fin d th e n e t flo w
o f th e v e cto r fie ld a cro ss C 1 .
Fo r 0  t  2  an d 0  s 
C1 :
x
1
( s , t ), y 1 ( s , t ), z 1 ( s , t ) 

2
:
C1

co s( s )(1  sin(8 s ))

  2 sin ( s ) co s( t ), sin( s ) sin ( t ),
s 
4
2

C 2 :  x 2 ( s , t ), y 2 ( s , t ), z 2 ( s , t ) 
  2 sin ( s ) cos( t ), sin( s ) sin ( t ), 0 
C2
Example 6: Using a Substitute Surface
When the Divergence is 0

Le t Fie ld( x , y , z )  z  y , z  x , x
2
.
Le t C b e th e b o u n d in g su rface o f th e so lid re g io n ,
th e u n io n o f th e cap , C 1 , an d th e e llip tic al b ase C 2 .
Fin d th e n e t flo w o f th e v e cto r fie ld acr o ss C 1 .
d iv Fie ld( x , y , z ) 
m
x
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C


n
y

p
z
 0
 d iv Fie ld d x
dy dz  0
R
S o th e n e t flo w o f th e v e cto r fie ld acro ss th e clo se d su rface C is 0 . H o w e v e r, th is
calcu latio n d o e s N O T im p ly th at th e n e t f lo w o f th e v e cto r fie ld acro ss C 1 o r C 2 is 0 .
Example 6: Using a Substitute Surface
When the Divergence is 0

Le t Fie ld( x , y , z )  z  y , z  x , x
2
.
Le t C b e th e b o u n d in g su rface o f th e so lid re g io n ,
th e u n io n o f th e cap , C 1 , an d th e e llip tic al b ase C 2 .
Fin d th e n e t flo w o f th e v e cto r fie ld acr o ss C 1 .
0 
 d iv Fie ld d x
dy dz
R

Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C

 Fie ld( x , y , z )  o u te rn o rm al d A   Fie ld( x , y , z )  o u t e rn o rm al d A
C2
So
 Fie ld( x , y , z )  o u te rn o rm al d A
C1
C1

 Fie ld( x , y , z )  o u te rn o rm al d A .
C2
S in ce C 2 is e asie r to w o rk w ith , w e 'll u se th is su b stitu te su rface in ste ad !
 Fie ld( x , y , z )  o u te rn o rm al d A
Example 3: Using a Substitute Surface
  
Fie ld( x( s , t ), y( s , t ), z( s , t ))  n o rm al( s , t ) d s d t
C2
2
0
 /2
0
C 2 :  x 2 ( s , t ), y 2 ( s , t ), z 2 ( s , t )    2 sin ( s ) co s( t ), sin( s ) sin ( t ), 0 
 x 2 y 2 z 2   x 2 y 2 z 2 
n o rm al( s , t )  
,
,
,
,



s

s

s

t

t

t

 


i
j
k
x 2
y 2
z 2
s
x 2
s
y 2
s
z 2
t
t
t
T h e se n o rm als p o in t in
j
k
th e co rre ct d ire ctio n
 2 co s( s ) co s( t )
co s( s ) sin ( t )
0
b e cau se fro m 0  s 
 2 sin ( s ) sin ( t )
sin ( s ) co s( t )
0
i
2
( 0 , 0 , 2 sin ( s ) co s( s ))
2
 0 i  0 j  ( 2 sin ( s ) c o s( s ) c o s ( t )  2 sin ( s ) c o s( s ) sin ( t ))k
 ( 0 , 0 , 2 sin ( s ) c o s( s ))
p o in ts u p o u t o f th e
e llip tical b ase .

2
,
Example 6: Using a Substitute Surface
When the Divergence is 0
Fie ld( x , y , z ) 
z  y,z  x,x 
2
C 2 :  x 2 ( s , t ), y 2 ( s , t ), z 2 ( s , t )    2 sin ( s ) co s( t ), sin( s ) sin ( t ), 0 
 Fie ld( x , y , z )  o u te rn o rm al d A
C2



2
 
0
 /2
0
2
Fie ld( x ( s , t ), y( s , t ), z( s , t ))  n o rm al( s , t ) d s d t
 /2
   sin( s )sin( t ),  2co s( t )sin( s ), 4 co s
0
0
2
 /2
 
0
0
3
2
2

( t )sin ( s )  ( 0 , 0 , 2 sin ( s ) co s( s )) d s d t
2
8 sin ( s ) c o s ( t ) c o s ( s ) d s d t
 2
T h e n e t flo w o f th e v e cto r fie ld acro ss C 1 is w ith th e d ire ctio n
o f th e n o rm al v e cto rs (d o w n to u p ).
Summary: Using a Substitute Surface When
the Divergence is 0
L e t C b e th e b o u n d in g su rfa c e o f a so lid re g io n su c h th a t C  C 1 U C 2
fo r tw o o p e n su rfa c e s C 1 a n d C 2 . L e t Fie ld (x , y ,z ) b e a v e c to r fie ld
w ith n o sin g u la ritie s c o n ta in e d w ith in C su c h th a t d iv Fie ld( x , y , z )  0
a w a y fro m sin g u la ritie s. T h e n :
 Fie ld( x , y , z )  o u te rn o rm al d A
C1

 Fie ld( x , y , z )  o u t e rn o rm al d A
C2
T h is allo w s u s to su b stitu te C 1 fo r C 2 o r v ice -v e rsa w h e n
co m p u tin g a su rface in te g ral. T rad e a cr azy su rface fo r
a sim p le r o n e !
N o te : Ju st b e cau se
 Fie ld( x , y , z )  o u te rn o rm a l d A
C
 0 , th at says n o th in g ab o u t C 1 o r C 2 .
VC.06: Flow Across When divField(x,y)=0
Le t d iv Fie ld( x , y ) 
m
x

n
y
 0 . H e re are so m e co n clu si o n s ab o u t th e n e t flo w
o f th e v e cto r fie ld acro ss v ario u s clo se d cu rv e s:
If C d o e sn 't co n tain an y sin g u laritie s, th e n
Ñ
  n( x , y )d x
 m ( x , y )d y  0 .
C
If C co n tain s a sin g u larity, th e n
Ñ
  n( x , y )d x
 m ( x , y )d y 
C
Ñ
  n( x , y )d x
 m ( x , y )d y
C1
fo r an y su b stitu te cu rv e C 1 co n tain in g th e sam e sin g u larity ( an d n o n e w e xtras).
If C co n tain s n sin g u laritie s, th e n
Ñ
  n( x , y )d x
C
 m ( x , y )d y 
Ñ
  n( x , y )d x
C1
 m ( x , y )d y  ... 
Ñ
  n( x , y )d x
Cn
fo r little circle s , C 1 , ..., C n , e n cap su latin g e ach o f th e se sin g u laritie s.
 m ( x , y )d y
VC.10: Flow Across When divField(x,y,z)=0
Let d ivField( x , y , z ) 
m
x

n
y

p
z
 0. H ere are so m e co n clu sio n s
ab o u t th e n et flo w o f th e vecto r field ac ro ss vario u s clo sed su rfaces:
If C d o e sn 't co n tain an y sin g u laritie s, th e n
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
 0.
C
If C co n tain s a sin g u larity, th e n
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A

C
Ò
 Fie ld( x , y , z )  o u t e rn o rm al d A
C1
fo r an y su b stitu te su rface C 1 co n tain in g th e sam e sin g u larity ( an d n o e x tras).
If C co n tain s n sin g u laritie s, th e n
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C

Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C1
 ... 
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
Cn
fo r little sp h e re s , C 1 , ..., C n , e n cap su latin g e ach o f th e se sin g u laritie s .
Example 7: Using a Substitute Surface With
Singularities (Details in Mathematica
Notebook)

y
x
z
Le t Fie ld( x , y , z )  
,
,
 ( x 4  y 4  z 4 )3 / 4 ( x 4  y 4  z 4 )3 / 4 ( x 4  y 4  z 4 )3 / 4

an d le t C b e th e b o u n d ary to th e re g io n p ictu re d at th e rig h t.
Fin d th e n e t flo w o f th e v e cto r fie ld acro ss C .
d ivField( x , y ) 
m
x

n
y

p
z
 0,
b u t w e h ave a sin g u larity at (0,0,0).
R ep lace th e su rface w ith a sm all
sp h ere cen tered at (0,0,0):
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C

2
 
0

0
Fie ld( x ( s , t ), y( s , t ), z( s , t ))  n o rm al( s , t ) d s d t



Example 7: Using a Substitute Surface With
Singularities (Details in Mathematica
Notebook)
Find outernorm al(s, t) :
Verify they truly are O UTERnorm als:
Example 7: Using a Substitute Surface With
Singularities (Details in Mathematica
Notebook)

y
x
z
Le t Fie ld( x , y , z )  
,
,
 ( x 4  y 4  z 4 )3 / 4 ( x 4  y 4  z 4 )3 / 4 ( x 4  y 4  z 4 )3 / 4

an d le t C b e th e b o u n d ary to th e re g io n p ictu re d at th e rig h t.
Fin d th e n e t flo w o f th e v e cto r fie ld acro ss C .
2
 
0

0
Fie ld( x ( s , t ), y ( s , t ), z( s , t ))  n o rm a l( s , t ) d s d t  1 9 .4 4 6
So th e flow of th e vector field across t h e w avy
su rface (an d th e sp h ere) is in sid e to ou tsid e.



For Monday…
• Surface area of curved surfaces
• Surface area of figures plotted on a curved
surface
• Surface integrals