VC.06
Gauss-Green Strikes Again!
(Day 1)
The Gauss-Green Formula
L e t R b e a re g io n in th e xy-p lan e w h o se b o u n d ary is p aram e te rize d
b y (x(t),y(t)) fo r t lo w  t  t h ig h . T h e n th e fo llo w in g fo rm u la h o ld s :
t h ig h
 
t low
 n m 
m  x( t ), y( t )  x '( t )  n  x( t ), y( t )  y ' ( t ) d t   

 dA
x
y 
R 

W ith th e p rop er in terp retation , w e can u se th is form u la
to h elp u s com p u te flow alon g /across m easu rem en ts!!
T h e b asic in te rp re tatio n o f G au ss-G re e n is th at it is a
c o rre sp o n d e n c e b e tw e e n a lin e in te g ral o f a c lo se d
c u rv e an d a d o u b le in te g ral o f th e in te rio r re g io n o f
th e c lo se d c u rv e . S o fo r th is to w o rk c o r re c tly, w e ju st
n e e d to m ak e su re th e v e c to r fie ld h as n o
sin g u laritie s in th e in te rio r re g io n !
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:
b
 Fie ld( x ( t ), y( t ))  ( x '( t ), y '( t ))d t
a
b

  m ( x ( t ), y( t ))x '( t )  n( x ( t ), y( 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( x , y ) d x
R
dy
Le t ro tFie ld( x , y ) 
n
x

m
y
.
Summary: 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
 C m ( x , y )d x  n( x , y )d y

 ro tFie ld( x , y ) 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 ]
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:
b
 Fie ld( x ( t ), y ( t ))  ( y '( t ),  x '( t ))d t
a
b

   n( x ( t ), y ( t )) x '( t )  m ( x ( t ), y ( t )) y '( 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( x , y ) d x
R
Le t d iv Fie ld( x , y ) 
dy
m
x

n
y
.
Summary: 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
 C  n( x , y )d x  m ( x , y )d y

 d iv Fie ld( x , y ) 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 ]
We Are Saving Singularities for
Tomorrow
Using Gauss-Green in this way is an amazing
computational tool provided our closed curves
do not encapsulate any singularities. So for
today, we’ll explore this material with no
singularities.
Tomorrow we’ll see how singularities spice things
up a bit…
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 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 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 ?
W h en d ivField (x,y)> 0 , th is is a so u rce o f n ew flu id .
A d d in g u p all so u rces w ill g ive yo u n et flo w o f th e
vecto r field acro ss th e cu rve fro m in sid e to o u tsid e.
T h is m e th o d w as n ic e b e c au se
it w as alm o st n o c o m p u tatio n !
Summary: The Divergence Locates Sources
and Sinks
Let C be a closed curve with a counterclockwise parameterization with no
singularities on the interior of the curve. Then:
If d iv Fie ld (x,y)> 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)< 0 fo r all p o in ts in C , t h 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 )  0 for all p oin ts in C, th en th e n et flow of
th e vector field across C is 0.
Summary: The Rotation Helps You Find
Clockwise/Counterclockwise Swirl
Let C be a closed curve with a counterclockwise parameterization with no
singularities on the interior of the curve. Then:
If ro tField (x,y)> 0 fo r all p o in ts in C , t h en all th ese p o in ts
ad d co u n terclo ckw ise sw irl an d th e n et f lo w o f th e vecto r
field alo n g C is fro m co u n terclo ckw ise.
If ro tFie ld (x,y)< 0 fo r all p o in ts in C , t h e n all o f th e se p o in ts
ad d clo ck w ise sw irl an d th e n e t flo w o f t h e v e cto r fie ld
acro ss C is clo ck w ise .
If ro tFie ld( x , y )  0 fo r all p o in ts in C , th e n th e se p o in ts h av e
n o sw irl, an d th e n e t flo w o f th e v e cto r fie ld alo n g C is 0
(irro tatio n al).
You Try One!

5
L e t Fie ld( x , y )  y ,  2 x

3

an d le t C b e a clo se d cu rv e g iv e n b y
2
C (t)  ( x( t ), y( t ))  sin ( t )  co s( t ), sin ( t )  co s( t )


 t  2.
2
Is th e flo w o f th e v e cto r fie ld alo n g th e cu rv e co u n te rclo ck w ise o r clo ck w ise ?
ro tFie ld( x , y ) 
Ñ
 C m ( x , y )d x  n( x , y )d y
n
x


m
y
fo r
2
 6 x  5y
 ro tFie ld( x , y ) d x
4
 0
dy  0
R
Sin ce rotField (x,y) is A LW A Y S n eg ative 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 n eg ative for an y closed cu rve.
T h at is, for A N Y closed cu rve, th e n et f low of th e vector field alon g
th e cu rve is clockw ise.
Example 2: Find the Net Flow of a Vector
Field ACROSS 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

Ñ
 C  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.
You Try One: Find the Net Flow of a Vector
Field ALONG 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 u re th e n e t flo w o f th e v e c to r fie ld
a lo n g th e c u rv e .
ro tFie ld( x , y ) 
n
x
m

Ñ
 C m ( x , y )d x  n( x , y )d y
 1  2x
y

 ro tFie ld( x , y ) d x
dy
R
4

5
  1  2x  dx
dy
1 2
 140
Positive. Th e n et flow of th e vector fie ld alon g ou r closed cu rve
is cou n terclockw ise.