16
VECTOR CALCULUS
VECTOR CALCULUS
16.4
Green’s Theorem
In this section, we will learn about:
Green’s Theorem for various regions and
its application in evaluating a line integral.
INTRODUCTION
Green’s Theorem gives the relationship
between a line integral around a simple closed
curve C and a double integral over the plane
region D bounded by C.
 We assume that D
consists of all points
inside C as well as
all points on C.
INTRODUCTION
In stating Green’s Theorem, we use
the convention:
 The positive orientation of a simple closed curve C
refers to a single counterclockwise traversal of C.
INTRODUCTION
Thus, if C is given by the vector function r(t),
a ≤ t ≤ b, then the region D is always on
the left as the point r(t) traverses C.
GREEN’S THEOREM
Let C be a positively oriented, piecewisesmooth, simple closed curve in the plane
and let D be the region bounded by C.
 If P and Q have continuous partial derivatives
on an open region that contains D, then
 Q P 
C P dx  Q dy  D  x  y  dA
Note
NOTATIONS
The notation

C
P dx  Qdy or

C
P dx  Qdy
is sometimes used to indicate that the line
integral is calculated using the positive
orientation of the closed curve C.
Note—Equation 1
NOTATIONS
Another notation for the positively oriented
boundary curve of D is ∂D.
So, the equation in Green’s Theorem can
be written as:
 Q P 
D  x  y  dA  D P dx  Q dy
GREEN’S THEOREM
Green’s Theorem should be regarded
as the counterpart of the Fundamental
Theorem of Calculus (FTC) for double
integrals.
GREEN’S THEOREM
Compare Equation 1 with the statement of
the FTC Part 2 (FTC2), in this equation:

b
a
F '( x) dx  F (b)  F (a)
In both cases,
 There is an integral involving derivatives
(F’, ∂Q/∂x, and ∂P/∂y) on the left side.
 The right side involves the values of the original
functions (F, Q, and P) only on the boundary of
the domain.
GREEN’S THEOREM
In the one-dimensional case, the domain
is an interval [a, b] whose boundary
consists of just two points, a and b.
SIMPLE REGION
The theorem is not easy to prove in general.
Still, we can give a proof for the special case
where the region is both of type I and type II
(Section 15.3).
 Let’s call such regions simple regions.
GREEN’S TH. (SIMPLE REGION)
Proof—Eqns. 2 & 3
Notice that the theorem will be proved if
we can show that:
P
P
dx


dA
C
D y
and
Q
C Q dy  D x dA
GREEN’S TH. (SIMPLE REGION)
Proof
We prove Equation 2 by expressing D as
a type I region:
D = {(x, y) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}
where g1 and g2 are continuous functions.
GREEN’S TH. (SIMPLE REGION)
Proof—Equations 4
That enables us to compute the double
integral on the right side of Equation 2 as:
b g 2 ( x ) P
P
D y dA  a g1 ( x ) y ( x, y) dy dx
b
  [ P( x, g 2 ( x))  P( x, g1 ( x))] dx
a
where the last step follows from the FTC.
GREEN’S TH. (SIMPLE REGION)
Proof
Now, we compute the left side of Equation 2
by breaking up C as the union of the four
curves C1, C2, C3, and C4.
GREEN’S TH. (SIMPLE REGION)
Proof
On C1 we take x as the parameter and write
the parametric equations as:
x = x, y = g1(x), a ≤ x ≤ b
 Thus,

C1
P ( x, y ) dx
b
  P ( x, g1 ( x)) dx
a
GREEN’S TH. (SIMPLE REGION)
Proof
Observe that C3 goes from right to left
but –C3 goes from left to right.
GREEN’S TH. (SIMPLE REGION)
Proof
So, we can write the parametric equations
of –C3 as: x = x, y = g2(x), a ≤ x ≤ b
 Therefore,

C3
P ( x, y ) dx
 
 C3
b
P ( x, y ) dx
   P ( x, g 2 ( x)) dx
a
GREEN’S TH. (SIMPLE REGION)
Proof
On C2 or C4 (either of which might reduce to
just a single point), x is constant.
 So, dx = 0
and

C2
P ( x, y ) dx
0
  P ( x, y ) dx
C4
GREEN’S TH. (SIMPLE REGION)
Proof
Hence,

C
P( x, y ) dx
  P( x, y ) dx   P( x, y ) dx   P ( x, y ) dx
C1
C3
C2
  P( x, y ) dx
C4
b
b
a
a
  P( x, g1 ( x)) dx   P ( x, g 2 ( x)) dx
GREEN’S TH. (SIMPLE REGION)
Proof
Comparing this expression with the one
in Equation 4, we see that:
P
C P( x, y) dx   D y dA
GREEN’S TH. (SIMPLE REGION)
Proof
Equation 3 can be proved in much the same
way by expressing D as a type II region.
 Then, by adding Equations 2 and 3,
we obtain Green’s Theorem.
 See Exercise 28.
GREEN’S THEOREM
Evaluate

C
Example 1
x dx  xy dy
4
where C is the triangular curve
consisting of the line segments
from (0, 0) to (1, 0)
from (1, 0) to (0, 1)
from (0, 1) to (0, 0)
GREEN’S THEOREM
Example 1
The given line integral could be evaluated
as usual by the methods of Section 16.2.
 However, that would involve setting up
three separate integrals along the three
sides of the triangle.
 So, let’s use Green’s Theorem instead.
GREEN’S TH. (SIMPLE REGION)
Example 1
Notice that the region D enclosed by C
is simple and C has positive orientation.
GREEN’S TH. (SIMPLE REGION)
Example 1
If we let P(x, y) = x4 and Q(x, y) = xy,
then
 Q P 
C x dx  xy dy  D  x  y  dA
4

1 1 x

0 0
1
( y  0) dy dx
 [ y ]
1
0 2

1
2
1
2 y 1 x
y 0
 (1  x)
0
2
dx
dx
3 1
  (1  x)  
0
1
6
1
6
GREEN’S THEOREM
Example 2
Evaluate
—
 (3y  e
sin x
C


) dx  7x  y 1 dy
where C is the circle x2 + y2 = 9.
 The region D bounded by C is
the disk x2 + y2 ≤ 9.
4
GREEN’S THEOREM
Example 2
 So, let’s change to polar coordinates after
applying Green’s Theorem:


sin x
4
(3y

e
)
dx

7x

y
 1 dy
—

C




4
sin x 
  
7x  y  1  (3y  e )  dA
x
x

D 

2
0

3
0
(7  3) r dr d
2
3
0
0
 4  d  r dr  36
GREEN’S THEOREM
In Examples 1 and 2, we found that
the double integral was easier to evaluate
than the line integral.
 Try setting up the line integral in Example 2
and you’ll soon be convinced!
REVERSE DIRECTION
Sometimes, though, it’s easier to evaluate
the line integral, and Green’s Theorem is
used in the reverse direction.
 For instance, if it is known that P(x, y) = Q(x, y) = 0
on the curve C, the theorem gives:
 Q P 
D  x  y  dA  C P dx  Q dy  0
no matter what values P and Q assume in D.
REVERSE DIRECTION
Another application of the reverse direction
of the theorem is in computing areas.
As the area of D is

D
P and Q so that:
1 dA, we wish to choose
Q P

1
x y
REVERSE DIRECTION
There are several possibilities:






P(x, y) = 0
P(x, y) = –y
P(x, y) = –½y
Q(x, y) = x
Q(x, y) = 0
Q(x, y) = ½x
Equation 5
REVERSE DIRECTION
Then, Green’s Theorem gives the following
formulas for the area of D:
A —
 x dy
C
 —
y
dx

C

1
2
—
 x dy  y dx
c
REVERSE DIRECTION
Example 3
Find the area enclosed by the ellipse
2
2
x
y


1
2
2
a
b
 The ellipse has parametric equations
x = a cos t, y = b sin t,
0 ≤ t ≤ 2π
REVERSE DIRECTION
Example 3
 Using the third formula in Equation 5,
we have:
A


1
2
1
2


 (a cos t )(b cos t ) dt  (b sin t )(a sin t ) dt
C
x dy  y dx
2
0
ab 2

dt   ab

2 0
UNION OF SIMPLE REGIONS
We have proved Green’s Theorem only
for the case where D is simple.
Still, we can now extend it to the case
where D is a finite union of simple regions.
UNION OF SIMPLE REGIONS
For example, if D is the region shown here,
we can write:
D = D1
where
D1 and D2
are both
simple.
D2
UNION OF SIMPLE REGIONS
The boundary of D1 is C1
C 3.
The boundary of D2 is C2
(–C3).
UNION OF SIMPLE REGIONS
So, applying Green’s Theorem to D1 and D2
separately, we get:

 Q P 
dA

P dx  Q dy   

C2
x y 
D1 

 Q P 
dA

P dx  Q dy   

(  C3 )
x y 
D2 
C1
C2
UNION OF SIMPLE REGIONS
If we add these two equations, the line
integrals along C3 and –C3 cancel.
So, we get:

C1
 Q P 
P dx  Q dy   

dA

C2
x y 
D 
 Its boundary is C = C1 C2 .
 Thus, this is Green’s Theorem for D = D1
D2.
UNION OF NONOVERLAPPING SIMPLE REGIONS
The same sort of argument allows us
to establish Green’s Theorem for any finite
union of nonoverlapping simple regions.
UNION OF SIMPLE REGIONS
Evaluate
Example 4
y
dx

3xy
dy
—

2
C
where C is the boundary of the semiannular
region D in the upper half-plane between
the circles x2 + y2 = 1 and x2 + y2 = 4.
UNION OF SIMPLE REGIONS
Example 4
Notice that, though D is not simple, the
y-axis divides it into two simple regions.
 In polar coordinates,
we can write:
D
= {(r, θ) | 1 ≤ r ≤ 2,
0 ≤ θ ≤π}
Example 4
UNION OF SIMPLE REGIONS
So, Green’s Theorem gives:

 2 
—
C y dx  3xy dy  D  x (3xy)  y ( y )  dA
2
  y dA
D


0

2
1
(r sin  ) r dr d

2
  sin  d  r dr
0
2
1
14
 [ cos ]0  r  
1
3

1
3
3
2
REGIONS WITH HOLES
Green’s Theorem can be extended
to regions with holes—that is, regions
that are not simply-connected.
REGIONS WITH HOLES
Observe that the boundary C of the region D
here consists of two simple closed curves
C1 and C2.
REGIONS WITH HOLES
We assume that these boundary curves are
oriented so that the region D is always on
the left as the curve C is traversed.
 So, the positive
direction is
counterclockwise
for C1 but
clockwise for C2.
REGIONS WITH HOLES
Let’s divide D into two regions D’ and D”
by means of the lines shown here.
REGIONS WITH HOLES
Then, applying Green’s Theorem to each
of D’ and D” , we get:
 Q P 
 Q P 
 Q P 
D  x  y  dA  D '  x  y  dA  D"  x  y  dA
  P dx  Q dy  
D '
D
P dx  Q dy
 As the line integrals along the common boundary
lines are in opposite directions, they cancel.
REGIONS WITH HOLES
Thus, we get:
 Q P 
D  x  y  dA  C1 P dx  Q dy  C2 P dx  Q dy
  P dx  Q dy
C
 This is Green’s Theorem for the region D.
REGIONS WITH HOLES
If
show that
Example 5
F(x, y) = (–y i + x j)/(x2 + y2)
∫C F · dr = 2π
for every positively oriented, simple closed
path that encloses the origin.
REGIONS WITH HOLES
Example 5
C is an arbitrary closed path that encloses
the origin.
Thus, it’s difficult to compute the given
integral directly.
REGIONS WITH HOLES
Example 5
So, let’s consider a counterclockwise-oriented
circle C’ with center the origin and radius a,
where a is chosen to be small enough that C’
lies inside C.
REGIONS WITH HOLES
Example 5
Let D be the region bounded by C
and C’.
 Then, its positively oriented boundary is C
(–C’).
Example 5
REGIONS WITH HOLES
So, the general version of Green’s Theorem
gives:

C
P dx  Q dy  
C '
P dx  Q dy
 Q P 
  

 dA
x y 
D 
2
2
 y2  x2
y x 
   2
 2
dA  0
2 2
2 2
(x  y ) (x  y ) 
D 
Example 5
REGIONS WITH HOLES
Therefore,

C
That is,

C
P dx  Q dy   P dx  Q dy
C'
F  dr   F  dr
C'
 We now easily compute this last integral
using the parametrization given by:
r(t) = a cos t i + a sin t j,
0 ≤ t ≤ 2π
REGIONS WITH HOLES
Example 5
Thus,

C
F  dr
  F  dr
C'
2
  F(r (t ))  r '(t ) dt
0

2
0
2
(a sin t )(a sin t )  (a cos t )(a cos t )
dt
2
2
2
2
a cos t  a sin t
  dt  2
0
GREEN’S THEOREM
We end by using Green’s Theorem
to discuss a result that was stated in
Section 16.3
THEOREM 6 IN SECTION 16.3
Proof
We’re assuming that:
 F = P i + Q j is a vector field on an open
simply-connected region D.
 P and Q have continuous first-order
partial derivatives.
P Q

throughout D

y x
THEOREM 6 IN SECTION 16.3
Proof
If C is any simple closed path in D and R is
the region that C encloses, Green’s Theorem
gives:
—
 F  dr  —
 P dx  Q dy
C
C
 Q P 
  
  dA
x y 
R 
  0 dA  0
R
THEOREM 6 IN SECTION 16.3
Proof
A curve that is not simple crosses itself at
one or more points and can be broken up
into a number of simple curves.
 We have shown that the line integrals of F
around these simple curves are all 0.
 Adding these integrals, we see that ∫C F · dr = 0
for any closed curve C.
THEOREM 6 IN SECTION 16.3
Proof
Thus, ∫C F · dr is independent of path in D
by Theorem 3 in Section 16.3.
It follows that F is a conservative vector field.