Green’s Theorem — Calculus III (MATH
2203)
S. F. Ellermeyer
November 2, 2013
Green’s Theorem gives an equality between the line integral of a vector
field (either a flow integral or a flux integral) around a simple closed curve,
, and the double integral of a function over the region, , enclosed by the
curve. A simple closed curve is a loop which does not intersect itself (as
pictured below).
Theorem 1 (Green’s Theorem) Suppose that  is a simple closed piecewise smooth curve in 2 and that  is the region enclosed by . Suppose
1
also that  is oriented counterclockwise — meaning that  is parameterized
in such a way that it is traced out counterclockwise by the parameterization.
Also suppose that
F ( ) =  ( ) i +  ( ) j
is a vector field such that  and  have continuous first order partial derivatives throughout  and on its boundary curve, . Then
¶
Z
ZZ µ


−
.
F · r =




Example 2 Let  be the unit circle oriented counterclockwise and let
F ( ) = −i + j.
We will verify that Green’s Theorem holds in this case.
First we will compute the flow of F around the boundary of  (oriented
counterclockwise). To do this we parameterize  in the usual way by
r () = cos () i + sin () j
0 ≤  ≤ 2.
Thus
r0 () = − sin () i + cos () j
and
This gives
F ( ()   ()) = − sin () i + cos () j.
Z

F · r =
Z
0
2
¢
¡ 2
sin () + cos2 ()  = 2.
Next we will compute the double integral on the right hand side of Green’s
Theorem using  = − and  = . We obtain
¶
ZZ µ
ZZ


(1 − (−1) 
−
 =




ZZ
1 
=2

= 2 × (Area of the unit circle)
= 2.
2
This shows that
Z

F · r =
ZZ µ


−



¶
 = 2.
Remark 3 We can also write the conclusion of Green’s Theorem as
¶
Z
ZZ µ


(  +  ) =
−





and this form is sometimes more convenient to use (as in the following example).
Example 4 Let  be the triangle with vertices at (0 0), (1 0), and (1 1)
oriented counterclockwise and let
F ( ) = −i + j.
We will verify that Green’s Theorem holds here.
3
To compute the line integral, we note that  = −1 and  = . The
curve  (broken down into three pieces) is pictured. On 1 we have
=
=0
 =   = 0
which gives us
Z
(  +  ) =
1
On 2 we have
Z
1
−1  = −1.
0
=1 =
 = 0  = 
which gives us
Z
(  +  ) =
2
Z
0
1
1
(1) ()  = .
2
On 3 we have
=
=
 =   = 
which gives us
Z
(  +  ) =
3
Z
(−1  +  )

Z 03
¡
¢
−1  + 2 
1
¶¯0
µ
1 3 ¯¯
= − +  ¯
3
¶1
µ
1
= 0 − −1 +
3
2
= .
3
=
We conclude that
Z

(  +  ) = −1 +
4
1 2
1
+ = .
2 3
6
Next we evaluate the double integral and see that we get the same result:
¶
ZZ µ
Z 1Z 


( − 0)  
−
 =


0
0

¶¯
Z 1µ
1 2 ¯¯
=
 ¯ 
2
0
0
Z 1
1 2
 
=
0 2
1
= .
6
Below is a picture of the triangle shown together with the vector field F ( ) =
−i + j.
Example 5 As another example, we use Green’s Theorem to evaluate
Z
¡ 3
¢
  − 3 

where  is the positively—oriented circle of radius 2.
5
Note that the region  is the interior of the disk of radius 2 centered at the
origin and that the functions  =  3 and  = −3 have partial derivatives
that are continuous on . By Green’s Theorem,
¶
ZZ µ
Z
¢
¡ 3


3
−

  −   =



ZZ
¡
¢
−32 − 3 2 
=

ZZ
¡ 2
¢
 +  2 .
= −3

In order to evaluate this double integral, it will be convenient to use polar
coordinates.
ZZ
Z 2 Z 2
¡ 2
¢
2
−3
2   
 +   = −3
0

= −3
= −3
Z
0
2
0
Z
2
0
¯=2
1 4 ¯¯


4 ¯=0
4 
= −24.
Therefore
Z

¢
¡ 3
  − 3  = −24.
Exercise 6 For  as given in the previous example, show directly (without
using Green’s Theorem) that
Z
¢
¡ 3
  − 3  = −24.

You will see that it is easier to use Green’s Theorem than to evaluate the line
integral directly.
Example 7 In this example we will evaluate the flux integral of
F ( ) = 2 i +  2 j
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around the (pictured) positively oriented curve that consists of the upper half
of the circle of radius 2 centered at the origin followed by the line segment
connecting the point (−2 0) to the point (2 0).
Since we are computing the flux of F across , then the line integral we
want to compute is
Z
¢
¡ 2
  −  2  .

This problem was given on Practice Exam 7. We can do it directly but it
is easier to use Green’s Theorem and evaluate the double integral instead.
Notice that since this is a flux integral of F, we need to be careful to apply
Green’s Theorem correctly. In particular we first rewrite the line integral we
want to compute as
Z
Z
¢
¢
¡ 2
¡ 2
2
  −   =
−  + 2 


so, in applying Green’s Theorem, we need to take  = − 2 and  = 2 . We
7
obtain
Z

¢
¡ 2
−  + 2  =
=
ZZ µ
ZZ
=2

0
Z
¶

(2 + 2) 

Z


−


Z
2
( cos () +  sin ())   
0

Z
2
2 (cos () + sin ())  
µZ 2
¶ µZ 
¶
2
=2
 
(cos () + sin ()) 
0
0
µ ¶
8
(sin () − cos ())|=
=2
=0
3
16
((0 − (−1)) − (0 − 1))
=
3
32
= .
3
Remark 8 The preceding example suggests how we can write Green’s Theorem in another form. The originally stated form of the theorem, called the
flow form, is
¶
Z
ZZ µ
Z


−
.
F · r =
(  +  ) =




=2
0
0

For a flux integral, we have
Z
Z
Z
F · n  =
(  −  ) =
(−  +  )



and Green’s Theorem, written in flux form, is thus
¶
Z
Z
ZZ µ

 (−)
F · n  =
(−  +  ) =
−






or
Z

F · n  =
Z

(  −  ) =
ZZ µ

8
 
+


¶
.
1
Application (Finding Area Using Green’s
Theorem)
We know that if  ⊆ 2 , then
Area of  =
ZZ
1 .

If we take  and  to be the functions
1
 ( ) = − 
2
1
 ( ) = ,
2
then

1
=

2
1

=− .

2
Thus
Area of  =
ZZ µ


−



¶

and by Green’s Theorem
Z
1
Area of  =
(  +  ) =
2

Z

(  −  )
where  is the boundary curve of . This remarkable formula allows us
to compute the area of a region using a line integral rather than a double
integral.
Example 9 We will use the formula
1
Area of  =
2
Z

(  −  )
to compute the area of the ellipse
2  2
+ 2 = 1.
2

9
To do this first note that the positively—oriented boundary of the ellipse is
the curve  given by
r () =  () i +  () j =  cos () i +  sin () j
0 ≤  ≤ 2.
For this curve we have
r () =


i + j = −  sin () i +  cos () j.


The area of the ellipse is thus
Z
1
Area of  =
(  −  )
2 
¶
Z µ
1 2


=
 ()
−  ()

2 0


Z
1 2
( cos ()  cos () −  sin () (− sin ())) 
=
2 0
Z 2
1
1 
= 
2
0
= .
Example 10 Suppose that we have a polygon with vertices at the points
(0  0 ), (1  1 ), (2  2 ),. . . ,(   ) where (   ) = (0  0 ) as pictured.
10
The area of the region, , enclosed by the polygon is
Z
1
Area of  =
(  −  )
2 
where  is the union of the line segments that make up the sides of . Thus
 Z
1X
(  −  ) .
Area of  =
2 =1 
R
In order to compute 1 (  −  ), we note that 1 is the line segment
joining the point (0  0 ) to the point (1  1 ). We can parameterize this
segment as
 = 0 + (1 − 0 ) 
 = 0 + (1 − 0 ) 
0≤≤1
and thus

= 1 − 0


= 1 − 0 .

11
This gives us
¶
Z
Z 1µ


 ()
(  −  ) =
−  ()



0
1
Z 1
=
((0 + (1 − 0 ) ) (1 − 0 ) − (0 + (1 − 0 ) ) (1 − 0 )) 
0
Z 1
=
(0 (1 − 0 ) − 0 (1 − 0 )) 
0
Z 1
=
(0 1 − 1 0 ) 
0
= 0 1 − 1 0 .
By similar reasoning we obtain
Z
(  −  ) = −1  −  −1

for all  = 1 2     . Therefore
1X
(−1  −  −1 ) .
Area of  =
2 =1

This formula allows us to compute the area of the region enclosed by a polygon
using only the coordinates of the vertices of the polygon.
Example 11 Let us find the area of the region enclosed by the polygon pictured here.
12
Using the formula derived in the previous example we obtain
1X
Area of  =
(−1  −  −1 )
2 =1
5
1
= [(0) (0) − (1) (0)
2
+ (1) (4) − (3) (0)
+ (3) (4) − (4) (4)
+ (4) (6) − (3) (4)
+ (3) (0) − (0) (6)]
1
= [0 + 4 − 4 + 12 + 0]
2
= 6.
Exercise 12 Find the area of the region enclosed by the polygon pictured
here.
13
2
Proof of Green’s Theorem
Rather than prove Green’s Theorem in its greatest generality, we will just
prove the theorem in the special case that  is the positively—oriented circle
of radius   0 centered at some point ( ) and the vector field
F ( ) =  ( ) i +  ( ) j
is such that  and  have continuous first partial derivatives on  and in
the enclosed region . We will actually prove that
ZZ
Z


  = −



and
Z

  =
ZZ

.


Adding these two equations together then gives Green’s Theorem.
To prove the first fact, we note that the circle, , has equation
( − )2 + ( − )2 = 2 .
14
The upper half of the circle has equation
q
 =  + 2 − ( − )2
and the lower half of the circle has equation
q
 =  − 2 − ( − )2 .
Therefore
¶
¶
µ
Z 
Z − µ
Z
q
q
2
2
2
2
  =
   +  − ( − )  +
   −  − ( − ) 

−

¶
µ
¶¶
Z µ µ
q
q
2
2
=
   − 2 − ( − ) −    + 2 − ( − )

−
and
ZZ
Z  Z +√2 −(−)2


−
 = −
 
√


− − 2 −(−)2

¶
µ
¶¶
Z µ µ
q
q
2
2
2
2
=−
   +  − ( − ) −    −  − ( − )

−
¶
µ
¶¶
Z µ µ
q
q
2
2
=
   − 2 − ( − ) −    + 2 − ( − )
.
−
This shows that
ZZ


  = −


Z

as promised.
To prove the second fact, we note that the right half of the circle has
equation
q
 =  + 2 − ( − )2
and the left half of the circle has equation
q
 =  − 2 − ( − )2 .
15
Therefore
¶
¶
Z  µ
Z − µ
Z
q
q
2
2
2
2
  =
  +  − ( − )    +
  −  − ( − )   
−


¶
µ
¶¶
Z µ µ
q
q
2
2
2
2
=
  +  − ( − )   −   −  − ( − )  

−
and
ZZ

 =


=
Z

−
Z

−
This shows that
Z
√
+
√
−
2 −(−)2
2 −(−)2

 

µ µ
¶
µ
¶¶
q
q
2
2
2
2
  +  − ( − )   −   −  − ( − )  

Z
  =

We conclude that
Z

(  +  ) =
ZZ

.


ZZ µ



−


¶
.
Although the above proof only proves Green’s Theorem in a special case, it
hints at the general idea of the proof.
Exercise 13 Prove Green’s Theorem in the special case that  is the rectangle  = [ ] × [ ].
16