MAC 2313
Green’s Theorem
Cal III
Sanchez
Summary 4-Green’s Theorem
Green’s theorem: Let C be a piecewise smooth simple closed curve that bounds
the region R in the plane. Suppose that the functions P(x,y) and Q(x,y) are
continuous and have continuous first-order partial derivatives on R. Then
 Q P 
Pdx

Qdy

 dA

 
 x y 
C
Proof:
1. Let R be represented by R  ( x, y) | f1 ( x)  y  f 2 ( x), a  x  b
and also by R  ( x, y) | g1 ( y)  x  g 2 ( y), c  y  d 
b
a
a
b
b
a
2.  P( x, y )dx   P( x, f1 ( x))dx   P( x, f 2 ( x)) dx
C
  P( x, f1 ( x)) dx   P( x, f 2 ( x))dx
a
a
b
y f2 ( x)
a
y  f1 ( X )
   P ( x, y )
  
R
d
b  y f2 ( x)
  
a


f1 ( X )
P( x, y ) 
dydx
y

P
dA
y
c
3.  Q( x, y )dy   Q( g 2 ( y ), y )dy   Q( g1 ( y )) dy
C
c
d
d
d
c
c
  Q( g 2 ( y ), y )dy   Q( g1 ( y )) dy
d
x g2 ( y )
c
x  g1 ( y )
  Q ( x, y )
 
R
d  y g2 ( y)
 
c


g1 ( y )
Q( x, y ) 
dxdy
x

Q
dA
x
 Q P 
Q
P
dA
dA  
dA   


x

y

x

y


C
R
R
R
Green’s Theorem, expresses a double integral over a region as a line integral along
the boundary of the region. This has many important consequences.
4.  P( x, y )dx  Q( x, y )dy  
-1-
Problem 1. Use Green’s theorem to evaluate the integral
 Pdx  Qdy where P(x,
C
y)=x-y and Q(x, y) =y, C is the boundary of the region between the x-axis and the
graph of y=sinx for 0  x  
Solution:
Y
Sin x
1

 sin x
 Pdx  Qdy   ( x  y)dx  ydy   (0  1)dA    dydx
C
C
R

0 0

  sin xdx   cos x 0  cos  cos 0  2
0
Problem 2. Use Green’s Theorem to evaluate the integral
 Pdx  Qdy if
C
P( x, y )  sin x  y and Q( x, y )  ln(1  y ), C is the circle x 2  y 2  9.
2
 Pdx  Qdy   sin
C



x  y dx  ln(1  y 2 ) dy   [0  1]dydx  
C
R

dydx  9
x  y 9
2
2
Corollary to Green’s Theorem. The area A of the
region R bounded by the piecewise smooth simple closed curve C is given by
1
A= C ydx  xdy   C ydx  C xdy
2
Proof:
1
1
1)   ydx  xdy    1  1}dA   dA  A
2C
2R
R
r
2)   ydx    ydx  0dy    0  1dA   dA  A
C
C
R
R
3)  xdy   0dx  xdy   1  0dA  A
C
C
R
-2-
Problem 3. Use the corollary to Green’s theorem to find the area of the region
between the x-axis and one arch of the cycloid with parametric equations
x=5(t-sint) and y=5(1-cost)
Y
Cycloid
C2
2
C1
2
0
0
2
A    ydx    ydx    ydx    odx   5(1  cos t )(5)(1  cos t )
C
C1
2
C2
2

2

1  cos 2t 

 25  1  cos t  dt  25  1  2 cos t  cos t dt  25  1  2 cos t 
dt
2


0
0
0
2
2
2
1
3
3

 25  t  2 sin t  sin 2t   25   2  75
4
2
2
0
Problem 4. Use the corollary to Green’s theorem to find the area of the region between the
yx
2
and
yx
3
Y
y  x2
y  x3
x
Area 
1
 xdy   x(3x
C C1C 2

3x 4 2 x

4
3
0
1
3
0

2
0
1
1
0
1
)dx   x(2 x)dx   3x dx   2 x 2 dx
3 2 1
 
4 3 12
-3-
3
0
Problem 5. Find the area bounded by one loop of the curve x=sin2t, y=sint
Use a calculator to graph the parametric curve x=sin2t, y=sint
Note:
x  sin 2t  x  2 sin t cost  x 2  4 sin 2 t cos2 t  x 2  4(sin 2 t )(1  sin 2 t )
 x 2  4 y 2 (1  y 2 )
For the y  int ercepts, x  0  4 y 2 (1  y 2 )  0  y  0 y  1 or y  1
Y
1
x
-1


2 cos3 t
Area   xdy   2 sin t cost costdt   2 cos t sin tdt  
3
C
0
0



2

0
2
4
cos3   cos3 0 
3
3
F  5 x y i  (7 x y ) j
2
3
3
2
Problem 6. Find the work done by the force
In moving a particle once around the triangle with vertices (0, 0), (3, 0) and ((0, 6)
6
y
3
x


Work   Pdx  Qdy   5 x 2 y 3 dx  7 x 3 y 2 dy   21x 2 y 2  15 x 2 y 2 dA
C
3 2 x 6

 6x
0
0
C
2
R
3
2 3 2 x6
y dydx   2 x y
2
0
0
3
dx   2 x 2  2 x  6 3 dx 
0
-4-
972
6
Vector form of Green’s Theorem or divergence theorem in the plane.
Let R be a bounded region with complete boundary C oriented in positive sense, and suppose F
 F  nds   divFdA
is a vector field defined on R and its boundary. Then
R
Proof:
dx dy
dx dy
i

j
 dt
 dt i  dt j
 dx dy
dt
1) T 
T 
T  i
j

ds
v
ds
ds
dt
 dy dx
2) Let N  i 
j
ds
ds
  dx dy dy dx
 

3) T  N 
  
 0, therefore T  N and N is normal to C
ds ds ds ds
 
dy
dx
 dy dx 
4) F  N  Pi  Qj    i 
j  P   Q
ds 
ds
ds
 ds
dx 
 dy
5)  F  nds    P   Q  ds   Pdy  Qdx
ds
ds 
C
C
C
6)  F  nds    Qdx  Pdy
C
C
Green 's Theorem

 P
Q 
R


  divF dA
R
Exercise 7. Verify the divergence theorem for R the interior of
And F=2xi+3yj
1) divF 
 P
  x   y dA    x 
R

Q 
dA
y 
x  y 1
2
2
P Q

 2  3  5   F  nds  divFdA   5dA  5
x y
R
Circle
 x  cos t
2) Let C : x 2  y 2  1 or 
, 0  t  2 , P  2 x, Q  3 y
 y  sin t
2
 F  nds   Pdy  Qdx   2 cost costdt  3 sin t  ( sin tdt )
C

C
2
 (2 cos
0
2
t 3 sin t )dt 
2
0
2
 2  sin
0
2

t dt 
2
0
2
1
5
5

  t  sin 2t    2  5
4
2
2
0
-5-

  2 
1  cos 2t 
dt
2

Exercise 8. Use the divergence theorem to evaluate the line integral
 (4 x  y)dx  (3x  2 y)dy ; C
the complete boundary of the annulus
C
1  x  y  4 in the positive sense.
2
2
P
4
 P  4x  y
P Q
x


divF


2

Q
x y
Q  3x  2 y
 2
y
2 2
2
2
2
P Q
4
(4 x  y )dx  (3x  2 y )dy 

dA 
2rdrd  r d  3d  6
x y
1




C
R
0 1
0


Definition: flux is the rate of flow of the fluid across C in the direction of
vector n
Exercise 9. Use Green’s theorem to calculate the outward flux = C
2
Where F=2xi+3yj and C is the ellipse
F  nds
2
x y
 1
9 4
3


P Q
flux   F  nds   divFdA  

dxdy  4

x

y
C
R
R
0
2 1
x2
9
 2  3dxdy
0
 /2
x2
x
 20  2 1  dx  40  1  sin 2  3 cos d , sin  
9
3
0
0
3
 /2
 120  cos d  120
0
2
 /2

0
 /2
1  cos 2
1
1

d  120    sin 2 
2
4
2
0
-6-
 30