Section 14.4: Green’s Theorem
Definition: A curve that does not intersect itself anywhere between its endpoints is called
a simple curve. The positive orientation of a simple closed curve C refers to a single
counterclockwise traversal of C.
Theorem: (Green’s Theorem)
Let C be a positively oriented, piecewise-smooth, 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
I
ZZ ∂Q ∂P
−
dA.
P dx + Qdy =
∂x
∂y
C
D
Note: Green’s Theorem gives a relationship between a line integral around a simple closed
curve C and a double integral over the plane region D bounded by C.
I
Example: Evaluate the line integral
xy 2 dx + 4x2 ydy, where C is the triangle with vertices
C
(0, 0), (2, 2), and (2, 4) with positive orientation.
Let P (x, y) = xy 2 and Q(x, y) = 4x2 y. By Green’s Theorem
ZZ I
∂Q ∂P
2
2
xy dx + 4x ydy =
−
dA
∂x
∂y
D
C
Z 2 Z 2x
=
(8xy − 2xy)dydx
0
x
Z 2 Z 2x
=
6xydydx
0
x
2x
Z 2
2
3xy dx
=
0
x
Z 2
=
9x3 dx
0
2
9 4 =
x
4 0
= 36.
I
−2y 3 dx + 2x3 dy, where C is the circle x2 + y 2 = 4.
Example: Evaluate the line integral
C
Let P (x, y) = −2y 3 and Q(x, y) = 2x3 . By Green’s Theorem
I
ZZ ∂Q ∂P
3
3
−
dA
−2y dx + 2x dy =
∂x
∂y
C
D
ZZ
=
6x2 + 6y 2 dA
Z 2π Z 2
=
6r3 drdθ
0
0
2
3 4 r = 2π
2
0
= 2π(24)
= 48π.
Example: Find the work done by the force F~ (x, y) = h7x, x3 + 3xy 2 i in moving a particle
√
from the point (−2, 0) along the x-axis to (2, 0), and then along the semicircle y = 4 − x2
to the starting point.
The work done is given by
I
W =
~
F~ · dR.
C
Let P (x, y) = 7x and Q(x, y) = x3 + 3xy 2 . By Green’s Theorem,
ZZ
(3x2 + 3y 2 )dA
W =
D
Z 2π
Z 2
=
3r3 drdθ
0
Z 20
= 2π
3r3 dr
0
2
3π 4 =
r
2 0
= 24π.
Note: Note that although the object is moved back to the starting point, the work done is
non-zero since the force field is not conservative.
Example: Find the work done by the force F~ (x, y) = hx2 + xy, xy 2 i in moving a particle from
(0, 0) along the x-axis to (1, 0), then along the line segment to (0, 1), and then back to the
origin along the y-axis.
The work done is given by
I
W =
~
F~ · dR.
C
2
2
Let P (x, y) = x + xy and Q(x, y) = xy . By Green’s Theorem,
ZZ
W =
(y 2 − x)dA
Z 1DZ 1−x
(y 2 − x)dydx
=
0
0
1−x
Z 1
1 3
=
dx
y − xy 0 3
0
Z 1
1
=
(1 − x)3 − x(1 − x)dx
3
Z0 1
=
(1 − 4x + 4x2 − x3 )dx
0
1
4 3 1 4 2
= x − 2x + x − x 3
4 0
1
.
=
12