17.4 Green’s Theorem
I
Notation The integral notation
is used when one wants to emphasize the curve C is closed. When the notation
C
I
is used, it is also assumed without saying that the curve C is counterclockwise oriented.
C
Theorem (Green’s Theorem)
Let C be a positively oriented(counterclockwise), 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 open region that contains D, then
Z Z I
P dx + Qdy =
C
D
∂Q ∂P
−
∂x
∂y
dA
(Remark 1) The Green’s Theorem applies to a region D with a hole(or holes) in it as well.
Suppose the region D is bounded by two circles C1 and C2 as in the left figure. Be aware that the inner
circle C2 is clockwise, not counterclockwise. By dividing the region D into D1 and D2 as in the right
figure, we can apply the Green’s Theorem on each D1 and D2 separately. In the following computation,
∂D1 and ∂D2 are boundaries of D1 and D2 with counterclockwise orientation. Then
Z Z
Z Z
Z Z
Qx − Py dA +
Qx − Py dA =
D
Qx − Py dA
D2
D1
I
=
I
P dx + Qdy +
∂D1
P dx + Qdy
∂D2
I
=
P dx + Qdy
C1 +C2
because along the shared boundaries between D1 and D2 , the two line integrals are opposite directions
and cancel each other.
(Remark 2) When F =< P (x, y), Q(x, y) >, in the next section we define the curl of F by

i
curlF = ∇ × F := det  ∂x
P
1
j
∂y
Q

k
∂z  = (∂x Q − ∂y P )k
0
Hence Green’s Theorem can be re-stated as
Z
Z Z
F · dr =
curlF · kdA
C
D
TheZGreen’s
Theorem implies that the total rotations in the region
Z
D(
curlF · kdA) is the same as sum of F along the closed
Inside D, each side shared by two adjacent
blocks has two opposite signed arrows canceling each other.
D
curve C that defines the boundary of D.
Example Evaluate the line integral using Green’s Theorem :
I
xydx + x2 dy, where C is the rectangle with vertices (0, 0), (3, 0), (3, 1), and (0, 1).
(a)
C
I
ydx − xdy, where C consists of the line segments from (0, 1) to (0, 0) and from (0, 0) to (1, 0) and
(b)
C
the parabola y = 1 − x2 from (1, 0) to (0, 1).
I
(Answer)
Z Z
2
(a)
Z
2
C
D
I
1
Z
Z
C
−2dydx = −
D
x=0
Z
1
xdydx =
x=0
1−x2
−2dA =
Z
xdA =
D
Z Z
ydx − xdy =
(b)
3
Z
(x )x − (xy)y dA =
xydx + x dy =
y=0
y=0
9
2
4
3
F · dr where F =< y 2 cos x, x2 + 2y sin x > and C is the triangle from
Example Use Green’s Theorem to evaluate
C
(0, 0) to (2, 6) to (2, 0) to (0, 0).
(Answer) Notice that the orientation of the path is clockwise (not counterclockwise). The answer is
Z
Z Z
I
−C
2
Z
3x
Qx − Py dA = −
P dx + Qdy = −
−
D
(2x + 2y cos x) − (2y cos x)dydx = 16
x=0
−yi + xj
, show that
x2 + y 2
encloses the origin.
y=0
Z
F · dr = 2π for every positively oriented simple closed path that
Example If F(x, y) =
C
−x2 + y 2
= Qx , we cannot apply the Fundamental Theorem of Line Integrals here.
(x2 + y 2 )2
Be reminded that the FTofLI applies only when F is a continuous vector field on a
open simply connected region that contains C. Any open simply connected region
that contains a curve enclosing the origin contains the origin. At origin, F is not
continuous.
(Answer) Notice that though Py =
The difficulty of the problem is in that the path is arbitrary containing the origin. To
solve this difficulty, we first show the claim :
Claim If Co is a counterclockwise oriented unit circle, then for any closed curve that encloses the origin,
I
I
F · dr =
F · dr
Co
C
(Proof of the claim) Let D be the region bounded by Co and C. The region D does not contain the origin and hence P and Q in F have continuous partial
derivatives in D.
By the Green’s Theorem,
I
F · dr
Z Z
Qx − Py dA
=
C−Co
D
Z Z
=
0dA = 0
D
I
I
F · dr +
C
F · dr
=
0
F · dr
=
−
−Co
I
I
C
I
F · dr
F · dr
=
−Co
Co
2
The proof of the claim is done.
Therefore, for any positively oriented simple closed path C, we have
I
I
F · dr =
F · dr,
Co : r(t) =< cos t, sin t >, 0 ≤ t ≤ 2π
C
Co
Z
2π
< − sin t, cos t > · < − sin t, cos t > dt
=
t=0
Z
2π
=
1dt = 2π
t=0
3