Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 6.4 Green’s Theorem (13.4)
We now discover the connection between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. Def: A simple closed curve is said to be positive oriented if it traversed counterclockwise. Green’s Thm: Let C be 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 ∂Q
∫ P dx + Q dy = ∫∫ ∂ x − ∂P∂ y dA C
D
Note: The circle on the line integral is sometime related to the positive oriented curve and sometime even drawn with an arrow on the circle. One views the Green’s theorem as a counterpart of Fundamental Theorem of Calculus ( ∫ F '( x ) dx = F (b) − F (a)) for double integral, i.e. ∫∫
b
a
∂Q
∂x
∫ P dx + Q dy . − ∂P
∂ y dA =
∂D
D
Proof: One shows the part of proof for 
∫ P dx = − ∫∫ ∂P∂ y dA by expressing the domain as a domain C
D
∂Q
of type I. The proof for the second part (not shown here), 
∫ Q dy = ∫∫ ∂ x dA , is similar but the C
D
domain should be expressed as a domain of type II. Thus, let D = {( x, y ) : a ≤ x ≤ b, g1 ( x ) ≤ y ≤ g2 ( x )} . From one side we have b g1( x )
− ∫∫ ∂P
∂ y dA = ∫
∫
a g2 ( x )
D
C :y=g (x) C C :y=g (x) 3
∂P
∂y
2
4
b
dy dx = ∫ P ( x, g2 ( x )) − P ( x, g1 ( x )) dx 1
1
a
From the other side, let C = C1 ∪ C2 ∪ C3 ∪ C4 as described in the figure, thus ∫ P dx = ∫
∫
P dx +
x,g1( x )
C
b
∫
P dx +
b,g1( b ) (1−t )+ b,g2 ( b ) t
b
a
a
b
a
b
P dx +
a,g2 ( a ) ( t−1)+ a,g1( a ) t
∫
P dx =
x,g2 ( x )
b
b
∫ P ( x, g ( x )) dx + ∫ P dx + ∫ P dx + ∫ P ( x, g ( x )) dx = ∫ P ( x, g ( x )) dx − ∫ P ( x, g ( x )) dx
1
2
a
Ex 4.
Evaluate
∫
∂D
∂D
2
2
(
2
0
)
(
= ∫ −4 y + 6 y dy = −2 y + 2 y
0
a
(x 2 − xy 3 ) dx + ( y 2 − 2xy) dy = where D is a square [ 0,2 ] × [ 0,2 ]
− xy 3 ) dx + ( y 2 − 2xy)dy = ∫
2
2
a
Q = ( y 2 − 2xy) ⇒ ∂Q
= −2 y ∂x
P = (x 2 − xy 3 ) ⇒ ∂∂ Py = −3xy 2
∫ (x
1
2
∫ ( −2 y + 3xy ) dx dy = ∫
2
2
0
3 y=2
y=0
2
0
) = −8 + 16 = 8
81 (−2xy +
3
2
x2 y2
x=2
x=0
) dy =
C 2
Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky One can verify it using 2
3
2
∫ (x − xy ) dx + ( y − 2xy)dy = ∫ + ∫ − ∫ − ∫ = 83 + ( −5 13 ) − 83 − ( −13 13 ) = 8 , where C j ’s are lines C1
C
C2
C3
C4
along the boundary starting from x-­‐axis and moving counterclockwise. 6.4.1 Extentions of Green’s Theorem

Let C be r ( t ) = t, t − t 2 , evaluate
Ex 5.
∫ (e
x
sin y − y 2 + x ) dx + ( e x cos y − cos(y 2 )) dy
C
The curve C isn’t a closed curve, since it is a curve y = x − x 2 therefore Green’s theorem cannot be used straightforward. Therefore we create another, closed curve C1 = C ∪ C2 for which Green’s theorem is defined. From the other side ∫ = ∫ + ∫ . and therefore ∫ = ∫ − ∫ . . The curve C2 C1
C
C2
C
C1
C2
can be arbitrary path connecting endpoints of C, but it would be wise to define it so that the integration of ∫ . is easy. In our case C is defined for x − x 2 > 0 ⇒ 0 ≤ x ≤ 1 therefore we rewrite it C2
as x + y = x or in polar coordinates r = cosθ ,0 ≤ θ ≤ π / 2 , which is a half circle. The easiest choice of curve C2 would be the line from 0 to 1 on x-­‐axis. Thus 2
2
∫ (e
C1
x
sin y − y 2 + x ) dx + ( e x cos y − cos(y 2 )) dy = ∫∫ e x cos y − e x cos y + 2y dA =
R
= ∫∫ 2y dA =
∫ (e
C
=
R
x
π /2 cosθ
∫ ∫
0
2r sin θ dr dθ =
2
0
π /2
∫
0
1
2 cos 3 θ sin θ
2u 3θ
2u 4θ
1
dθ = ∫
dθ =
=
3
3
3⋅ 4 0 6
0
1
and so
sin y − y + x ) dx + ( e cos y − cos(y )) dy =
∫ (e
2
x
x
2
sin y − y 2 + x ) dx + ( e x cos y − cos(y 2 )) dy −
C1
∫ (e
x
sin y − y 2 + x ) dx + ( e x cos y − cos(y 2 )) dy = C2
1
1
1
1
d
d
1
1 1
1
= − ∫ ( et sin 0 − 0 2 + t ) t dt − ∫ ( et cos 0 − cos(0 2 )) 0 dt = − ∫ t dt = − = −
6 0
dt
dt
6 0
6 2
3
0
Another example of smart use of Green’s theorem is for A = ∫∫ dA by
Ex 6.
D
choosing
∂Q
∂x
− ∂P
∂y = 1 ,
for example P ( x, y ) = 0 and Q ( x, y ) = x which gives
A = ∫∫ dA = 
∫ x dy .
D
C
The following theorem provides a way to use Greens theorem for more general domains, even for a domain with holes. Thm: Let D be a domain. Rewrite D as union of subdomains, e.g. D = D1 ∪ D2 , let ∂D = C1 ∪ C2 and C3 = D1 ∩ D2 , i.e. D1 = C1 ∪ C3 and D2 = C2 ∪ ( −C3 ) . ∂P
Now ∫∫ ∂Q
∂ x − ∂ y dA =
D
∫
C1 ∪C3
P dx + Q dy +
∫
C2 ∪( −C3 )
P dx + Q dy = ∫ P dx + Q dy + ∫ P dx + Q dy =
C1
82 C2
∫
C1 ∪C2
P dx + Q dy Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky Ex 7. Evaluate ∫ y 3 dx − x 3 dy where C are the two circles of radius 2
C
and radius 1 centered at the origin with positive orientation.
By Green’s theorem: 2π 2
3
3
2
2
2
2
3
∫ y dx − x dy = −3∫∫ x + y dA − 3∫∫ x + y dA = −3 ∫ ∫ r dr dθ =
C
D1
D2
2 4 − 14 )
(
r4
45
= −3⋅ 2π ⋅
= −3π
=− π
4 1
2
2
0 1
2
83