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18
FUNDAMENTAL
THEOREMS OF
VECTOR ANALYSIS
18.1 Green’s Theorem
(ET Section 17.1)
Preliminary Questions
1.
Which vector feld
F
is being integrated in the line integral
I
x
2
dy
−
e
y
dx
?
SOLUTION
The line integral can be rewritten as
H
−
e
y
+
x
2
. This is the line integral oF
F
=
D
−
e
y
,
x
2
E
along
the curve.
2.
Draw a domain in the shape oF an ellipse and indicate with an arrow the boundary orientation oF the boundary curve.
Do the same For the annulus (the region between two concentric circles).
The orientation on
C
is counterclockwise, meaning that the region enclosed by
C
lies to the leFt in traversing
C
.
C
±or the annulus, the inner boundary is oriented clockwise and the outer boundary is oriented counterclockwise. The
region between the circles lies to the leFt while traversing each circle.
3.
The circulation oF a gradient vector feld around a closed curve is zero. Is this Fact consistent with Green’s Theorem?
Explain.
Green’s Theorem asserts that
Z
C
F
·
d
s
=
Z
C
Pdx
+
Qdy
=
ZZ
D
µ
∂
Q
∂
x
−
∂
P
∂
y
¶
dA
(1)
IF
F
is a gradient vector feld, the cross partials are equal, that is,
∂
P
∂
y
=
∂
Q
∂
x
⇒
∂
Q
∂
x
−
∂
P
∂
y
=
0(
2
)
Combining (1) and (2) we obtain
R
C
F
·
d
s
=
0. That is, Green’s Theorem implies that the integral oF a gradient vector
feld around a simple closed curve is zero.
4.
Which oF the Following vector felds possess the Following property: ±or every simple closed curve
C
,
Z
C
F
·
d
s
is
equal to the area enclosed by
C
?
(a) F
= h−
y
,
0
i
(b) F
= h
x
,
y
i
(c) F
=
±
sin
(
x
2
),
x
+
e
y
2
®
By Green’s Theorem,
Z
C
F
·
d
s
=
D
µ
∂
Q
∂
x
−
∂
P
∂
y
¶
dx dy
(1)
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View Full DocumentSECTION
18.1
Green’s Theorem
(ET Section 17.1)
637
D
C
We compute the curl of each one of the given Felds.
(a)
Here,
P
=−
y
and
Q
=
0, hence
∂
Q
∂
x
−
∂
P
∂
y
=
0
−
(
−
1
)
=
1. Therefore, by (1),
Z
C
F
·
d
s
=
ZZ
D
1
dx dy
=
Area
(
D
)
(b)
We have
P
=
x
and
Q
=
y
, therefore
∂
Q
∂
x
−
∂
P
∂
y
=
0
−
0
=
0. By (1) we get
Z
C
F
·
d
s
=
D
0
=
0
6=
Area
(
D
)
(c)
In this vector Feld we have
P
=
sin
(
x
2
)
and
Q
=
x
+
e
y
2
. Therefore,
∂
Q
∂
x
−
∂
P
∂
y
=
1
−
0
=
1
.
By (1) we obtain
Z
C
F
·
d
s
=
D
1
=
Area
(
D
).
Exercises
1.
Verify Green’s Theorem for the line integral
I
C
xydx
+
ydy
,where
C
is the unit circle, oriented counterclockwise.
SOLUTION
Step 1.
Evaluate the line integral. We use the parametrization
γ
(
θ
)
= h
cos
,
sin
i
,0
≤
≤
2
π
of the unit circle. Then
dx
sin
d
,
dy
=
cos
d
and
+
=
cos
sin
(
−
sin
d
)
+
sin
cos
d
=
³
−
cos
sin
2
+
sin
cos
±
d
The line integral is thus
Z
C
+
=
Z
2
0
³
−
cos
sin
2
+
sin
cos
±
d
=
Z
2
0
−
cos
sin
2
d
+
Z
2
0
sin
cos
d
sin
3
3
¯
¯
¯
¯
2
0
−
cos 2
4
¯
¯
¯
¯
2
0
=
0(
1
)
x
y
C
D
Step 2.
Evaluate the double integral. Since
P
=
xy
and
Q
=
y
,wehave
∂
Q
∂
x
−
∂
P
∂
y
=
0
−
x
x
We compute the double integral in Green’s Theorem:
D
µ
∂
Q
∂
x
−
∂
P
∂
y
¶
=
D
−
xdxdy
D
638
CHAPTER 18
FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS
(ET CHAPTER 17)
The integral of
x
over the disk
D
is zero, since by symmetry the positive and negative values of
x
cancel each other.
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 Fall '08
 Estrada
 Calculus

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