(4.20)
C
where A = ( x( a), y( a)) and B = ( x( b), y( b)) are the endpoints of C. Thus, the line integral is independent of the path between its endpoints, since it depends only on the values of F at
those endpoints.
Proof: By definition of
f
C · dr, we have
b
f · dr =
P( x( t), y( t)) x ′( t) + Q( x( t), y( t)) y′( t) dt C
a
b
∂F dx
∂F d y
∂F
∂F
=
+
dt (since ∇ F = f ⇒
= P and
= Q)
a
∂x dt
∂y dt
∂x
∂y
b
=
F ′( x( t), y( t)) dt (by the Chain Rule in Theorem 4.4)
a
b
= F( x( t), y( t))
= F( B) − F( A)
a
by the Fundamental Theorem of Calculus.
QED
148
CHAPTER 4. LINE AND SURFACE INTEGRALS
Theorem 4.5 can be thought of as the line integral version of the Fundamental Theorem
of Calculus. A real-valued function F( x, y) such that ∇ F( x, y) = f( x, y) is called a potential for f. A conservative vector field is one which has a potential.
Example 4.5. Recall from Examples 4.2 and 4.3 in Section 4.1 that the line integral ( x 2
C
+
y 2) dx + 2 xy d y was found to have the value 13 for three different curves C going from the
3
point (0, 0) to the point (1, 2). Use Theorem 4.5 to show that this line integral is indeed path
independent.
Solution: We need to find a real-valued function F( x, y) such that
∂F
∂F
= x 2 + y 2 and
= 2 xy .
∂x
∂y
Suppose that ∂F
x 3
∂x = x 2 + y 2, Then we must have F ( x, y) = 1
3
+ xy 2 + g( y) for some function
g( y). So ∂F
∂ y = 2 x y + g ′( y) satisfies the condition ∂F
∂ y = 2 x y if g ′( y) = 0, i.e. g( y) = K , where K
is a constant. Since any choice for K will do (why?), we pick K = 0. Thus, a potential F( x, y)
for f( x, y) = ( x 2 + y 2)i + 2 xyj exists, namely
1
F( x, y) =
x 3 + xy 2 .
3
Hence the line integral
( x 2
C
+ y 2) dx + 2 xy d y is path independent.
Note that we can also verify that the value of the line integral of f along any curve C going
from (0, 0) to (1, 2) will always be 13 , since by Theorem 4.5
3
1
1
13
f · dr = F(1,2) − F(0,0) =
(1)3 + (1)(2)2 − (0 + 0) =
+ 4 =
.
C
3
3
3
A consequence of Theorem 4.5 in the special case where C is a closed curve, so that the
endpoints A and B are the same point, is the following important corollary:
Corollary 4.6. If a vector field f has a potential in a region R, then
f· dr = 0 for any closed
C
curve C in R (i.e.
∇ F · dr = 0 for any real-valued function F( x, y)).
C
Example 4.6. Evaluate
x dx + y d y for C : x = 2cos t, y = 3sin t, 0 ≤ t ≤ 2 π.
C
Solution: The vector field f( x, y) = x i + yj has a potential F( x, y):
∂F
1
= x ⇒ F( x, y) =
x 2 + g( y) ,so
∂x
2
∂F
1
= y ⇒ g ′( y) = y ⇒ g( y) =
y 2 + K
∂y
2
4.2 Properties of Line Integrals
149
1
1
for any constant K , so F( x, y) = x 2 + y 2 is a potential for f( x, y). Thus,
2
2
x dx + y d y =
f · dr = 0
C
C
by Corollary 4.6, since the curve C is closed (it is the ellipse x 2
4 + y 2
9 = 1).
Exercises
A
1. Evaluate
( x 2 + y 2) dx + 2 xy d y for C : x = cos t, y = sin t, 0 ≤ t ≤ 2 π.
C
2. Evaluate
( x 2 + y 2) dx + 2 xy d y for C : x = cos t, y = sin t, 0 ≤ t ≤ π.
C
3. Is there a potential F( x, y) for f( x, y) = yi − x j? If so, find one.
4. Is there a potential F( x, y) for f( x, y) = x i − yj? If so, find one.
5. Is there a potential F( x, y) for f( x, y) = xy 2 i + x 3 yj? If so, find one.
B
6. Let f( x, y) and g( x, y) be vector fields, let a and b be constants, and let C be a curve in R2.
Show that
( a f ± b g) · dr = a
f · dr ± b
g · dr .
C
C
C
7. Let C be a curve whose arc length is L. Show that
1 ds
C
= L.
8. Let f ( x, y) and g( x, y) be continuously differentiable real-valued functions in a region R.
Show that
f ∇ g · dr = −
g ∇ f · dr
C
C
for any closed curve C in R. ( Hint: Use Exercise 21 in Section 2.4. )
9. Let f( x, y) = − y i + x j for all ( x, y) = (0,0), and C : x = cos t, y = sin t, 0 ≤ t ≤ 2 π.
x 2+ y 2
x 2+ y 2
(a) Show that f = ∇ F, for F( x, y) = tan−1( y/ x).
(b) Show that
f · dr = 2 π. Does this contradict Corollary 4.6? Explain.
C
C
10. Let g( x) and h( y) be differentiable functions, and let f( x, y) = h( y)i + g( x)j. Can f have a potential F( x, y)? If so, find it. You may assume that F would be smooth. ( Hint: Consider
the mixed partial derivatives of F. )
150
CHAPTER 4. LINE AND SURFACE INTEGRALS
4.3 Green’s Theorem
We will now see a way of evaluating the line integral of a smooth vector field around a
simple closed curve. A vector field f( x, y) = P( x, y)i + Q( x, y)j is smooth if its component functions P( x, y) and Q( x, y) are smooth. We will use Green’s Theorem (sometimes called
Green’s Theorem in the plane) to relate the line integral around a closed curve with a double
integral over the region inside the curve:
Theorem 4.7. (Green’s Theorem) Let R be a region in R2 whose boundary is a simple
closed curve C which is piecewise smooth. Let f( x, y) = P( x, y)i + Q( x, y)j be a smooth vector field defined on both R and C. Then
∂Q
∂P
f · dr =
−
d A ,
(4.21)
C
∂x
∂y
R
where C is traversed so that R is always on the left side of C.
Proof: We will prove the theorem in the case for a simple region R, that is, where the
boundary curve C can be written as C = C
in two distinct ways:
1 ∪ C 2
C
( x) from the point X to the point X
(4.22)
1 = the curve y = y 1
1
2
C
( x) from the point X to the point X ,
(4.23)
2 = the curve y = y 2
2
1
where X and X are the points on C farthest to the left and right, respectively; and
1
2
C
( y) from the point Y to the point Y
(4.24)
1 = the curve x = x 1
2
1
C
( y) from the point Y to the point Y ,
(4.25)
2 = the curve x = x 2
1
2
where Y and Y are the lowest and highest points, respectively, on C. See Figure 4.3.1.
1
2
y
y = y 2( x)
d
Y 2
X 2 x = x 2( y)
x = x 1( y) X 1
R
C
Y 1
c
y = y 1( x)
x
a
b
Figure 4.3.1
Integrate P( x, y) around C using the representation C = C
given by (4.23) and (4.24).
1 ∪ C 2
4.3 Green’s Theorem
151
Since y = y ( x) along C (as x goes from a to b) and y
( x) along C (as x goes from b to
1
1
= y 2
2
a), as we see from Figure 4.3.1, then we have
P( x, y) dx =
P( x, y) dx +
P( x, y) dx
C
C 1
C 2
b
a
=
P( x, y ( x)) dx
P( x, y ( x)) dx
1
+
2
a
b
b
b
=
P( x, y ( x)) dx
P( x, y ( x)) dx
1
−
2
a
a
b
= −
( P( x, y ( x))
( x))
2
− P( x, y 1
) dx
a
b
y= y 2( x)
= −
P( x, y)
dx
a
y= y 1( x)
b
y 2( x) ∂P( x, y)
= −
d y dx (by the Fundamental Theorem of Calculus)
a
y
∂y
1( x)
∂P
= −
d A .
∂y
R
Likewise, integrate Q( x, y) around C using the representation C = C
given by (4.25)
1 ∪ C 2
and (4.26). Since x = x ( y) along C (as y goes from d to c) and x
( y) along C (as y goes
1
1
= x 2
2
from c to d), as we see from Figure 4.3.1, then we have
Q( x, y) d y =
Q( x, y) d y +
Q( x, y) d y
C
C 1
C 2
c
d
=
Q( x ( y), y) d y
Q( x ( y), y) d y
1
+
2
d
c
d
d
= −
Q( x ( y), y) d y
Q( x ( y), y) d y
1
+
2
c
<