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12.4 Independence of Path and Potential Theory 385

for all (x, y) except the origin. This is a vector field in the plane with the origin removed, with
−y x
f (x, y) = and g(x, y) = .
x + y 2 x + y 2
2
2
Routine integrations would appear to derive the potential function

x
ϕ(x, y) =−arctan .
y
However, this potential is not defined for all (x, y).
If we restrict (x, y) to the right quarter plane x > 0, y > 0, then ϕ is indeed a potential
function and F is conservative in this region. However, suppose we attempt to consider F over the
set D consisting of the entire plane with the origin removed. Then ϕ is not a potential function.

Further, F is not conservative over D because F · dR is not independent of path in D.Tosee
C
this, we will evaluate this integral over two paths from (1,0) to (−1,0), shown in Figure 12.9.
First, let C 1 be the half-circle given by x =cos(θ), y =sin(θ) for 0≤θ ≤π. This is the upper
2
2
half of the circle x + y = 1. Then
π
F · dR = [(−sin(θ))(−sin(θ)) + cos(θ)(cos(θ))]dθ
C 1 0
π
= dθ = π.
0
Next let C 2 be the half-circle from (1,0) to (−1,0) given by x =cos(θ), y =−sin(θ) for 0≤θ ≤
2
2
π. This is the lower half of the circle x + y = 1 and
π
F · dR = [sin(θ)(−sin(θ)) + cos(θ)(−cos(θ))]dθ
C 2 0
π
=− dθ =−π.
0

In this example, C F · dR depends not only on the vector field, but also on the curve, and the
vector field cannot be conservative over the plane with the origin removed.

In attempting a converse of the test of Theorem 12.5, Example 12.14 means that we must
place some condition on the set D over which the vector field is defined. This leads us to define
aset D of points in the plane to be a domain if it satisfies two conditions:
1. If P is a point in D, then there is a circle about P that encloses only points of D.
2. Between any two points of D there is a path lying entirely in D.

y

C 1

x
(–1, 0) (1, 0)

C 2

FIGURE 12.9 Two paths of integration
in Example 12.14.

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October 14, 2010 14:53 THM/NEIL Page-385 27410_12_ch12_p367-424