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12.4 Independence of Path and Potential Theory 383

with β(z) an as yet unknown function of z.Wenow have
2
z
3
3
2
ϕ(x, y, z) = x yz + α(y, z) = x yz + ye + β(z)
and we have only to determine β(z). For this use the third equation, 12.7, to write
∂ϕ
3
z
3
z
= 2x yz + ye = 2x yz + ye + β (z).

∂z
2
z
3
This forces β (z)=0, so β(z)=k, any number. With ϕ(x, y, z)= x yz + ye +k for any number

k (which we can choose to be 0), we have F =∇ϕ and ϕ is a potential function for F.

This enables us to easily evaluate F · dR. If, for example, C is a path from (0,0,0) to
C
(−1,3,−2), then

−2
F · dR = ϕ(−1,3,−2) − ϕ(0,0,0) =−12 + 3e .
C

And if C is a closed path, then F · dR = 0.
C
This method for finding a potential function for a function of two variables was seen
previously in solving exact differential equations (Example 1.12).
There are nonconservative vector fields.
EXAMPLE 12.12
x
Let F = yi + e j, a vector field in the plane. If this is conservative, there would be a potential
function ϕ(x, y) such that
∂ϕ ∂ϕ
x
= y and = e .
∂x ∂y
Integrate the first with respect to x, thinking of y as fixed, to get

ϕ(x, y) = ydx = xy + α(y).
But then we would have to have
∂ϕ x ∂

= e = (xy + α(y)) = x + α (y).
∂y ∂y
This would make α depend on x. This is impossible, since α(y) was the “constant” of integration
with respect to x. F has no potential and is not conservative.
If a vector field is conservative, we may be able to find a potential function by integration.
But in general, integration is an ineffective way to determine if a vector field is conservative, one
problem being that we cannot integrate every function. The following test is simple to apply for
vector fields defined over a rectangle in the plane. We will extend this test to a three dimensional
version later when we have Stokes’s theorem.

THEOREM 12.5 Test for a Conservative Field in the Plane

Let f and g be continuous in a region D of the plane bounded by a rectangle having its sides
parallel to the axes. Then F(x, y) = f (x, y)i + g(x, y)j is conservative on D if and only if, for
all (x, y) in D,
∂g ∂ f
= .
∂x ∂y
Proof In one direction the proof is a simple differentiation. If F is conservative on D, then
F =∇ϕ. Then

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October 14, 2010 14:53 THM/NEIL Page-383 27410_12_ch12_p367-424