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384 CHAPTER 12 Vector Integral Calculus

∂ϕ ∂ϕ
f (x, y) = and g(x, y) =
∂x ∂y
so
2
2
∂g ∂ ϕ ∂ ϕ ∂ f
= = = .
∂x ∂x∂y ∂y∂x ∂y
A proof of the converse is outlined in Problem 22.

EXAMPLE 12.13
Often Theorem 12.5 is used in the following form: if
∂g ∂ f
=
∂x ∂y
then f (x, y)i + g(x,)j is not conservative. As an example of the use of this test, consider
2
x
2
F(x, y) = (2xy + y)i + (2x y + e y)j = f (x, y)i + g(x, y)j.
This is continuous over the entire plane, hence on any rectangular region. Compute
∂g ∂ f
x
= 4xy + e y and = 4xy + 1.
∂x ∂y
These partial derivatives are not equal throughout any rectangular region, so F is not conservative.
If we attempted to find a potential function ϕ by integration, we would begin with
∂ϕ ∂ϕ
2
x
2
= 2xy + y and = 2x y + e y.
∂x ∂y
Integrate the first equation with respect to x to obtain
2
2
ϕ(x, y) = x y + xy + α(y),
in which α(y) is the “constant” of integration with respect to x. But then
∂ϕ 2 2 x
= 2x y + x + α (y) = g(x, y) = 2x y + e y.

∂y
This requires that
x

α (y) = ye ,
and then α (y) would depend on x, not y, a contradiction. Thus F has no potential function, as

we found with less effort using the test of Theorem 12.5.
In special regions (rectangular), existence of a potential function for f (x, y)i + g(x, y)j
implies that
∂g ∂ f
= .
∂x ∂y
We can ask whether equality of these partial derivatives implies that f i + gj has a potential
function. This is a subtle question, and the answer depends not only on the vector field, but on
the set D over which this field is defined. The following example demonstrates this.

EXAMPLE 12.14

Let
−y x
F(x, y) = i + j
2
2
x + y 2 x + y 2

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October 14, 2010 14:53 THM/NEIL Page-384 27410_12_ch12_p367-424