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386 CHAPTER 12 Vector Integral Calculus

For example, the interior of a solid rectangle is a domain and the entire plane is a domain.
The right quarter plane consisting of points (x, y) with x >0 and y >0 is also a domain. However,
if we include parts of the axes, considering points (x, y) with x ≥ 0 and y ≥ 0, the resulting set
is not a domain. For example, (0,1) is in this set, but no circle about this point can contain only
points with nonnegative coordinates, violating condition (1) for a domain.
A domain D is called simply connected if every simple closed path in D encloses only points
of D. In Example 12.14, the plane with the origin removed is not simply connected, because a
closed path about the origin encloses a point (the origin) not in the set.
Now we can improve Theorem 12.5 to obtain a necessary and sufficient condition for a
vector field to be conservative.

THEOREM 12.6

Let F= f i+ gj be a vector field defined over a simply connected domain D in the plane. Suppose
f and g are continuous and that ∂ f/∂y and ∂g/∂x are continuous. Then F is conservative on D
if and only if
∂ f ∂g
= . (12.8)
∂y ∂x
Thus, under the given conditions, equality of these partials is both necessary and sufficient
for the vector field to have a potential function.
In Example 12.14, the components of F satisfy equation (12.8), but the set (the plane with
the origin removed), is not simply connected, so the theorem does not apply. In that example we
saw that there is no potential function for F over the entire punctured plane.
In 3-space there is a similar test for a vector field to be conservative, with adjustments to
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accommodate the extra dimension. A set S of points in R is a domain if it satisfies the following
two conditions:
1. If P is a point in S, then there is a sphere about P that encloses only points of S.
2. Between any two points of S there is a path lying entirely in S.

For example, the interior of a cube is a domain.
Furthermore, S is simply connected if every simple closed path in S is the boundary of a
surface in S.
With this notion of simple connectivity in 3-space, we can state a three-dimensional version
of Theorem 12.6.

THEOREM 12.7

3
Let F be a vector field defined over a simply connected domain S in R . Then F is conservative
on S if and only if
∇× F = O.

3
Thus, the conservative vector fields in R are those with zero curl. These are the irrotational
vector fields. We will prove this theorem in Section 12.9.1 when we have Stokes’s theorem.
With the right perspective, these tests in 2-space and 3-space can be combined into a single
test. Given F = f i + gj in the plane, define

G(x, y, z) = f (x, y)i + g(x, y)j + 0k

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October 14, 2010 14:53 THM/NEIL Page-386 27410_12_ch12_p367-424