Mathematics of Physics and Engineering

156 The Three Theorems

has a continuous gradient in G. Then the following three conditions are

equivalent:

(i) The field F has the path independence property: §c F • dr = 0 for every

simple closed piece-wise smooth curve C in G.

(ii) The field F has a potential: there exists a function f so that F = V/

everywhere in G.

(Hi) The field F is irrotational: curl F = 0 everywhere in G.

EXERCISE 3.2.8. (a)c Prove the theorem. Hint: for (i) => (ii), see Theorem

3.1.1. For (ii) =>• (Hi), use (3.1.30). For (Hi) =>• (i) use Stokes's Theorem. (b)B

Verify that the vector field F(x, y, z) = (-y/(x^2 +y^2 ))i+(x/(x^2 +y^2 ))j+OK

is irrotational in the domain {(x, y, z) : x^2 +y^2 > 0} (direct computation) but

does not have the path independence property there (integrate over a unit circle

in the (I, j) plane, centered at the origin). Explain why the result is consistent

with the above theorem (note that the function is not defined anywhere on the

z axis).

Assume that the vector field F satisfies the conditions of Stoke's Theo-
rem everywhere in M^3. Then, by (3.2.11), the value of J J curlF • dcr does
S
not depend on the particular surface S, as long as the boundary of S is
the given simple closed piece-wise smooth curve C. This conclusion is con-
sistent with Gauss's Theorem. Indeed, let S\ and 5*2 be two orientable
piece-wise smooth surfaces with the same boundary C and no other points
in common, and denote by G the region enclosed by Si and S2 (draw a
picture). The boundary dG of G is the union of S\ and S2. If the vec-
tor field F is sufficiently smooth so that curlF satisfies the conditions of
Gauss's Theorem, then, by (3.2.11), ff curl F-dcr = J//div(curlF)dV = 0,
dG G
where the last equality follows from (3.1.30) on page 139, and therefore
// curl F -dcr = JJ curl F • da, if the orientations of both Si and S2 agree
Si S 2
with the orientation of C.
By now we have seen enough connections between Green's, Gauss's, and
Stokes's Theorems to suspect that there should be a way to unify the three
into one formula. Such a formula does indeed exist:

/ UJ= I duj, (3.2.12)

JdM JM

where dM is the boundary of the set M, and du is a certain derivative
of the expression u> being integrated on the left-hand side. Making a pre-