Examples of Path-Dependent Integrals: Vector Fields with Non-Vanishing Curl
Here is a vector function (xyz, xyz, xyz) for which the curl does not vanish anywhere, except the origin
These are the conditions that the curl is zero:
There is only one point where this occurs:
Let's evaluate the integral of the vector potential ( ) for any curve that wraps around a cylinder of radius R with an axis that coincides with the z-axis
Any curve that wraps around the cylinder can be parameritized as (x(t), y(t), z(t)) = (R cos(t), R sin(t), A (t)) where
(t) =
(t + 2π) and in particular
(0) =
(2π).
Therefore d= (-R sin(t), R cos(t),
(t)) dt = (-y(t), x(t), A
(t)) dt
The integrand for an integral of "VectorFunction" around such a curve is (written in terms of an arbitrary P(t):
The integral depends on the choice of P(t)
Let's introduce some specific periodic functions for P. Note how the value of the integral changes as the path changes:
However, here is curious result which shows that some special paths can ``accidentally'' have zero integrals : let P(t) = cos(n t),
Apparently, the symmetry of the vector function causes cancelation (note results for P = Sin and P=Cos, differ by a minus sign)
But, why doesn't n=1 give us the correct result above? Note that the denominator goes to zero as n→1
Created by Wolfram Mathematica 6.0 (24 September 2007) | ![]() |