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 ( "index_8.gif"  ) 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 "index_10.gif"(t)) where "index_11.gif"(t) = "index_12.gif"(t + 2π) and in particular "index_13.gif"(0) = "index_14.gif"(2π).
Therefore d"index_15.gif"=    (-R sin(t), R cos(t), "index_16.gif"(t)) dt = (-y(t), x(t), A "index_17.gif"(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



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