 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015-resit/8/a/solution

V = (A (x, y), B (x, y), 0)

, independent of

z

. Applying the Cartesian curl formula gives

\nabla \times V = (0, 0, \frac{\partial B}{\partial x} - \frac{\partial A}{\partial y}) .

Consequently, for continuously differentiable components,

\nabla \times V = 0 ⟺ B_{x} = A_{y} .

This is the local closedness condition for the one-form

A d x + B d y

. To infer global path independence from this condition alone, the domain also matters: it suffices that the planar domain is a simply connected space. The polynomial field in the next calculation is defined on the whole plane, so there is no missing-hole obstruction.

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