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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2022/9/b/i/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2022 / 9 / b / i / Solution by

Codex  0 2026-10-25

For a sufficiently smooth vector field on an oriented surface

S

with compatible boundary orientation, Stokes theorem states

\oint_{\partial S} F \cdot d r = \iint_{S} (\nabla \times F) \cdot n d S .

(1)

Apply it to

F = (P, Q, 0)

in the plane. The closed line integral is the area integral of

Q_{x} - P_{y}

. Thus

P_{y} = Q_{x}

makes it vanish whenever a spanning surface lies within the smooth domain. In a simply connected domain this implies path independence: join one path from

A

B

to the reverse of another, and their difference is a zero closed line integral.

The domain qualification is essential. On the punctured plane,

P = - y / (x^{2} + y^{2})

Q = x / (x^{2} + y^{2})

satisfy

P_{y} = Q_{x}

, but the line integral around the unit circle is

2 π

. A disk spanning that circle passes through the singularity. Hence the printed implication requires an appropriate domain assumption; zero curl alone is not a global path independence theorem.

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