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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-3/12c/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3 / 12C / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

\partial_{i} F_{j} = \partial_{j} F_{i}

. Contracting this symmetric pair with the antisymmetric Levi-Civita symbol gives

(\nabla \times F)_{k} = ϵ_{kij} \partial_{i} F_{j} = 0.

(1)

Every circle bounds its planar disk, and the vector field is smooth on all of

R^{3}

, including that disk. Stokes theorem therefore gives for either orientation

\oint_{C} F \cdot d x = \int_{D} (\nabla \times F) \cdot n d S = 0.

(2)

The global domain is essential to this argument; the same local symmetry condition on a domain with a hole need not make every closed-curve line integral vanish.

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