 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/1/b/solution

Take the closed surface with its outward orientation. For sufficiently smooth fields defined throughout the solid, the divergence theorem and the identity that the divergence of a curl is zero give

I = ∭_{V} \nabla \cdot (\nablaΩ + \nabla \times U) d V = ∭_{V} ΔΩ d V .

Thus the curl term carries no net outward flux, and

I

is independent of

U

. The hole in the solid torus does not invalidate the divergence theorem; singularities of

U

inside the integration region would invalidate the smoothness argument and must be excluded.

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