 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/11/a/ii/solution

ϕ

agree. Therefore

\nabla \times v = \nabla \times \nabla ϕ = 0

everywhere, and

\int_{Γ_{2}} (\nabla \times v) \cdot d x = 0 .

Unlike part (i), this integrand does not involve the problematic auxiliary field.

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