 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/1/b/solution

Take any constant vector

a

. Its curl is zero, so part (a) and the divergence theorem give

a \cdot \int_{V} \nabla \times F d V = - \int_{V} \nabla \cdot (a \times F) d V = - \int_{S} (a \times F) \cdot n d S,

where

n

is the outward unit normal vector. The scalar triple product identity rewrites the last integrand as

a \cdot (F \times n)

. Since

a

is arbitrary, equality of these scalar products implies the vector identity

\int_{V} \nabla \times F d V = - \int_{S} F \times n d S = \int_{S} n \times F d S .

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