Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 329 / 1 / b

Table of contents
- Solution b

Solution

Put

S (x) = x \cdot A \cdot x

. Since

u = - S r^{- 5} x

and

\nabla \times x = 0

, the product rule gives

ω = \nabla \times u = \nabla (- S r^{- 5}) \times x .

For the symmetric stresslet tensor

A

,

\nabla (S r^{- 5}) = 2 r^{- 5} A \cdot x - 5 S r^{- 7} x .

The final term is parallel to

x

and drops out of the cross product, leaving

ω (x) = \frac{2 x \times ( A \cdot x )}{r ^{5}} .

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