Put $S(\mathbf x)=\mathbf x\cdot\mathbf A\cdot\mathbf x$. Since $\mathbf u=-Sr^{-5}\mathbf x$ and $\nabla\times\mathbf x=0$, the <product rule> gives
$$
\boldsymbol\omega=\nabla\times\mathbf u
=\nabla(-Sr^{-5})\times\mathbf x.
$$
For the symmetric <stresslet> tensor $\mathbf A$,
$$
\nabla(Sr^{-5})=2r^{-5}\mathbf A\cdot\mathbf x-5Sr^{-7}\mathbf x.
$$
The final term is parallel to $\mathbf x$ and drops out of the <cross product>, leaving
$$
\boxed{\boldsymbol\omega(\mathbf x)=\frac{2\mathbf x\times(\mathbf A\cdot\mathbf x)}{r^5}.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-329/1/b/solution