Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 6 / c / Solution

Use the vector triple product expression from part (a). Put $s=\Omega \cdot x$ and $r^2=x\cdot x$. Since $u=\Omega +\beta x$,Expanding $W=(u\cdot u)\Omega -(u\cdot \Omega )u$ and canceling the zeroth-order terms givesThe first bracket is odd in $x$ and integrates to zero over the centred ball. With $\alpha =4\pi a^5/15$, the Cartesian moments of a three-dimensional ball giveThereforeThe resulting vector is parallel to $\Omega$, as the rotational symmetry of the centred domain requires.