= Solution
Use the <vector triple product> expression from part (a). Put $s=\Omega\cdot x$ and $r^2=x\cdot x$. Since $u=\Omega+\beta x$,
$$
u\cdot u=\Omega^2+2\beta s+\beta^2r^2,\qquad u\cdot\Omega=\Omega^2+\beta s.
$$
Expanding $W=(u\cdot u)\Omega-(u\cdot\Omega)u$ and canceling the zeroth-order terms gives
$$
\boxed{W(x)=\beta\big[(\Omega\cdot x)\Omega-\Omega^2x\big]+\beta^2\big[r^2\Omega-(\Omega\cdot x)x\big].}
$$
The 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> give
$$
\int_Vr^2\,dV=3\alpha,\qquad\int_V(\Omega\cdot x)x_i\,dV=\Omega_j\int_Vx_jx_i\,dV=\alpha\Omega_i.
$$
Therefore
$$
\boxed{\int_VW_i\,dV=2\alpha\beta^2\Omega_i=\frac{8\pi a^5\beta^2}{15}\Omega_i.}
$$
The resulting <vector> is parallel to $\Omega$, as the rotational symmetry of the centred domain requires.
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