= Solution
The <Papkovich–Neuber representation> writes a homogeneous <Stokes flow> in terms of a <harmonic function>[harmonic vector field] $\boldsymbol\Phi$ and a harmonic scalar $\chi$ as
$$
\mu\mathbf u=\boldsymbol\Phi-\frac12\nabla(\mathbf x\mathbin\cdot\boldsymbol\Phi)+\nabla\chi,
\qquad
p=-\nabla\mathbin\cdot\boldsymbol\Phi.
$$
For translation, rotational symmetry and decay at infinity restrict the trial harmonic fields to the fundamental harmonic $1/r$ and its directional derivatives contracted with $\mathbf U$. For rotation, the only decaying isotropic axial-vector field with the required boundary value is proportional to $\boldsymbol\Omega\times\mathbf x/r^3$. Matching the <no-slip boundary condition> $\mathbf u=\mathbf U+\boldsymbol\Omega\times\mathbf x$ at $r=a$ gives the superposition of the <translating sphere in Stokes flow> and the <rotating sphere in Stokes flow>:
$$
\boxed{
\mathbf u=
\frac{3a}{4r}\left(\mathbf I+\frac{\mathbf x\mathbf x}{r^2}\right)\mathbf U
+\frac{a^3}{4r^3}\left(\mathbf I-3\frac{\mathbf x\mathbf x}{r^2}\right)\mathbf U
+\frac{a^3}{r^3}\boldsymbol\Omega\times\mathbf x,
\qquad
p=\frac{3\mu a}{2r^3}\mathbf U\mathbin\cdot\mathbf x .}
$$
Each term decays at infinity, and direct substitution at $r=a$ gives the prescribed rigid velocity.
When $\mathbf U=0$, the pressure is constant and may be set to zero. Differentiating the rotational velocity and using the <Newtonian fluid stress tensor> gives
$$
\boxed{
\boldsymbol\sigma
=-\frac{3\mu a^3}{r^5}
\left[(\boldsymbol\Omega\times\mathbf x)\mathbf x
+\mathbf x(\boldsymbol\Omega\times\mathbf x)\right].}
$$
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