= Solution
Substitution gives
$$
\boxed{u_R=U\left(\frac{3a}{2R}-\frac{a^3}{2R^3}\right)\cos\theta,
\qquad u_\theta=-U\left(\frac{3a}{4R}+\frac{a^3}{4R^3}\right)\sin\theta,
\qquad u_\phi=0.}
$$
Rotational symmetry about the translation axis forbids a transverse force. Linearity of the Stokes equations and boundary conditions makes the force linear in $U$; dissipation makes the force on the translating sphere oppose its motion. Thus \b[the drag is axial and proportional to $U$], without computing the stress integral. Dimensional analysis further gives scale $\mu aU$; the exact single-sphere coefficient is $6\pi$.
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