= Solution
Write $D=\partial_nA_n$ and use repeated-index summation. Because every $A_i$ is a <harmonic function>,
$$
\partial_iu_i=D-\partial_kA_k-x_k\nabla^2A_k=0,
$$
and
$$
\nabla^2u_i=-2\partial_k\partial_iA_k=-2\partial_iD,\qquad \partial_ip=-2\mu\partial_iD.
$$
Thus $\mu\nabla^2u_i-\partial_ip=0$, the incompressible <Stokes equation>. Moreover
$$
\partial_ju_i=\partial_jA_i-\partial_iA_j-x_k\partial_i\partial_jA_k.
$$
The antisymmetric first two terms cancel in the sum with $\partial_iu_j$, giving the <Newtonian fluid> <Cauchy stress tensor>
$$
\boxed{\sigma_{ij}=-p\delta_{ij}+\mu(\partial_ju_i+\partial_iu_j)=2\mu(\delta_{ij}D-x_k\partial_i\partial_jA_k).}
$$
For $A_i=V_i(1-a/(2r))$, harmonic outside the origin,
$$
\partial_iA_k=\frac{aV_kx_i}{2r^3},\quad u_i=V_i\left(1-\frac a{2r}\right)-\frac{ax_i(\mathbf V\cdot\mathbf x)}{2r^3},\quad p=-\frac{\mu a(\mathbf V\cdot\mathbf x)}{r^3}.
$$
At $r=a$ this gives $\mathbf u=\tfrac12[\mathbf V-(\mathbf V\cdot\mathbf n)\mathbf n]$, where $\mathbf n=\mathbf x/a$, so the normal velocity is zero. Also
$$
\partial_i\partial_jA_k=\frac{aV_k}{2}\left(\frac{\delta_{ij}}{r^3}-\frac{3x_ix_j}{r^5}\right),\qquad \sigma_{ij}=\frac{3\mu a(\mathbf V\cdot\mathbf x)x_ix_j}{r^5}.
$$
Hence on the surface,
$$
\boxed{\sigma_{ij}n_j=\frac{3\mu}{a}(\mathbf V\cdot\mathbf n)n_i,}
$$
a purely normal <traction>, appropriate to a shear-free sphere rather than a no-slip solid sphere. Finally rotational symmetry gives $\int_{S^2}n_in_j\,d\Omega=(4\pi/3)\delta_{ij}$, so the fluid's force on the sphere is
$$
\boxed{\mathbf F=\int_{r=a}\boldsymbol\sigma\mathbf n\,dS=4\pi\mu a\mathbf V.}
$$
This is the <clean-bubble Stokes drag> in the far-field velocity $\mathbf V$.
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