= Solution
Use the <force flux at infinity in Stokes flow>. With an outward normal from a large sphere, the <force> on the body is $\mathbf F=\int_{S_R}\sigma\mathbf n\,dS$. Let $C=3a/4$. The leading far-field <velocity> and <pressure> are
$$
u_i=C\left(\frac{U_i}{r}+\frac{(\mathbf U\cdot\mathbf x)x_i}{r^3}\right)+O(r^{-3}),\qquad
p=2\mu C\frac{\mathbf U\cdot\mathbf x}{r^3}.
$$
Their <rate-of-strain tensor> is
$$
e_{ij}=C(\mathbf U\cdot\mathbf x)\left(\frac{\delta_{ij}}{r^3}-\frac{3x_ix_j}{r^5}\right)+O(r^{-4}).
$$
In the <Newtonian fluid stress tensor> $\sigma_{ij}=-p\delta_{ij}+2\mu e_{ij}$, the isotropic pieces cancel. Consequently,
$$
\sigma\mathbf n=-\frac{6\mu C}{R^2}(\mathbf U\cdot\mathbf n)\mathbf n+O(R^{-4}).
$$
The <second spherical moment integral> $\int n_in_j\,d\Omega=(4\pi/3)\delta_{ij}$ follows from rotational symmetry and $\sum_i n_i^2=1$. Since $dS=R^2d\Omega$, taking $R\to\infty$ gives the <Stokes drag law>:
$$
\boxed{\mathbf F=-6\mu C\frac{4\pi}{3}\mathbf U=-6\pi\mu a\mathbf U.}
$$
The omitted dipole contributes a surface integral of order $R^{-2}$ and vanishes in this limit. The sign refers to the fluid's drag on the moving sphere.
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