= Solution
Let $\mathbf F$ denote the applied <force>, and take the normal $\mathbf n_S$ outward from the solid. At leading order the <Stokes drag law> and the <torque> on a <rotating sphere in Stokes flow> give
$$
\boxed{\mathbf U_0=\frac{\mathbf F}{6\pi\mu a},\qquad\boldsymbol\Omega_0=\mathbf0.}
$$
The exact <no-slip boundary condition> and the exact resultant-load conditions are
$$
\begin{gathered}
\mathbf u(\mathbf x)=\mathbf U+\boldsymbol\Omega\times\mathbf x\quad(\mathbf x\in S),\qquad\mathbf u\to\mathbf0\quad(r\to\infty),\\
\int_S\boldsymbol\sigma\mathbf n_S\,dS=-\mathbf F,\qquad
\int_S\mathbf x\times(\boldsymbol\sigma\mathbf n_S)\,dS=\mathbf0.
\end{gathered}
$$
The <stress> condition specifies the total <force> and <torque>, not a pointwise prescribed <traction>.
For the <boundary perturbation of a nearly spherical particle>, evaluate the no-slip condition at $\mathbf x=(a+\varepsilon f)\mathbf n$. Taylor expansion gives
$$
\mathbf u_0(a\mathbf n)+\varepsilon\left[\mathbf u_1(a\mathbf n)+f\partial_r\mathbf u_0(a\mathbf n)\right]
=\mathbf U_0+\varepsilon\left[\mathbf U_1+\boldsymbol\Omega_1\times(a\mathbf n)\right]+O(\varepsilon^2).
$$
There is no $f\boldsymbol\Omega_0\times\mathbf n$ term because $\boldsymbol\Omega_0=0$. Therefore
$$
\boxed{\mathbf u_1=\mathbf U_1+\boldsymbol\Omega_1\times\mathbf x-f\partial_r\mathbf u_0\quad(r=a).}
$$
To expand the integral conditions correctly, use <surface independence of Stokes force and torque integrals>. The <Stokes equation> implies $\nabla\cdot\boldsymbol\sigma=0$, and symmetry of the <stress tensor> implies that the angular-momentum flux is divergence-free as well. Evaluate the exact <force> and <torque> on a fixed sphere of radius $R$ enclosing the perturbed particle. There are then no geometric terms: the $O(\varepsilon)$ integrals of $\boldsymbol\sigma_1$ on that fixed sphere vanish because the external <force> and <torque> have no $O(\varepsilon)$ corrections. Each perturbation field solves the homogeneous exterior Stokes equations, so the <divergence theorem> transfers those integrals back to $r=a$. Hence
$$
\boxed{\int_{r=a}\boldsymbol\sigma_1\mathbf n\,dS=\mathbf0,\qquad
\int_{r=a}\mathbf x\times(\boldsymbol\sigma_1\mathbf n)\,dS=\mathbf0.}
$$
If the integrals are expanded directly on the displaced surface, terms involving the displaced evaluation of $\boldsymbol\sigma_0$, the changed normal and the changed area do appear individually. Their sum vanishes by the same surface-independence identity. Thus their absence in the final equations is a cancellation, not permission to omit geometric terms independently.
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