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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 giveThe exact no-slip boundary condition and the exact resultant-load conditions areThe 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 $\mathbf{x}=(a+\epsilon f)\mathbf{n}$. Taylor expansion givesThere is no $f{\mathbf{\Omega }}_0\times \mathbf{n}$ term because ${\mathbf{\Omega }}_0=0$. Therefore

To expand the integral conditions correctly, use surface independence of Stokes force and torque integrals. The Stokes equation implies $\nabla \cdot \mathbf{\sigma }=0$, 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 $O(\epsilon )$ integrals of ${\mathbf{\sigma }}_1$ on that fixed sphere vanish because the external force and torque have no $O(\epsilon )$ corrections. Each perturbation field solves the homogeneous exterior Stokes equations, so the divergence theorem transfers those integrals back to $r=a$. HenceIf the integrals are expanded directly on the displaced surface, terms involving the displaced evaluation of ${\mathbf{\sigma }}_0$, 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.