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The fluid is outside the body, so its normal on the inner boundary is $-n^{\prime }$. Let $t=\sigma n^{\prime }$ be traction on the body's surface from the fluid. For an exterior point and a flow vanishing at infinity, the boundary representation readsFor $r=|y|\gg a$, expand the single layer through its first moment and the double layer through its zeroth moment:The next single-layer and double-layer terms are both $O(r^{-3})$ at fixed body size and boundary data. Differentiating $J$ givesHence the dipole terms combine intoThe antisymmetric part obeys $(\Sigma -{\Sigma }^T)y=G\times y$ for $G_j=-{ϵ}_{j\alpha \beta }{\Sigma }_{\alpha \beta }$. For the symmetric trace-free tensor $S$, $y\cdot My=y\cdot Sy+(\mathrm{Tr}M)r^2/3$. Also $\mathrm{Tr}\Sigma -\mathrm{Tr}M=2\mu \mathrm{Tr}U=2\mu Q$. Substitution givesThis derives the Stokes far-field multipoles of a deforming body. $F$ is the force exerted by the body on the fluid, the negative of the integrated fluid traction. $G$ is its torque on the fluid: $G=-\int x\times tdS$. Their kernels are the Stokeslet and rotlet. $Q=\int u\cdot n^{\prime }dS$ is the rate of increase of the body's volume, by the Reynolds transport theorem. It gives a radial source flow with volume flux $Q$. A volume-preserving body has $Q=0$. The trace-free tensor $S$ gives the stresslet, which can remain nonzero for a force-free, torque-free swimmer.