= Solution
Use the same constant <dynamic viscosity> $\mu$ for both incompressible <Stokes flows>, and let $\mathbf f^{(a)}$ be <body force> per unit volume. Their equations and <Newtonian fluid stress tensor> are
$$
\partial_j\sigma^{(a)}_{ij}+f^{(a)}_i=0,\qquad
\partial_i u^{(a)}_i=0,\qquad
\sigma^{(a)}_{ij}=-p^{(a)}\delta_{ij}+2\mu e^{(a)}_{ij},
$$
where $e^{(a)}_{ij}=(\partial_i u^{(a)}_j+\partial_j u^{(a)}_i)/2$ is the <strain-rate tensor>. Consider the cross-work flux
$$
Q_j=u_i^{(2)}\sigma^{(1)}_{ij}-u_i^{(1)}\sigma^{(2)}_{ij}.
$$
By the product rule,
$$
\partial_jQ_j
=(\partial_j u_i^{(2)})\sigma^{(1)}_{ij}
-(\partial_j u_i^{(1)})\sigma^{(2)}_{ij}
-u_i^{(2)}f_i^{(1)}+u_i^{(1)}f_i^{(2)}.
$$
The pressure terms vanish by <incompressibility>. Symmetry of the <stress tensor> removes antisymmetric <velocity gradient> contributions, leaving $2\mu e^{(1)}:e^{(2)}$ in both cross terms; they cancel. Apply the <divergence theorem>, with $\mathbf n$ outward from $V$, and rearrange:
$$
\boxed{\begin{aligned}
\int_V\mathbf f^{(1)}\cdot\mathbf u^{(2)}\,dV
+\int_{\partial V}\mathbf u^{(2)}\cdot\boldsymbol\sigma^{(1)}\mathbf n\,dS
={}&\int_V\mathbf f^{(2)}\cdot\mathbf u^{(1)}\,dV\\
&+\int_{\partial V}\mathbf u^{(1)}\cdot\boldsymbol\sigma^{(2)}\mathbf n\,dS.
\end{aligned}}
$$
This is the <Lorentz reciprocal theorem with body forces>. It depends on reciprocal linear viscous response in the same fluid; arbitrary unrelated constitutive laws would not supply the cross-term cancellation.
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