= Solution
Take $\mathbf F,\mathbf G$ to be the force and couple exerted by the body on the fluid, equivalently the external force and couple needed to maintain its motion. This fixes the sign for a positive <hydrodynamic resistance matrix>. If one uses the fluid's force on the body, both resultants have the opposite sign.
For two <Stokes flows> $(\mathbf u,\boldsymbol\sigma)$ and $(\widetilde{\mathbf u},\widetilde{\boldsymbol\sigma})$ in the same fluid domain $\mathcal D$, with zero body force, the <Lorentz reciprocal theorem for Stokes flow> is
$$
\boxed{\int_{\partial\mathcal D}\mathbf u\cdot\widetilde{\boldsymbol\sigma}\mathbf n\,dS
=\int_{\partial\mathcal D}\widetilde{\mathbf u}\cdot\boldsymbol\sigma\mathbf n\,dS.}
$$
Here $\mathbf n$ points out of the fluid. To see the identity, take the divergence of the difference of the two cross-work fluxes. The stress divergences vanish, while <incompressibility> and symmetry of the <Newtonian fluid stress tensor> reduce the remaining terms to $2\mu(\mathbf e:\widetilde{\mathbf e}-\widetilde{\mathbf e}:\mathbf e)=0$. The <divergence theorem> proves the result.
On the body, the <no-slip boundary condition> is $\mathbf u=\mathbf U+\boldsymbol\Omega\times\mathbf r$, and the far-field contribution vanishes for decaying exterior flows. Reciprocity becomes
$$
\mathbf U\cdot\widetilde{\mathbf F}+\boldsymbol\Omega\cdot\widetilde{\mathbf G}
=\widetilde{\mathbf U}\cdot\mathbf F+\widetilde{\boldsymbol\Omega}\cdot\mathbf G.
$$
With $q=(\mathbf U,\boldsymbol\Omega)^T$ and $(\mathbf F,\mathbf G)^T=\mathsf Rq$, this is $q^T\mathsf R\widetilde q=\widetilde q^T\mathsf Rq$ for all pairs, so $\mathsf R^T=\mathsf R$. The power identity gives
$$
q^T\mathsf Rq=\mathbf F\cdot\mathbf U+\mathbf G\cdot\boldsymbol\Omega
=2\mu\int_{\mathcal D}\mathbf e:\mathbf e\,dV\ge0.
$$
Equality would force zero strain throughout the connected exterior domain, hence a rigid fluid motion. Decay at infinity makes that motion zero, and the <no-slip boundary condition> then gives $q=0$. This proves the <symmetry and positivity of a rigid-body resistance matrix>:
$$
\boxed{\mathsf R^T=\mathsf R,\qquad q^T\mathsf Rq>0\text{ for }q\ne0.}
$$
For the <helix>, assume $0<\phi<\pi/2$ and use cylindrical unit vectors. Its arclength element and tangent are
$$
\frac{ds}{d\theta}=\left|\frac{d\mathbf X}{d\theta}\right|=b\sec\phi,
\qquad\mathbf t=\frac{d\mathbf X}{ds}=\cos\phi\,\mathbf e_\theta+\sin\phi\,\mathbf e_z.
$$
Integrating to $\theta=L\cos\phi/b$ gives $\boxed{\text{total arc length}=L}$. In a combined axial translation and rotation, $\mathbf V=U\mathbf e_z+b\Omega\mathbf e_\theta$. The <slender-body force density>, or local <resistive-force theory>, gives
$$
\mathbf f=C\left[\mathbf V-\frac12\mathbf t(\mathbf t\cdot\mathbf V)\right],
\qquad C=\frac{4\pi\mu}{|\log\epsilon|}.
$$
Writing $s_\phi=\sin\phi$, $c_\phi=\cos\phi$, its axial and azimuthal components are
$$
f_z=C\left[(1-s_\phi^2/2)U-bs_\phi c_\phi\Omega/2\right],\qquad
f_\theta=C\left[b(1-c_\phi^2/2)\Omega-s_\phi c_\phi U/2\right].
$$
The axial couple density is $bf_\theta$. Integration over arclength gives the <axial resistance matrix of a slender helix>:
$$
\boxed{\begin{pmatrix}F_z\\G_z\end{pmatrix}
=\begin{pmatrix}A&B\\B&D\end{pmatrix}\begin{pmatrix}U\\\Omega\end{pmatrix},\quad
A=\frac{CL}{2}(1+c_\phi^2),\quad
B=-\frac{CLb}{2}s_\phi c_\phi,\quad
D=\frac{CLb^2}{2}(1+s_\phi^2).}
$$
Thus pure translation gives $(F_z,G_z)=(AU,BU)$, and pure rotation gives $(F_z,G_z)=(B\Omega,D\Omega)$. The mixed coefficients agree, as reciprocity requires, and $AD-B^2=C^2L^2b^2/2>0$. The sign of $B$ follows the handedness in the parametrization; reversing handedness reverses propulsion.
For the <helical microswimmer with a spherical head>, the diagram's rotation relation is $\Omega_0=\Omega-\omega$. Neutral buoyancy and the absence of external forcing make the whole swimmer <force-free> and <torque-free>. Neglecting interactions between its parts gives
$$
(A_0+A)U+B\Omega=0,\qquad
BU+D\Omega+D_0(\Omega-\omega)=0.
$$
With $\Delta=(A_0+A)(D_0+D)-B^2>0$, solving this pair gives
$$
\boxed{U=-\frac{BD_0\omega}{\Delta},\qquad
\Omega=\frac{(A_0+A)D_0\omega}{\Delta},\qquad
\Omega_0=-\frac{(A_0+A)D-B^2}{\Delta}\omega.}
$$
The head counterrotates, providing the reaction to the flagellum's rotation. The figure fixes this relative rotation convention; it is duplicated and corrupted in the local TeX.
For a very large head, the coefficient regime is $A_0\gg A$ and $D_0\gg D$. Consequently
$$
\boxed{U\simeq-\frac{B\omega}{A_0},\qquad\Omega\simeq\omega\quad(a\gg L).}
$$
The large translational resistance $A_0$ makes the <microswimmer> slow even though the flagellum rotates almost at the motor rate.
For a very small head, $A_0\ll A$ and $D_0\ll D$. The appropriate denominator retains the translation-rotation coupling:
$$
\boxed{U\simeq-\frac{BD_0\omega}{AD-B^2},\qquad
\Omega\simeq\frac{AD_0\omega}{AD-B^2}\quad(a\ll(Lb^2)^{1/3}).}
$$
Both tend to zero with $D_0\propto a^3$, while $\Omega_0\simeq-\omega$. The motor mainly rotates the low-resistance head, and produces little flagellar motion relative to the fluid.
In the intermediate coefficient regime $A_0\ll A$ and $D_0\gg D$, the head supplies a strong rotational reaction with little translational penalty. Then
$$
\boxed{U\simeq-\frac{B\omega}{A}
=\omega b\frac{\sin\phi\cos\phi}{1+\cos^2\phi},\qquad\Omega\simeq\omega.}
$$
The head radius has dropped out. To find the <optimal pitch of a helical microswimmer>, put $z=\tan\phi$. The dimensionless speed becomes $z/(z^2+2)$ and has derivative $(2-z^2)/(z^2+2)^2$. Hence, for $\omega>0$ with the chosen handedness,
$$
\boxed{\tan\phi_{\mathrm{opt}}=\sqrt2,\qquad
\phi_{\mathrm{opt}}\simeq54.74^\circ,\qquad U_{\max}=\frac{\omega b}{2\sqrt2}.}
$$
The angle is measured from the horizontal plane. An intermediate pitch combines axial and transverse tangent directions, allowing anisotropic drag to convert rotation into translation; either extreme removes that coupling.
The geometric head-size regimes quoted in the paper hold with the logarithmic slenderness factor treated as fixed. More precisely, $A_0\ll A$ requires $a\ll L(1+\cos^2\phi)/(3|\log\epsilon|)$, while $D_0\gg D$ requires $a^3\gg Lb^2(1+\sin^2\phi)/(4|\log\epsilon|)$. These coefficient inequalities state the validity of the intermediate approximation when the logarithm is quantitatively important. The many-turn and local-drag assumptions of <slender-body theory> also remain in force.
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