= Solution
Use reference cylindrical coordinates $(R,\Theta,Z)$ and current coordinates $(r,\theta,z)$. The prescribed map has $r=\lambda^{-1/2}R$, $\theta=\Theta+\alpha Z$ and $z=\lambda Z$. In the corresponding orthonormal cylindrical bases, differentiate physical line elements to obtain the <deformation gradient>
$$
F=\begin{pmatrix}\lambda^{-1/2}&0&0\\0&\lambda^{-1/2}&\alpha R\lambda^{-1/2}\\0&0&\lambda\end{pmatrix},\qquad J=1.
$$
The radial direction is an eigenvector of the <right Cauchy-Green deformation tensor> with eigenvalue $\lambda^{-1}$. Its other block is
$$
\begin{pmatrix}\lambda^{-1}&\alpha R\lambda^{-1}\\\alpha R\lambda^{-1}&\lambda^2+\alpha^2R^2\lambda^{-1}\end{pmatrix}.
$$
Its trace is $s=\lambda^{-1}(1+\alpha^2R^2)+\lambda^2$ and its determinant is $\lambda$. Thus its eigenvalues solve $z^2-sz+\lambda=0$. The <principal stretches> are the positive square roots of these eigenvalues:
$$
\boxed{\lambda_1^2=\lambda^{-1},\qquad \lambda_{2,3}^2=\tfrac12\left(s\pm\sqrt{s^2-4\lambda}\right)}.
$$
Their product is one, independently checking <incompressibility>.
The <strain energy density> $W$ is per reference volume, so stored energy per initial cylinder height is $\mathcal E(\alpha,\lambda)=2\pi\int_0^A W(\lambda_1,\lambda_2,\lambda_3)R\,dR$. The end rotation is $\alpha H$ and the current height is $\lambda H$, giving applied power per initial height $M\dot\alpha+N\dot\lambda$. For quasistatic reversible loading, equate this with $\dot{\mathcal E}$ for independent changes of $\alpha,\lambda$. This proves the <torque and axial force of a twisted incompressible cylinder> formulas
$$
\boxed{M=2\pi\frac{\partial}{\partial\alpha}\int_0^A WR\,dR,\qquad N=2\pi\frac{\partial}{\partial\lambda}\int_0^A WR\,dR}.
$$
In the second derivative hold $\alpha$, the twist per initial length, fixed. Holding twist per current length instead would mix the generalized forces.
For the <Mooney-Rivlin solid>, compute invariants directly from $F$ rather than differentiating individual eigenvalue square roots:
$$
I_1=\lambda^2+2\lambda^{-1}+\alpha^2R^2\lambda^{-1},\qquad I_2=\lambda^{-2}+2\lambda+\alpha^2R^2\lambda^{-2}.
$$
The second identity follows from $I_2=\operatorname{tr}(F^TF)^{-1}$ when $J=1$. Retain the paper's minus sign on its $\mu_2$ term. Integrating $R$ and $R^3$ gives
$$
\mathcal E=\frac{\pi A^2}{2}\{\mu_1(\lambda^2+2\lambda^{-1}-3)-\mu_2(\lambda^{-2}+2\lambda-3)\}+\frac{\pi A^4\alpha^2}{4}(\mu_1\lambda^{-1}-\mu_2\lambda^{-2}).
$$
Consequently
$$
\boxed{M=\frac{\pi A^4\alpha}{2}(\mu_1\lambda^{-1}-\mu_2\lambda^{-2})},
$$
$$
\boxed{N=\pi A^2\{\mu_1(\lambda-\lambda^{-2})-\mu_2(1-\lambda^{-3})\}+\frac{\pi A^4\alpha^2}{4}(-\mu_1\lambda^{-2}+2\mu_2\lambda^{-3})}.
$$
As checks, untwisted unit stretch gives $M=N=0$, and setting $\mu_2=0$ recovers the <neo-Hookean solid> formulas. A convention with a positive second-invariant coefficient corresponds to replacing this $\mu_2$ by its negative.
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