= Solution
{c}
Using $a=A$ and the end-face <shear stress> from part (c), the printed <torque> normalization gives
$$
M(\tau)=2\tau\int_0^A\left[\mu_1+\mu_2(3+\tau^2r^2)^n\right]r^3\,dr.
$$
The <binomial theorem> expands the remaining <integral> into powers of $r$:
$$
(3+\tau^2r^2)^n
=\sum_{k=0}^n\binom nk3^{n-k}\tau^{2k}r^{2k}.
$$
Integrating each term gives an expression valid also at $\tau=0$:
$$
\boxed{M(\tau)=\frac{\mu_1A^4}{2}\tau
+\mu_2\sum_{k=0}^n\binom nk
\frac{3^{n-k}A^{2k+4}}{k+2}\tau^{2k+1}.}
$$
For example, $n=1$ gives $M=(\mu_1+3\mu_2)A^4\tau/2+\mu_2A^6\tau^3/3$. The <torque> is an odd <polynomial> in the twist per unit reference length, as expected when reversing the twist reverses the end-face <shear stress>.
For a physical end-face <torque> $N$, integration over the azimuth normally gives
$$
N=\int_0^{2\pi}\!\int_0^A T_{\theta z}r^2\,dr\,d\theta=2\pi M.
$$
The original PDF explicitly defines $M$ without this $2\pi$, so the boxed answer retains its convention. Since $\mathbf T$ is the <Cauchy stress tensor>, its shear component here is spatial $T_{\theta z}$; the printed capital $Z$ in $T_{\theta Z}$ does not require a change to a <nominal stress tensor>.
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