= Solution
The <polar shear stress in a Newtonian fluid> is
$$
\sigma_{r\theta}
=\mu\left[
r\frac\partial{\partial r}\left(\frac{u_\theta}{r}\right)
+\frac1r\frac{\partial u_r}{\partial\theta}
\right].
$$
On $r=a$, no slip holds for every $\theta$, so $u_\theta=0$ and $\partial_\theta u_r=0$. Therefore
$$
\sigma_{r\theta}(a,\theta)
=\mu\,\partial_ru_\theta(a,\theta)
=-\mu\psi_{rr}(a,\theta).
$$
Differentiating the solution from part (ii) gives
$$
\psi_{rr}(a,\theta)=\gamma-2\gamma\cos2\theta,
$$
and hence
$$
\boxed{
\sigma_{r\theta}(a,\theta)
=-\mu\gamma+2\mu\gamma\cos2\theta
}.
$$
The moment arm and line element are both $a$, so the torque exerted by the fluid on the disk per unit axial length is
$$
\begin{aligned}
\mathcal T_z
&=a^2\int_0^{2\pi}\sigma_{r\theta}(a,\theta)\,d\theta\\
&=\boxed{-2\pi\mu\gamma a^2}.
\end{aligned}
$$
For $\gamma>0$ the negative sign denotes the clockwise sense of the ambient shear rotation. This agrees with the <hydrodynamic torque on a fixed disk in planar shear>.
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