= Solution
The <rate-of-strain tensor> is $e_{ij}=\frac12(\partial_ju_i+\partial_iu_j)$. For an incompressible <Newtonian fluid>, the <stress tensor> is $\sigma_{ij}=-p\delta_{ij}+2\mu e_{ij}$. Dotting the momentum equation with velocity and using the product rule gives the local mechanical-energy balance
$$
\partial_t(\tfrac12\rho|u|^2)+\partial_j(\tfrac12\rho|u|^2u_j)=\partial_j(u_i\sigma_{ij})-\sigma_{ij}\partial_ju_i
$$
without body forces. Write $\partial_ju_i=e_{ij}+\omega_{ij}$, with $\omega_{ij}$ antisymmetric. The pressure contribution is $-p\partial_iu_i=0$, and $e_{ij}\omega_{ij}=0$ by symmetry. Thus the remaining loss is $2\mu e_{ij}e_{ij}\ge0$, proving \b[the local viscous dissipation rate is $2\mu e_{ij}e_{ij}$ per unit volume].
For the oscillating boundary, take $u=(U(y,t),0,0)$ and constant pressure. The nonlinear advection vanishes, leaving $U_t=\nu U_{yy}$, $\nu=\mu/\rho$. Seek the real part of $V e^{i\omega t-\gamma y}$; decay at infinity selects $\gamma=(1+i)/\delta$, $\delta=\sqrt{2\nu/\omega}$. After transients,
$$
\boxed{U(y,t)=V e^{-y/\delta}\cos(\omega t-y/\delta).}
$$
Only $e_{xy}=e_{yx}=U_y/2$ is nonzero, so the dissipation is $\mu U_y^2$. Its time average is $\mu V^2\delta^{-2}e^{-2y/\delta}$: the two sine and cosine contributions each average to one half. Integrating over $y>0$ gives
$$
\boxed{\left\langle\frac{\text{dissipation}}{\text{boundary area}}\right\rangle=\frac{\mu V^2}{2\delta}=V^2\sqrt{\frac{\mu\rho\omega}{8}}.}
$$
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