= Solution
The <traction> $\mathbf t(\mathbf n)$ is force per unit area across an oriented surface with unit normal $\mathbf n$. The stress component $\sigma_{ij}$ is its $i$th component on a surface normal to the $j$th coordinate axis. <Cauchy's stress theorem> states that this dependence is linear:
$$
\boxed{t_i(\mathbf n)=\sigma_{ij}n_j.}
$$
For the boundary of a <material volume>, take the outward normal and the <traction> exerted by the surrounding fluid. Newton's law and the <divergence theorem> give
$$
\frac d{dt}\int_{V(t)}\rho u_i\,dV
=\int_{\partial V(t)}\sigma_{ij}n_j\,dS
=\int_{V(t)}\partial_j\sigma_{ij}\,dV.
$$
Constant <density> and <incompressibility> make the transport-theorem integrand on the left $\rho Du_i/Dt$. The volume is arbitrary, so
$$
\boxed{\rho\frac{Du_i}{Dt}=\partial_j\sigma_{ij}.}
$$
Substitute the Newtonian constitutive law. With constant viscosity,
$$
\partial_j\sigma_{ij}=-\partial_i p+\mu\partial_j\partial_j u_i
+\mu\partial_i(\partial_j u_j)=-\partial_i p+\mu\nabla^2u_i.
$$
Hence
$$
\boxed{\rho(\mathbf u_t+(\mathbf u\cdot\nabla)\mathbf u)
=-\nabla p+\mu\nabla^2\mathbf u.}
$$
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