= Solution
The same <Maxwell equations> give
$$
\partial_t\mathbf u^B=\frac12\nabla\times\partial_t\mathbf B
=-\frac12\nabla\times(\nabla\times\mathbf E)=\frac12\nabla^2\mathbf E,
$$
where the last equality uses $\nabla\cdot\mathbf E=0$. Combining with part (ii),
$$
\partial_t(u_i^E+u_i^B)=\frac12\nabla^2E_i-\frac{c^2}{2}\nabla^2B_i.
$$
The divergences of the symmetric gradients satisfy
$$
\partial_jS^E_{ij}=\frac12(\partial_i\partial_jE_j+\partial_j\partial_jE_i)=\frac12\nabla^2E_i,
\qquad \partial_jS^B_{ij}=\frac12\nabla^2B_i.
$$
Thus the local identity is $\partial_t(u_i^E+u_i^B)=\partial_j(S^E_{ij}-c^2S^B_{ij})$. Integrate over the fixed volume and apply the <divergence theorem> componentwise:
$$
\boxed{\frac\partial{\partial t}\int_V(u_i^E+u_i^B)\,dV
=\oint_A(S^E_{ij}-c^2S^B_{ij})\,dA_j}.
$$
Here $dA_j=n_jdA$ uses the outward <normal vector>.
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