= Solution
Multiply <equilibrium> by the <displacement> and integrate. <Integration by parts> and the <traction> condition give
$$
0=\int_B\mathbf u\cdot(\nabla\cdot T+\rho\mathbf g)\,dV
=\int_{\partial B}\mathbf u\cdot\boldsymbol\sigma\,dS-\int_BT:\nabla\mathbf u\,dV+\int_B\rho\mathbf u\cdot\mathbf g\,dV.
$$
The <stress> is symmetric, so it contracts only with the symmetric part of $\nabla\mathbf u$, namely $e$. <Hooke's law> then gives $T:e=\lambda(\operatorname{tr}e)^2+2\mu e:e=2W$. Hence
$$
\boxed{\int_B\rho\mathbf u\cdot\mathbf g\,dV=\int_B2W\,dV-\int_{\partial B}\mathbf u\cdot\boldsymbol\sigma\,dS.}
$$
The identity is the linear-elastic work balance: twice the stored quadratic energy equals the body-force and boundary-<traction> work paired with the final <displacement>.
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