Solution

Let us first look at the first term of equation~(1) and apply the divergence theorem (2). Thus,

${\displaystyle {\begin{aligned}\int _{\partial B}e_{ijk}~x_{j}~n_{l}~\sigma _{lk}~dS&=\int _{\partial B}n_{l}~(e_{ijk}~x_{j}~\sigma _{lk})~dS\\&=\int _{B}{\frac {\partial {(e_{ijk}~x_{j}~\sigma _{lk})}}{\partial x_{l}}}~dV\\&=\int _{B}\left(e_{ijk}~{\frac {\partial x_{j}}{\partial x_{l}}}~\sigma _{lk}+e_{ijk}~x_{j}~{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ijk}~\delta _{jl}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\\&=\int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV\end{aligned}}}$

Plugging this back into equation~(1) gives

${\displaystyle \int _{B}\left(e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}{\frac {\partial \sigma _{lk}}{\partial x_{l}}}\right)~dV+\int _{B}\rho ~e_{ijk}~x_{j}~b_{k}~dV={\frac {d}{dt}}\left(\int _{B}\rho ~e_{ijk}~x_{j}~v_{k}~dV\right)}$

Therefore, bringing all terms to the left hand side,

${\displaystyle {\text{(5)}}\qquad \int _{B}\left[e_{ilk}~\sigma _{lk}+e_{ijk}~x_{j}\left({\frac {\partial \sigma _{lk}}{\partial x_{l}}}+\rho ~b_{k}-{\frac {d}{dt}}\left(\rho ~v_{k}\right)\right)\right]~dV=0}$

Using the equilibrium equations~(3), equation~(5) reduces to

${\displaystyle {\text{(6)}}\qquad \int _{B}e_{ilk}~\sigma _{lk}~dV=0}$

Since this holds for any ${\displaystyle B}$, we have

${\displaystyle {\text{(7)}}\qquad e_{ilk}~\sigma _{lk}=0}$

If you work this expression out, you will see that ${\displaystyle \sigma _{ij}=\sigma _{ji}}$. Hence, the stress tensor is symmetric.