= Solution
The product rule gives
$$
\partial_j(T_{ij}v_i)=(\partial_jT_{ij})v_i+T_{ij}\partial_jv_i.
$$
Integrating and applying the divergence theorem proves the identity.
Here $\operatorname{div}T=4(x,y,z)$ and $v=x(x,y,z)$, so
$$
\int_V\operatorname{div}T\cdot v\,dV
=4\int_Vx(x^2+y^2+z^2)\,dV=\frac73.
$$
On the boundary, $(Tn)\cdot v=x(x\cdot n)(x^2+y^2+z^2-1)$. The nonzero contributions from the faces $x=1,y=1,z=1$ are respectively $2/3,5/12,5/12$, totaling $3/2$.
Finally, direct contraction gives
$$
T_{ij}\partial_jv_i=2x(x^2+y^2+z^2)-4x,
$$
whose <integral> is $-5/6$. Thus the right side is $3/2-(-5/6)=7/3$, equal to the left side.
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