= Solution
Choose the coordinate vectors $b=e_i$ and $c=e_j$ in the assumed bilinear identity. It gives $T_{ij}=0$ for each pair of indices, hence \b[the tensor is zero].
Let $\Omega$ be the enclosed region. Apply the <divergence theorem> to the vector <field> $(b\cdot x)c$. Since $b,c$ are constant,
$$
\nabla\cdot[(b\cdot x)c]=b\cdot c.
$$
Therefore
$$
\boxed{\int_S(b\cdot x)(c\cdot n)\,dS=V b\cdot c.}
$$
Writing the left side as $b_i c_j\int_S x_i n_j\,dS$ and using the initial tensor argument gives the <boundary integral of position and normal components>
$$
\boxed{\int_S x_i n_j\,dS=V\delta_{ij}.}
$$
Contract with the <Kronecker delta> to obtain $\int_S x\cdot n\,dS=3V$. Contract with the <Levi-Civita symbol> to obtain
$$
\left(\int_S x\times n\,dS\right)_k
=\epsilon_{kij}V\delta_{ij}=0.
$$
Thus the requested integrals are \b[$3V$ and the zero vector], respectively. These applications require the usual sufficiently regular closed boundary and outward-normal orientation of the divergence theorem.
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