# Divergence theorem, but with a second order tensor formed via dyadic product

I'm currently working through a chapter on tensor analysis. One of the end-of-chapter questions asks me to show that, given a vector field $$\mathbf{v}$$ and a second order tensor field $$\mathbf{S}$$, $$\int_{\partial B} (\mathbf{S}\mathbf{n})\otimes\mathbf{v}\ dA = \int_{B}(\nabla \cdot\mathbf{S})\otimes\mathbf{v} + \mathbf{S}\nabla\mathbf{v}^{T} dV$$ Here, $$B$$ is the volume over which the integral is taken, with $$\partial B$$ representing its boundary. The vector $$\mathbf{n}$$ represents the outward normal field on $$\partial B$$. The question directs me to use the divergence theorem, which for second order tensors is $$\int_{\partial B} \mathbf{S}\mathbf{n}\ dA = \int_{B}(\nabla \cdot\mathbf{S})\ dV$$

This leads me to believe that, in some way, I have to rewrite $$(\mathbf{Sn})\otimes\mathbf{v}$$ in a way that $$\mathbf{v}$$ is included in the bracket, with $$\mathbf{n}$$ outside of it.

Is there any standard identity for this kind of representation? How does the transpose of the vector gradient of $$\textbf{v}$$ come about?

In index notation, the integrand is a matrix but it can also be written as the normal vector times a third-order tensor, i.e. \eqalign{ M &= (S\cdot n)\otimes v \quad&\implies\quad M_{ij} = S_{ik} n_k v_j \\ &= n\cdot T \quad&\implies\quad T_{kij} = S_{ik}v_j \\ } where $$T$$ is a third-order tensor which can be used in the divergence theorem \eqalign{ Q &= \int_{\partial B} n\cdot T\;\,dA = \int_B \left(\nabla\cdot T\right)\,dV \\ Q_{ij} &= \int_B \big(\partial_k T_{kij}\big)\,dV \\ &= \int_B \big(\partial_k S_{ik}\big)v_j + S_{ik}\big(\partial_k v_j\big)\;dV \\ Q &= \int_B \left(\nabla\cdot S^T\right)\otimes v + S\cdot\left(\nabla\otimes v\right)\;\;dV \\ }