Micromechanics of composites/Proof 4

Question
Let $$\partial{\Omega}$$ be a surface. Let $$\mathbf{x}$$ be the position vector of points on the surface and let $$\mathbf{t}$$ be a vector field that are defined on $$\partial{\Omega}$$. If

\int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} = \mathbf{0} $$ show that

\int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{t}~\text{dA} = \int_{\partial{\Omega}} \mathbf{t}\otimes\mathbf{x}~\text{dA} ~. $$

Proof
If we assume a Cartesian basis, we can write the given relation in index notation as

\int_{\partial{\Omega}} e_{ijk}~x_j~t_k~\text{dA} = 0 $$ where $$e_{ijk}$$ is the Levi-Civita (permutation) symbol. Since $$e_{ijk}$$ is does not depend upon the position we can write

e_{ijk} \int_{\partial{\Omega}} ~x_j~t_k~\text{dA} = 0 ~. $$ Define

A_{jk} := \int_{\partial{\Omega}} x_j~t_k~\text{dA} ~. $$ Then,

e_{ijk} A_{jk} = 0 ~. $$ Expanding, we get

e_{i11} A_{11} + e_{i12} A_{12} + e_{i13} A_{13} + e_{i21} A_{21} + e_{i22} A_{22} + e_{i23} A_{23} + e_{i31} A_{31} + e_{i32} A_{32} + e_{i33} A_{33} = 0 \qquad i=1,2,3 ~. $$ Expanding further, we get three equations

\begin{align} e_{112} A_{12} + e_{113} A_{13} + e_{121} A_{21} + e_{123} A_{23} + e_{131} A_{31} + e_{132} A_{32} & = 0 \\ e_{212} A_{12} + e_{213} A_{13} + e_{221} A_{21} + e_{223} A_{23} + e_{231} A_{31} + e_{232} A_{32} & = 0 \\ e_{312} A_{12} + e_{313} A_{13} + e_{321} A_{21} + e_{323} A_{23} + e_{331} A_{31} + e_{332} A_{32} & = 0 \end{align} $$ or,

\begin{align} e_{123} A_{23} + e_{132} A_{32} & = 0 \\ e_{213} A_{13} + e_{231} A_{31} & = 0 \\ e_{312} A_{12} + e_{321} A_{21} & = 0 \end{align} $$ or,

A_{23} = A_{32} ~; A_{13} = A_{31} ~; A_{12} = A_{21} ~. $$ Therefore,

\int_{\partial{\Omega}} x_2~t_3~\text{dA} = \int_{\partial{\Omega}} x_3~t_2~\text{dA} = \int_{\partial{\Omega}} t_2~x_3~\text{dA} ~; \int_{\partial{\Omega}} x_1~t_3~\text{dA} = \int_{\partial{\Omega}} x_3~t_1~\text{dA} = \int_{\partial{\Omega}} t_1~x_3~\text{dA}~; \int_{\partial{\Omega}} x_1~t_2~\text{dA} = \int_{\partial{\Omega}} x_2~t_1~\text{dA} = \int_{\partial{\Omega}} t_1~x_2~\text{dA} ~. $$ Also, by symmetry,

\int_{\partial{\Omega}} x_1~t_1~\text{dA} = \int_{\partial{\Omega}} t_1~x_1~\text{dA} ~; \int_{\partial{\Omega}} x_2~t_2~\text{dA} = \int_{\partial{\Omega}} t_2~x_2~\text{dA} ~; \int_{\partial{\Omega}} x_3~t_3~\text{dA} = \int_{\partial{\Omega}} t_3~x_3~\text{dA} ~. $$ Therefore we may write,

\int_{\partial{\Omega}} x_j~t_k~\text{dA} = \int_{\partial{\Omega}} t_j~x_k~\text{dA} ~. $$ Reverting back to direct tensor notation, we get

{ \int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{t}~\text{dA} = \int_{\partial{\Omega}} \mathbf{t}\otimes\mathbf{x}~\text{dA} ~. } $$