Micromechanics of composites/Proof 5

Surface-volume integral relation 2
Let $$\Omega$$ be a body and let $$\partial{\Omega}$$ be its surface. Let $$\mathbf{n}$$ be the normal to the surface. Let $$\mathbf{v}$$ be a vector field on $$\Omega$$. Show that

\int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA}~. $$

 Proof:

Recall that

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\cdot\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} $$ where $$\boldsymbol{S}$$ is any second-order tensor field on $$\Omega$$. Let us assume that $$\boldsymbol{S} = \boldsymbol{\mathit{1}}$$. Then we have

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{\mathit{1}}\cdot\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{\mathit{1}} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{\mathit{1}})]~\text{dV} $$ Now,

\boldsymbol{\mathit{1}}\cdot\mathbf{n} = \mathbf{n} ~; \boldsymbol{\nabla} \bullet \boldsymbol{\mathit{1}} = \mathbf{0} ~; \boldsymbol{A}\cdot\boldsymbol{\mathit{1}} = \boldsymbol{A} $$ where $$\boldsymbol{A}$$ is any second-order tensor. Therefore,

\int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} ~. $$ Rearranging,

{ \int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA} \qquad \square } $$