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,

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