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,

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

← Proof 4

Micromechanics of composites

Proof 6 →