# Continuum mechanics/Relations between surface and volume integrals

## Surface-volume integral relation 1

Let ${\displaystyle \Omega }$  be a body and let ${\displaystyle \partial {\Omega }}$  be its surface. Let ${\displaystyle \mathbf {n} }$  be the normal to the surface. Let ${\displaystyle \mathbf {v} }$  be a vector field on ${\displaystyle \Omega }$  and let ${\displaystyle {\boldsymbol {S}}}$  be a second-order tensor field on ${\displaystyle \Omega }$ . Show that

${\displaystyle \int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}~.}$

Proof:

Recall the relation

${\displaystyle {\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]=\mathbf {a} \cdot [\{{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})\}\cdot \mathbf {b} ]~.}$

Integrating over the volume, we have

${\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\Omega }\mathbf {a} \cdot [\{{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})\}\cdot \mathbf {b} ~{\text{dV}}~.}$

Since ${\displaystyle \mathbf {a} }$  and ${\displaystyle \mathbf {b} }$  are constant, we have

${\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\mathbf {a} \cdot \left[\left\{\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}\right\}\cdot \mathbf {b} \right]~.}$

From the divergence theorem,

${\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet \mathbf {u} ~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {u} \cdot \mathbf {n} ~{\text{dA}}}$

we get

${\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\partial {\Omega }}[(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]\cdot \mathbf {n} ~{\text{dA}}~.}$

Using the relation

${\displaystyle [(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]\cdot \mathbf {n} =\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]}$

we get

${\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]~{\text{dA}}~.}$

Since ${\displaystyle \mathbf {a} }$  and ${\displaystyle \mathbf {b} }$  are constant, we have

${\displaystyle \int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\mathbf {a} \cdot \left[\left\{\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}\right\}\cdot \mathbf {b} \right]~.}$

Therefore,

${\displaystyle \mathbf {a} \cdot \left[\left\{\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}\right\}\cdot \mathbf {b} \right]=\mathbf {a} \cdot \left[\left\{\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}\right\}\cdot \mathbf {b} \right]~.}$

Since ${\displaystyle \mathbf {a} }$  and ${\displaystyle \mathbf {b} }$  are arbitrary, we have

${\displaystyle {\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}}\qquad \qquad \qquad \square }$

## Surface-volume integral relation 2

Let ${\displaystyle \Omega }$  be a body and let ${\displaystyle \partial {\Omega }}$  be its surface. Let ${\displaystyle \mathbf {n} }$  be the normal to the surface. Let ${\displaystyle \mathbf {v} }$  be a vector field on ${\displaystyle \Omega }$ . Show that

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

Proof:

Recall that

${\displaystyle \int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}}$

where ${\displaystyle {\boldsymbol {S}}}$  is any second-order tensor field on ${\displaystyle \Omega }$ . Let us assume that ${\displaystyle {\boldsymbol {S}}={\boldsymbol {\mathit {1}}}}$ . Then we have

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

Now,

${\displaystyle {\boldsymbol {\mathit {1}}}\cdot \mathbf {n} =\mathbf {n} ~;~~{\boldsymbol {\nabla }}\bullet {\boldsymbol {\mathit {1}}}=\mathbf {0} ~;~~{\boldsymbol {A}}\cdot {\boldsymbol {\mathit {1}}}={\boldsymbol {A}}}$

where ${\displaystyle {\boldsymbol {A}}}$  is any second-order tensor. Therefore,

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

Rearranging,

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