# Continuum mechanics/Relations between surface and volume integrals
## Surface-volume integral relation 1

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 $\Omega$ and let ${\boldsymbol {S}}$ be a second-order tensor field on $\Omega$. Show that
    $\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\cdot \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}~.$
Proof:
Recall the relation
    ${\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]=\mathbf {a} \cdot [\{{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})\}\cdot \mathbf {b} ]~.$
Integrating over the volume, we have
    $\int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\Omega }\mathbf {a} \cdot [\{{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})\}\cdot \mathbf {b} ~{\text{dV}}~.$
Since $\mathbf {a}$ and $\mathbf {b}$ are constant, we have
    $\int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\mathbf {a} \cdot \left[\left\{\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}\right\}\cdot \mathbf {b} \right]~.$
From the divergence theorem,
    $\int _{\Omega }{\boldsymbol {\nabla }}\bullet \mathbf {u} ~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {u} \cdot \mathbf {n} ~{\text{dA}}$
we get
    $\int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\partial {\Omega }}[(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]\cdot \mathbf {n} ~{\text{dA}}~.$
Using the relation
    $[(\mathbf {v} \bullet \mathbf {a} )({\boldsymbol {S}}\bullet \mathbf {b} )]\cdot \mathbf {n} =\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]$
we get
    $\int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\int _{\partial {\Omega }}\mathbf {a} \cdot [\{\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )\}\cdot \mathbf {b} ]~{\text{dA}}~.$
Since $\mathbf {a}$ and $\mathbf {b}$ are constant, we have
    $\int _{\Omega }{\boldsymbol {\nabla }}\bullet [(\mathbf {v} \cdot \mathbf {a} )({\boldsymbol {S}}\cdot \mathbf {b} )]~{\text{dV}}=\mathbf {a} \cdot \left[\left\{\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}\right\}\cdot \mathbf {b} \right]~.$
Therefore,
    $\mathbf {a} \cdot \left[\left\{\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}\right\}\cdot \mathbf {b} \right]=\mathbf {a} \cdot \left[\left\{\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}\right\}\cdot \mathbf {b} \right]~.$
Since $\mathbf {a}$ and $\mathbf {b}$ are arbitrary, we have
    ${\int _{\partial {\Omega }}\mathbf {v} \otimes ({\boldsymbol {S}}^{T}\bullet \mathbf {n} )~{\text{dA}}=\int _{\Omega }[{\boldsymbol {\nabla }}\mathbf {v} \cdot {\boldsymbol {S}}+\mathbf {v} \otimes ({\boldsymbol {\nabla }}\bullet {\boldsymbol {S}}^{T})]~{\text{dV}}}\qquad \qquad \qquad \square$

## Surface-volume integral relation 2

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