# 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,

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