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,

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