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Up to date as of January 14, 2010

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Surface-volume integral relation 1

Let Ω 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 Ω and let \boldsymbol{S} be a second-order tensor field on Ω. 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 Ω 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 Ω. 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 Ω. 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 }

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