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 $𝐧$ be the normal to the surface. Let $𝐯$ be a vector field on $\Omega$ and let $𝑺$ be a second-order tensor field on $\Omega$. Show that

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

Proof:

Recall the relation

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

Integrating over the volume, we have

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

Since $𝐚$ and $𝐛$ are constant, we have

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

From the divergence theorem,

${\int }_{\Omega }\mathrm{\nabla }\bullet 𝐮\text{dV}={\int }_{\partial \Omega }𝐮\cdot 𝐧\text{dA}$

we get

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

Using the relation

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

we get

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

Since $𝐚$ and $𝐛$ are constant, we have

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

Therefore,

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

Since $𝐚$ and $𝐛$ are arbitrary, we have

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

← Curl of a gradient of a vector

Continuum mechanics

Leibniz formula →

Surface-volume integral relation 2

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

${\int }_{\Omega }\mathrm{\nabla }𝐯\text{dV}={\int }_{\partial \Omega }𝐯\otimes 𝐧\text{dA}.$

Proof:

Recall that

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

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

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

Now,

$1\cdot 𝐧=𝐧;\mathrm{\nabla }\bullet 1=\mathrm{𝟎};𝑨\cdot 1=𝑨$

where $𝑨$ is any second-order tensor. Therefore,

${\int }_{\partial \Omega }𝐯\otimes 𝐧\text{dA}={\int }_{\Omega }\mathrm{\nabla }𝐯\text{dV}.$

Rearranging,

${\int }_{\Omega }\mathrm{\nabla }𝐯\text{dV}={\int }_{\partial \Omega }𝐯\otimes 𝐧\text{dA}◻$

← Curl of a gradient of a vector

Continuum mechanics

Leibniz formula →