Micromechanics of composites/Proof 5

Surface-volume integral relation 2

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

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

Proof:

Recall that

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

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

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

Now,

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

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

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

Rearranging,

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

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