$\int_{V}\partial_{j} T_{ij}dV = \int_{\partial V}T_{ij}dS_{j} $ is the divergence theorem for a second rank tensor. I need to show that this is true.I tried to mimic the proof for "the normal" divergence theorem but couldn't succed:
$\int_{V}\vec{∇}\cdot \vec{F}dV =\int_{\partial V}\vec{F}\cdot d\vec{S}$,
$V=[(x,y,z)|(x,y) \in D, f(x,y)<z<g(x,y)]$,
$\int_{V}(\partial_{x} F_{x}+\partial_{y} F_{y}+\partial_{z} F_{z})dxdydz$ this is where it stopps. How does one interpet this to a second rank tensor?
Does any one know a way to tackle this problem? Thank you for answers.
Best Answer
First note that $\int_\Omega \vec{\nabla} \cdot \vec{F} \, dV = \oint_{\partial \Omega} \vec{F} \cdot d\vec{S}$ can be written as $\int_\Omega \partial_i F_i \, dV = \oint_{\partial\Omega} F_i dS_i.$
Now let $\vec A = (A_i)$ be a constant vector field. Then $$ A_i \int_\Omega \partial_{j} T_{ij} \, dV = \int_\Omega \partial_j (A_i T_{ij}) \, dV = \oint_{\partial \Omega} A_i T_{ij} \, dS_j = A_i \oint_{\partial \Omega} T_{ij} \, dS_j. $$ This is valid for all constant $\vec A,$ e.g. for $\vec A = e_1, e_2, e_3$ which gives that $$\int_\Omega \partial_j T_{ij} \, dV = \oint_{\partial\Omega} T_{ij} \, dS_j$$ for $i=1,2,3,$ i.e. the formula is valid for all $i.$