Divergence theorem with rank 2 tensor.

$\int_{V}\partial_{j} T_{ij}dV = \int_{\partial V}T_{ij}dS_{j}$ 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.

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 $i.$
• Thank you for answer, but I don't really understand the argument. How come you make them $\int_\Omega \partial_j (A_i T_{ij}) \, dV = \oint_{\partial \Omega} A_i T_{ij} \, dS_j$ in the first place, and why would it matter if it is valid for all constant vectors? – 123 Jul 19 '18 at 20:04
• That equality is just $\int_\Omega \vec{\nabla} \cdot \vec{F} \, dV = \oint_{\partial \Omega} \vec{F} \cdot d\vec{S}$ for $F_j = A_i T_{ij},$ and you know from before that this is true. – md2perpe Jul 19 '18 at 20:29
• If $A_i u_i = A_i v_i$ for all constant $A_i$ then $u_i = v_i$ (for all $i$). This is because you can take $\vec A = e_j,$ the unit vector field in the $j$th coordinate direction. But since $A_i = (e_j)_i = \delta_{ij}$ (Kronecker delta), you then get $u_j = v_j$ for that $j$. But since you can to that for any $j$ you get $u_j = v_j$ for all $j$. – md2perpe Jul 19 '18 at 20:36
• Yes, the product $A_i T_{ij}$ is a rank 1 tensor on which we can apply our already known theorem/formula. – md2perpe Jul 19 '18 at 20:49