I'm unsure how to apply the divergence theorem correctly on a volume integral of following type of object: $\partial_{a} \partial_{b} \partial_{c} T^{a b c}$. The volume integral is over a $n$-dimensional Euclidean space. For clarity, let $(x_1,x_2,x_3,x_4)$ be the coordinate in this space. Now the partial derivatives $\partial_a = (\partial_{x_1},\partial_{x_2},\partial_{x_3},\partial_{x_4})$, same for $\partial_b$ and $\partial_c$ and $T^{a b c} = T^{a b c}(x_1,x_2,x_3,x_4)$. $$ \int_{R^n} \mathrm{d}^n V \partial_{a} \partial_{b} \partial_{c} T^{a b c} $$ and I'm supposed to write this as an integral over the surface of $R^n$. Initially I thought the answer was $$ \int_{S^{n-1}} \mathrm{d}^{n-1} S n_a \partial_{b} \partial_{c} T^{a b c} \quad [1] $$ but I'm uncertain. Especially, don't know of any alternatives, but it seems odd I'd have to choose which one of the contractions to remove, which makes me think I made a mistake. On the other hand I don't really see a way to prove this interpretation.
Say that $x$ has dimensions $x^{1}$, so that $\partial_a$ has dimensions $x^{-1}$. Then $\mathrm{d}^n V$ has dimensions $x^{n}$. Assume that $T^{a b c}$ has dimensions $x^{t}$. Then the result of the volume integral has dimensions $x^{n + t - 3}$. The same arguments, assuming $[n_a] = x^0$ give $x^{n-1 + t - 2} = x^{n + t - 3}$ as the dimensions for $[1]$, so they match.
Footnote: I'm a physics student. Mathematical explanation is appreciated, but please take it slowly.
Provided that $T^{abc}$ is smooth enough, $\partial_a \partial_b \partial_c T^{abc}=\partial_b \partial_a \partial_c T^{abc}$, and so on, so in fact only the symmetric part of $T^{abc}$ survives, and $$ \partial_a \partial_b \partial_c T^{abc} = \partial_a \partial_b \partial_c T^{(abc)}, $$ where the brackets take the standard meaning of symmetrisation, i.e., average over all the permutations of the indices. In particular, this means that when you apply the divergence theorem, the choice of $a$ as the index is irrelevant: you find that $$ \int n_a \partial_b \partial_c T^{abc} = \int n_a \partial_b \partial_c T^{(abc)}, $$ and so $n_a \partial_a \partial_b T^{(abc)} = n_b \partial_a \partial_c T^{(bac)}=n_b \partial_a \partial_c T^{(abc)}$, by relabelling and using the symmetry.