Integrate over boundary and partial differentiation wrt r and n When solving a problem, suppose I am in $\Bbb R^3$ and I want to integrate over $B_R\setminus B_\epsilon$ with respect to $x$. 
My first question is that when using the Green's formula (Integration by parts), for the term which integrats over the boundary, I should have $$\int_{S_R}...dS+\int_{S_\epsilon}...dS \text{ or } \int_{S_R}...dS-\int_{S_\epsilon}...dS \text{ ?}$$  I think the '+' one is correct because the region of $B_R\setminus B_\epsilon$ has two boundaries?
And my second question is that, for $S_\epsilon$, let $||x||=r$, then I would have the external normal $n=\frac{-x}{||x||}$ (Since for $S_\epsilon$, the normal vector is pointing inwards or is it due to other reasons?), then I would have the derivative along the external normal $$\partial_n=\sum_{j=1}^{3}n_j\partial_{x_j}=-\sum_{i=1}^{3}\frac{x_i\partial_{x_i}}{||x||}=-\partial_r$$
I am confused about why the last equal sign holds? Could someone explain more details for me? And does the result $\partial_n=-\partial_r$ hold for higher dimensions?
I greatly appreciate any help and advice!
 A: Yes, you can take the plus sign in $\int_{S_R} + \int_{S_\epsilon}$, provided that you choose the normal vector $n$ to point outwards from the region $B_{R} \ \backslash \ B_{\epsilon}$. So on the $S_R$ surface, the normal vector should point away from the origin, but on the $S_\epsilon$ surface, the normal vector should point towards the origin, as you correctly identified.
For your second question, we can write $r$ as $$ r = \sqrt{x_1^2 + x_2^2 + x_3^2}.$$ Using the chain rule, we get
$$ \frac{\partial}{\partial x_i} = \frac{\partial r}{\partial x_i } \frac{\partial}{\partial r} = \frac{x_i}{\sqrt{x_1^2+x_2^2+x_3^2}}\frac{\partial}{\partial r} = \frac{x_i}{r} \frac{\partial}{\partial r}$$
for each  $i \in \{ 1,2,3 \}$.
But then,
$$ \sum_{i=1}^3 \frac{x_i}{|| x_i ||} \frac{\partial}{\partial x_i}  = \sum_{i=1}^3 \frac{x_i}{r} \frac{\partial}{\partial x_i} = \sum_{i=1}^3\frac{x_i^2}{r^2} \frac{\partial}{\partial r} = \frac{x_1^2 + x_2^2 + x_3^2}{r^2} \frac{\partial}{\partial r} = \frac{\partial}{\partial r},$$
as required.
Hopefully it is clear that this derivation of $\sum_i \frac{x_i}{|| x_i ||} \frac{\partial}{\partial x_i} = \frac{\partial}{\partial r}$ works in any number of dimensions.
