...if inside the value $(\nabla , \mathbf A ) = 0$, and on the surface of the value $ A_{n} = 0$.
So, I need to prove that
$$ \int \mathbf A \, dV = 0. \tag{.0} $$
I tried to prove it by the next way. $(\nabla , \mathbf A ) = 0$ means that the lines of the $ \mathbf A$-field are closed, and $ A_{n} = 0$ on the surface means that all the lines are inside the value and on the surface. So it means that the integral can be transformed into a number of integrals looks as
$$ \oint I \, d \mathbf l, \tag{.1} $$
where I is the analogue of closed current. So all of these integral's are equal to 0.
But I have some doubts about correctness of the transition from $\int \mathbf A \, dV$ to (.1) and I need to prove (.0) more strictly. Can you help me?
Consider the vector field $x\mathbf{A}$.
By the divergence theorem we have that
$$0 = \int_{\partial \Omega} x\mathbf{A}\cdot n = \int_{\Omega} \operatorname{div} (x\mathbf{A}) = \int_{\Omega} (A_x + x\operatorname{div} \mathbf{A}) = \int_{\Omega} A_x,$$
and similarly for the other components of $A$.