I saw recently the follow gradient definition
$$\nabla\phi = \lim_{\Delta\text{Vol}\to0}\frac{1}{\Delta\text{Vol}} \int_{\partial\text{Vol}}\phi\ \text{d}\vec{A}$$
I can't get that definition using $\lim$ and $\int_{\partial\text{Vol}}$ to get a differential operator. $\phi$ is a scalar function.
Can someone help me to understand? Regards!
OK, so to fix ideas I'll do this in $3$-space, but it works in any dimension.
Fix $\mathbf a$ and let $B(\mathbf a,r)$ be a ball of radius $r$ centered at $\mathbf a$. (You can use other shapes, too, if you so desire, but they do have to shrink down to the point $\bf a$ in a reasonable manner.) Let's consider the first coordinate of your equation. To get $\frac{\partial\phi}{\partial x}(\mathbf a)$ I want to take the first coordinate of the right-hand side, so I consider $$\iint_{\partial B(\mathbf a,r)} (\phi\vec i)\cdot d\vec A.$$ (So we're going to consider the flux of the vector field $\vec F = \phi\vec i$ across the surface. The divergence of $\vec F$ is $\partial\phi/\partial x$.) By the Divergence Theorem, this integral is equal to $$\iiint_{B(\mathbf a,r)} \frac{\partial\phi}{\partial x}\,dV \approx \frac{\partial\phi}{\partial x}(\mathbf a)\text{vol}(B(\mathbf a,r)),$$ by continuity. Thus, to get the correct statement in what you posted, you need to take the limit of the integral divided by the volume of the region. (Obviously, if you do not do that, the RHS just goes to $0$.) That is, the correct statement is $$\nabla\phi(\mathbf a) = \lim_{r\to 0}\frac 1{\text{vol}(B(\mathbf a,r))}\iint_{\partial B(\mathbf a,r)} \phi\,d\vec A.$$