Integral of divergence equal to divergence of integral? Just as the heading reads...is the integral of the divergence of a vector field equal to the divergence of the integral of a vector field? 
$\int\nabla\cdot\vec U dz = 0$
same as
$\frac\partial\partial_x \int u(x,y,z)  dz +\frac\partial\partial_y \int v(x,y,z)  dz +\frac\partial\partial_z \int w(x,y,z)  dz =0$
Is this statement true? The integral is arbitrary and can be taken over any coordinate. Basically I have to integrate the divergence of a vector field and I'm not sure if I can simply move the divergence operator outside the integral?
OR should it read:
$\int \frac\partial\partial_x u(x,y,z)  dz + \int \frac\partial\partial_y  v(x,y,z)  dz + \int \frac\partial\partial_z w(x,y,z)  dz =0$
EDIT: Tried to make it more clear by example. The limits of integration cover the full scope of the z-coordinate which is bounded from say A to B.

For terms like $\frac{\partial}{\partial x} \int u(x,y,z) \mathrm dz$ you could apply the Leibniz integral rule, see https://en.wikipedia.org/wiki/Leibniz_integral_rule.
