Infinite plane gravity: what is "mass density per unit area"? Recently I learned that the gravity of an infinity plane is independent of the distance from that plane. In fact it is
$$g = 2\pi G \sigma$$
where $\sigma$ is "the mass density of the plane per unit area". 
I am struggling to understand what this actually means. I do understand mass density (per volume), but "per area"? Would this not always be zero?
Looking for example at a $2\,mm$ thick sheet of copper, where copper has a mass density of $\rho_{\text{Cu}} =8.92 \,g/cm³ $. What then is the $\sigma$ on the surface of the sheet? Is it just (at least approximately) the stacked density on each surface point, i.e. $\sigma = w\cdot\rho$ where $w=0.2\,cm$ is the thickness of the plate? 
What if the plate is not negligibly thick but, say $w=1\,km$?
Edit: removed reference to finite plane, some comments may no longer apply.
 A: Let’s start with 1D.
If you were to buy climbing rope, you might want to ask the vendor: “how much does a meter of rope weight?”. 
This question is the same as asking what is the mass per unit length of the rope. Yes, you might care what the actual density (mass per unit volume) of the rope is, but since the main variable will be the length, you abstract the girth away.
mass per unit length = cross sectional area * density
Same for the 2D case. Say you are buying material to build a sail. You want to know what is the mass per unit area because it is a more relevant piece of information than the density.
So you see, we are not talking about exactly 1 or 2 dimensional objects (which should have a density of 0) but of situations in which only one or two dimensions of the object in question matter.
In the case of an infinite plane, the way we derive that formula is to look how much mass is inside an (imagined) cylinder straddling the plane, where the bases of the cylinder (the two disks) are perpendicular to the plane. That mass will always be proportional to the area of the base, and completely independent of the height of the cylinder (can you see why?). So we don’t care about the density (or the thickness) of the plane itself.
A: If the sheet had non-zero thickness $\tau$, you'd have $\sigma=\rho\tau$. Now just take $\tau\to 0^+$ while holding $\sigma$ fixed.
