Electrostatics: Charge distribution and Dirac's delta, lots of questions

[fluidistic]

The problem was to give the charge distribution of a uniformly charged disk of radius R centered at the origin and perpendicular to the z-axis.
The professor explained us how to do so (though many parts were unclear to me and I couldn't copy everything that was on the blackboard because we were too much students and I couldn't see the bottom part of the blackboard).
For a punctual charge, the general form of the charge distribution is \rho (\vec x) = q \delta (x) \delta (y) \delta (z).
For a charged plane: \rho (\vec x) = q \delta (z) and for a charged line: \rho (x,y,z)=q \delta (x) \delta (y).

However for a finite (in extension) plane perpendicular to the z-axis, \rho (\vec x) = f(x,y)\delta (z). The professor told us that the Dirac's delta "reduces" the dimension (though I don't understand why/how. If you have any comment on his, this would be nice). That's why for a point-like charge we need 3 of these deltas, 1 for a plane and 2 for a line.
The purpose of the function f in the above example is to make the final value coincide with what it should be; that's what I understood.
Using cylindrical coordinates (\rho, z, \phi), for the uniformly charged disk of radius R we have that \rho (\vec x)=\delta (z) \theta (\rho -R) f(\rho) where \theta (\rho - R) = 1 for \rho \leq R and 0 for \rho \geq R. Thus the theta function is a constraint that delimit the disk of radius R.
Then I don't really know what the professor has done, some integration I believe and he reached a final result I couldn't copy.
In spherical coordinates (r, \theta, \phi ) and for the same problem, he reached that \rho (\vec x) =\frac{Q}{\pi R^2 r} \delta (\theta - \frac{\pi}{2}) \hat \theta (r-R) which looks different from the result he reached in cylindrical coordinates but it must be the same.
If I understand well the final result, it has no sense to ask for the charge distribution in a particular point in \mathbb{R}^3. However it makes sense to ask for the charge distribution in any surface element; though I don't know how to proceed. I think I should do an integration or so, but I'm not sure and an example would be welcome.
So the final answer to the question is a function that is ill defined because for example in the origin it is not defined nor at the boundaries although there are charges there.
And if I'm not wrong, if I do integrate the final answer over an area (or volume?!) A, then I should get \sigma _0 A ? Namely the amount of charges enclosed into this small area? Is this right?
Why is the function "ill defined" as I call it? Is it because we DO NOT assume charges like points but like a property of a non zero space extension?

If I have the problem "What is the charge distribution of a uniformly charged cube?", I'd rather answer \rho _0 for -a \leq x \leq a, -a \leq y \leq a, -a \leq y \leq a. 0 everywhere else.
My answer wouldn't be ill defined and I could give you the charge distribution in any particular point in space, unlike the method used in more complicated problems.

 

[Dale]

A disk is a cylinder whose height has been "flattened". So, if you can come up with the expression for a cube, then you can probably come up with the expression for a cylinder. Then, simply take the part of the expression of the cylinder that sets the height, and replace it with the delta function.

 

[fluidistic]

A disk is a cylinder whose height has been "flattened". So, if you can come up with the expression for a cube, then you can probably come up with the expression for a cylinder. Then, simply take the part of the expression of the cylinder that sets the height, and replace it with the delta function.

Thanks for this insight.
I use cylindrical coordinates (I can't remember the variables so I take (\rho , \theta , z) ).
0\leq \rho \leq a , 0 \leq \theta \leq 2 \pi , 0\leq z \leq b for a cylinder. I replace z by \delta (z-z_0) and I get a disk perpendicular to the z-axis which crosses it at z=z_0 and with radius a.
Is this good? How would you personally write it? I feel my description of the cylinder is kind of "the ugly way".

 

[Dale]

Well, your description is fine, but if you want to make it prettier then you can define a couple of useful functions.

1) The unit step function:
u(x)=\int_{-\infty}^x{\delta(X)dX}

2) The boxcar function:
B_{a,b}(x)=u(x-a)-u(x-b)

Then your charge distribution becomes
\frac{Q}{A}B_{0,a}(\rho) B_{0,2\pi}(\theta)\delta(z-z_0)