Hemispherical Bowl : how is it possible? I was reading up about how to extrusion print an overhang greater than 45 degrees, when something weird stuck me.

I see lots and lots of 3D extrusion printed bowls that have perfectly printed hemispherical surfaces. However, given that there is some deformation expected as part of the print process, how is it possible to get a perfect hemispherical  bowl - where both inner and outer surfaces are perfectly hemispherical?

Note: I am not talking about surface smoothness, but about the shape of the bowl itself. I understand that surface smoothing will make the bowl look great, but it will not fix a deformation in the shape of a print.
There are several factors playing together. For example orientation, printer &amp; slicer settings and more.
Reminder
First of all, not all overhangs of greater than 45° need support. Many printers manage up to 60°, even 70° is not unheard of - with the right settings. Pretty much all printers manage tiny 90° overhangs.
U-Bowls (open side up)
Let's look at bowls that shape like a U - the dome is at the bottom. With a flat contact surface only the next areas need support, and with PVA or other easily removable support structures on the rise, it is no problem to use support and leave no trace. With support, there is no sagging, so the bowl gets its &quot;perfect&quot; look even though there are overhangs and support used.
n-bowls (open side down)
The other way around - dome on top - is probably the more &quot;smooth&quot; one, using different tricks to get it set.
The outer wall now sits on the infill like the inner we had before, perfect, but we used that for the inner wall before. But what about the inner one?! Well, we need no support for the sides till we reach, let's say 55° angle. Can we go further without support? Yes... if we are tricksy!

Print the walls from the center to the outer wall, as then the inner walls stand on the lower layer and can hold the new, neighboring, floating wall
Have more than 2 walls! Why? Well, simple math: You go up by $z$, and out by $(n-1)\times d$ where $n$ is the number of walls. We have a maximum overhang angle of $\tan^{-1}(z^{-2}\times((n-0.5)\times d)^2)$ before our walls don't stand on the wall below by more than a half wall. As one can see, $z$ and $d$ are fixed in that calculation while a higher factor $n$ directly increases the value of the angle. As long as there is support for the wall (like the neighboring, already printed wall), you get some kind of dome.
Print thinner layers. If you decrease $z$, you also increase the maximum angle that has one wall standing on half a supporting wall.
Print slow. Printing slower allows the material to cool and harden while still somewhat held up by the cohesion to the filament that comes from the nozzle. This can support a higher overhang angle.
Cooling. Together with slow printing, you might want to use an extra strong cooling solution to speed up the solidification. Aim well.
Print narrow1 domes. Why is it much easier to print more narrow domes than wider domes? Part of the answer is how much geometry does not need to be supported by itself, another part is the weight of the full structure: A narrow dome spans a smaller distance. This means it has less area of a high angle to the last layer, and the overhangs, in general, are shorter. This can result in better printing.


1 What counts as narrow is printer and filament dependent. It is no problem to print small spherical hollows in a print if they are just small enough as the overhang then is just short enough so it can carry itself. At some point it gets too much though.