I only have a basic understanding of the principles of stone sculpting
and any insights are appreciated.
Use your imagination. What do you want it to look like?
D&D 5e is not reality simulation, nor does it attempt to be. Difficult terrain is ground that is hard to walk on, an area where you are slowed down by trying to move through it. This could be a pile of boulders, a swamp, mud, or just very uneven ground where no two steps are on the same level.
Assuming the caster is skilled enough to smooth the stone within the hour, such that it is no longer difficult terrain, what happens when the spell ends?
The magical effect ends. Whatever was there becomes what it was before the cantrip was cast. The duration of the magical effect is one hour.
How exactly was the stone warped in the first place?
By magic.
How does normal terrain become less difficult terrain?
By magic.
Would the effect be ascetically pleasing or would it look sloppy?
That depends upon the interaction between the player and the DM.
Use your imagination. What do you want it to look like? This is where a DM's ruling is appropriate. Tell the DM what you want it to look like. The flow of events in D&D (p. 3, Basic Rules) is:
How to Play
1. The DM describes the environment.
2. The players describe what they want to do.
3. The DM narrates the results of the adventurers’ actions.
You describe what you want it to look like. The DM rules on that with your input considered.
The philosophy that the game started with still applies. D&D 5e has tried to unify the D&D fan base, in some cases reaching back to first principles ... When Imagination Was The Only Rule ... the theme behind Rob Kuntz'1 projects.
Action required: use your imagination and describe what efforts you are putting into this bit of magic-assisted sculpture, and what you are trying to make it look like. Give the DM something to work with so that he doesn't have to do a lot of work to arrive at a ruling. (As a DM of some experience, I promise you, such efforts are appreciated).
1 Who is Rob Kuntz and what does he have to do with D&D 5e? The front page of the rule book says that the game is:
Based on the original D&D game created by E. Gary Gygax and Dave Arneson, with Brian Blume, Rob Kuntz, James Ward, and Don Kaye
Under the default rules there are no squares and the caster is free to pick any point in 3D space; the 5-foot radius then extends from that point and if any creature is touched then it is affected.
I assume you are using the Playing on a Grid variant on page 192 of the Player's Handbook. In which case your caster still decides which point in 3d space to target and your DM decides which squares are affected.
Alternatively, you can use the variant to the variant rule on p. 251 of the Dungeon Master's Guide which says:
Choose an intersection of squares or hexes as the
point of origin of an area of effect, then follow its rules
as normal. If an area of effect is circular and covers at
least half a square, it affects that square.
So your Moonbeam affects the 4 squares that touch the intersection. A 15 foot radius fireball affects 32 squares (a 6x6 area with the corners missing).
Best Answer
The range is 10 feet
Range is not changed by grid combat. Whether you use a grid or not, if the spell or effect has a range of 10 feet, it has a range of 10 feet. The only difference with a grid is that you target the corner of a square with the spell, but it still needs to be in range. Here is how grids change how you can place the area of effect (DMG p. 251):
So here, you only an affect the ground within 10 feet or yourself. If there is no ground within 10 feet, you cannot affect it. If you fly higher, you cannot affect the ground. The game has pretty iffy consideration of vertical space in combat, but this is not one of the ambiguous cases.
How many squares do you affect?
You affect the ground in within a range of 10 feet , which will be measured from yourself (or around a corner of your square).
You can use the math approach of this answer to calculate how many squares are affected. Since the area of effect here is circular around a corner of your square, you calculate the intersection of the sphere around that point with the grid on the floor depending on your height, and each square that is covered by more than 50% is affected.
Since the floor is only touched at a single point if you are 10 foot up, that will affect no squares. If you are 0 feet up, it will affect the normal radius. So we really only need to calculate what squares are affected if you are 5 feet up.
Here (other than in the schematic depiction above to illustrate the idea) h is 5 feet, and r is 10 feet, so r' is $$ r' = \sqrt{r^2 - h^2} = r' = \sqrt{10^2 - 5^2} \approx 8.66 $$
Generally, if the center of a square is within range, more than half of the square will be covered - it is not quite exact but easier check (you can use the detail math in the linked answer if you want to know exactly). Here are the approximate distances in feet of the centerpoint of each square around the corner directly underneath you:
From this you can see that both for the green area, and for the yellow ara, the center points are well under the range for r' = 8.66, and all of these areas would be affected if you are five feet up. That incidentally is the same area, as if you were on the ground (because then, you have the full 10 feet radius around the corner, but the red square centers are a good bit further out than 10 feet, so those squares are still not affected).
So, you affect the same plus-shaped area if you are on level G or 2, and no area if you are on level 3 or 4.
(This assumes that the grid has the same dimensions vertically, i.e. each level is 5 feet high, so one on level G is at 0 feet, on level 2 at 5 feet, on level 3 at 10 feet and on level 4 at 15 feet above ground with the space they fly in. This is however not required to work this out. You just need to decide how high above the ground the caster is, i.e. where the spot to center the 10 feet radius is).