I am writing to attempt an answer to your question about multiplying the area of a circle (1-dimension, call it S1 for surface 1) times the volume of an n-ball (call it Vn) and obtaining the volume of S(n+1). That this is true is a major step in the derivation of the actual volume of Vn. To understand why this is true, it helps to understand what a gamma function is. Most people know what a factorial is.
I am not going into the math of the torus, but that combined with the above convinces me that multiplying the volume of an n-ball by the length of the circle does imply a rotation to produce the cover of a higher dimension ball. This raises another interesting point. I’m sure you know that the size of the unit n-ball increases up to five dimensions and then decreases for higher dimensions. That is interesting, but partly a function of the metric used. If one takes a one-yard ruler and uses it as a physical radius to rotate through n-dimensions the hypersphere does appear to shrink after five dimensions, but if one uses the same ruler, but calls it a three-foot ruler, the shrinkage starts at dimensions much higher than five. If one calls it a 36-inch ruler, the dimension at which the hyperspheres start to shrink is still higher. However, if one puts an n-ball in a cube so that it touches each wall of the cube, the diminishing of the volume of the n-ball relative to the cube while increasing the dimensions is metric free. The relative volume decreases with every increase in dimensions starting with a square and a circle. At two dimensions the volume (area) of a square is 4 and that of circle is 3.14159 (π) given a diameter of 2. The ratio is 1.2732. At three dimensions the volume of the cube is 8 and the volume of a 3-ball is 4.1887. The ratio is 1.9099. The ratio keeps getting larger for higher dimensions. In this case the ratios don’t change with a change in the length of the diameter. A similar situation exists with the covers, but the increase in the ratio is slower.
So now I come to the reason for the last paragraph. What is the ratio of the volumes of a 3-d cover to a 3-d ball and more the ratio of a spiral built on a 3-d cover to a 3-d ball? That is the ratio of your universe to a 3-d-ball universe. The volume of the 3-d cover is larger than that of the 3-d ball by a factor of (π(n+1))/2. The problem with the spiral is somewhat trickier but it looks like your universe is 1.57 times as large. I’ll explain. According to Archimedes the area marked off by a spiral in a circle (see below) is 1/3 of the whole circle.
When one rotates by multiplying by S1, one obtains the spiral plaid on the
3-d cover of a 4-d ball. Its volume should be 1/3 of the whole cover or 2/3*πx R3 = 6.58 *R3. The 3-d ball has a volume of 4/3*π x R3 = 4.19*R3. Then 6.58/4.19 = 1.57 = π/2. Note, since R3 is in both the numerator and denominator, I left it out of the ratio . Therefore your universe is apparently 1.57 times the size it would be if it was a simple 3-d ball.
The letter has become long. I’ll close. Hope my comments on you universe etc. are interesting.
PS Magda says “Hi” and says she’s surprised you haven’t asked how we got back so fast.
PPS Artemy wants to email you so the two of you can set a date for her to visit.