I'm wondering how to scale up the surface area of a sphere of 6 million meters in radius, into a hypersphere of similar radius (i.e. a Hyper Earth). I would also like to know the ratio. I would like to know the basic value in 4th dimension, but knowing values for 5th, 6th, and higher would be useful too. I attempted using something like pi^2*r^3 after badly integrating numbers that probably shouldn't be integrated, but then I got confused, because the units produced different surface areas. like if the unit is 1 earth radius, then taking it to the 3rd, 4th, or 5th power doesn't change the output, and if the units are km, they produce a smaller change than if the units are meters, or nanometers. I got so lost on this issue that I decided to post here.
Wikipedia articles are Gibberish, not informative. from what I got out of that they imagined Pi from thin air and then applied it to S_n(R) = S_n R^n and n-sphere in (n + 1)-dimensional Euclidean... which from what I am reading says n = 4 since thats my happy dimension to start with, and 4+1 = 5. now I stick 5 in the n place, and get Surface area of a 6e6m radius sphere is 6e6^5, which is about 68 on a decibel scale, times 5, which is about 340, or 1e34 times s sub 4 which for some unknown reason in 3 dimensions becomes 4 so uhh.. 8pi*1e34? Now you know that's not the answer. So how about a more sincere reply to my question than "go to wikipedia" ?
It's gibberish if you don't understand it. In reality, the Earth is a 3-ball (volume is 3 dimensional) and a 2-sphere (surface area is 2 dimensional). Now this is gibberish!
I think the surface area is always the derivative (wrt R) of the volume, and the volume of a 4-ball is π^2R^4/2, I believe. so the surface area of a 4 sphere is lets see: 2π^2R^3?? see my notes for the epsilon geometry camp for bright 9 year olds, last pages. http://www.math.uga.edu/~roy/camp2011/10.pdf