What is the surface area and volume (or equivalent in 4D) of a hypersphere?

Question

I was just curious since I never took a differential geometry course how one would calculate this.

As for the volume of a hypersphere, I was thinking one would integrate the surface area of a hypersphere from radius=0 to radius=R (similar to how the volume of a sphere is derived from integrating the surface area of a sphere), but in 4D would it become a multiple integral since there is a new dimension to be considered?

Answer 1

I suppose you do know this
In mathematics, a hypersphere is a sphere which has dimension 3 or higher
The set of all points on this sphere has dimension n − 1, so it is called the (n − 1)-sphere and is denoted . It may be written as (x1,x2,...,xn) where

The hyperdimensional volume of the space which a (n − 1)-sphere encloses (the n-ball) is

_
V=II^n/2. R^n/Γn/2=1 II-pie


If the dimension, , is not limited to integral values, the hypersphere volume is a continuous function of with a global maximum for the unit sphere in "dimension" n = 5.2569464... where the "volume" is 5.277768...

The hypercube circumscribed around the unit n-sphere has an edge length of 2 and hence a volume of 2n; the ratio of the volume of the hypersphere to its circumscribed hypercube decreases monotonically as the dimension increases.


HEY I KNOW ONLY THE VOLUME

Answer 2

You're right that the volume of any sphere is just the integral:

V(R) = \int_0^R S(r)dr

or in other words the surface S(R) is the derivative of V(R) with respect to R, so knowing one immediately gives the other. For a sphere in a space with an even number of dimensions 2m:

V = (pi^m/m!)R^(2m) = (pi^2/2)R^4 (in 4 dimensions)

so

S = 2m*(pi^m/m!)R^(2m-1) = 2 pi^2 R^3 (in 4 dimensions)

******

Just a sketch of how to prove it for a four dimensional sphere (ball): The sphere is defined by:

x^2+y^2+z^2+w^2 = R^2

where x,y,z,w are the 4 coordinates. Add up the volumes of each slice of the ball of thickness dx from x=-R to x=+R. The intersection of an x=constant, plane with the ball is a *three dimensional* sphere of volume (4/3)pi*r^3 where:

r^2 = y^2 + z^2 + w^2 = R^2 - x^2
--> r = sqrt(R^2-x^2)

So the total volume of the ball in four dimensions is the integral:

\int (x=-R to +R) V(r) dx =
\int (x=-R to +R) (4/3)pi (R^2-x^2)^(3/2) dx
= (pi^2/2)R^4

Answer 3

that could be the heaviest solids if weight and shape have been the identity characteristics. Take 2 products of same weight and shape. The densest one, which incorporate plutonium might have least quantity. particularly recently 2 new aspects have been chanced on at Berkeley - 116 and 118 (pointed out by employing their form of protons). yet once you're finding for a shapewise answer then it would be a flat sheet. The flatter, the extra helpful to fulfill your requirement to cut back quantity to floor. and that i declare it does no longer matter what 2 dimensional shape it took so long because it grew to become into flat. As a thought test, think of that an merchandise might desire to be flattened all the way down to a million molecule thick. you've gotten all molecules uncovered and the section may be the part of a million molecule cases the form of molecules interior the article. So the two dimensional shape does no longer matter. Flatness is the main.

Answer 4

Volume Of Hypersphere

Answer 5

Vol. of Sphere = 4/3 (pi)r^2
Take integral= [4(pi)(r^3)] / 9 + c

Answer 6

think of dimension and how to calc:

e.g. circle -- area is pi * r * r

do integral to get volume: it is (4/3) * pi * r * r * r

so hyper shere would be another integral of that