 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Spheres in 4-D (Posted on 2003-12-20) Derive the formula for the 4-D volume of a hypersphere.

 See The Solution Submitted by Brian Smith Rating: 3.2000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 3 of 6 | The hypersphere can be considered as a stack of solid (filled-in) spheres (just as a sphere can be considered a stack of filled-in circles), for purposes of calculating (hyper-)volume, where there approach infinitely many infinitely thin (in the 4th dimension) spheres.

If the radius of the hypersphere is 1, and it's centered on the origin, we can integrate along the x-axis. The radius of each 3-dimensional sphere is r = √(1-x^2). Each sphere has volume (4/3)pi. (Later we'll scale up the results by the fourth power of the radius of the hypersphere.)

We need ∫{-1 to 1} (4 pi/3) (1-x^2)^3/2 dx, as we need the cube of the radius of each sphere.

Going to Wolfram's integrator, we find the integral is:

(4 pi/3) [(√(1-x^2))(5x/8 - x^3/4) + 3 arcsin(x)/8]{-1 to 1}

With the substitutions and subtraction, this comes out to

(4 pi/3) (3 arcsin(1)/8 - 3 arcsin(-1)/8)

or

8 pi/3 (3 arcsin(1)/8)

or pi (pi/2) = pi^2/2

So the hypervolume is pi^2 R^4 / 2 where R is the radius of the hypersphere.
 Posted by Charlie on 2003-12-20 14:10:20 Please log in:

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