Find the "volume" of the "four-dimensional sphere"

Question

Find the "volume" of the "four-dimensional sphere"

x² + y² + z² + w² = a²

by evaluating

Answer 1

(I) Relationship between the volume of the 4D sphere and the suggested integral:

The volume of the 4D sphere is the same as the suggested integral, but

the bounds should be

from -a to a instead of 0 to a,

from -√(a²-x²) to +√(a²-x²) instead of from 0 to +√(a²-x²),

and so on.

By symmetry each of the suggested integrals computes half of its actual contribution to the sphere.

Therefore the whole 4D sphere is 16 times the suggested integral.

(II) Intermediate result to be used later:

For any r ≥ 0 the definite integral from 0 to r

∫√(r²-x²)dx = πr²/4 (II)

The reason is that the integral represents the area of the quarter-circle of radius r (in the first quadrant).

(III) The suggested integral over dw:

The indefinite

∫dw = w

Therefore the given definite integral is

√(a²-x²-y²-z²)

(IV) The suggested integral over dz:

The definite integral from 0 to √(a²-x²-y²)

∫√(a²-x²-y²-z²)dz = π(a²-x²-y²)/4

This is by applying the above result (II) with r = √(a²-x²-y²)

(V) The suggested integral over y:

Let's first do the indefinite integral

F(y) = ∫π(a²-x²-y²)/4 dy

∫π(a²-x²-y²)/4 dy =

π/4 ∫(a²-x²-y²)dy =

π/4 (∫(a²-x²)dy - ∫y²dy) =

π/4 ((a²-x²)y - y³/3)

The definite integral is then

F(√(a²-x²)) - F(0) =

(a²-x²)√(a²-x²) - (√(a²-x²))³ / 3 - 0 =

(2/3)(a²-x²)^3/2

Therefore the complete definite integral over y is

(π/6)(a²-x²)^3/2

(VI) The suggested integral over x:

Let's first do the indefinite integral

∫(a²-x²)^3/2dx =

∫(a²-x²)(a²-x²)^1/2dx =

∫a²(a²-x²)^1/2dx - ∫x²(a²-x²)^1/2dx

Ignoring the first integral for the time being, we do the second one by parts using

∫uv' = uv - ∫u'v

where

u = x, hence u' = 1

v' = x(a²-x²)^1/2, hence v = (-1/3)(a²-x²)^3/2

Continuing

∫a²(a²-x²)^1/2dx - ∫x²(a²-x²)^1/2dx =

a²∫(a²-x²)^1/2dx + (x/3)(a²-x²)^3/2 - ∫(1/3)(a²-x²)^3/2

Now consider the whole equation we have derived so far

∫(a²-x²)^3/2dx = a²∫(a²-x²)^1/2dx + (x/3)(a²-x²)^3/2 - ∫(1/3)(a²-x²)^3/2

From that

(4/3)∫(a²-x²)^3/2dx = a²∫(a²-x²)^1/2dx + (x/3)(a²-x²)^3/2

∫(a²-x²)^3/2dx = (3/4)a²∫(a²-x²)^1/2dx + (x/4)(a²-x²)^3/2 (1)

Now consider the definite integral from 0 to a of ∫(a²-x²)^3/2d

The first half of the sum in (1) becomes

(3/4)a²(πa²/4) using the result (I) with r = a

The second half of the sum (1) is 0 because it is 0 for both x = a and x = 0.

Thus the whole suggested integral becomes

(π/6)(3/4)a²(πa²/4) = π²a⁴/32

Using (I) the whole 4D sphere has volume π²a⁴/2