Volume of Ball in R⁴

Question

How do you calculate the (four-dimensional) volume V = r⁴⋅π²/2 of the ball B(r) = {(x₁,x₂,x₃,x₄) ∈ R⁴: x₁²+x₂²+x₃²+x₄² ≤ r²}?

Answer 1

We know that the formula in Rⁿ is V_n(r)= C_n·rⁿ for some constant C_n.

The recursion is V_n(r) = ∫^r_-r V_n-1(√(r²-x²)) dx.

Using the trigonometric substitution x = r sin(θ):

I_n = ∫^r_-r C_n-1(r²-x²)^(n-1)/2 dx = 2C_n-1rⁿ ∫₀^π/2 cosⁿ(θ) dθ

The reduction formula ∫ cosⁿ(θ) dθ = (1/n) cos^n-1 x sin x - ((n-1)/n) ∫ cos^n-2(θ) dθ gives:

∫₀^π/2 cosⁿ(θ) dθ = (n-1)!!/n!! for odd n, or π(n-1)!!/n!! for even n. Using gamma functions, it is:

∫₀^π/2 cosⁿ(θ) dθ = (Γ((n+1)/2)√π) / (2Γ(n/2+1))

C_n = 2C_n-1(Γ((n+1)/2)√π) / (Γ(n/2+1)) = 2C_n-2 π/n with C₁ = 2 and C₂ = π.

This gives C₃ = 4π/3, which is well known, and C₄ = π²/2.