Riley hobson bence · Problem 6.23

Multiple integrals

Mathematics — Methods · multiple-integrals-volume-recurrence

Problem

(a) Let

R

be a real positive number and define

K_m

K_m = \int_{-R}^{R} \left(R^2 - x^2\right)^m dx.

Show, using integration by parts, that

K_m

satisfies the recurrence relation

(2m + 1)K_m = 2mR^2 K_{m-1}.

(b) For integer

n

, define

I_n = K_n

and

J_n = K_{n+1/2}

. Evaluate

I_0

and

J_0

directly and hence prove that

I_n = \frac{2^{2n+1}(n!)^2 R^{2n+1}}{(2n+1)!} \qquad \text{and} \qquad J_n = \frac{\pi(2n+1)! R^{2n+2}}{2^{2n+1} n!(n+1)!}.

V_n(R)

is defined by

V_0(R) = 1,

V_n(R) = \int_{-R}^{R} V_{n-1}\!\left(\sqrt{R^2 - x^2}\right) dx, \qquad n \geq 1.

Prove by induction that

V_{2n}(R) = \frac{\pi^n R^{2n}}{n!}, \qquad V_{2n+1}(R) = \frac{\pi^n 2^{2n+1} n! R^{2n+1}}{(2n+1)!}.

(d) For interest, ((i)) show that

V_{2n+2}(1) < V_{2n}(1)

and

V_{2n+1}(1) < V_{2n-1}(1)

for all

n \geq 3

; ((ii)) hence, by explicitly writing out

V_k(R)

for

1 \leq k \leq 8

(say), show that the `volume' of the totally symmetric solid of unit radius is a maximum in five dimensions.

Intuitive explanation

The recurrence relation is the bridge between different powers of the polynomial. Use integration by parts strategically: differentiate the power $(R^2 - x^2)^m$ to reduce the exponent, and integrate $dx$ to create a factor of $x$ . This will generate a term involving $K_{m-1}$ and a boundary term that vanishes—rearranging gives your recurrence. Once you have the relation, you can climb from base cases $I_0$ and $J_0$ (which you compute directly) to general formulas by repeatedly applying it, keeping track of the factorial and power patterns that emerge.

Intuitive explanation

Related theory

Multiple Integrals

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