11A (Vector Calculus)

Solution

For coordinates

x = x (q)

, the Jacobian is

J_{ij} = \partial x_{i} / \partial q_{j}

. Expanding after separating

x_{1} = r cos θ_{1}

from the remaining coordinates gives

∣ J_{n} ∣ = r sin^{n - 2} θ_{1} ∣ J_{n - 1} ∣.

Consequently

d V = r^{n - 1} \prod_{j = 1}^{n - 2} sin^{n - 1 - j} θ_{j} d r d θ_{1} \dots d θ_{n - 2} d ϕ,

and on

r = R

the surface element is obtained by omitting

d r

and replacing

r

R

The ball volume is

V_{n} (R) = \frac{π ^{n /2} R ^{n}}{Γ ( n /2 + 1 )} .

For

n = 2 m

this is

π^{m} R^{2 m} / m!

; for

n = 2 m + 1

it is

2^{2 m + 1} m! π^{m} R^{2 m + 1} / (2 m + 1)!

Differentiating the ball volume with respect to

R

gives

A_{n - 1} (R) = \frac{2 π ^{n /2} R ^{n - 1}}{Γ ( n /2 )} = \frac{n}{R} V_{n} (R) .

i \neq = j

R^{2} A_{n - 1} (R)

. Therefore

\int_{r = R} x_{i} x_{j} d S = δ_{ij} \frac{R ^{2} A _{n - 1} ( R )}{n} = δ_{ij} \frac{π ^{n /2} R ^{n + 1}}{Γ ( n /2 + 1 )} .

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