Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 6 / b

Take the centre of the ball as the origin. An orthogonal change of variables leaves both this domain and its volume measure invariant. Consequently its Cartesian moment integrals are isotropic tensors. Equivalently, inversion

x \mapsto - x

immediately makes every odd moment zero.

There is no nonzero invariant rank-one vector, so

\int_{V} x_{i} d V = 0.

\int_{V} x_{i} x_{j} d V = α δ_{ij}

. Contract the indices and use spherical coordinates:

3 α = \int_{V} r^{2} d V = 4 π \int_{0}^{a} r^{4} d r = \frac{4 π a ^{5}}{5} .

α = \frac{4 π a ^{5}}{15}, \int_{V} x_{i} x_{j} d V = \frac{4 π a ^{5}}{15} δ_{ij}, \int_{V} x_{i} x_{j} x_{k} d V = 0.

For the last identity, one may also use the general rotation-invariant rank-three tensor

c ϵ_{ijk}

: the moment is fully symmetric in its indices, whereas the Levi-Civita symbol is antisymmetric, forcing

c = 0

The origin convention is necessary. For a ball centred at a nonzero vector

c

, its first moment would instead be

(4 π a^{3} /3) c_{i}

, so the displayed claims implicitly refer to coordinates measured from its centre.

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