Tensors for Physics

8.3 Volume Integrals, Gauss 133

integrals to be evaluated are of the form

∫

V

...d^3 r= 2 π

∫a 2

a 1

r^2 dr

∫θmax

0

sinθdθ...= 2 π

∫a 2

a 1

r^2 dr

∫ 1

ζmin

dζ..., (8.59)

where the integrand...is assumed to be independent of the polar angleφ. Further-
more,ζ=cosθandζmin=cosθmaxare used.
First, the volume is computed with the help of (8.59), with the integrand 1. The
result is

V=

2 π

3

(a 23 −a 13 )( 1 −ζmin). (8.60)

The volume of a sphere with radiusais recovered from this expression witha 2 =a,
a 1 =0 andζmin=cosπ=−1. The mass of the cap isM=ρ 0 V.
Due to the uniaxial symmetry, the vectorRis parallel (or anti-parallel) to the unit
vectoru. The calculation of the center of mass is simplified with the help of this
argument. The ansatz

Rμ=cuμ, (8.61)

is made. The coefficientcis inferred from the scalar multiplication of this equation
withuμ, thusMc=MuμRμ. The vectoruis constant, so it can be put inside the
integral (8.58)usedforMRμ. Then the integrand isuμrμ=rcosθ=rζ. With the
help of (8.59),

Vc=

2 π

8

(a^42 −a^41 )( 1 −ζmin^2 ) (8.62)

is obtained. Clearly, forθmax=π, corresponding toζmin=−1, the coefficientc
is zero. As expected, in this case, the center of mass coincides with the geometric
center. For a half-sphere, with radiusa=a 2 ,a 1 =0, andζmin=0, on the other
hand, the center of mass

R=

3

8

au (8.63)

is shifted “upwards”.
The casea 1 =a,a 2 =a+δa, with 0<δaacorresponds to thin shell
structure with thicknessδa. Then the factors(a 23 −a 13 )and(a 24 −a 14 ), occurring in
(8.60) and (8.62), reduce to 3a^2 δaand 4a^3 δa, respectively. As a consequence,

R=

1

2

au (8.64)

is found for the hollow hemisphere.