In order to solve this integral we must multiply and divide by the constant a to give the differential the correct form and then we factor the integrand into the square of the sine and the sine multiplied by the differential. We thus get

(Eq’n 1)

We can put that result into a more convenient form by factoring out a cosine and carrying out the obvious algebra;

(Eq’n 2)

Example: Angular Momentum of a Thin Spherical Shell

We have formed a material of density
ρ
into a thin spherical shell of radius R and thickness T<<R, so for the total
mass of the shell we have, to a good approximation, M=ρ4πR^{2}T.
The shell rotates as a solid body about an axis that passes through its center,
rotating such that a point on the sphere’s equator moves with a longitudinal
speed of W. We now want to organize those data into a description of the shell’s
angular momentum L.

To set up the calculation we divide the shell into
minuscule rings with planes perpendicular to the shell’s axis. Each ring has
thickness T, width Rdθ,
and length 2πr
(in which r=Rsinθ,
with θ
measured from 0 at the poles to π/2
at the equator). That ring spins with longitudinal speed w=Wsinθ
and carries angular momentum dL=wrdM=ρTRdθ(2πr)wr=2πρTR^{3}Wsin^{3}θdθ.
To calculate the total angular momentum of the shell we integrate that
description from θ=0
to θ=π/2
to get the angular momentum of one hemisphere and then double the result to get
the angular momentum of the whole shell. Thus we get

(Eq’n A-1)

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