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    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=ρ4πR2T. 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=2πρTR3Wsin3θdθ. 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)

abefab

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