A spherical shell of radius #R# and uniformly charged with charge #Q# is rotating about its axis with frequency #f#. Find the magnetic moment of the sphere?

Uniformly charged spherical shell of radius #R# carries a total charge #=Q#
Hence it has surface charge density

#sigma=Q/(4πR^2)#

It rotates about its axis with frequency#=f#
#:.# Its angular velocity #omega=2pif#
Suppose the angular velocity #vecomega=omegahatz#

To find the magnetic moment of spinning shell we can divide
it into infinitesimal charges.
Using spherical polar coordinates #(rho, phi, theta)# with origin at the center of the spherical shell, we consider infinitesimal area #dS# of a circular ring of infinitesimal width and of radius #r# located at a distance #R# from the origin on the surface of sphere.
Here #vecr=Rsinthetahatr# and coordinate #rho=R#, is constant for each elemental area.

#dq=sigmadS=sigmaR^2sinthetad thetadphi#

Current in the ring is given by

#dI=(dq)xxf=(sigmaR^2sinthetad thetadphi)xxomega/(2pi)#
#=>dI=(omegasigmaR^2sinthetad thetadphi)/(2pi)#

Now the magnetic dipole moment of ring is given by the expression in terms of position vector #vecr# and current density #vecJ# as

#dvecm=1/2int_"ring"vecrxxvecJ#
#=>dvecm=1/2dIint_"ring"vecrxxdvecl#
where #dvecl# is element length of the ring.

Line integral becomes equal to the circumference of ring#=2piRsintheta#

#:.dvecm=dI(piR^2sin^2theta)hatz#

Inserting value of #dI# we get

#dvecm=(omegasigmaR^2sinthetad thetadphi)/(2pi)(piR^2sin^2theta)hatz#
#=>dvecm=(vecomegasigmaR^4sin^3thetad thetadphi)/2#

Total magnetic moment is integral with respect to both variables within respective limits

#vecm=(vecomegasigmaR^4)/2int_0^pisin^3thetad thetaint_0^(2pi)dphi#
#=>vecm=(vecomegasigmaR^4)/2xx4/3xx2pi#
#=>vecm=4/3pisigmaR^4vecomega#

Rewriting in terms of charge #Q#

#vecm=4/3pi(Q/(4πR^2))R^4vecomega#
#vecm=Q/3R^2vecomega#

.-.-.-.-.-.-.-.-

Reference figure for #dS#

Using online integral calculator

#int_0^pisin^3thetad theta=|cos^3theta/3-costheta|_0^pi=4/3#