根据x,y,z的对等性所以∫x^2dS=∫y^2dS=∫z^2dS ∫xdS=∫ydS=∫zdS所以∫xdS=∫ydS=∫zdS=(1/3)∫(x+y+z)dS=0所以原积分=∫x^2dS+2∫ydS+∫zdS=∫x^2dS=(1/3)∫(x^2+y^2+z^2)dS=(1/3)∫dS=2π/3
