求区域a^2
1个回答

由于对称性,知道质心的y0=z0=0

设质心M(x0,0,0)

x0=∫∫∫xdV / ∫∫∫dV

=∫∫∫rsinφcosθ r^2sinφdrdθdφ /[(1/2)(4πb^3-4πa^3)/3]

=∫(-π/2->π/2)dθ∫(0->π)dφ∫(a->b)[r^3(sinφ)^2cosθ]drdθdφ / [2(πb^3-πa^3)/3]

=[π(b^4-a^4)/4] / [2(πb^3-πa^3)/3]

=3(b^4-a^4)/[8(b^3-a^3)]

所以质心M(3(b^4-a^4)/[8(b^3-a^3)],0,0)