M.I. of sphere along tangent

Let the center be the origin point, the z-axis be parallel to the rotational axis, and take the spherical coordinate system, then the inertia momentum=
∫[0,π]∫[0,2π]∫[0,R] k*[R^2-ρRsinφcosθ+(ρsinφ)^2] *ρ^2 sinφ dρ dθ dφ
=k*2π∫[0,π]∫[0,R] [R^2+(ρsinφ)^2]*ρ^2sinφ dρ dθ dφ
=k*2πR^5∫[0,π] [(1/3)+(1/5)(sinφ)^2]sinφ dφ
= 2πkR^5*(14/15)
= (28/15)πR^5*M/[(4πR^3)/3]
= (7/5)MR^2

Note:(1) k is the density of the solid.
(2) Another method, by shifting theroem of I
I(L)=MR^2+I(O)=MR^2+(2/5)MR^2=(7/5)MR^2

2009-12-24 00:25:01 補充：
Sorry!
∫[0,π]∫[0,2π]∫[0,R] k*[R^2-ρRsinφcosθ+(ρsinφ)^2] *ρ^2 sinφ dρ dθ dφ
=k*2π∫[0,π]∫[0,R] [R^2+(ρsinφ)^2]*ρ^2 sinφ dρ dφ

這裏可以幫到你
http://actionte.ht18.1t39.com/yahoo.com.hk/hk/auction/178987536

Moment of inertia around solid sphere center:
I(c)=(2/5)MR
Moment of inertia around parallel displaced R axis:
I(d)=I(c) + MR =(2/5)MR+MR=(7/5)MR