设R-r=Rcosx
r=R(1-cosx)=0~R,x=0~π/2,
dr=Rsinxdx
原式=θ∫(0,π/2)Rsinx.R(1-cosx) Rsinxdx
=θR³∫(0,π/2)(sin²x-sin²xcosx)dx
= θR³[(1/2)∫(0,π/2)(1-cos2x)dx-∫(0,π/2)sin²xdsinx ]
= θR³[(1/2)(x-0.2sin2x)| (0,π/2) -(1/3)(sin³x )| (0,π/2)]
= θR³[(1/2)(π/2) -(1/3)]
= θR³[π/4 -1/3]