设圆柱 底面半径为r,高为h,【放置为截面方程:y^2+(x-r)^2 =r^2,0≤x≤2r】
则截面矩形 长为h,宽为 2√(r^2-(r-x)^2) = 2√(2rx-x^2) ,厚度为 dx
dV = h *2√(2rx-x^2)*dx
V=∫(0,2r) h2√(r^2-(r-x)^2) dx
令:r-x=rcost ,x=r-rcost ,dx=rsint ,0≤t≤π
= 2h ∫(0,2r) √(r^2-(r-x)^2) dx
= 2h ∫(0,π) r^2sin^2t dx
= 2h r^2 ∫(0,π) [1-cos2t]/2 dx
= 2h r^2 {[t-sin2t/2]/2 |[0,π]}
= 2h r^2*π/2
= h*πr^2