此題需要討論.設(shè)圓柱體桶的底面半徑為R,高為H.并設(shè)傾斜角度為θ.
當0≤θ≤tan^(-1)(H/2R)時,倒出的水的總體積為1/3*π*R^2*2Rtanθ=2π/3*R^3*tanθ；
當θ＞tan^(-1)(H/2R)時,剩下的水的體積為
1/3*{[π/2+sin^(-1)(Hcotθ/R-1)]*R^2+(Hcotθ-R)√[R^2-(Hcotθ-R)^2]}*H
則倒出來的水的總體積為
π*R^2*H-H/3*{[π/2+sin^(-1)(Hcotθ/R-1)]*R^2+(Hcotθ-R)√[R^2-(Hcotθ-R)^2]}
因此,本題中,桶每傾斜一度,能倒出的水占整桶水的體積的百分比是變化的,隨著傾斜角度的變化而變化.如,當0≤θ≤tan^(-1)(H/2R)時,命θ=1°,得倒出百分比為2π/3*R^3*tan1°/(π*R^2*H)=2Rtan1°/(3H)=1.1636709R/H%；命θ=2°,得倒出百分比為2π/3*R^3*(tan2°-tan1°)/(π*R^2*H)=2R(tan2°-tan1°)/(3H)=1.1643803R/H%.由于當θ=10°時,tanθ/(θ*π/180)=1.010279181≈1,故若0≤θ≤10°≤tan^(-1)(H/2R)時,每傾斜一度所能倒出的水占整桶水體積的百分比基本不變化,約為2π/3*R^3/(π*R^2*H)*tanθ/(θ*π/180)*π/180=πR/(270H)=0.011635528R/H=
1.1635528R/H%.當然,誤差不超過1.028%.
當然,精確的解是：
當0≤θ≤tan^(-1)(H/2R)時,每傾斜到θ°,能倒出的水占整桶水體積的百分比為2π/3*R^3*[tanθ°-tan(θ°-1°)]/(π*R^2*H)=2R*[tanθ°-tan(θ°-1°)]/(3H)*100%；
當θ＞tan^(-1)(H/2R)時,每傾斜到θ°,能倒出的水占整桶水體積的百分比為
{π*R^2*H-H/3*{[π/2+sin^(-1)(Hcotθ/R-1)]*R^2+(Hcotθ-R)√[R^2-(Hcotθ-R)^2]}-
π*R^2*H-H/3*{[π/2+sin^(-1)(Hcot(θ-1)/R-1)]*R^2+(Hcot(θ-1)-R)√[R^2-(Hcot(θ-1)-R)^2]}}/(π*R^2*H)={[sin^(-1)(Hcot(θ-1)/R-1)-sin^(-1)(Hcotθ/R-1)]*R^2+(Hcot(θ-1)-R)√[R^2-(Hcot(θ-1)-R)^2]-(Hcotθ-R)√[R^2-(Hcotθ-R)^2]}/(3π*R^2)*100%