此题需要讨论.设圆柱体桶的底面半径为R,高为H.并设倾斜角度为θ.

　　当0≤θ≤tan^(-1)(H/2R)时,倒出的水的总体积为1/3*π*R^2*2Rtanθ=2π/3*R^3*tanθ；

　　当θ＞tan^(-1)(H/2R)时,剩下的水的体积为

　　1/3*{[π/2+sin^(-1)(Hcotθ/R-1)]*R^2+(Hcotθ-R)√[R^2-(Hcotθ-R)^2]}*H

　　则倒出来的水的总体积为

　　π*R^2*H-H/3*{[π/2+sin^(-1)(Hcotθ/R-1)]*R^2+(Hcotθ-R)√[R^2-(Hcotθ-R)^2]}

　　因此,本题中,桶每倾斜一度,能倒出的水占整桶水的体积的百分比是变化的,随着倾斜角度的变化而变化.如,当0≤θ≤tan^(-1)(H/2R)时,命θ=1°,得倒出百分比为2π/3*R^3*tan1°/(π*R^2*H)=2Rtan1°/(3H)=1.1636709R/H%；命θ=2°,得倒出百分比为2π/3*R^3*(tan2°-tan1°)/(π*R^2*H)=2R(tan2°-tan1°)/(3H)=1.1643803R/H%.由于当θ=10°时,tanθ/(θ*π/180)=1.010279181≈1,故若0≤θ≤10°≤tan^(-1)(H/2R)时,每倾斜一度所能倒出的水占整桶水体积的百分比基本不变化,约为2π/3*R^3/(π*R^2*H)*tanθ/(θ*π/180)*π/180=πR/(270H)=0.011635528R/H=

　　1.1635528R/H%.当然,误差不超过1.028%.

　　当然,精确的解是：

　　当0≤θ≤tan^(-1)(H/2R)时,每倾斜到θ°,能倒出的水占整桶水体积的百分比为2π/3*R^3*[tanθ°-tan(θ°-1°)]/(π*R^2*H)=2R*[tanθ°-tan(θ°-1°)]/(3H)*100%；

　　当θ＞tan^(-1)(H/2R)时,每倾斜到θ°,能倒出的水占整桶水体积的百分比为

　　{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%