圆柱体横放,容量随高度增加的关系式!
1个回答

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