Let a solid of revolution be formed by revolving the region enclosed by the curve y = f(x) ,the x-axis and the vertical lines x =a and x =b about the x-axis.
We divide the solid of revolution into a large number n of very small elements of circular discs of mass Δm= ρ π y^2 Δx with moment of inertia about the x-axis ΔI = (y^2 /2) Δm = (ρ π y^4 /2) Δx where ρ is the density in mass per unit volume.
圖片參考：http://imgcld.yimg.com/8/n/HA05475666/o/701109020061413873452710.jpg
小弟有一問題,為何ΔI = (y^2 /2) Δm 入面 個y^2 要除2? solid of revolution,solid cylinder 等等都一樣除2 , equilateral triangular lamina 就係除3
i.e. ΔI = (x^2 /3) Δm ......希望能知道概念來由.......(此問題煩了我一星期,不得不上來問問)......................
可能係小弟學有所忘或道行不足,希望有識之士賜教... thank you very much~!!
更新1:
先感謝你的回答, 但未能對問題提出正解.... 我的問題係moment of inertia of a solid of revolution中為何ΔI = (y^2 /2) Δm ? 以我的想法,I = mr^2 的話,那solid of revolution 唔係應該同disc一樣都係 ΔI = y^2 Δm 架咩....?.........thanks..........
更新2:
我明白了 a_a 原來 disc 同thin rod 係拆solid , lamina之類的基礎,用佢地自身的moment of inertia 砌入新 ΔI 再用 upper 同lower limit 去in 返佢地,搵出總和(=solid 的moment of inertia ) =] ..........!!!!!!! 唔好意思T_T 係我未明箇中關連............真係唔該晒你 數學知識長 =>
