本文章由 WyOJ Shojo 从洛谷专栏拉取，原发布时间为 2020-03-31 16:03:01

把圆锥沿高分成$\color{orange}n$份

每份高 $\color{brown}=\frac{\color{red}h}{\color{orange}n}$

第$\color{green}k$份半径 $\color{brown}=\frac{\color{green}k\color{blue}r}{\color{orange}n}$

第$\color{green}k$份底面积 $\color{brown}=\frac{\color{purple}\pi\color{green}k^2\color{blue}r^2}{\color{orange}n^2}$

第$\color{green}k$份体积 $\color{brown}=\frac{\color{purple}\pi\color{red}h\color{green}k^2\color{blue}r^2}{\color{orange}n^3}$

总体积 $\color{orange}n$ 份

$=\color{brown}\sum\limits_{{\color{green}k}=\color{pink}1}^{\color{orange}n}\frac{\color{purple}{\pi}\color{red}h\color{green}k^2\color{blue}r^2}{\color{orange}n^3}$

$\color{brown}=\frac{\color{purple}{\pi}\color{red}h\color{pink}(1^2+2^2+3^2+4^2+...+\color{orange}{n}^2)\color{blue}r^2}{\color{orange}n^3}$

$\color{brown}∵\color{pink}1^2+2^2+3^2+4^2+..\color{orange}.+n^2\color{brown}=\frac{\color{orange}n(n+1)(2n+1)}{\color{pink}6}$

$\color{brown}∴$ 总体积 $\color{orange}n$ 份

$\color{brown}=\frac{\color{purple}{\pi}\color{red}h\color{blue}r^2\color{orange}n(n+1)(2n+1)}{\color{pink}6\color{orange}n^3}$

$\color{brown}=\frac{\color{purple}{\pi}\color{red}h\color{blue}r^2\color{orange}(1+\frac{1}{n})(2+\frac{1}{n})}{\color{pink}6}$

$\color{brown}∵$ 当$\color{orange}n$越来越大，总体积越接近圆锥体积，$\frac{\color{pink}1}{\color{orange}n}$趋于$\color{pink}0$

$\color{brown}∴\lim\limits_{\color{orange}n\color{brown}\to\color{purple}∞}\frac{\color{purple}{\pi}\color{red}h\color{blue}r^2\color{orange}(1+\frac{1}{n})(2+\frac{1}{n})}{\color{pink}6}=\frac{\color{purple}{\pi}\color{red}h\color{blue}r^2}{\color{pink}3}$

因为圆柱体积 $\color{brown}=\color{tan}S\color{red}h\color{black}\color{brown}=\color{purple}{\pi}\color{red}h\color{blue}r^2$

所以圆锥体积是与它等底等高的圆柱体积的$\color{pink}\frac{1}{3}$，也就是$\Large\color{pink}\frac{1}{3}\color{tan}S\color{red}h$。