圆锥体积公式推导

cxm1024 · 2020-03-31 16:03:01 · 个人记录

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。