Question Number 227054 by Spillover last updated on 28/Dec/25
$${A}\:{Segment}\:{of}\:{a}\:{sphere}\:{has}\:{radius}\:{r} \\ $$$${and}\:{maximum}\:{height}\:{h}.{Prove}\:{that} \\ $$$${its}\:{volume}\:\frac{\boldsymbol{\pi{h}}}{\mathrm{6}}\left(\boldsymbol{{h}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{r}}^{\mathrm{2}} \right) \\ $$
Answered by fantastic2 last updated on 29/Dec/25
Commented by fantastic2 last updated on 29/Dec/25
![AC=h โตAO=Rโh (Rโh)^2 +r^2 =R^2 ...i โ2Rh=h^2 +r^2 โR=((h^2 +r^2 )/(2h)) AB=โ โดOB=Rโh+โ (Rโh+โ)^2 +x^2 =R^2 โดR^2 +h^2 +โ^2 โ2Rh+2Rโโ2hโ+x^2 =R^2 +h^2 โ2Rh+r^2 [from i] โx^2 =r^2 โโ^2 +2โ(hโR) dV=ฯx^2 dโ โดV=โซ_0 ^h ฯ(r^2 โโ^2 +2โ(hโR))dโ =ฯ(r^2 hโ(h^3 /3)+h^3 โRh^2 ) =ฯh(r^2 โ(h^2 /3)+h^2 โ((h^2 +r^2 )/2)) =((ฯh)/6)(6r^2 โ2h^2 +6h^2 โ3h^2 โ3r^2 ) =(1/6)ฯh(h^2 +3r^2 )](https://www.tinkutara.com/question/Q227062.png)
$${AC}={h} \\ $$$$\because{AO}={R}โ{h} \\ $$$$\left({R}โ{h}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} ={R}^{\mathrm{2}} …{i} \\ $$$$\Rightarrow\mathrm{2}{Rh}={h}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}{h}} \\ $$$${AB}=\ell \\ $$$$\therefore{OB}={R}โ{h}+\ell \\ $$$$\left({R}โ{h}+\ell\right)^{\mathrm{2}} +{x}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\therefore{R}^{\mathrm{2}} +{h}^{\mathrm{2}} +\ell^{\mathrm{2}} โ\mathrm{2}{Rh}+\mathrm{2}{R}\ellโ\mathrm{2}{h}\ell+{x}^{\mathrm{2}} ={R}^{\mathrm{2}} +{h}^{\mathrm{2}} โ\mathrm{2}{Rh}+{r}^{\mathrm{2}} \left[{from}\:{i}\right] \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={r}^{\mathrm{2}} โ\ell^{\mathrm{2}} +\mathrm{2}\ell\left({h}โ{R}\right) \\ $$$${dV}=\pi{x}^{\mathrm{2}} {d}\ell \\ $$$$\therefore{V}=\int_{\mathrm{0}} ^{{h}} \pi\left({r}^{\mathrm{2}} โ\ell^{\mathrm{2}} +\mathrm{2}\ell\left({h}โ{R}\right)\right){d}\ell \\ $$$$=\pi\left({r}^{\mathrm{2}} {h}โ\frac{{h}^{\mathrm{3}} }{\mathrm{3}}+{h}^{\mathrm{3}} โ{Rh}^{\mathrm{2}} \right) \\ $$$$=\pi{h}\left({r}^{\mathrm{2}} โ\frac{{h}^{\mathrm{2}} }{\mathrm{3}}+{h}^{\mathrm{2}} โ\frac{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\frac{\pi{h}}{\mathrm{6}}\left(\mathrm{6}{r}^{\mathrm{2}} โ\mathrm{2}{h}^{\mathrm{2}} +\mathrm{6}{h}^{\mathrm{2}} โ\mathrm{3}{h}^{\mathrm{2}} โ\mathrm{3}{r}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\pi{h}\left({h}^{\mathrm{2}} +\mathrm{3}{r}^{\mathrm{2}} \right) \\ $$$$ \\ $$
Commented by Spillover last updated on 29/Dec/25
$${thanks} \\ $$