Question Number 187921 by Mingma last updated on 23/Feb/23
Answered by HeferH last updated on 23/Feb/23
$${V}_{\mathrm{1}} −{V}_{\mathrm{2}} \:=\:{V}_{\mathrm{3}} \\ $$
Answered by mr W last updated on 23/Feb/23
Commented by mr W last updated on 23/Feb/23
$$\frac{{h}_{\mathrm{1}} }{{h}_{\mathrm{1}} +{h}}=\frac{{r}}{{R}}\:\Rightarrow{h}_{\mathrm{1}} =\frac{{rh}}{{R}−{r}} \\ $$$${big}\:{cone}: \\ $$$${V}_{{big}} =\frac{\pi{R}^{\mathrm{2}} \left({h}+{h}_{\mathrm{1}} \right)}{\mathrm{3}}=\frac{\pi{R}^{\mathrm{3}} {h}}{\mathrm{3}\left({R}−{r}\right)} \\ $$$${small}\:{cone}: \\ $$$${V}_{{small}} =\frac{\pi{r}^{\mathrm{2}} {h}_{\mathrm{1}} }{\mathrm{3}}=\frac{\pi{r}^{\mathrm{3}} {h}}{\mathrm{3}\left({R}−{r}\right)} \\ $$$${frustum}: \\ $$$${V}={V}_{{big}} −{V}_{{small}} =\frac{\pi{R}^{\mathrm{3}} {h}}{\mathrm{3}\left({R}−{r}\right)}−\frac{\pi{r}^{\mathrm{3}} {h}}{\mathrm{3}\left({R}−{r}\right)} \\ $$$$\:\:=\frac{\pi{h}\left({R}^{\mathrm{3}} −{r}^{\mathrm{3}} \right)}{\mathrm{3}\left({R}−{r}\right)}=\frac{\pi{h}\left({R}^{\mathrm{2}} +{Rr}+{r}^{\mathrm{2}} \right)}{\mathrm{3}} \\ $$
Commented by Mingma last updated on 23/Feb/23
Perfect!
Commented by otchereabdullai last updated on 24/Feb/23
$${supper}\:{prof}\:{W} \\ $$
Commented by Humble last updated on 01/Mar/23
$${superb}! \\ $$
Commented by Humble last updated on 01/Mar/23
$${superb}! \\ $$
Answered by mr W last updated on 23/Feb/23
Commented by mr W last updated on 23/Feb/23
![y=r+(((R−r)x)/h) dV=πy^2 dx V=∫_0 ^h πy^2 dx=π∫_0 ^h [r^2 +((2r(R−r)x)/h)+(((R−r)^2 x^2 )/h^2 )]dx =π[r^2 x+((r(R−r)x^2 )/h)+(((R−r)^2 x^3 )/(3h^2 ))]_0 ^h =π[r^2 h+((r(R−r)h^2 )/h)+(((R−r)^2 h^3 )/(3h^2 ))] =πh[r^2 +r(R−r)+(((R−r)^2 )/3)] =πh[((3r^2 +3rR−3r^2 +R^2 −2Rr+r^2 )/3)] =((πh(R^2 +Rr+r^2 ))/3) ✓](https://www.tinkutara.com/question/Q187942.png)
$${y}={r}+\frac{\left({R}−{r}\right){x}}{{h}} \\ $$$${dV}=\pi{y}^{\mathrm{2}} {dx} \\ $$$${V}=\int_{\mathrm{0}} ^{{h}} \pi{y}^{\mathrm{2}} {dx}=\pi\int_{\mathrm{0}} ^{{h}} \left[{r}^{\mathrm{2}} +\frac{\mathrm{2}{r}\left({R}−{r}\right){x}}{{h}}+\frac{\left({R}−{r}\right)^{\mathrm{2}} {x}^{\mathrm{2}} }{{h}^{\mathrm{2}} }\right]{dx} \\ $$$$\:=\pi\left[{r}^{\mathrm{2}} {x}+\frac{{r}\left({R}−{r}\right){x}^{\mathrm{2}} }{{h}}+\frac{\left({R}−{r}\right)^{\mathrm{2}} {x}^{\mathrm{3}} }{\mathrm{3}{h}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{{h}} \\ $$$$\:=\pi\left[{r}^{\mathrm{2}} {h}+\frac{{r}\left({R}−{r}\right){h}^{\mathrm{2}} }{{h}}+\frac{\left({R}−{r}\right)^{\mathrm{2}} {h}^{\mathrm{3}} }{\mathrm{3}{h}^{\mathrm{2}} }\right] \\ $$$$\:=\pi{h}\left[{r}^{\mathrm{2}} +{r}\left({R}−{r}\right)+\frac{\left({R}−{r}\right)^{\mathrm{2}} }{\mathrm{3}}\right] \\ $$$$\:=\pi{h}\left[\frac{\mathrm{3}{r}^{\mathrm{2}} +\mathrm{3}{rR}−\mathrm{3}{r}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{Rr}+{r}^{\mathrm{2}} }{\mathrm{3}}\right] \\ $$$$\:=\frac{\pi{h}\left({R}^{\mathrm{2}} +{Rr}+{r}^{\mathrm{2}} \right)}{\mathrm{3}}\:\checkmark \\ $$
Commented by Mingma last updated on 23/Feb/23
Nice solution, sir!
Commented by Mingma last updated on 23/Feb/23
Nice sir!