Question Number 225356 by Ari last updated on 22/Oct/25
Commented by Ari last updated on 22/Oct/25
$${find}\:{volu}\mathrm{8}{e}\:{of}\:{the}\:{glass} \\ $$
Answered by mr W last updated on 23/Oct/25
Commented by mr W last updated on 23/Oct/25
![V=V_1 −V_2 V_1 =πr_2 ^2 h V_2 =(((r_2 −r_1 )h)/2)×2π(r_2 −((r_2 −r_1 )/3)) =((π(r_2 −r_1 )(2r_2 +r_1 )h)/3) ⇒V=πh[r_2 ^2 −(((r_2 −r_1 )(2r_2 +r_1 ))/3)] =((π(r_1 ^2 +r_1 r_2 +r_2 ^2 )h)/3) =((π(3^2 +3×5+5^2 )×9)/3)=147π](https://www.tinkutara.com/question/Q225368.png)
$${V}={V}_{\mathrm{1}} −{V}_{\mathrm{2}} \\ $$$${V}_{\mathrm{1}} =\pi{r}_{\mathrm{2}} ^{\mathrm{2}} {h} \\ $$$${V}_{\mathrm{2}} =\frac{\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right){h}}{\mathrm{2}}×\mathrm{2}\pi\left({r}_{\mathrm{2}} −\frac{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} }{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:=\frac{\pi\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)\left(\mathrm{2}{r}_{\mathrm{2}} +{r}_{\mathrm{1}} \right){h}}{\mathrm{3}} \\ $$$$\Rightarrow{V}=\pi{h}\left[{r}_{\mathrm{2}} ^{\mathrm{2}} −\frac{\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)\left(\mathrm{2}{r}_{\mathrm{2}} +{r}_{\mathrm{1}} \right)}{\mathrm{3}}\right] \\ $$$$\:\:\:\:\:\:\:\:=\frac{\pi\left({r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} \right){h}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\pi\left(\mathrm{3}^{\mathrm{2}} +\mathrm{3}×\mathrm{5}+\mathrm{5}^{\mathrm{2}} \right)×\mathrm{9}}{\mathrm{3}}=\mathrm{147}\pi \\ $$
Commented by mr W last updated on 23/Oct/25
Commented by mr W last updated on 23/Oct/25
$${in}\:{my}\:{approach},\:{V}_{\mathrm{1}} \:{is}\:{the}\:{cylinder} \\ $$$${with}\:{radius}\:{r}_{\mathrm{2}} \:{and}\:{height}\:{h}.\:{V}_{\mathrm{2}} \:{is} \\ $$$${the}\:{volume}\:{of}\:{the}\:{orange}\:{area}\: \\ $$$${rotated}\:{about}\:{y}−{axis}. \\ $$
Commented by mr W last updated on 23/Oct/25
$${see}\:{Q}\mathrm{158237} \\ $$
Answered by fantastic last updated on 23/Oct/25
Commented by fantastic last updated on 23/Oct/25
![H=9 (2/( 9))=(x/z)⇒x=((2z)/( 9)) r_dz =3+((2z)/( 9)) V_r_dz =π(3+((2z)/( 9)))^2 dz V=∫_0 ^( 9) π(3+((2z)/( 9)))^2 dz =π∫_0 ^( 9) (9+((4z^2 )/(81))+12(z/( 9)))dz =π(9∫_0 ^( 9) dz+(4/(81))∫_0 ^( 9) z^2 dz+((12)/( 9))∫_0 ^( 9) z dz) =π[9z+(4/(81))×(z^3 /3)+((12)/( 9))×(z^2 /2)]_0 ^9 =π(9×9+(4/( 81))×(((9)^3 )/3)+(6/( 9))×(9)^2 ) =π(81+12+54) =147πcm^3 ≈461.814cm^3](https://www.tinkutara.com/question/Q225377.png)
$${H}=\mathrm{9} \\ $$$$\frac{\mathrm{2}}{\:\mathrm{9}}=\frac{{x}}{{z}}\Rightarrow{x}=\frac{\mathrm{2}{z}}{\:\mathrm{9}} \\ $$$${r}_{{dz}} =\mathrm{3}+\frac{\mathrm{2}{z}}{\:\mathrm{9}} \\ $$$${V}_{{r}_{{dz}} } =\pi\left(\mathrm{3}+\frac{\mathrm{2}{z}}{\:\mathrm{9}}\right)^{\mathrm{2}} {dz} \\ $$$${V}=\int_{\mathrm{0}} ^{\:\mathrm{9}} \pi\left(\mathrm{3}+\frac{\mathrm{2}{z}}{\:\mathrm{9}}\right)^{\mathrm{2}} {dz} \\ $$$$=\pi\int_{\mathrm{0}} ^{\:\mathrm{9}} \left(\mathrm{9}+\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{81}}+\mathrm{12}\frac{{z}}{\:\mathrm{9}}\right){dz} \\ $$$$=\pi\left(\mathrm{9}\int_{\mathrm{0}} ^{\:\mathrm{9}} {dz}+\frac{\mathrm{4}}{\mathrm{81}}\int_{\mathrm{0}} ^{\:\mathrm{9}} {z}^{\mathrm{2}} {dz}+\frac{\mathrm{12}}{\:\mathrm{9}}\int_{\mathrm{0}} ^{\:\mathrm{9}} {z}\:{dz}\right) \\ $$$$=\pi\left[\mathrm{9}{z}+\frac{\mathrm{4}}{\mathrm{81}}×\frac{{z}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{12}}{\:\mathrm{9}}×\frac{{z}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{9}} \\ $$$$=\pi\left(\mathrm{9}×\mathrm{9}+\frac{\mathrm{4}}{\:\mathrm{81}}×\frac{\left(\mathrm{9}\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{6}}{\:\mathrm{9}}×\left(\mathrm{9}\right)^{\mathrm{2}} \right) \\ $$$$=\pi\left(\mathrm{81}+\mathrm{12}+\mathrm{54}\right) \\ $$$$=\mathrm{147}\pi{cm}^{\mathrm{3}} \approx\mathrm{461}.\mathrm{814}{cm}^{\mathrm{3}} \\ $$$$ \\ $$
Commented by fantastic last updated on 23/Oct/25
$${how}\:{did}\:{you}\:{get}\:{V}_{\mathrm{2}} \:{sir}? \\ $$$${pls}\:{tell} \\ $$
Answered by TonyCWX last updated on 23/Oct/25
$${r}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3} \\ $$$${R}=\frac{\mathrm{10}}{\mathrm{2}}=\mathrm{5} \\ $$$${h}=\mathrm{9} \\ $$$$ \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}\pi{h}\left({R}^{\mathrm{2}} +{Rr}+{r}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\pi\left(\mathrm{9}\right)\left(\mathrm{5}^{\mathrm{2}} +\mathrm{5}\left(\mathrm{3}\right)+\mathrm{3}^{\mathrm{2}} \right) \\ $$$$=\mathrm{147}\pi\:{cm}^{\mathrm{3}} \\ $$$$ \\ $$$$\begin{array}{|c|}{{Volume}\:{of}\:{the}\:{cup}\:=\:\mathrm{147}\pi\:{cm}^{\mathrm{3}} }\\\hline\end{array} \\ $$
Answered by jjthomson93 last updated on 24/Oct/25
$$\:{V}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\pi{h}\left({R}^{\mathrm{2}} +{rR}+{r}^{\mathrm{2}} \right) \\ $$$$\:{where}\:{r}\:=\:\mathrm{6}\:{cm},\:{R}\:=\:\mathrm{10}\:{cm}\:{and} \\ $$$$\:{h}\:=\:\mathrm{9}\:{cm} \\ $$$$\:{V}\:=\:\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{9}×\pi\left(\mathrm{10}^{\mathrm{2}} +\mathrm{6}×\mathrm{10}+\mathrm{6}^{\mathrm{2}} \right) \\ $$$$\:=\:\mathrm{3}×\pi×\mathrm{196} \\ $$$$\:=\:\mathrm{588}\pi \\ $$$$\:\approx\:\mathrm{1847}.\mathrm{3}\:{cm}^{\mathrm{3}} \\ $$
Commented by fantastic last updated on 24/Oct/25
$${bro}\:{thats}\:{diameter}\:{not} \\ $$$${radius}! \\ $$