Question Number 170412 by Mastermind last updated on 23/May/22
$${If}\:{part}\:{of}\:{the}\:{curve}\:{y}={x}+\mathrm{1}\:{from}\:{x}=\mathrm{1} \\ $$$${and}\:{x}=\mathrm{2}\:{is}\:{rotated}\:{completely}\:{about} \\ $$$${the}\:{y}−{axis}.\:{find}\:{the}\:{volume}\:{of}\:{the} \\ $$$${solid}\:{form}. \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by aleks041103 last updated on 23/May/22
$${The}\:{resulted}\:{shape}\:{is}\:{a}\:{trincated}\:{cone}. \\ $$$${r}_{\mathrm{1}} =\mathrm{1},\:{r}_{\mathrm{2}} =\mathrm{2} \\ $$$${h}=\mid{y}_{\mathrm{2}} −{y}_{\mathrm{1}} \mid=\mathrm{1} \\ $$$$\Rightarrow{V}=\frac{\mathrm{1}}{\mathrm{3}}\pi{h}\left({r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{2}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\pi\left(\mathrm{1}+\mathrm{4}+\mathrm{2}\right)=\frac{\mathrm{7}\pi}{\mathrm{3}} \\ $$$$\Rightarrow{V}=\frac{\mathrm{7}\pi}{\mathrm{3}} \\ $$
Commented by Mastermind last updated on 24/May/22
$$ \\ $$$${Thanks} \\ $$
Answered by MikeH last updated on 23/May/22
![V = π∫_a ^b y^2 dx ⇒ V = π∫_1 ^2 (x^2 +2x+1)dx V = π[(x^3 /3) + x^2 +x]_1 ^2 V = ((19π)/3) cubic units](https://www.tinkutara.com/question/Q170439.png)
$${V}\:=\:\pi\int_{{a}} ^{{b}} {y}^{\mathrm{2}} \:{dx} \\ $$$$\Rightarrow\:{V}\:=\:\pi\int_{\mathrm{1}} ^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$$${V}\:=\:\pi\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+\:{x}^{\mathrm{2}} +{x}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${V}\:=\:\frac{\mathrm{19}\pi}{\mathrm{3}}\:\mathrm{cubic}\:\mathrm{units} \\ $$
Commented by Mastermind last updated on 24/May/22
$$ \\ $$$${Thank}\:{you}\: \\ $$$$ \\ $$