Question Number 134587 by bobhans last updated on 05/Mar/21
$$\mathrm{Precalculus}\: \\ $$How do I find the volume of a solid obtained by rotating the region bounded by x=(y−3)^2 and x=4 about y=1?
Commented by EDWIN88 last updated on 05/Mar/21
Commented by EDWIN88 last updated on 05/Mar/21
![Vol = 2π∫_1 ^( 5) (y−1)(4−(y−3)^2 ) dy Vol=2π∫_1 ^( 5) (y−1)(−y^2 +6y−5)dy Vol=2π∫_1 ^( 5) (−y^3 +7y^2 −11y+5)dy Vol=2π [−(y^4 /4)+((7y^3 )/3)−((11y^2 )/2)+5y ]_1 ^5 Vol=2π [−(((624)/4))+((7(124))/3)−((11(24))/2)+20] Vol=2π(−156+((868)/3)−132+20 ) Vol=2π(((868)/3)−((804)/3))=((128π)/3)](https://www.tinkutara.com/question/Q134590.png)
$$\mathrm{Vol}\:=\:\mathrm{2}\pi\int_{\mathrm{1}} ^{\:\mathrm{5}} \left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{4}−\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} \right)\:\mathrm{dy} \\ $$$$\mathrm{Vol}=\mathrm{2}\pi\int_{\mathrm{1}} ^{\:\mathrm{5}} \left(\mathrm{y}−\mathrm{1}\right)\left(−\mathrm{y}^{\mathrm{2}} +\mathrm{6y}−\mathrm{5}\right)\mathrm{dy} \\ $$$$\mathrm{Vol}=\mathrm{2}\pi\int_{\mathrm{1}} ^{\:\mathrm{5}} \left(−\mathrm{y}^{\mathrm{3}} +\mathrm{7y}^{\mathrm{2}} −\mathrm{11y}+\mathrm{5}\right)\mathrm{dy} \\ $$$$\mathrm{Vol}=\mathrm{2}\pi\:\left[−\frac{\mathrm{y}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{7y}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{11y}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{5y}\:\right]_{\mathrm{1}} ^{\mathrm{5}} \\ $$$$\mathrm{Vol}=\mathrm{2}\pi\:\left[−\left(\frac{\mathrm{624}}{\mathrm{4}}\right)+\frac{\mathrm{7}\left(\mathrm{124}\right)}{\mathrm{3}}−\frac{\mathrm{11}\left(\mathrm{24}\right)}{\mathrm{2}}+\mathrm{20}\right] \\ $$$$\mathrm{Vol}=\mathrm{2}\pi\left(−\mathrm{156}+\frac{\mathrm{868}}{\mathrm{3}}−\mathrm{132}+\mathrm{20}\:\right) \\ $$$$\mathrm{Vol}=\mathrm{2}\pi\left(\frac{\mathrm{868}}{\mathrm{3}}−\frac{\mathrm{804}}{\mathrm{3}}\right)=\frac{\mathrm{128}\pi}{\mathrm{3}} \\ $$