Question Number 133533 by Raxreedoroid last updated on 22/Feb/21
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{that}\:\mathrm{is}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curves} \\ $$$${y}={x}^{\mathrm{3}} ,{y}=\mathrm{8},{x}=\mathrm{0},\:\mathrm{rotated}\:\mathrm{about}\:{x}=\mathrm{9} \\ $$
Answered by bemath last updated on 23/Feb/21
![V=π∫_0 ^8 (9−(y)^(1/3) )^2 dy =π∫_0 ^8 (81−18y^(1/3) +y^(2/3) )dy =π [81y−((27)/2)y^(4/3) +(3/5)y^(5/3) ]_0 ^8 = 8π [81−((27)/2)(8^(1/3) )+(3/5)(8)^(2/3) ] = 8π [81−27+((12)/5) ] = 8π[ 54+ ((12)/5) ] = 451π (1/5)](https://www.tinkutara.com/question/Q133559.png)
$$\mathrm{V}=\pi\underset{\mathrm{0}} {\overset{\mathrm{8}} {\int}}\left(\mathrm{9}−\sqrt[{\mathrm{3}}]{\mathrm{y}}\:\right)^{\mathrm{2}} \:\mathrm{dy} \\ $$$$\:\:\:\:=\pi\underset{\mathrm{0}} {\overset{\mathrm{8}} {\int}}\left(\mathrm{81}−\mathrm{18y}^{\mathrm{1}/\mathrm{3}} +\mathrm{y}^{\mathrm{2}/\mathrm{3}} \right)\mathrm{dy} \\ $$$$\:\:\:\:=\pi\:\left[\mathrm{81y}−\frac{\mathrm{27}}{\mathrm{2}}\mathrm{y}^{\mathrm{4}/\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{5}}\mathrm{y}^{\mathrm{5}/\mathrm{3}} \:\right]_{\mathrm{0}} ^{\mathrm{8}} \\ $$$$\:\:\:=\:\mathrm{8}\pi\:\left[\mathrm{81}−\frac{\mathrm{27}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{1}/\mathrm{3}} \right)+\frac{\mathrm{3}}{\mathrm{5}}\left(\mathrm{8}\right)^{\mathrm{2}/\mathrm{3}} \:\right] \\ $$$$\:\:\:=\:\mathrm{8}\pi\:\left[\mathrm{81}−\mathrm{27}+\frac{\mathrm{12}}{\mathrm{5}}\:\right] \\ $$$$\:\:\:=\:\mathrm{8}\pi\left[\:\mathrm{54}+\:\frac{\mathrm{12}}{\mathrm{5}}\:\right] \\ $$$$\:\:=\:\mathrm{451}\pi\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$