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Question Number 133533 by Raxreedoroid last updated on 22/Feb/21

$$\mathrm{Find}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{region}\mathrm{that}\mathrm{is}\mathrm{bounded}\mathrm{by}\mathrm{the}\mathrm{curves}$$$$y=x^3,y=8,x=0,\mathrm{rotated}\mathrm{about}x=9$$

Answered by bemath last updated on 23/Feb/21

$$\mathrm{V}=\pi \underset{0}{\stackrel{8}{\int }}{(9-\sqrt[3]{\mathrm{y}})}^2\mathrm{dy}$$$$=\pi \underset{0}{\stackrel{8}{\int }}(81-{18\mathrm{y}}^{1/3}+{\mathrm{y}}^{2/3})\mathrm{dy}$$$$=\pi {[81\mathrm{y}-\frac{27}{2}{\mathrm{y}}^{4/3}+\frac{3}{5}{\mathrm{y}}^{5/3}]}_0^8$$$$=8\pi [81-\frac{27}{2}\left(8^{1/3}\right)+\frac{3}{5}{\left(8\right)}^{2/3}]$$$$=8\pi [81-27+\frac{12}{5}]$$$$=8\pi [54+\frac{12}{5}]$$$$=451\pi \frac{1}{5}$$

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