Question Number 134038 by Raxreedoroid last updated on 27/Feb/21
Answered by EDWIN88 last updated on 27/Feb/21
![V=2π∫_0 ^2 (5−x)(8−x^3 ) dx V=2π∫_0 ^2 (40−5x^3 −8x+x^4 )dx V=2π [ 40x−((5x^4 )/4)−4x^2 +(x^5 /5) ]_0 ^2 V= 2π [ 80−20−16+((32)/5) ] V= 2π (44+((32)/5)) = 2π(((220+32)/5))=((504π)/5)](Q134042.png)
$$\mathrm{V}=\mathrm{2}\pi\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\left(\mathrm{5}−\mathrm{x}\right)\left(\mathrm{8}−\mathrm{x}^{\mathrm{3}} \right)\:\mathrm{dx} \\ $$$$\:\mathrm{V}=\mathrm{2}\pi\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\left(\mathrm{40}−\mathrm{5x}^{\mathrm{3}} −\mathrm{8x}+\mathrm{x}^{\mathrm{4}} \right)\mathrm{dx} \\ $$$$\:\mathrm{V}=\mathrm{2}\pi\:\left[\:\mathrm{40x}−\frac{\mathrm{5x}^{\mathrm{4}} }{\mathrm{4}}−\mathrm{4x}^{\mathrm{2}} +\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}\:\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\mathrm{V}=\:\mathrm{2}\pi\:\left[\:\mathrm{80}−\mathrm{20}−\mathrm{16}+\frac{\mathrm{32}}{\mathrm{5}}\:\right] \\ $$$$\:\mathrm{V}=\:\mathrm{2}\pi\:\left(\mathrm{44}+\frac{\mathrm{32}}{\mathrm{5}}\right)\:=\:\mathrm{2}\pi\left(\frac{\mathrm{220}+\mathrm{32}}{\mathrm{5}}\right)=\frac{\mathrm{504}\pi}{\mathrm{5}} \\ $$
Commented by Raxreedoroid last updated on 27/Feb/21
$$\mathrm{Thanks}\:\mathrm{sir}... \\ $$