Question Number 136910 by leena12345 last updated on 27/Mar/21
$${use}−{cylindrical}−{shells}−{to}−{find}−{volume} \\ $$$${y}={x}^{\mathrm{3}} /{y}=\mathrm{8}/{about}−{x}=\mathrm{9} \\ $$$$ \\ $$
Answered by liberty last updated on 27/Mar/21
![Vol = 2π∫_0 ^( 8) (8−x)x^3 dx = 2π∫_0 ^( 8) (8x^3 −x^4 ) dx =2π [ 2x^4 −(x^5 /5) ]_0 ^8 =2π (x^4 (2−(x/5))]_0 ^8 =2π.8^4 .(2/5) = ((4π.8^4 )/5) = ((16,384π)/5)](https://www.tinkutara.com/question/Q136939.png)
$$\mathrm{Vol}\:=\:\mathrm{2}\pi\int_{\mathrm{0}} ^{\:\mathrm{8}} \left(\mathrm{8}−\mathrm{x}\right)\mathrm{x}^{\mathrm{3}} \:\mathrm{dx}\: \\ $$$$=\:\mathrm{2}\pi\int_{\mathrm{0}} ^{\:\mathrm{8}} \left(\mathrm{8x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{4}} \right)\:\mathrm{dx} \\ $$$$=\mathrm{2}\pi\:\left[\:\mathrm{2x}^{\mathrm{4}} −\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}\:\right]_{\mathrm{0}} ^{\mathrm{8}} \\ $$$$=\mathrm{2}\pi\:\left(\mathrm{x}^{\mathrm{4}} \left(\mathrm{2}−\frac{\mathrm{x}}{\mathrm{5}}\right)\right]_{\mathrm{0}} ^{\mathrm{8}} \\ $$$$=\mathrm{2}\pi.\mathrm{8}^{\mathrm{4}} .\frac{\mathrm{2}}{\mathrm{5}}\:=\:\frac{\mathrm{4}\pi.\mathrm{8}^{\mathrm{4}} }{\mathrm{5}}\:=\:\frac{\mathrm{16},\mathrm{384}\pi}{\mathrm{5}} \\ $$