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Question Number 141191 by Eric002 last updated on 17/May/21
find the volume of the solid generated  when the region bounded by the y=x  y=x+2, x=2 and x=4 revolved about the x-axis
$${find}\:{the}\:{volume}\:{of}\:{the}\:{solid}\:{generated} \\ $$$${when}\:{the}\:{region}\:{bounded}\:{by}\:{the}\:{y}={x} \\ $$$${y}={x}+\mathrm{2},\:{x}=\mathrm{2}\:{and}\:{x}=\mathrm{4}\:{revolved}\:{about}\:{the}\:{x}-{axis} \\ $$
Answered by ajfour last updated on 17/May/21
 V=π∫_2 ^(  4) {(x+2)^2 −x^2 }dx      =π(2x^2 +4x)∣_2 ^4      =π(24+8)= 32π
$$\:{V}=\pi\int_{\mathrm{2}} ^{\:\:\mathrm{4}} \left\{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \right\}{dx} \\ $$$$\:\:\:\:=\pi\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}\right)\mid_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$\:\:\:=\pi\left(\mathrm{24}+\mathrm{8}\right)=\:\mathrm{32}\pi\: \\ $$

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