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Question Number 163223 by ZiYangLee last updated on 05/Jan/22

The arc of parabola y=−x^2 +9 in 0<x<3  is revolved about the line y=c and 0<c<9  to generate a solid.  Find the value of c that minimizes the   volume of the solid.

$$\mathrm{The}\:\mathrm{arc}\:\mathrm{of}\:\mathrm{parabola}\:{y}=−{x}^{\mathrm{2}} +\mathrm{9}\:\mathrm{in}\:\mathrm{0}<{x}<\mathrm{3} \\ $$ $$\mathrm{is}\:\mathrm{revolved}\:\mathrm{about}\:\mathrm{the}\:\mathrm{line}\:{y}={c}\:\mathrm{and}\:\mathrm{0}<{c}<\mathrm{9} \\ $$ $$\mathrm{to}\:\mathrm{generate}\:\mathrm{a}\:\mathrm{solid}. \\ $$ $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{c}\:\mathrm{that}\:\mathrm{minimizes}\:\mathrm{the}\: \\ $$ $$\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}. \\ $$

Answered by TheSupreme last updated on 05/Jan/22

f(x)=−x^2 +9−c  V=2π∫_(−3) ^3 f(x)^2 dx=2π∫_(−3) ^3 (9−c)^2 +x^4 −2x^2 (9−c)dx=  =2π[6(9−c)^2 +2(3^5 /5)−2(3^3 /3)(9−c)]  (∂V/∂c)=2π[−12(9−c)+12]=0  2π[−12∙9+12c+12]=0  12c=12∙8  c=8

$${f}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{9}−{c} \\ $$ $${V}=\mathrm{2}\pi\int_{−\mathrm{3}} ^{\mathrm{3}} {f}\left({x}\right)^{\mathrm{2}} {dx}=\mathrm{2}\pi\int_{−\mathrm{3}} ^{\mathrm{3}} \left(\mathrm{9}−{c}\right)^{\mathrm{2}} +{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{9}−{c}\right){dx}= \\ $$ $$=\mathrm{2}\pi\left[\mathrm{6}\left(\mathrm{9}−{c}\right)^{\mathrm{2}} +\mathrm{2}\frac{\mathrm{3}^{\mathrm{5}} }{\mathrm{5}}−\mathrm{2}\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{3}}\left(\mathrm{9}−{c}\right)\right] \\ $$ $$\frac{\partial{V}}{\partial{c}}=\mathrm{2}\pi\left[−\mathrm{12}\left(\mathrm{9}−{c}\right)+\mathrm{12}\right]=\mathrm{0} \\ $$ $$\mathrm{2}\pi\left[−\mathrm{12}\centerdot\mathrm{9}+\mathrm{12}{c}+\mathrm{12}\right]=\mathrm{0} \\ $$ $$\mathrm{12}{c}=\mathrm{12}\centerdot\mathrm{8} \\ $$ $${c}=\mathrm{8} \\ $$

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