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

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

Answered by TheSupreme last updated on 05/Jan/22

$$f\left(x\right)=-x^2+9-c$$ $$V=2\pi {\int }_{-3}^{3}f{\left(x\right)}^{2}dx=2\pi {\int }_{-3}^3{(9-c)}^2+x^4-2x^2(9-c)dx=$$ $$=2\pi [6{(9-c)}^2+2\frac{3^5}{5}-2\frac{3^3}{3}(9-c)]$$ $$\frac{\partial V}{\partial c}=2\pi [-12(9-c)+12]=0$$ $$2\pi [-12\cdot 9+12c+12]=0$$ $$12c=12\cdot 8$$ $$c=8$$

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