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Question Number 130208 by liberty last updated on 23/Jan/21

$$\mathrm{The}\mathrm{loop}\mathrm{of}\mathrm{curve}{2\mathrm{ay}}^2=\mathrm{x}{(\mathrm{x}-\mathrm{a})}^2$$$$\mathrm{revolves}\mathrm{about}\mathrm{straight}\mathrm{line}$$$$\mathrm{y}=\mathrm{a}.\mathrm{Find}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{solid}$$$$\mathrm{generated}.$$

Answered by benjo_mathlover last updated on 23/Jan/21

$$\mathrm{The}\mathrm{given}\mathrm{curve}\mathrm{is}{2\mathrm{ay}}^2=\mathrm{x}{(\mathrm{x}-\mathrm{a})}^2...\left(1\right)$$$$\mathrm{the}\mathrm{curve}\left(1\right)\mathrm{is}\mathrm{symetrical}\mathrm{about}\mathrm{the}\mathrm{x}-\mathrm{axis}$$$$\mathrm{and}\mathrm{the}\mathrm{loop}\mathrm{lies}\mathrm{between}\mathrm{x}=0\mathrm{and}\mathrm{x}=\mathrm{a}.$$$$\mathrm{Differentiating}\left(1\right)\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x},\mathrm{we}\mathrm{get}$$$$4\mathrm{ay}\left(\mathrm{dy}/\mathrm{dx}\right)=2\mathrm{x}(\mathrm{x}-\mathrm{a})+{(\mathrm{x}-\mathrm{a})}^2={3\mathrm{x}}^2-4\mathrm{ax}+{\mathrm{a}}^2$$$$\left(\mathrm{dy}/\mathrm{dx}\right)=0\mathrm{when}{3\mathrm{x}}^2-4\mathrm{ax}+{\mathrm{a}}^2=0\mathrm{or}\mathrm{when}$$$$\mathrm{x}=\frac{\mathrm{a}}{3}\mathrm{which}\mathrm{gives}\mathrm{from}\left(1\right),\mathrm{y}=\frac{\mathrm{a}\sqrt{2}}{3\sqrt{3}}.$$$$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{loop}\mathrm{A}=2{\int }_0^{\mathrm{a}}\mathrm{y}\mathrm{dx}=2{\int }_0^{\mathrm{a}}\frac{(\mathrm{x}-\mathrm{a})\sqrt{\mathrm{x}}}{\sqrt{2\mathrm{a}}}\mathrm{dx}$$$$\mathrm{A}=\sqrt{\frac{2}{\mathrm{a}}}{\int }_0^{\mathrm{a}}({\mathrm{x}}^{3/2}-{\mathrm{ax}}^{1/2})\mathrm{dx}=\frac{4}{15}\sqrt{2}{\mathrm{a}}^2$$$$\therefore \mathrm{by}\mathrm{Pappus}\mathrm{theorem}\mathrm{the}\mathrm{required}\mathrm{volume}$$$$=2\pi \mathrm{a}\times \mathrm{A}=\frac{8}{15}\sqrt{2}\pi {\mathrm{a}}^3....................◊$$

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