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Question Number 169860 by Shrinava last updated on 11/May/22

$$\mathrm{y}={\mathrm{x}}^2+1$$$$\mathrm{y}=0$$$$\mathrm{x}=-1$$$$\mathrm{x}=2$$$$\mathrm{find}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{object}\mathrm{obtained}$$$$\mathrm{by}\mathrm{rotating}\mathrm{the}\mathrm{figure}\mathrm{bounded}\mathrm{by}\mathrm{lines}$$$$\mathrm{around}\mathrm{the}\mathrm{abscissa}\mathrm{axis}$$

Answered by mr W last updated on 11/May/22

Commented by mr W last updated on 11/May/22

$$tryingwithoutintegralcalculation.$$$$V_1=2\pi \times 3\times 1\times 0.5=3\pi$$$$V_2=2\pi \times \frac{1\times 1}{3}\times (1+\frac{1}{4})=\frac{5\pi }{6}$$$$V_3=2\pi \times \frac{2\times 4}{3}\times (1+\frac{4}{4})=\frac{32\pi }{3}$$$$\Rightarrow V=\mathrm{\Sigma }=\frac{29\pi }{2}$$

Commented by Shrinava last updated on 12/May/22

$$\mathrm{Perfect}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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