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Question Number 190940 by Spillover last updated on 14/Apr/23

$$\mathrm{The}\mathrm{parametric}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{curve}\mathrm{are}$$$$\mathrm{x}={3\mathrm{t}}^2\mathrm{and}\mathrm{y}=3\mathrm{t}-{\mathrm{t}}^2.$$$$\mathrm{Find}\mathrm{the}\mathrm{volume}\mathrm{generated}$$$$\mathrm{when}\mathrm{the}\mathrm{plane}\mathrm{bounded}\mathrm{by}\mathrm{the}\mathrm{curve}$$$$,\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{and}\mathrm{the}\mathrm{ordinates}$$$$\mathrm{corresponding}\mathrm{to}$$$$\mathrm{t}=0\mathrm{and}\mathrm{t}=2\mathrm{rotates}\mathrm{about}\mathrm{the}\mathrm{y}-\mathrm{axis}$$

Answered by MikeH last updated on 16/Apr/23

$$V=\pi {\int }_{x_1}^{x_2}{y}^{2}dx$$$$\Rightarrow V=\pi {\int }_{t_1}^{t_2}{y}^2\left(\frac{dx}{dt}\right)dt$$$$\frac{dx}{dt}=6t\mathrm{and}V=\pi {\int }_0^2{(3t-t^2)}^{2}6tdt$$$$\Rightarrow V=\pi {\int }_0^2(6t^5-36t^4+54t^3)dt$$$$V=49.6\pi \mathrm{cubic}\mathrm{units}.$$$$$$

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