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Question Number 190940 by Spillover last updated on 14/Apr/23
The parametric equation of a curve  are  x=3t^2  and y=3t−t^2 .  Find the volume generated  when the plane bounded by the curve  ,the x−axis and the ordinates   corresponding to   t=0   and t=2  rotates about the y−axis
$$\mathrm{The}\:\mathrm{parametric}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{curve}\:\:\mathrm{are} \\ $$$$\mathrm{x}=\mathrm{3t}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{y}=\mathrm{3t}−\mathrm{t}^{\mathrm{2}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{generated} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve} \\ $$$$,\mathrm{the}\:\mathrm{x}−\mathrm{axis}\:\mathrm{and}\:\mathrm{the}\:\mathrm{ordinates}\: \\ $$$$\mathrm{corresponding}\:\mathrm{to}\: \\ $$$$\mathrm{t}=\mathrm{0}\:\:\:\mathrm{and}\:\mathrm{t}=\mathrm{2}\:\:\mathrm{rotates}\:\mathrm{about}\:\mathrm{the}\:\mathrm{y}−\mathrm{axis} \\ $$
Answered by MikeH last updated on 16/Apr/23
V = π∫_x_1  ^x_2  y^2 dx   ⇒ V = π∫_t_1  ^t_2  y^2 ((dx/dt))dt   (dx/dt) = 6t and V = π∫_0 ^2 (3t−t^2 )^2 6t dt  ⇒ V = π∫_0 ^2 (6t^5 −36t^4 +54t^3 )dt   V = 49.6 π cubic units.
$${V}\:=\:\pi\int_{{x}_{\mathrm{1}} } ^{{x}_{\mathrm{2}} } {y}^{\mathrm{2}} {dx}\: \\ $$$$\Rightarrow\:{V}\:=\:\pi\int_{{t}_{\mathrm{1}} } ^{{t}_{\mathrm{2}} } {y}^{\mathrm{2}} \left(\frac{{dx}}{{dt}}\right){dt}\: \\ $$$$\frac{{dx}}{{dt}}\:=\:\mathrm{6}{t}\:\mathrm{and}\:{V}\:=\:\pi\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{3}{t}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{6}{t}\:{dt} \\ $$$$\Rightarrow\:{V}\:=\:\pi\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{6}{t}^{\mathrm{5}} −\mathrm{36}{t}^{\mathrm{4}} +\mathrm{54}{t}^{\mathrm{3}} \right){dt}\: \\ $$$${V}\:=\:\mathrm{49}.\mathrm{6}\:\pi\:\mathrm{cubic}\:\mathrm{units}. \\ $$$$ \\ $$

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