Question Number 136076 by physicstutes last updated on 18/Mar/21
$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:\mathrm{curve}\:{y}\:=\:\mathrm{3}\:+\:\mathrm{2}{x}\:−{x}^{\mathrm{2}} ,\:\mathrm{the}\:{x}−\mathrm{axis} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:{y}\:=\:\mathrm{3}. \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{generated}\:\mathrm{when}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{is} \\ $$$$\mathrm{rotated}\:\mathrm{completely}\:\mathrm{about}\:\mathrm{the}\:\mathrm{line}\:{y}\:=\:\mathrm{3} \\ $$
Answered by mr W last updated on 18/Mar/21
$${Area}\:{A}=\frac{\mathrm{2}}{\mathrm{3}}×\left(\mathrm{4}×\mathrm{4}−\mathrm{2}×\mathrm{1}\right)=\frac{\mathrm{28}}{\mathrm{3}} \\ $$
Commented by mr W last updated on 18/Mar/21
Commented by mr W last updated on 18/Mar/21
![volume V x^2 −2x−3+y=0 (x_2 −x_1 )^2 =(x_2 +x_1 )^2 −4x_1 x_2 =2^2 −4(−3+y)=4(4−y) x_2 −x_1 =2(√(4−y)) dV=2π(3−y)(x_2 −x_1 )dy V=∫_0 ^3 2π(3−y)(x_2 −x_1 )dy =4π∫_0 ^3 (3−y)(√(4−y))dy =4π∫_3 ^0 (3−y)(√(1+3−y))d(3−y) =4π∫_0 ^3 u(√(1+u))du =4π×(2/(15))[(u+1)^(3/2) (3u−2)]_0 ^3 =4π×((2×58)/(15)) =((464π)/(15))](https://www.tinkutara.com/question/Q136097.png)
$${volume}\:{V} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}+{y}=\mathrm{0} \\ $$$$\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({x}_{\mathrm{2}} +{x}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{3}+{y}\right)=\mathrm{4}\left(\mathrm{4}−{y}\right) \\ $$$${x}_{\mathrm{2}} −{x}_{\mathrm{1}} =\mathrm{2}\sqrt{\mathrm{4}−{y}} \\ $$$${dV}=\mathrm{2}\pi\left(\mathrm{3}−{y}\right)\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right){dy} \\ $$$${V}=\int_{\mathrm{0}} ^{\mathrm{3}} \mathrm{2}\pi\left(\mathrm{3}−{y}\right)\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right){dy} \\ $$$$\:\:=\mathrm{4}\pi\int_{\mathrm{0}} ^{\mathrm{3}} \left(\mathrm{3}−{y}\right)\sqrt{\mathrm{4}−{y}}{dy} \\ $$$$\:\:=\mathrm{4}\pi\int_{\mathrm{3}} ^{\mathrm{0}} \left(\mathrm{3}−{y}\right)\sqrt{\mathrm{1}+\mathrm{3}−{y}}{d}\left(\mathrm{3}−{y}\right) \\ $$$$\:\:=\mathrm{4}\pi\int_{\mathrm{0}} ^{\mathrm{3}} {u}\sqrt{\mathrm{1}+{u}}{du} \\ $$$$\:\:=\mathrm{4}\pi×\frac{\mathrm{2}}{\mathrm{15}}\left[\left({u}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{3}{u}−\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{3}} \\ $$$$\:\:=\mathrm{4}\pi×\frac{\mathrm{2}×\mathrm{58}}{\mathrm{15}} \\ $$$$\:\:=\frac{\mathrm{464}\pi}{\mathrm{15}} \\ $$