Question Number 87298 by ajfour last updated on 03/Apr/20
$${If}\:{y}=\mathrm{sin}\:{x}\:,\:\:{x}=\mathrm{0}\:{to}\:{x}=\mathrm{2}\pi\:{is} \\ $$$${revolved}\:{about}\:{the}\:{x}-{axis},\:{find} \\ $$$${the}\:{surface}\:{of}\:{the}\:{solid}\:{of} \\ $$$${revolution}. \\ $$
Answered by ajfour last updated on 04/Apr/20
![A=2∫_0 ^( π) (2πy)ds ds=dx×sec θ tan θ=(dy/dx) = cos x A=4π∫_0 ^( π) (sin x)((√(1+cos^2 x)) dx) let cos x=t ⇒ −sin xdx=dt A=4π∫_(−1) ^( 1) (√(1+t^2 )) dt =8π∫_0 ^( 1) (√(1+t^2 )) dt = 8π{(t/2)(√(1+t^2 ))+(1/2)ln ∣t+(√(1+t^2 ))∣ }∣_0 ^1 = 8π[(1/( (√2)))+(1/2)ln (1+(√2))] A=4𝛑[(√2)+ln (1+(√2))].](https://www.tinkutara.com/question/Q87332.png)
$${A}=\mathrm{2}\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{2}\pi{y}\right){ds} \\ $$$$\:\:{ds}={dx}×\mathrm{sec}\:\theta \\ $$$$\:\:\:\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}\:=\:\mathrm{cos}\:{x} \\ $$$${A}=\mathrm{4}\pi\int_{\mathrm{0}} ^{\:\pi} \left(\mathrm{sin}\:{x}\right)\left(\sqrt{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx}\right) \\ $$$$\:\:\:{let}\:\:\mathrm{cos}\:{x}={t}\:\:\Rightarrow\:\:−\mathrm{sin}\:{xdx}={dt} \\ $$$$\:{A}=\mathrm{4}\pi\int_{−\mathrm{1}} ^{\:\:\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$$\:\:\:\:=\mathrm{8}\pi\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$$\:\:\:=\:\mathrm{8}\pi\left\{\frac{{t}}{\mathrm{2}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid{t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\mid\:\right\}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:=\:\mathrm{8}\pi\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right] \\ $$$$\:\:\boldsymbol{{A}}=\mathrm{4}\boldsymbol{\pi}\left[\sqrt{\mathrm{2}}+\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right]. \\ $$$$ \\ $$
Commented by mr W last updated on 04/Apr/20
$${you}\:{are}\:{right}\:{sir}!\:{thanks}! \\ $$
Commented by ajfour last updated on 04/Apr/20
$${You}\:{might}\:{like}\:{Q}.\mathrm{87296}\:{Sir}. \\ $$