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Question Number 25201 by tawa tawa last updated on 06/Dec/17

Answered by ajfour last updated on 06/Dec/17

y=(6/(3x−2))   ⇒  (dy/dx)∣_(x=2) =−((18)/((3x−2)^2 ))∣_(x=2)      (dy/dx)∣_(x=2) =−(9/8) .  V=∫_1 ^( 2) 2πxydx        =2π∫_1 ^( 2) ((2×3dx)/(3x−2))=4πln ∣3x−2∣_(x=1) ^(x=2)    V =4πln 4 .

$${y}=\frac{\mathrm{6}}{\mathrm{3}{x}−\mathrm{2}}\:\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\mid_{{x}=\mathrm{2}} =−\frac{\mathrm{18}}{\left(\mathrm{3}{x}−\mathrm{2}\right)^{\mathrm{2}} }\mid_{{x}=\mathrm{2}} \\ $$$$\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\mathrm{2}} =−\frac{\mathrm{9}}{\mathrm{8}}\:. \\ $$$${V}=\int_{\mathrm{1}} ^{\:\mathrm{2}} \mathrm{2}\pi{xydx}\: \\ $$$$\:\:\:\:\:=\mathrm{2}\pi\int_{\mathrm{1}} ^{\:\mathrm{2}} \frac{\mathrm{2}×\mathrm{3}{dx}}{\mathrm{3}{x}−\mathrm{2}}=\mathrm{4}\pi\mathrm{ln}\:\mid\mathrm{3}{x}−\mathrm{2}\mid_{{x}=\mathrm{1}} ^{{x}=\mathrm{2}} \\ $$$$\:{V}\:=\mathrm{4}\pi\mathrm{ln}\:\mathrm{4}\:. \\ $$

Commented by tawa tawa last updated on 06/Dec/17

I appreciate sir. God bless you.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by mrW1 last updated on 06/Dec/17

ajfour sir, please check:  it should be a rotation about x−axis.    V=∫_1 ^( 2) πy^2 dx   ⇒V=π∫_1 ^( 2) ((36)/((3x−2)^2 ))dx   ⇒V=12π∫_1 ^( 2) (1/((3x−2)^2 ))d(3x)   ⇒V=12π[−(1/(3x−2))]_1 ^2   ⇒V=12π((1/1)−(1/4))  ⇒V=9π

$${ajfour}\:{sir},\:{please}\:{check}: \\ $$$${it}\:{should}\:{be}\:{a}\:{rotation}\:{about}\:{x}−{axis}. \\ $$$$ \\ $$$${V}=\int_{\mathrm{1}} ^{\:\mathrm{2}} \pi{y}^{\mathrm{2}} {dx}\: \\ $$$$\Rightarrow{V}=\pi\int_{\mathrm{1}} ^{\:\mathrm{2}} \frac{\mathrm{36}}{\left(\mathrm{3}{x}−\mathrm{2}\right)^{\mathrm{2}} }{dx}\: \\ $$$$\Rightarrow{V}=\mathrm{12}\pi\int_{\mathrm{1}} ^{\:\mathrm{2}} \frac{\mathrm{1}}{\left(\mathrm{3}{x}−\mathrm{2}\right)^{\mathrm{2}} }{d}\left(\mathrm{3}{x}\right)\: \\ $$$$\Rightarrow{V}=\mathrm{12}\pi\left[−\frac{\mathrm{1}}{\mathrm{3}{x}−\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow{V}=\mathrm{12}\pi\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\Rightarrow{V}=\mathrm{9}\pi \\ $$

Commented by tawa tawa last updated on 06/Dec/17

Yes 9π is in the options. God bless you sir.

$$\mathrm{Yes}\:\mathrm{9}\pi\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{options}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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