Volume of rotation, y-axis

1. The problem statement, all variables and given/known data



Hello!



English is not my native language so I hope the terminology is right.



Q:

Find the volume generated by the curve y=1/x+2, positive x- and y-axis and the line x=1.

Calculate the volume obtained by rotation around the:

a) x-axis

b) y-axis



2. Relevant equations

The text book use this one:

[tex]Vx= \pi \int_{a}^{b}(f(x))^{2} dx[/tex]



3. The attempt at a solution

a) I got this one right:



[tex]Vx = \pi \int_{0}^{1}\frac{1}{(x+2)^{2}} dy= \pi \left[ -\frac{1}{x+2} \right]= \pi(\frac{1}{3} - \frac{1}{2}) = \frac{\pi}{6}[/tex]



b) I can't get this straight and I'm not sure about the upper-/lower-limits:



[tex] y(1)= \frac{1}{1+2}=\frac{1}{3}\\y(0)=\frac{1}{0+2}=\frac{1}{2}.[/tex]



X as a function of y:



[tex]x(y)=\frac{1}{y}-2[/tex]



We have:



[tex]x(y)=\frac{1}{y}-2\\Vy = \pi \int_{1/2}^{1/3}(\frac{1}{y}-2)^{2} dy = 4\pi \left[y-\frac{1}{4y}-lny \right] = 4\pi ((\frac{1}{2}-\frac{1}{2}-ln\frac{1}{2}) - (\frac{1}{3}-\frac{3}{4}-ln\frac{1}{3}))=\\= 4\pi (-\frac{1}{3}+\frac{3}{4}+ln\frac{1}{3}-ln\frac{1}{2}) = 4\pi (\frac{5}{12}+ln\frac{2}{3}) [/tex]



Stock here! I have tested the limits 0 to 1, 0 to 1/3 och 1/3 to ½. Anybody have any clue?







The answer is:

Spoiler

[tex]4\pi (\frac{1}{2}+ ln\frac{2}{3}) v.e[/tex]