Volume of Rotation around the y-axis for y=1/x+2 and x=1

[Hacca]

Homework Statement

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

Homework Equations

The textbook use this one:
[tex]Vx= \pi \int_{a}^{b}(f(x))^{2} dx[/tex]

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]

 

[Panphobia]

I would have to say that the shell method would be easier here,
2*∏*∫(SHELL RADIUS)*(SHELL HEIGHT) dx
Of course the limits of integration would be [0,1]

 

[STEMucator]

It would be convenient to use vertical line segments in this case in order to avoid setting up multiple integrals. So:

##r_{in} = x - dx##
##r_{out} = x##
##height = \frac{1}{x+2}##

##dV = 2\pi(\frac{r_{in} + r_{out}}{2})(r_{out} - r_{in})(height)##

 

[Hacca]

I would have to say that the shell method would be easier here,
2*∏*∫(SHELL RADIUS)*(SHELL HEIGHT) dx
Of course the limits of integration would be [0,1]

It would be convenient to use vertical line segments in this case in order to avoid setting up multiple integrals. So:

##r_{in} = x - dx##
##r_{out} = x##
##height = \frac{1}{x+2}##

##dV = 2\pi(\frac{r_{in} + r_{out}}{2})(r_{out} - r_{in})(height)##

Thanks,
I have never seen this methods. The textbook & teacher only use pi(f(x))² and Washer method. Should it not be possiable with disc/washer?

With [0,1] I got:
[tex]Vy = 4\pi ((1-\frac{1}{4*1}-ln1) - (\frac{0}{0}-\frac{1}{4*0}-ln\frac{0}{0})) = 4\pi (1-\frac{1}{4}) = 4\pi (3/4) [/tex]
Isn't the parentheses with 0 undefined, so I can't really use lower limit 0?

 

[STEMucator]

The method I'm using is simply a generalized version of the disk and washer methods. Makes it so you only have to remember one formula really.

Using the method I get 1.18796 as the answer.

 

[Panphobia]

Using the shell method I got the same answer as Zondrina, but using your integral Hacca, I evaluated the integral on my own, and on wolfram, and both came to the same conclusion of 0.0448.