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Calculating the volume of a function that rotates around the x-axis

  • Thread starter jakobs
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  • #1
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Homework Statement



y=(1/x) is rotating around the x-axis.
find the volume for the figure between x=1 and x=2

Homework Equations





The Attempt at a Solution


V=pi y2 = pi*(1/x)2

V= 1∫2 pi(1/x)2dx wich gives V= 1∫2 pi(1/x2)dx

A primitive function for (1/x2) is -(1/x)

V=1∫2 pi((-(1/2))-(-(1/1)))
How do I go from here?
 

Answers and Replies

  • #2
Bacle2
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Why are you multiplying in your last line:

V=∫21 (pi)...

Once you have the antiderivative , you evaluate the antiderivative at the endpoints

2,1 respectively, and that is the value of the integral.
 
  • #3
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Sorry, I posted wrong, don't know what happened.

but pi*(-(1/2))=-1,57
pi*(-(1/1))=-3,14

-1,57-(-3,14)= 1,57

is this the right answer?
 
  • #4
CAF123
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Yes, but a nicer answer would have been [itex] \frac{\pi}{2} [/itex] units3.
 
  • #5
Zondrina
Homework Helper
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Homework Statement



y=(1/x) is rotating around the x-axis.
find the volume for the figure between x=1 and x=2

Homework Equations





The Attempt at a Solution


V=pi y2 = pi*(1/x)2

V= 1∫2 pi(1/x)2dx wich gives V= 1∫2 pi(1/x2)dx

A primitive function for (1/x2) is -(1/x)

V=1∫2 pi((-(1/2))-(-(1/1)))
How do I go from here?
Easily stated you want :

[itex]2π\int_{1}^{2}\frac{1}{x^2}dx[/itex]

You could have also used :

[itex]π\int_{1}^{2}1dx[/itex]
 
  • #6
CAF123
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Easily stated you want :

[itex]2π\int_{1}^{2}\frac{1}{x^2}dx[/itex]
The OP correctly identified the volume of revolution to be [tex] V = \pi\int_{1}^{2} \frac{1}{x^2} dx. [/tex]
You could have also used :

[itex]π\int_{1}^{2}1dx[/itex]
What mathematical reasoning gives this?
 
  • #7
Zondrina
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The OP correctly identified the volume of revolution to be [tex] V = \pi\int_{1}^{2} \frac{1}{x^2} dx. [/tex]


What mathematical reasoning gives this?
There are two formulas I remember from Stewart calc 4th edition which highlight the formulas to use when calculating volumes of solids.

[tex]\pi\int_{a}^{b}xf(x)dx[/tex]

[tex]2\pi\int_{a}^{b}f^2(x)dx[/tex]
 
Last edited:
  • #8
CAF123
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There are two formulas I remember from Stewart calc 4th edition which highlight the formulas to use when calculating volumes of solids.

[tex]\pi\int_{a}^{b}xf(x)dx[/tex]

[tex]2\pi\int_{a}^{b}f^2(x)dx[/tex]
I have come across the method of cylindrical shells, where for some rotation about y axis, [tex] V = 2\pi \int_{a}^{b} x f(x) dx [/tex]
I was wondering if someone could attempt the problem in the OP using cylindrical shells, because I am always getting either π or π/4, not the required π/2.
 
  • #9
vela
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Easily stated you want :

[itex]2π\int_{1}^{2}\frac{1}{x^2}dx[/itex]

You could have also used :

[itex]π\int_{1}^{2}1dx[/itex]
There are two formulas I remember from Stewart calc 4th edition which highlight the formulas to use when calculating volumes of solids.

[tex]\pi\int_{a}^{b}xf(x)dx[/tex]

[tex]2\pi\int_{a}^{b}f^2(x)dx[/tex]
You've remembered incorrectly.
 
  • #10
vela
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I have come across the method of cylindrical shells, where for some rotation about y axis, [tex] V = 2\pi \int_{a}^{b} x f(x) dx [/tex]
I was wondering if someone could attempt the problem in the OP using cylindrical shells, because I am always getting either π or π/4, not the required π/2.
Show us what you tried. You should have had to split it up into two integrals.
 
  • #11
CAF123
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I split it up into 2 regions: one integral using y limits from 1/2 to 1 and another from 0 to 1/2.
For the first region: [tex] V = 2\pi \int_{\frac{1}{2}}^{1} (\frac{1}{2} + y)(\frac{1}{y} - 1) dy ,[/tex] where the shell radius is 1/2 + y and it's height is 1/y - 1.

Then the region using y limits 0 to 1/2 was a rectangular region:
[tex] V = 2\pi\int_{0}^{\frac{1}{2}} y dy, [/tex] where the height is a constant 1 unit.

Doing the integral does not give me the correct answer. Where am I going wrong?
Many thanks.
 
  • #12
vela
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The shell radius in the first integral should simply be y.
 
  • #13
CAF123
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The shell radius in the first integral should simply be y.
Perfect, I must have thought that I was rotating about the line y = 1/2 at my first attempt, but I see why it is simply y: y is the distance from the axis of rotation, which is the x axis. I now have the correct answer.
Thanks again.
 
  • #14
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You could have also used :

[itex]π\int_{1}^{2}1dx[/itex]
It would be a good idea for you to review formulas that you "remember" to see if they are actually valid. The one above has nothing to do with this problem.
 

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