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

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations



    3. 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?
     
  2. jcsd
  3. Sep 2, 2012 #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.
     
  4. Sep 2, 2012 #3
    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?
     
  5. Sep 2, 2012 #4

    CAF123

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    Yes, but a nicer answer would have been [itex] \frac{\pi}{2} [/itex] units3.
     
  6. Sep 2, 2012 #5

    Zondrina

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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]
     
  7. Sep 3, 2012 #6

    CAF123

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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?
     
  8. Sep 3, 2012 #7

    Zondrina

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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]
     
    Last edited: Sep 3, 2012
  9. Sep 5, 2012 #8

    CAF123

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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.
     
  10. Sep 5, 2012 #9

    vela

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    You've remembered incorrectly.
     
  11. Sep 5, 2012 #10

    vela

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    Show us what you tried. You should have had to split it up into two integrals.
     
  12. Sep 5, 2012 #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.
     
  13. Sep 5, 2012 #12

    vela

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    The shell radius in the first integral should simply be y.
     
  14. Sep 5, 2012 #13

    CAF123

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    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.
     
  15. Sep 5, 2012 #14

    Mark44

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    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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