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Volume from rotating function about y axis

  1. Jun 27, 2009 #1
    y=1/x, y=0, x=1,x=1, rotate around y axis;

    curious how to set this up using washers
     
  2. jcsd
  3. Jun 27, 2009 #2
    Re: volumes

    It's been a long time since I have done this, but I think this should be done with disks & I believe you meant x=1 and y=1 ?
     
  4. Jun 27, 2009 #3

    HallsofIvy

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    Re: volumes

    You have "x= 1, x= 1". Surely, that's not what you meant! I am going to assume the limits are x= a, x= b, with a< b.

    Because you have constant x limits (I assume), and are rotating around the y-axis, "washers" is not the best way to do this. The simplest method would be shells.

    But since you ask:

    There is no graph below y= 1/b. But you have y= 0 as the lower limit for the figure. So, for y= 0 to 1/a, this is a cylinder. its volume is [itex]\pi(b^2- a^2)(1/a)[/itex], the area between the two circles, of radii a and b, times it height, 1/a.

    For y= 1/a to 1/b, draw a horizontal line from line from x= (a, y) to (1/y, y) (Since y= 1/x, x= 1/y). Rotated about the y-axis that gives a "washer". Its area is the difference between the areas of the two circles: [itex]\pi((1/y)^2- a^2)[/itex]. The volume of that "washer" is that area times its thickness, "dy": [itex]\pi\int_{y= 1/a}^{1/b} ((1/y)^2- a^2)dy[/itex]. Don't forget to add that first volume to get the volume of the entire thing.
     
  5. Jun 27, 2009 #4

    tiny-tim

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    Hi nameVoid! :smile:

    I'm not sure what you mean :redface:, but the general technique is to cut it into slices with thickness dx … then the volume of each slice is (area)dx, and so the total volume is ∫(area)dx. :wink:
     
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