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Cylindrical shell

  1. Aug 6, 2006 #1
    find volume:

    [tex]y=x^2, y=0, x=1, x=2[/tex] about x=1

    I found the height to be x^2

    and circumference to be 2pi(1-x)

    So [tex]V= \int_1^2 2\pi(1-x)(x^2)dx[/tex]

    This is not giving me the right answer.
  2. jcsd
  3. Aug 6, 2006 #2
    I don't think the (1-x) is correct, I think it should just be x.
  4. Aug 6, 2006 #3
    hrrmm..still seems to be giving me wrong answer
  5. Aug 7, 2006 #4
    Oh I didn't see the about x=1 part, so I think I'm wrong that it should only be x sorry about that. Do you know what the right answer should be because it seems to me that what you had originally should be correct unless I'm forgetting something important about the shell method.

    Edit: I see the problem now you have (1-x) representing the radius, but if you have this then at x=2 the shell has a radius of negative 1 and this is obviously incorrect, so try the problem again but with (x-1) instead of (1-x) and you should be able to get the right answer.
    Last edited: Aug 7, 2006
  6. Aug 7, 2006 #5
    I dont think that is right, wouldnt that just be a vector?...it should be 16pi/15
  7. Aug 7, 2006 #6
    What do you mean when you say a vector? And unless I'm completely wrong when I'm remembering how to do the shell method 16pi/15 doesnt seem right I did this problem using (x-1) as a radius and shifting the graph of y=x[sup2[/sup] so that I could just use x as a radius and I am coming up with the same answer in both cases neither of which is 16pi/15.
    Last edited: Aug 7, 2006
  8. Aug 7, 2006 #7
    The radius of -1 just represents direction(probly wrong)?Well, I found another example where curve is rotated about x=2, they also take the radius to be 2-x. Your way makes sense to me, but why would they do examples this way...i try the method in the example and get a wrong answer.
  9. Aug 7, 2006 #8
    The -1 definitely matters since the integral will be of different sign then if you integrate using (x-1) as the radius. I understand you're difficulty and it has been a while since I've done a problem with the shell method so I've been trying to review this and I found a problem in my book that does the same thing as yours using 2-x as the radius when revolving around x=2 but there the integral is from 0 to 1 and the radius is always positive, so I think what you need to do about finding an expression for the radius is to look at what the radius should be at the points of integration relative to the axis of rotation and then try and come up with an expression that makes sense so that the radius is always positive.
  10. Aug 7, 2006 #9


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    "Direction" is not relevant. x-1 is the distance from the given x to the line x=1 about which you are rotating: x is larger than 1 and distance is always positive. The area of a "thin" cylinder from r to r+ dr is [itex]2\pi r dr[/itex]. Here r= x-1. The volume is that multiplied by the height, y= x2. The volume of the thin cylinder is [itex]2\pi(1-x)x^2dx[/itex] so the entire volume is
    [tex]2\pi\int_1^2 (x-1)x^2 dx[/itex]
  11. Aug 7, 2006 #10
    hrmm..Ok Answer in the back must be wrong!
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