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Quick question about volume of revolution

  1. Jun 15, 2013 #1
    Hi, this isn't a specific question but say you had a function y = x - 1 and you were told that the region from x = 1 to x = 3 was rotated 2pi radians and were asked to find the volume of revolution formed.

    My question is, would this volume of revolution be the same if they said it was rotated 4pi radians, or 3pi radians? i.e. would I still use [itex] \pi \int_1^3 (x-1)^2 \ dx [/itex] or would I use 2pi if it was rotated 4pi radians, or 3pi/2 if it was rotated 3pi radians. As how I see it, no "new" volume would be formed after rotating a full 2pi.
     
  2. jcsd
  3. Jun 15, 2013 #2

    HallsofIvy

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    Using the standard definition of "volume", yes, going around [itex]4\pi[/itex] rather than [itex]2\pi[/itex] just goes over the same region twice but does not change the volume.
     
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