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Revolution Volume by Cylinder Shells

  1. Nov 12, 2006 #1
    Rotate about x-axis the region enclosed by y=e^x, y=1/x, x=1 and x=2. I can do the problem with the rings method but I don't how to even set up the integral to solve by the shells method. Help? Thanks
  2. jcsd
  3. Nov 13, 2006 #2


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    The two curves cross well before x= 1 so you have a region in which y=ex is the upper curve and y= 1/x is the lower. That means that the length of your cylinder is exx- 1/x. The radius is x so you cylinder will have a surface area, for each x, of [itex]2\pi x(e^x- 1/x)[/itex]. Multiply that by the "thickness", dx, to find the differential of volume.
  4. Nov 13, 2006 #3
    Thanks, but since the region enclosed by the boundaries given is rotated about the x-axis, then, doesn't it mean that the radius of the cylinder is y? I'm a little confused. ??
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