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Volume of solid curves

  1. Apr 22, 2006 #1

    Could someone please give me an idea on how to go about this problem

    Find the volume of the curve genereated by revolving the area between the curve y =(cos x)/x and the x axis in the interval pie/6 to pie/2

    Thanks a lot..
  2. jcsd
  3. Apr 22, 2006 #2

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  4. Apr 22, 2006 #3


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    Better than a formula is to think through it: Draw a line from any point on the x-axis up to the curve. As the curve is rotated around the x-axis that line sweeps out a disk of radius y= cos(x)/x. It's area is [itex]\pi y^2[/itex] and if we imagine that as a very shallow cylinder of height dx (the height of the disk is in the x direction) its volume is [itex]\pi y^2 dx[/itex].
    The volume of the whole thing is a sum of those volumes (a Riemann sum) and becomes the integral Tinaaa said:
    [tex]\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}y^2 dx[/tex]
    Since y= cos(x)/x, put that in and integrate.

    Was this really for a PRE-Calculus course?
  5. Apr 22, 2006 #4
    @ Hallofivy

    Thanks a lot. Actually I couldnt think up where to put this post so i just dropped it in Pre calculus.
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