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Disk and Shell Method (Ignore my other thread)

  1. Aug 12, 2008 #1
    Reducing it to three questions, because I'm pretty confident on the others.
    9. Find the volume of the solid generated by revolving about the line x = -1, the region bounded by the curves y = -x^2 + 4x - 3 and y = 0.
    ---
    I graphed everything, and then translated the graph 1 to the right making it y=-(x-1)^2+4(x-1)-3 and y=0 rotated around the y-axis because this is easier for me.

    So then, I should be able to do this with the shell method, and get [tex]\int_2^4 2x \pi (-(x-1)^2+4(x-1)-3)\,dx[/tex]

    I don't think I'm right because this is negative
    ---

    10. Consider the region in the xy-plane between x = 0 and x=pi/2 bounded by y = 0 and y = sin x. Find the volume of the solid generated by revolving this region about the x-axis.
    stuck here too

    Honestly, I just am posting this because my answer seems to simple
    [tex]\int_0^\pi/2 sin^2x\,dx [/tex]

    11. Let R be the region bounded by y = 1/x, y = x^2, x = 0, and y = 2. Suppose R is revolved around the x-axis. Set up but do not evaluate the integrals for the volume of rotation using: a) the method of cylindrical shells; b) the method of circular disks.
    Honestly don't know where to start
     
    Last edited: Aug 12, 2008
  2. jcsd
  3. Aug 12, 2008 #2

    tiny-tim

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    Hi myanmar! :smile:

    9. No, it is positive … for example, if x = 3, the bracket is -4 + 8 - 3 = 1. :smile:

    10. I assume you mean [tex]\int_0^{\pi/2} \pi sin^2x\,dx [/tex] ?

    Yes, that's fine! :smile:

    11. Don't know how to help you on this … you just have to draw the diagram carefully, and decide what the endpoints are of each slice.
     
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