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Homework Help: Revolution about y=-1

  1. Sep 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid obtained by rotating the region
    bounded by the given curves about the specified line. Sketch the
    region, the solid, and a typical disk or washer.

    y=sinx, y=cosx, 0≤ x≤∏/4, About y=-1

    2. Relevant equations

    3. The attempt at a solution
    Tried 2 ways, shell method and washer method..

    [tex]\pi \int_0^\frac{\pi}{4}(cosx-sinx)^{2}\,dx - \pi \int_0^\frac{\pi}{4}(-1)^{2}\,dx[/tex]
    [tex]\pi \int_0^\frac{\pi}{4}cos^{2}x-2sinxcosx+sin^{2}x\,dx - \pi \int_0^\frac{\pi}{4}1\,dx[/tex]
    ^Hard integral

    [tex]2\pi \int_0^\frac{\pi}{4}(x+1)(cosx-sinx)\,dx[/tex]
    [tex]2\pi \int_0^\frac{\pi}{4}xcosx+cosx-xsinx-sinx\,dx[/tex]
    [tex]2\pi (sinx+cosx)|[/tex] (From 0 to ∏/4, not sure how to do this in latex)
    (And so on)

    Thanks for any help.
    Last edited: Sep 13, 2012
  2. jcsd
  3. Sep 13, 2012 #2
    You might want to relook at the way you set up the problem.

    Your first integral is some finite value, let's say 1 (I know that isn't right, but follow the idea). Now imagine, instead of the line y = -1, it is rotated about y = -100. Your volume should be bigger, right? But the way you have the equation, it's getting smaller.

    So, for starters, imagine instead you rotate about the line y = 0. What would each individual part look like (the Cos part minus the sin part). Now shift the radius by 1 larger (since you are rotating about y=-1). What is the radius of the outer diameter as you go from 0 to pi/4?
  4. Sep 13, 2012 #3


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    The result is the same as subtracting,
    the volume generated by rotating the region bounded by y=sin(x), 0≤ x≤∏/4, about y=-1
    the volume generated by rotating the region bounded by y=cos(x), 0≤ x≤∏/4, about y=-1 .
  5. Sep 13, 2012 #4
    @SammyS, OK, but it's still a hard integral.
    Last edited: Sep 13, 2012
  6. Sep 13, 2012 #5
    Ah I see, I thought it was becoming a cylinder..
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