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Trig integrals and finding volume.

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the volume obtained by rotating the following curves bounded by:

    y=sinx y=0 pi/2≤ x ≤ pi

    2. Relevant equations

    I know I have to use the cylindrial disk method so ∫2pi(x)f(x)dx.

    3. The attempt at a solution

    I did the following ∫(pi/2 to pi) 2pi (x) sinxdx
    I factor out the 2pi and im left with ∫(x)(sinx)(dx) always from pi/2 to pi
    I then used integration by part with u=x du=dx and dv=sinx v=-cosx

    therefore uv -∫vdu --> -(xcos(x)) (from pi/2 to pi) - ∫-cos(x)dx from pi/2 to pi

    However my answer key shows a completely different work / answer.. I don't know how to approach the question differently
     
  2. jcsd
  3. Mar 13, 2013 #2
    Which axis are you rotating about?
    Also if this is about the x-axis, all you have to do is use:
    [tex]\int_a^b \pi f(x)^{2}dx[/tex]
     
    Last edited: Mar 13, 2013
  4. Mar 13, 2013 #3
    Wait but I thought a cylindrical would be 2pi(x) f(x) dx . X for the radius and f(x) for the height. And yes it was about the x-axis
     
  5. Mar 13, 2013 #4

    SteamKing

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    What's the area of a disk?
     
  6. Mar 13, 2013 #5
    It would be pi (x)^2. And this was a stupid question. I meant cylindrical shell... So my teacher used the disk method. How would I be able to decide between cylindrical shell, the disk and washer method :/ ( I mean washer there would have to be a "hole" of some sort no?)
     
  7. Mar 13, 2013 #6
    Nope you can use the shell or disk method anytime, sometimes (like exams) there is only one way to do it because the other way will make the integral really nasty.
     
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