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Volume of rotation - my answer seems off a bit

  1. May 12, 2009 #1
    Find the area of the given function, rotated about the y axis

    The Area below the line y=2 and above y=sin(x) from [tex]0-\pi[/tex]

    I did this rotated around the X AXIS no problem,

    I found the area of the disk to be [tex]\pi(4-sin^{2}[/tex](x))

    The volume around the X AXIS to be[tex]\frac{7\pi^{2}}{2}[/tex]



    3. The attempt at a solution

    For the Y axis I have used this:[tex]\int^{\pi}_{0} 2\pi xf(x)dx[/tex]
    Plugging in:

    [tex]\int^{\pi}_{0} 2\pi x(2-sin(x))dx[/tex]

    I solved this to be [tex]2\pi^{3}-2\pi^{2}[/tex]


    Does this look right?
     
  2. jcsd
  3. May 12, 2009 #2

    Mark44

    Staff: Mentor

    Yes.
     
  4. May 12, 2009 #3
    Really? That is awesome!

    Any second opinions?
     
  5. May 12, 2009 #4

    Mark44

    Staff: Mentor

    OK.
    No.:smile:
     
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