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Volume of solid of revolution - y axis.

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid of revolution when we rotate the area limited by the x axis and the function f(x) = 1 - cosx where x e [0, 2∏] once around the y-axis?

    3. The attempt at a solution

    In my notes I have the following equation:

    V = ∫ 2∏x f(x) dx

    If I put in my limits (upper limit 2∏, lower limit 0) and my function I get the following:

    V = 2∏ ∫x(1-cos(x)) dx

    V = 2∏ ∫x - xcos(x) dx

    V = 2∏[[itex]\frac{x^2}{2}[/itex] - (xsin(x)+cos(x))]

    V = 2∏ [[itex]\frac{x^2}{2}[/itex] - (∏sin(∏) + cos(∏)] - 2∏ [[itex]\frac{0^2}{2}[/itex] - (0sin(0) + cos(0)]

    Since ∏ sin(∏) = 0, cos(∏) = -1 , 0sin(0) = 0 and cos(0) = 1 I get the following:

    V = 2∏ ([itex]\frac{∏^2}{2}[/itex]) - 2∏ + 1

    V = ##2∏^3## - 4∏ + 2

    Is this correct? Am I using the correct formulas/equations?

    Please let me know if there is something I need to explain better. Any help and guiding is massively appreciated. Thanks.
     
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2

    PeroK

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    Check your integration by parts: should be -cos(x).
     
  4. Nov 17, 2013 #3
    Sloppy mistake by me, there should of course be parenthesis around that equation.

    V = 2∏[[itex]\frac{x^2}{2}[/itex] - (xsin(x)+cos(x))]

    That slipped past me when I was writing off of my notes - therefore unless I again have overlooked something (which I've double checked I haven't), that mistake was a one off - I've solved the rest of the problem as if there were parenthesis around (xsin(x)+cos(x)). Therefore I get - (-1) -> +1 in my answer which I later multiply by 2.

    Does it look ok besides that error?

    Thanks for your input!
     
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