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Evaluate the definite integral for the area of the surface.

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

    y=(x3/6) + (1/2x), [1,2]


    2. Relevant equations
    2π∫[r(x)](1+[f'(x)2])


    3. The attempt at a solution
    First I found the derivative.

    f'(x)= (x2/2) + (1/2x2)dx​

    And since y is a function of x, r(x) is

    r(x)= (x3/6) + (1/2x)​

    Then I plug everything in and get

    2π∫ [(x3/6) + (1/2x)] * {1 + [(x2/2) + (1/2x2)]2}1/2}dx​

    And then I'm stuck. The book tells me that I am suppose to get

    2π∫ [(x3/6) + (1/2x)] * [(x2/2) + (1/2x2)]dx​

    But I have no idea how they got that. Specifically, I don't know how they got rid of the radical...
     
  2. jcsd
  3. Jun 2, 2010 #2
    For starters you derived it wrong, the derivative of f(x) is f'(x)=x2/2-1/2x2. Maybe this is the problem. Unless you just wrote your f(x) wrong.

    Edit: Yes that is the problem, use the correct f'(x), and then expand f'(x)2 and put everything over a common denominator, and then you can make an equation which is being squared that is equal to that, and then the root cancels out the squares.
     
    Last edited: Jun 2, 2010
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