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Math Help - Arc length

  1. #1
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    Arc length

    Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x axis.

    y = \frac {x^3}{6} + \frac {1}{2x}

    Now, to find the derivative is easy.
     y \prime = \frac {x^2}{2} - \frac {1}{2x^2}
    Not sure why the solution manual shows a - instead of a plus!

    Anyways, at this point we have the form
     1 + (y \prime)^2 = ( \frac {x^2}{2} - \frac {1}{2x^2})^2
    which i can follow along with..

    What is going on here? because if i squared everything i would get
    \frac {x^4}{4} + \frac {1}{4x^2}

    However, the next step i dont see how the solutions manual got it..?

    S = 2 \pi \int (\frac {x^3}{6} + \frac {1}{2x}) ( \frac {x^2}{2} + \frac {1}{2x^2}) dx



    and why are they attaching y prime on the end? you dont need to explain the 2 pi
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  2. #2
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    Re: Arc length

    Quote Originally Posted by icelated View Post
    Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x axis.

    y = \frac {x^3}{6} + \frac {1}{2x}

    Now, to find the derivative is easy.
     y \prime = \frac {x^2}{2} - \frac {1}{2x^2}
    Not sure why the solution manual shows a - instead of a plus!

    Anyways, at this point we have the form
     1 + (y \prime)^2 = ( \frac {x^2}{2} - \frac {1}{2x^2})^2
    which i can follow along with..

    What is going on here? because if i squared everything i would get
    \frac {x^4}{4} + \frac {1}{4x^2}

    no ... you're forgetting the middle term

    However, the next step i dont see how the solutions manual got it..?

    S = 2 \pi \int (\frac {x^3}{6} + \frac {1}{2x}) ( \frac {x^2}{2} + \frac {1}{2x^2}) dx



    and why are they attaching y prime on the end? you dont need to explain the 2 pi
    y = \frac{1}{6}x^3 + \frac{1}{2} x^{-1}

    y' = \frac{1}{2}x^2 - \frac{1}{2}x^{-2} = \frac{x^2}{2} - \frac{1}{2x^2}

    ... see why the second term of the derivative is negative?

    (y')^2 = \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2 = \frac{x^4}{4} - 2\frac{x^2}{2} \cdot \frac{1}{2x^2} + \frac{1}{4x^4} = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}

    A = 2\pi \int_a^b y\sqrt{1 + (y')^2} \, dx

    note ...

    1+(y')^2 =

    1 + \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4} =

    \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} =

    \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2

    A = 2\pi \int_a^b \left(\frac {x^3}{6} + \frac {1}{2x}\right) \cdot \sqrt{\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2} \, dx

    you never did mention the limits of integration ... finish it.
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  3. #3
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    Re: Arc length

    wow, perfect... Yeah, i limits of integration...
    Thank you so much!!!!!!!
    I guess i just need more practice..
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