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Math Help - Surface Area Problem - 3

  1. #1
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    Surface Area Problem - 3

    Find the area of the surface by rotating the curve about the x axis

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

    \dfrac{1}{2} \leq x \leq 1

    y' = \dfrac{3}{2}x^{2} - 6x^{-2}

    S = \int_{1/2}^{1} 2\pi(y) \sqrt{1 + (\dfrac{3}{2}x^{2} - 6x^{-2})^{2}} dx

    Squared

    (\dfrac{3}{2}x^{2} - 6x^{-2})(\dfrac{3}{2}x^{2} - 6x^{-2}) = \dfrac{9}{2}x^{4} -18x + 36x^{-4}

    S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{1 + (\dfrac{9}{2}x^{4} -18x + 36x^{-4}   )} dx

    Next move?
    Last edited by Jason76; July 19th 2014 at 11:59 PM.
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  2. #2
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    Re: Surface Area Problem - 3

    $\displaystyle \begin{align*} \left( \frac{3}{2}x^2 - 6x^{-2} \right) ^2 = \frac{9}{4}x^4 - 18 + 36x^{-4} \end{align*}$, not $\displaystyle \begin{align*} \frac{9}{4}x^4 - 18x + 36x^{-4} \end{align*}$.

    Once you have fixed this up in your integrand, get a common denominator and simplify.
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    Re: Surface Area Problem - 3

    Find the area of the surface by rotating the curve about the x axis

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

    \dfrac{1}{2} \leq x \leq 1

    y' = \dfrac{3}{2}x^{2} - 6x^{-2}

    S = \int_{1/2}^{1} 2\pi(y) \sqrt{1 + (\dfrac{3}{2}x^{2} - 6x^{-2})^{2}} dx

    Squared

    (\dfrac{3}{2}x^{2} - 6x^{-2})(\dfrac{3}{2}x^{2} - 6x^{-2}) = \dfrac{9}{2}x^{4} -18 + 36x^{-4}

    S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{1 + \dfrac{9}{2}x^{4} -18 + 36x^{-4}   )} dx

    You could multiply everything by 2 to get rid of the fraction

    S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{2+ 18x^{4} -36 + 72x^{-4}   )} dx ???

    S = 2\pi \int_{1/2}^{1} (\dfrac{1}{2}x^{3} + 6x^{-1}) \sqrt{18x^{4} -34 + 72x^{-4}   )} dx ???
    Last edited by Jason76; July 21st 2014 at 12:25 AM.
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