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Math Help - Surface Area Problem - 4

  1. #1
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    Surface Area Problem - 4

    Find the surface area, considering the curve is being rotated about the y axis.

    y = 3rd root of x

    y = x^{1/3}

    y^{3} = x

    x = y^{3}

    2 \leq y \leq 3

    x' = 3y^{2}

    S = \int_{2}^{3} 2\pi(x) \sqrt{1 + (3y^{2})^{2}} dy

    S = 2\pi\int_{2}^{3} (y^{3}) \sqrt{1 + (9y^{4})} dy

    Next move, probably trig?
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  2. #2
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    Re: Surface Area Problem - 4

    A little bit of simplification first, then yes, a trigonometric substitution

    $\displaystyle \begin{align*} S &= 2\pi \int_2^3{ y^3 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, \mathrm{d}y } \\ &= \frac{\pi}{3} \int_2^3{ y^2 \, \sqrt{ 1 + \left( 3y ^2 \right) ^2 } \, 6y\, \mathrm{d}y } \\ &= \frac{\pi}{9} \int_2^3{ 3y^2 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, 6y \, \mathrm{d}y } \end{align*}$

    So now substitute $\displaystyle \begin{align*} 3y^2 = \tan{ \left( \theta \right) } \implies 6y \, \mathrm{d}y = \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta \end{align*}$ and notice that when $\displaystyle \begin{align*} y = 2, \theta = \arctan{ \left( 12 \right) } \end{align*}$ and when $\displaystyle \begin{align*} y = 3, \theta = \arctan{ \left( 27 \right) } \end{align*}$, the integral becomes

    $\displaystyle \begin{align*} \frac{\pi}{9} \int_2^3{ 3y^2 \, \sqrt{ 1 + \left( 3y^2 \right) ^2 } \, 6y\, \mathrm{d}y } &= \frac{\pi}{9} \int_{\arctan{ \left( 12 \right) } } ^{\arctan{ \left( 27 \right) } } { \tan{ \left( \theta \right) } \, \sqrt{ 1 + \tan^2{ \left( \theta \right) } } \, \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta } \end{align*}$

    Go from here.
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    Re: Surface Area Problem - 4

    Do you have to use trig sub stuff on this one?
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    Re: Surface Area Problem - 4

    Find the surface area, considering the curve is being rotated about the y axis.

    y = 3rd root of x

    y = x^{1/3}

    y^{3} = x

    x = y^{3}

    2 \leq y \leq 3

    x' = 3y^{2}

    S = \int_{2}^{3} 2\pi(x) \sqrt{1 + (3y^{2})^{2}} dy

    S = 2\pi\int_{2}^{3} (y^{3}) \sqrt{1 + (9y^{4})} dy

    Let's not use trig

    S = 2\pi\int_{2}^{3} (y^{3}) {1 + (9y^{4})^{1/2} dy

    S = 2\pi\int_{2}^{3} (y^{3}) {u)^{1/2} dy

    u = 1 + 9y^{4}

    du = 36y^{3} dy

    \dfrac{1}{36}du = y^{3} dy

    S = 2\pi\int_{2}^{3} \dfrac{1}{36} (u)^{1/2} dy ?? On the right track here?

    = 2\pi \dfrac{1}{36}\dfrac{2}{3}(u)^{3/2}


    = 2\pi \dfrac{1}{54}(u)^{3/2} evaluated at upper bound 3 and lower bound 2

    = 2\pi \dfrac{1}{54}(1 + 9y^{4})^{3/2} evaluated at upper bound 3 and lower bound 2

    [2\pi \dfrac{1}{54}(1 + 9(3)^{4})^{3/2}] - [2\pi \dfrac{1}{54}(1 + 9(2)^{4})^{3/2}]
    Last edited by Jason76; July 20th 2014 at 11:51 PM.
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