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Math Help - Surface Area Problem

  1. #1
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    Surface Area Problem

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

    y = x^{3}

    0 \leq x \leq 3

    y' = 3x^{2}

    S =    \int_{0}^{3} 2\pi (x^{3}) \sqrt{1 + (3x^{2})^{2}} dx

    S =   2\pi \int_{0}^{3}  (x^{3})  \sqrt{1 + 9x^{4}} dx

    S =   2\pi  \int_{0}^{3}(x^{3})     \sqrt{1 + 9x^{4}} dx

    Use trig stuff here?
    Last edited by Jason76; July 19th 2014 at 11:01 PM.
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  2. #2
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    Re: Surface Area Problem

    You also need to convert the $\displaystyle \begin{align*} x^3 \end{align*}$ from your trigonometric substitution.
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    Re: Surface Area Problem

    Quote Originally Posted by Prove It View Post
    You also need to convert the $\displaystyle \begin{align*} x^3 \end{align*}$ from your trigonometric substitution.
    What do you mean about the x^{3} and how would you go about trig sub. considering we have 1 + 9x^{4} This going to the 4th power not 2nd as required to use a\tan\theta
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    Re: Surface Area Problem

    Since you have edited your original post I can't really state what we originally had. Anyway, you have correctly gotten to $\displaystyle \begin{align*} S = \int_0^3{ 2\pi\left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x} \end{align*}$. From here I would do...

    $\displaystyle \begin{align*} \int_0^3{ 2\pi \left( x^3 \right) \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } &= 2\pi \int_0^3{ x^3 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, \mathrm{d}x } \\ &= \frac{\pi}{3} \int_0^3{ x^2 \, \sqrt{1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \\ &= \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } \end{align*}$

    Now I would substitute $\displaystyle \begin{align*} 3x^2 = \tan{ \left( \theta \right) } \implies 6x\,\mathrm{d}x = \sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$, and also note that when $\displaystyle \begin{align*} x = 0, \theta = 0 \end{align*}$ and when $\displaystyle \begin{align*} x = 3, \theta = \arctan{ \left( 27 \right) } \end{align*}$, the integral becomes

    $\displaystyle \begin{align*} \frac{\pi}{9} \int_0^3{ 3x^2 \, \sqrt{ 1 + \left( 3x^2 \right) ^2 } \, 6x \, \mathrm{d}x } &= \frac{\pi}{9} \int_0^{\arctan{ \left( 27 \right) } } { \tan{ \left( \theta \right) } \, \sqrt{ 1 + \tan^2{ \left( \theta \right) } } \, \sec^2{ \left( \theta \right) } \, \mathrm{d}\theta } \end{align*}$

    Can you go from here?
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    Re: Surface Area Problem

    Quote Originally Posted by Jason76 View Post
    Use trig stuff here?
    You don't need a trigonometric substitution.
    Use the substitution \sqrt{1+9x^4}=t\Rightarrow 1+9x^4=t^2\Rightarrow 36x^3dx=2tdt\Rightarrow x^3dx=\frac{1}{18}tdt.
    You can continue from here.
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    Re: Surface Area Problem

    Quote Originally Posted by red_dog View Post
    You don't need a trigonometric substitution.
    Use the substitution \sqrt{1+9x^4}=t\Rightarrow 1+9x^4=t^2\Rightarrow 36x^3dx=2tdt\Rightarrow x^3dx=\frac{1}{18}tdt.
    You can continue from here.
    Yes you DO need a trigonometric substitution here. When you resubstitute in for the x^3 term, you will end up with another $\displaystyle \begin{align*} \sqrt{ a^2 + t^2} \end{align*}$ form.
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    Re: Surface Area Problem

    This problem needs u substitution, not trig sub. It's just like an arc length problem before. Same numbers.

    y = x^{3}

    0 \leq x \leq 3

    y' = 3x^{2}

    S =    \int_{0}^{3} 2\pi (x^{3}) \sqrt{1 + (3x^{2})^{2}} dx

    S =   2\pi \int_{0}^{3}  (x^{3})  \sqrt{1 + 9x^{4}} dx

    S =   2\pi  \int_{0}^{3}(x^{3})     \sqrt{1 + 9x^{4}} dx

    S =   2\pi  \int_{0}^{3}(x^{3})     (1 + 9x^{4})^{1/2} dx

    S =   2\pi  \int_{0}^{3}(x^{3})     (u)^{1/2} dx

    u = 1 + 9x^{4}

    du = 36x^{3} dx

    \dfrac{1}{36} = x^{3} dx

    =   2\pi     \dfrac{1}{54}(u)^{3/2} evaluated at upper bound 3 and lower bound 0

    =   2\pi     \dfrac{1}{54}(1 + 9x^{4})^{3/2} evaluated at upper bound 3 and lower bound 0

    [2\pi     \dfrac{1}{54}(1 + 93)^{4})^{3/2}] - [2\pi     \dfrac{1}{54}(1 + 9(0)^{4})^{3/2}]
    Last edited by Jason76; July 21st 2014 at 12:17 AM.
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