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Thread: Surface Area Problem - 2

  1. #1
    MHF Contributor Jason76's Avatar
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    Surface Area Problem - 2

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

    $\displaystyle y = \sqrt{1 +5x}$

    $\displaystyle 1 \leq x \leq 7$

    $\displaystyle y = (1 + 5x)^{1/2}$

    $\displaystyle y' = (u)^{1/2} du$

    $\displaystyle u = 1 + 5x$

    $\displaystyle du = 5 dx$

    $\displaystyle y' = \dfrac{5}{2}(1 + 5x)^{-1/2}$

    $\displaystyle S = \int_{1}^{7} 2\pi[(1 + 5x)^{1/2}]\sqrt{1 + [(\dfrac{5}{2})(1 + 5x)^{-1/2}]^{2}] dx$

    $\displaystyle S = \int_{1}^{7} 2\pi[(1 + 5x)^{1/2}]\sqrt{1 + (\dfrac{25}{4})(1 + 5x)^{-1} dx$

    $\displaystyle S = \int_{1}^{7} 2\pi[(1 + 5x)^{1/2}]\sqrt{1 + (\dfrac{25}{4})(\dfrac{1}{1 + 5x}) dx$

    $\displaystyle S = \int_{1}^{7} 2\pi[(1 + 5x)^{1/2}]\sqrt{1 + (\dfrac{25}{4 + 20x}) dx$

    Next move?
    Last edited by Jason76; Jul 19th 2014 at 10:31 PM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Re: Surface Area Problem - 2

    $\displaystyle S=\2\pi\int_1^7\sqrt{1+5x}\cdot\displaystyle\sqrt{ \1+\frac{25}{4(1+5x)}}dx=\\=2\pi\int_1^7\sqrt{1+5x }\cdot\displaystyle\frac{\sqrt{20x+29}}{2\sqrt{1+5 x}}dx=\pi\int_1^7\sqrt{20x+29}dx$
    Can you continue?
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  3. #3
    MHF Contributor Jason76's Avatar
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    Re: Surface Area Problem - 2

    I have no idea what is going on in the post above. What's going on in the top line?
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  4. #4
    MHF Contributor red_dog's Avatar
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    Re: Surface Area Problem - 2

    Quote Originally Posted by Jason76 View Post
    I have no idea what is going on in the post above. What's going on in the top line?
    $\displaystyle \left(1+5x\right)^{\frac{1}{2}}=\sqrt{1+5x}$
    and $\displaystyle 1+\left[\frac{5}{2}\left(1+5x\right)^{-\frac{1}{2}}\right]^2=1+\frac{25}{4}\left(\frac{1}{\sqrt{1+5x}}\right )^2=1+\frac{25}{4}\cdot\frac{1}{1+5x}=\frac{20x+29 }{1+5x}$
    Then $\displaystyle \sqrt{1+5x}\cdot\sqrt{\frac{20x+29}{1+5x}}=\sqrt{1 +5x}\cdot\frac{\sqrt{20x+29}}{\sqrt{1+5x}}=\sqrt{2 0x+9}$
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