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Math Help - Surface Area Problem - 2

  1. #1
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    Surface Area Problem - 2

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

    y = \sqrt{1 +5x}

    1 \leq x \leq 7

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

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

    u = 1 + 5x

    du = 5 dx

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

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

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

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

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

    Next move?
    Last edited by Jason76; July 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

    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
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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?
    \left(1+5x\right)^{\frac{1}{2}}=\sqrt{1+5x}
    and 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 \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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