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Math Help - [SOLVED] Surface Area

  1. #1
    Junior Member superman69's Avatar
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    [SOLVED] Surface Area

    Find the surface area of the surface of revolution generated by revolving the graph , around the x-axis.

    This is what I got so far I don't know what to do next.
    Attached Thumbnails Attached Thumbnails [SOLVED] Surface Area-msp62419805b9efe3g18bb00002004d5gia929fc3c.gif  
    Last edited by superman69; October 21st 2009 at 11:00 PM. Reason: error
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by superman69 View Post
    Find the surface area of the surface of revolution generated by revolving the graph , around the x-axis.

    This is what I got so far I don't know what to do next.
    I'm not sure how you got your result ...

    If y = f(x) then the surface area produced by revolution of the curve around the x-axis is:

    A = 2\pi \int_a^b\left(y \sqrt{1+(y')^2}\right)dx

    With your example you get:

    A = 2\pi \int_0^8\left(x^3 \sqrt{1+9x^4}\right)dx

    Use integration by substitution:

    u = 1+9x^4~\implies~\dfrac{du}{dx}=36x^3\implies~du=36  x^3 \cdot dx

    Your integral becomes:

    A = 2\pi \cdot \dfrac1{36} \int_0^8\left(36x^3 \sqrt{1+9x^4}\right)dx = \dfrac \pi{18} \int u^{\frac12} du

    Can you take it from here?
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  3. #3
    Junior Member superman69's Avatar
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    yes thank you
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