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Math Help - Surface Area of a Revolution

  1. #1
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    Surface Area of a Revolution

    Find the surface area given by rotating the following curve about the y-axis.

    y= \frac{x^4}{4} + \frac{1}{8x^2} 0\leq x \leq 3

    \frac{dy}{dx}=x^3-\frac{1}{4x^3}

    (\frac{dy}{dx})^2=x^6-\frac{1}{2x^6}-\frac{1}{2}

    s=\int^3_0 2\pi*x*\sqrt{1+x^6-\frac{1}{2x^6}-\frac{1}{2}}

    s=\int^3_0 2\pi*x*\sqrt{x^6-\frac{1}{2x^6}+\frac{1}{2}}

    That is as far as I can get. Can someone tell me how to take this integral? I need to take it just using algebraic manipulation and u-substitution if necessary. Thanks
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  2. #2
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    Whats going on here? I couldn't get the preview function to work but I thought it was just broken or something. I'm using LaTex the way I always have...am I doing it wrong now? Did something change?
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  3. #3
    MHF Contributor

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    LaTeX hasn't been working for about a week. Hadn't you noticed?
    A number of new "buttons" apparently connected to LaTeX that don't showing up properly seem to been added. I suspect that they tried to add something to make it easier to write LaTeX and blew up in the forum's face!

    You appear to have written
    y=\frac{x^4}{4} + \frac{1}{8x^2} 0\leq x \leq 3

    \frac{dy}{dx}=x^3-\frac{1}{4x^3}

    (\frac{dy}{dx})^2=x^6-\frac{1}{2x^6}-\frac{1}{2}

    s=\int^3_0 2\pi*x*\sqrt{1+x^6-\frac{1}{2x^6}-\frac{1}{2}}

    s=\int^3_0 2\pi*x*\sqrt{x^6-\frac{1}{2x^6}+\frac{1}{2}}

    Your first problem is that
    \frac{1}{4x^3}
    squared is
    \frac{1}{16x^2}
    not
    [tex]\frac{1}{2x^6}

    Now, for the rest-
    That's a very common trick for this kind of problem- just because it is so hard to find f so that
    \sqrt{1- \left(\frac{df}{dx}\right)^2}
    is easy to integrate.
    Notice that the only difference between your last line and your third is that the last line has
    \sqrt{1+ \frac{1}{2}+ \frac{1}{16x^2}}
    while the third line has
    \sqrt{1+ \frac{1}{2}- \frac{1}{16x^2}}
    is that the middle term has changed from "-" to "+".

    That is, it is just the difference between (a- b)^2= a^2- 2ab+ b^2 and a^2+ 2ab+ b^2= (a+ b)^2.
    Last edited by HallsofIvy; April 30th 2011 at 10:51 AM.
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