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Math Help - [SOLVED] Surface Area - check my work?

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Surface Area - check my work?

    Find the area of the surface obtained by rotating the following curve about the x-axis:

    y = x^3, 0 \leq x \leq 3

    Here is my work, which I'm pretty sure is correct, but wrong in Webassign:

    2\pi \int\limits^{3}_{0} (x^3) \sqrt{1+9x^4} dx

    \frac{\pi}{18} \int\limits^{729}_{1} \sqrt{u} du

    = \frac{\pi}{18} \biggl[\frac{2}{3} u^{\frac{3}{2}} \biggr]^{729}_{1}

    = \frac{\pi}{27} \biggl[729^{\frac{3}{2}} - 1 \biggr]

    Can someone please check my work and let me know where the error is?
    Thanks!! Molly
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Find the area of the surface obtained by rotating the following curve about the x-axis:

    y = x^3, 0 \leq x \leq 3

    Here is my work, which I'm pretty sure is correct, but wrong in Webassign:

    2\pi \int\limits^{3}_{0} (x^3) \sqrt{1+9x^4} dx

    \frac{\pi}{18} \int\limits^{7{\color{red}30}}_{1} \sqrt{u} du

    = \frac{\pi}{18} \biggl[\frac{2}{3} u^{\frac{3}{2}} \biggr]^{7{\color{red}30}}_{1}

    = \frac{\pi}{27} \biggl[7{\color{red}30}^{\frac{3}{2}} - 1 \biggr]

    Can someone please check my work and let me know where the error is?
    Thanks!! Molly
    Note my corrections in red.
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