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Math Help - Stumped on an integral.

  1. #1
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    Stumped on an integral.

    Been beating my head against the wall on an integral for awhile now....

    The question is this:

    Compute the surface area generated by revolving the curve \[f(x) = 2\sqrt x ,\,\,{\text{for 1}} \leqslant x \leqslant 4\] about the x-axis.

    I took the 1st derivative of the equation:

    \[f'(x) = \frac{1}{x}\]

    Easy enough. Then I went into my integral:

    \[2\pi \int_{\,1}^{\,4} {2\sqrt x \sqrt {1 + \frac{1}{{{x^2}}}} } dx\]

    That didn't look very pretty, and I didn't seem to be getting anywhere trying to solve it (no u-sub, and integration by parts didn't seem to be working), so I tried to manipulate the equation several times to see if a different variation would work...

    \[4\pi \int_{\,1}^{\,4} {\sqrt {x + \frac{1}{x}} } dx\]

    and

    \[4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {\left( {{x^2} + 1} \right)} dx\]

    and

    \[4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} dx\]

    and

    \[4\pi \int_{\,1}^{\,4} {x\sqrt {\left( {\frac{{{x^2} + 1}}{{{x^3}}}} \right)} } dx\]

    But I can't find one to work out. My professor stated on the assignment that the integral can be done exactly, otherwise I would have approximated it a long time ago.

    Would one of you be able to possibly hint where I'm going wrong, or which form you'd take?
    Last edited by mr fantastic; March 9th 2011 at 02:09 AM. Reason: Title.
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  2. #2
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    Here's a similar example for your consideration.

    Areas of Surfaces of Revolution
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by Malaclypse View Post
    Been beating my head against the wall on an integral for awhile now....

    The question is this:

    Compute the surface area generated by revolving the curve \[f(x) = 2\sqrt x ,\,\,{\text{for 1}} \leqslant x \leqslant 4\] about the x-axis.

    I took the 1st derivative of the equation:

    \[f'(x) = \frac{1}{x}\]

    Easy enough. Then I went into my integral:

    \[2\pi \int_{\,1}^{\,4} {2\sqrt x \sqrt {1 + \frac{1}{{{x^2}}}} } dx\]

    That didn't look very pretty, and I didn't seem to be getting anywhere trying to solve it (no u-sub, and integration by parts didn't seem to be working), so I tried to manipulate the equation several times to see if a different variation would work...

    \[4\pi \int_{\,1}^{\,4} {\sqrt {x + \frac{1}{x}} } dx\]

    and

    \[4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {\left( {{x^2} + 1} \right)} dx\]

    and

    \[4\pi \int_{\,1}^{\,4} {\left( {\sqrt {\frac{1}{x}} } \right)} \sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} dx\]

    and

    \[4\pi \int_{\,1}^{\,4} {x\sqrt {\left( {\frac{{{x^2} + 1}}{{{x^3}}}} \right)} } dx\]

    But I can't find one to work out. My professor stated on the assignment that the integral can be done exactly, otherwise I would have approximated it a long time ago.

    Would one of you be able to possibly hint where I'm going wrong, or which form you'd take?
    Your calculation of f'(x) is wrong.
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  4. #4
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    The derivative of ln(x) is \frac{1}{x}, not the derivative of x^{1/2}. To differentiate that, use the power rule.
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  5. #5
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    Well now I feel like an idiot. Sure enough, I forgot to place the radical in the denominator when I performed it and didn't give it second thought. 1 / sqrt(x), NOT 1/x. I even looked at it several times. Urgh. Sorry all, and thanks.

    That makes the integral nice and easy. What a waste of 3 hours.
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