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Math Help - Tough integral (for me)

  1. #1
    s3a
    s3a is offline
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    Tough integral (for me)

    The initial question was (if anyone is interested):
    Find the area of the surface obtained by rotating the curve
    from to about the -axis.

    but I am stuck at the integral mentioned above. Having just checked. I do not get the right answer when I use math software so I guess I did something wrong prior to getting this integral.

    Here is the integral I am finding tough to do:
    integrate 4*pi*sqrt(x)*sqrt(&# x28;sqrt(x) + 2)/sqrt(x)), from 0 to 2 - Wolfram|Alpha

    I am probably not using the right integral since using math software gives me the wrong answer so any help would be greatly appreciated!

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    The initial question was (if anyone is interested):

    Find the area of the surface obtained by rotating the curve

    from to about the -axis.

    but I am stuck at the integral mentioned above. Having just checked. I do not get the right answer when I use math software so I guess I did something wrong prior to getting this integral.

    Here is the integral I am finding tough to do:
    integrate 4*pi*sqrt(x)*sqrt(&# x28;sqrt(x) + 2)/sqrt(x)), from 0 to 2 - Wolfram|Alpha

    I am probably not using the right integral since using math software gives me the wrong answer so any help would be greatly appreciated!

    Thanks in advance!
    S=2\pi\int_a^bf(x)\sqrt{1+(f'(x))^2}\,dx

    f(x)=\sqrt{4x}=2x^{\frac{1}{2}}

    f'(x)=x^{-\frac{1}{2}}

    S=2\pi\int_0^2 2\sqrt{x}\sqrt{1+\frac{1}{x}}\,dx
    Last edited by ione; March 24th 2010 at 03:07 PM.
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