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Homework Help: Volume of a Solid-Revolution About X-Axis

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the region bounded by the curves y= (x)/(1-x) , x= 0, x=(1/2), y=0.
    Calculate the volume of the solid that is created when this region is revolved about the x-axis.

    2. Relevant equations

    3. The attempt at a solution
    This is the work I have so far, but it seems to be giving me undefined answers, and I know this can't possibly be right.

    V= [itex] \int_{0}^{\frac{1}{2}} [\frac{x}{1-x}]^{2}\, dx [/itex] Whole thing is squared, by the way.
    ∏ * [itex] \int_{0}^{\frac{1}{2}} [\frac{x}{1-x}]^{2}\, dx [/itex]
    ∏ * [itex] (\left. x + \frac{1}{(1-x)} + 2 ln (x-1))\right|_{0}^{\frac{1}{2}} [/itex]

    Would appreciate any help on this as I've been working on it for hours and can't seem to figure out what's going on. Thank you.
  2. jcsd
  3. Aug 4, 2013 #2


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    I believe your integral is correct though you might need to change 2ln(x-1) to ln(x-1)2 to not get an undefined answer.
  4. Aug 4, 2013 #3


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    Use this :

    ##2ln(x-1) = ln(x^2-2x+1)## and it will evaluate properly.
  5. Aug 4, 2013 #4
    So now I get:

    pi * (x + 1/(1-x) + ln(x-1)^(2) | 0 and 1/2 being lower and upper limits of integration respectively
    pi(0.5 + 2 + ln(1/4)) - pi (0 + 1 + 0)
    5/2*(pi) + ln(1/4)*pi - pi
    3/2*(pi) + ln(1/4)*pi

    Is this correct? Thank you !
  6. Aug 4, 2013 #5


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    I agree with you answer.
    In your initial post, the error is that ##\int\frac {dx}x = \ln(|x|)##.
    Fwiw, you can avoid that complication and the partial fractions by substituting w = 1-x.
  7. Aug 5, 2013 #6


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    Also don't forget that there's a factor [itex]\pi[/itex] missing. The volume is
    [tex]V=\pi \int_{x_0}^{x_1} \mathrm{d} x \; [f(x)]^2.[/tex]
    This is easily seen from the geometric derivation of the integral, which sums infinitesimal cylindrical discs of height [itex]\mathrm{d} x[/itex] and radius [itex]f(x)[/itex].
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