Volume of a Solid-Revolution About X-Axis

[Justabeginner]

Homework Statement

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.

Homework Equations

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

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.

 

[rock.freak667]

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.

 

[STEMucator]

Use this :

$2ln(x-1) = ln(x^2-2x+1)$ and it will evaluate properly.

 

[Justabeginner]

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 !

 

[haruspex]

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.

 

[vanhees71]

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