

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Find the volume of the region bounded by $y= \frac{\ln{x}}{x^2}$, $y=0$, $x=1$, and $x=2$ when revolved about the y-axis. Indicate which method to use.

I think i would be using vertical slices. Which might mean I have to use the shell method. But I'm just not sure where to begin. I have drawn my graph but now I'm stuck. Any help would be appreciated.

share|cite|improve this question
up vote 2 down vote accepted

Yes, the Method of Shells looks like the natural one to use. Look at a thin vertical slice, "at" $x$, and of width "$dx$." Rotate it around the $y$-axis. This generates a shell of volume approximately $2\pi x y\,dx$. Here $y=\frac{\ln x}{x^2}$. "Add up" (integrate) from $x=1$ to $x=2$. So our volume is $$\int_1^2 2\pi x\frac{\ln x}{x^2}\,dx.$$ Do the obvious cancellation. Then, for the integration, let $u=\ln x$.

Remark: Doing it by taking slices perpendicular to the $y$-axis is very unpleasant to set up. A sketch of the curve will show you why. We end up wanting to integrate with respect to $y$, and it looks as if we have to take the equation $y=\frac{\ln x}{x^2}$, and solve for $x$ in terms of $y$, a hopeless task.

In fact one can work around the issue by making an appropriate change of variable. But the details are not easy.

share|cite|improve this answer
Argh, beat me by 20 seconds! ;) – Neal Apr 18 '13 at 3:20
Two answers are better than one. And our styles undoubtedly differ. – André Nicolas Apr 18 '13 at 3:31

Your Answer


