# Improper integrals and solids of revolution

1. Feb 18, 2013

### phosgene

1. The problem statement, all variables and given/known data

Let n>1/2 and consider the function

$f(x)=x^{-n}$ for $x\in[1,∞)$

Calculate the volume of the solid generated by rorating f(x) about the x-axis, showing all details of your working.

2. Relevant equations

Since it is rotated about the x-axis, its axis of symmetry is the x-axis, and by slicing up the solid of revolution into circles perpendicular to the x-axis, the volume is given by

$V=lim_{t→∞}∫^{t}_{1}A(x)dx$
$A= \pi r^{2}=\pi(x^{-n})^{2}$

So

$V=lim_{t→∞}∫^{t}_{1}\pi(x^{-n})^{2}dx$

3. The attempt at a solution

$V= \pi lim_{t→∞}∫^{t}_{1}(x^{-n})^{2}dx$

$= \pi lim_{t→∞}∫^{t}_{1}(x^{-2n})dx$

$= \pi lim_{t→∞} \frac{x^{-2n+1}}{-2n+1}|^{t}_{1}$

$= \pi lim_{t→∞} \frac{1}{-2n+1}(t^{-2n+1}-1)$

$= \frac{\pi}{-2n+1}$

So a few questions. First, am I right? Second, should I do another substitution with the limit and turn it into the limit as s goes to 0 where t=1/s? Appreciate any help :)

2. Feb 18, 2013

### Dick

You are almost right. But if you put a number n>1/2 into that shouldn't you get something positive? And taking your limit works precisely because (-2n+1)<0, right? Just make sure you are clear on that. No need to do the limit another way.

3. Feb 19, 2013

### phosgene

Oops, you're right. I messed up the last step and forgot to multiply by the negative 1. Thanks :)