1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Improper integrals and solids of revolution

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Let n>1/2 and consider the function

    [itex]f(x)=x^{-n}[/itex] for [itex] x\in[1,∞)[/itex]

    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

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

    So

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

    3. The attempt at a solution

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


    [itex]= \pi lim_{t→∞}∫^{t}_{1}(x^{-2n})dx[/itex]

    [itex]= \pi lim_{t→∞} \frac{x^{-2n+1}}{-2n+1}|^{t}_{1}[/itex]

    [itex]= \pi lim_{t→∞} \frac{1}{-2n+1}(t^{-2n+1}-1)[/itex]

    [itex]= \frac{\pi}{-2n+1}[/itex]

    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. jcsd
  3. Feb 18, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Feb 19, 2013 #3
    Oops, you're right. I messed up the last step and forgot to multiply by the negative 1. Thanks :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook