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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]A= \pi r^{2}=\pi(x^{-n})^{2}[/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


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    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 :)
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