Describe the solid generated by the integral

  1. I've been told this is a trick question, but I don't understand why:

    How would I describe the solid generated by 2∏∫ x/(1+x2)dx on [0,2]

    How I would do it I would rewrite the intergal as 2∏∫ x * 1/(1+x2)dx and apply substitution.

    I would then use the volume of disks method and integrate the integral about the new bounds. Is this the right way to go about it?
     
    Last edited: Oct 4, 2012
  2. jcsd
  3. LCKurtz

    LCKurtz 8,173
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    Your title line says "describe the solid" generated by the integral, it doesn't say to calculate its volume. Which do you really intend to do? The integral itself calculates the volume under a certain curve which is revolved about the y axis using the shell, not the disk, method. And a u-substitution would be appropriate, if the problem is to calculate the volume.
     
  4. Consider the formula for a solid of revolution involving rotating the function f(x) around the x-axis. That formula is:

    [tex]V=\int_a^b \pi \left(f(x)\right)^2 dx[/tex]

    Now, how could you re-write your integral in that form to represent a solid of revolution?
     
  5. uart

    uart 2,751
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    I wouldn't call it so much a "trick" question as an ambiguous one.

    Firstly, the question doesn't appear to be asking you to actually evaluate the integral, but merely to describe the solid for which the volume calculation would generate this integral.

    The ambiguous part is "the solid". If the question were to ask for "a solid" then I'd consider it a legitimate question.

    Even if we restrict ourselves to volumes of rotation then I can think of at least two different solids that could lead to that integral. Perhaps you could find them both as an exercise.

    That is, find:

    1. A volume of rotation around the "y" axis, calculated using the method of shells.

    2. A volume of rotation around the "x" axis, calculated using the method of discs.

    Edit: I posted the above before seeing either of the two previous replies (due to slow typing and distractions :smile:). I agree however with both of the above replies. The form on the integral expression lends itself most naturally to a method of shells rotation (because of the [itex]2 \pi[/itex] out front) as LCKurtz noted. There is however no real reason why we couldn't include a [itex]\sqrt{2}[/itex] in the function being evaluated by the disc method as per jackmels approach, resulting in the same given integral. This is why I call it an ambiguous question that could have been better worded.
     
    Last edited: Oct 4, 2012
  6. SammyS

    SammyS 7,886
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    It's not all that clear as to what question you're asking.

    Are we viewing the integral [itex]\displaystyle 2\pi\int_{0}^{2}\frac{x}{1+x^2}\,dx [/itex] as giving the volume of a solid formed by rotating the area bounded by some function and the coordinate axes about the y-axis?

    What is it that you are supposed to say about this integral or do with this integral?
     
  7. I don't feel it's a trick question at all. I'm getting an A on this one I'm sure:

    [tex]2\pi \int_0^2 \frac{x}{1+x^2}dx=\int_0^2 \pi \left(\sqrt{\frac{2x}{1+x^2}}\right)^2dx[/tex]

    and so the solid is that rotated around the x-axis. Now put mine on top of the stack when you hand them out to class.

    Of course I'm not a HH soooooo, watch it.
     
  8. uart

    uart 2,751
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    As I said above, that's one possible "answer". There's also a different function for which using the method of shells and rotation about the "y" gives the same integral.

    What about this one. The volume below the function [itex]f(x,y) = \frac{x}{x^2+1}[/itex], and above the rectangle in the x,y plane given by [itex]0 \leq x \leq 2[/itex] and [itex]0 \leq y \leq 2 \pi[/itex].
     
    Last edited: Oct 4, 2012
  9. LCKurtz, that's exactly what I was looking for. I apologise to the others if I the question was too ambiguous, the textbook just worded it this way.

    Nevertheless, thanks everyone!
     
  10. LCKurtz

    LCKurtz 8,173
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    You do understand, don't you, the point the others are making? My answer is the one that "leaps out at you" given the way the integral was written. But the truth is that just knowing the integral is insufficient to give a unique answer to the question. It is a bit like questions you see on aptitude tests like what is the next term in 1,3,5,7,9,11,? Everyone "knows" the answer is 13 but there is actually no objective reason to assume that. Similarly, as we have seen, there is more than one correct answer to your question. If this is a problem you are going to hand in, it might be interesting to give Jackmell's example of a solid rotated about the x axis if you are up for a possible discussion with the teacher. :uhh:
     
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