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Volume of Revolution with tan(x)

  1. Aug 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Let R be the region between y=tan(x) and the x-axis from x=0 to x=pi/2. Find the volume of the solid formed when R is revolved around the y-axis.


    2. Relevant equations
    Please try to solve this problem using elementary calculus. The textbook is an elementary calculus textbook. Anything involving complex analysis or polylogarithmic functions is outside the scope of my understanding.


    3. The attempt at a solution
    First, we try the washer method. we have:
    [itex] \int_0^\infty \pi*((\pi/2)^2-(\arctan y)^2)\,dy [/itex]

    I do not know how to integrate [itex]arctan^2(y)[/itex]. Heck, the textbook didn't even explain the integration of inverse trigonometric functions, let alone [itex]arctan^2(y)[/itex]. I did find out that the integral of arctan(y) can be found via integration by parts, but I think the integral of [itex]arctan^2(y)[/itex] involves complex exponents, which is outside the scope of the textbook.

    Second, we try the shell method. we have:
    [itex] \int_0^{\pi/2} 2*\pi*xtanx\ dx [/itex]

    And I fail to integrate x*tan(x) dx. Again, polylogarithmic functions is outisde the scope of my understanding.

    Please help?
     
    Last edited: Aug 5, 2013
  2. jcsd
  3. Aug 5, 2013 #2
    There is a vertical asymptote at x = [itex]\pi[/itex]/2. In this case the area and the volume of revolution are both divergent.

    Both are improper integrals:
    Area = [itex]\int^{\pi/2}_{0} tan \space x \space dx[/itex]
    Volume = [itex]\pi \int^{\pi/2}_{0} tan^{2} x \space dx[/itex]

    Junaid Mansuri
     
  4. Aug 5, 2013 #3
    Thanks for the reply. But I think there are two things:
    First, just because the area under the curve diverges (which can be easily shown) does not mean the volume diverges. the curve 1/x is one such example. I must show that the volume diverges by integrating, which is what's trumping me.

    Second, I believe your formula for the volume is for the revolution around the x-axis. The question asks for the revolution around the y-axis, which make things a lot more complex.

    Any ideas?
     
  5. Aug 5, 2013 #4
    You're right. The are and volume I mentioned are all relative to the x-axis. For rotation about the y-axis, the integral would be quite complex. You can take a look at it on Wolfram Alpha.

    Your integrals do look correct though.

    -J
     
  6. Aug 6, 2013 #5

    haruspex

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    It's probably one of those definite integrals that can be solved without having to solve the indefinite version. Played around with it some, but no breakthroughs.
     
  7. Aug 6, 2013 #6

    vanhees71

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    To integrate [itex]\tan^2 x[/itex] note that
    [tex]\tan^2 x=\frac{\sin^2 x}{\cos^2 x}=\frac{1-\cos^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}-1[/tex]
    and
    [tex]\frac{\mathrm{d}}{\mathrm{d} x} \tan x=\frac{1}{\cos^2 x}.[/tex]
     
  8. Aug 6, 2013 #7

    haruspex

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    Sure, but that's not the problem here. The integrand is more like [itex]x^2\tan^2 x[/itex]
     
  9. Aug 6, 2013 #8

    vanhees71

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    Argh. Ok, I've not read the original question carefully enough. One has to find the volume of revolution arount the [itex]y[/itex] axis. This means, of course, one has to evaluate
    [tex]V=\pi \int_0^{\infty} \mathrm{d} y \arctan^2 y,[/tex]
    which of course cannot exist in the first place (as one sees from the geometry of the problem).

    Also the body formed when rotating around the [itex]x[/itex] axis has infinite volume. Perhaps the original question was for [itex]x \in [0,\pi/4][/itex]? Then, however, the question for revolving around the [itex]y[/itex]-axis leads to polylogarithmic functions (according to Mathematica), which I'd not call "elementary" at all ;-).
     
    Last edited: Aug 6, 2013
  10. Aug 6, 2013 #9

    haruspex

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    No, that's still not it. It's the volume generated from the region under the curve, so as it says in the OP it's [tex]V=\pi \int_0^{\infty} \mathrm{d} y ((\frac \pi 2)^2 - \arctan^2 y)[/tex], or [tex]V=\pi \int_0^{\frac \pi 2} \mathrm{d} x ((\frac \pi 2)^2 - x^2 ) \sec^2 x=\pi \int_0^{\frac \pi 2} \mathrm{d} x ( \pi - x ) x \sin^{-2} x[/tex]
    That still looks unbounded to me, but it isn't quite as obvious.
     
  11. Aug 6, 2013 #10
    I understand your first formula, which is what was in the original post. However, I could not see the reasoning behind the second formula you've posted-it seems that you are using x as the variable of integration? In that case, we would be using the shell method, which has the form ## 2\pi \int_0^{\pi/2} xtanx\,dx ##.
     
  12. Aug 6, 2013 #11

    haruspex

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    Yes, the second formula switches the integration variable to x, but it still uses the washer decomposition. Just plug in y = tan(x) in the first formula. The dy becomes sec2(x)dx. To get the last formula, replace x everywhere by (π/2)-x.
     
  13. Aug 6, 2013 #12
    For this problem there is no need to evaluate the integrals from the washer or shell method. Keep in mind that if the area of a region is infinite, then the volume produced by that region when revolved around an axis will also be infinite. All you need to do is show that the area of the region is infinite, which is easy in this case.
     
    Last edited: Aug 6, 2013
  14. Aug 7, 2013 #13

    haruspex

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    Need not be true when the axis lies in or at one edge of the region. Consider 0 <= y < 1/x, 0 <= x <= 1, rotated around y axis.
     
  15. Aug 7, 2013 #14
    Ah, you're right. For this problem I was considering the following where A is the area of the region being rotated, V is the volume of revolution around the y axis, and a is the value of x where the region begins greater than zero:
    [tex]V\approx\sum_{i=1}^{n}A\frac{2\pi a}{n}[/tex]
    That is for large values of n. It enough to show that if a>0 and A is infinite, V will also be infinite. It isn't valid for when a=0, which I did I overlook. My mistake.
     
    Last edited: Aug 7, 2013
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