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Solids of Revolution

  1. Dec 20, 2011 #1
    Problem
    The area between [tex]y=x^{-2}[/tex] and x=1 & y=e is rotating around the y-axis. What is the volume?

    Attempt
    [tex]\pi\left( r_{outer\mbox{}} \right)^{2}\; -\; \pi \left( r_{inner\mbox{}} \right)^{2} \; \; \; \delta y[/tex].
    [tex]\frac{1}{x^{2}}=y\; gives\; \frac{1}{y}=x^{2}\; and\; r=\sqrt{y}[/tex]
    [tex]V=\pi \int_{1}^{e}{\frac{1}{y}-1\; dy}[/tex]
     
  2. jcsd
  3. Dec 20, 2011 #2
    Can I assume that the question means the area above the x axis?

    Anyway, firstly you must draw the graph. Are you trying to find the volume of the cylinder then subtract the volume not included in the region? If so isn't the x=1 the outer radius? And so then rather the inner radius should then be (1-x)

    I find it easier and a nicer integral to just sum the volumes - so calculate the volume generated by the cylinder from y=0 to y=1, and then add to this the volume generated by the curve from y=1 to y=e. Hopefully this helps
     
  4. Dec 20, 2011 #3
    Yes, that would be more effective. I am, however, trying to learn this formula, and would like to use it on this problem.

    It is indeed the area over the x-axis. The upper bound is e, the right side is x=1, and the third line is x^-2. If you see what I mean…

    I don't follow when you say "If so isn't the x=1 the outer radius? And so then rather the inner radius should then be (1-x)"
     
  5. Dec 20, 2011 #4
    Sorry for lag reply. Oh my god I've just spent the last 30 min scribbling furiously away at my desk because I was getting different answers from 2 different methods and didn't know why. When I did your way I got it wrong because the algebra was more complicated. So YEAH I really recommend just doing it the simplest way.

    I shall now guess what you're trying to do. Are you trying to find the volume of the cylinder from y=0 to e made by a rotation, and subtract from this the volume generated by the unwanted area (between y=1/x^2, y=e and x=1)?

    Okay, if so, what you've done is switch the 1 and 1/y. When you draw the diagram (draw a nice big one!), you will see that x=1 is the OUTER radius, and the inner radius is x.
    So an infinitesimal thickness disc would have volume [tex]dV = \pi (1^2-x^2)dy[/tex]. Now since [tex]x=\frac{1}{\sqrt{y}}[/tex], then simply complete the integral, subtract this from the area of the cylinder, and the answer should be [tex]2\pi[/tex] cubic units!
     
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