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Homework Help: Volume of the solid of revolution

  1. Jul 12, 2007 #1
    Find the volume of the solid of revolution obtained when the region under the graph of

    [tex]f(x) = \left( \frac{1}{x} \right) e^\frac{1}{x} [/tex]

    from x = 1 to x = 6

    2. Relevant equations

    [tex] \pi \int (f(x))^2 dx [/tex]

    3. The attempt at a solution

    Ok, the equation I gave above should be that of a definite integral with a=1 and b=6 (If anyone can tell me how to write that in Latex it would be much appreciated)

    So, the volume is

    [tex]\pi \int_1^6 \left( \frac{1}{x} \right) \left( e^\frac{1}{x} \right)^2 dx[/tex]

    So, we can simplyfy this to

    [tex]\pi \int_1^6 (x)^{-2} \times e^\frac{2}{x} dx[/tex]

    Now I'm a bit stuck as to where to go from here. Do I use the integration by parts method? I think I'm getting bogged down in unnecessary calculations. Can someone give me a hint or point me in the right direction?
    Last edited: Jul 12, 2007
  2. jcsd
  3. Jul 12, 2007 #2
    Try setting u = 1/x or something like that.
  4. Jul 12, 2007 #3


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    Science Advisor

    ?? You didn't finish your sentence! Between what limits? Rotated around what axis?

    So you are rotating around the x-axis?

    And only the region between x= 1 and x= 6?

    No, If [itex]f(x)= \frac{1}{x}e^{1/x}[/itex] then both [itex]f^2(x)= \frac{1}{x^2}e^{2/x}[/itex]

    Okay, good. Now you have first x squared also. By the way, the code to put the limits of integration in is "\int_1^6". In other words, treat the lower limit as a subscript and the upper limit as a superscript on the integral sign.

    Looks to me like the substitution u= 2/x should work nicely.
  5. Jul 12, 2007 #4
    Many thanx for your time and help..... I've managed to solve using substitution as you suggested.
    Last edited: Jul 12, 2007
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