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Homework Help: Volume generated

  1. Jul 19, 2006 #1
    A question here:
    Given two curves
    y=e^x
    y=1+2e^(-x)

    The region in the first quadrant that is bounded by the y-axis and these two curves is rotated through one complete revolution about the x-axis. Calculate the exact volume of the solid generated.

    My problem is, in the first quadrant, y=1+2e^(-x) seems touching the x-axis at [tex]x=\infty[/tex], so how do we find the volume?
     
    Last edited: Jul 19, 2006
  2. jcsd
  3. Jul 19, 2006 #2
    You sure you copied the problem correctly?
     
  4. Jul 19, 2006 #3
    Thanks, you are right that the question has not been copied correctly. I have changed
    y=1+2e^(x)
    to
    y=1+2e^(-x)
    Please refer to the original question again. Very sorry for any inconvenience caused.
     
  5. Jul 19, 2006 #4

    HallsofIvy

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    Science Advisor

    The two curves cross at (0,3), of course, and the region under the two curves is symmetric about the y-axis. However, if that really is the correct formula, because y goes to 1 as x goes to [itex]\infty[/itex], and as x goes to [itex]-\infty[/itex], the volume generated contains an infinitely long cylinder of radius 1 and so is not finite.
     
  6. Jul 19, 2006 #5

    benorin

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    Homework Helper

    A start

    The two curves are [tex]y=e^{x}[/tex] and [tex]y=1+2e^{-x}[/tex] which intersect when [tex]e^{x}=1+2e^{-x}[/tex] multiply by e^x to get [tex]e^{2x}-e^{x}-2=0[/tex] so by the quadratic formula we have [tex]e^{x}=2[/tex] or [tex]x = \log {2}[/tex] so the curves meet at the point (log 2, 2). The other boundary is the y-axis so the bounded area is now finite (see attached plot) and to be rotated about the x-axis, so do an integral :smile: . --Ben
     

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