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Volumes of revolution

  1. Feb 24, 2006 #1

    Hootenanny

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    If the finite region bounded by the curve [itex] y = \text{e}^{x} +1 [/itex], the y-axis and the line [itex] x = \ln2 [/itex] is rotatated around the x-axis by [itex] 360^{\circ} [/itex] show that the volume of the solid formed is:
    [tex] \frac{\pi}{2} (7 + \ln4 ) [/tex]
    I did the intergral and got:
    [tex] V = \pi \left[ (\text{e}^{4} + 2\text{e}^{2} +1) - (1 + 2 + 1) \right] [/tex]
    But I can't see how I can manipulate it to get the required answer. :confused: Any help would be much appreciated.
     
    Last edited: Feb 24, 2006
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  3. Feb 24, 2006 #2

    TD

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    You won't be able to rewrite it like that since they aren't the same.
    Perhaps you could show how you got that?
     
  4. Feb 24, 2006 #3

    Hootenanny

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    [tex]V = \int_{0}^{\ln2} \pi\left[ f(x) \right]^2 \;\; dx[/tex]
    [tex]V = \pi \int_{0}^{\ln2} \left[ e^{x} +1 \right]^2 \;\; dx = \pi \int_{0}^{\ln2} e^{2x} + 2e^{x} +1 \;\; dx[/tex]
    [tex]V = \pi \left[ (e^{4} + 2e^{2} +1) - (1 + 2 + 1) \right][/tex]
     
  5. Feb 24, 2006 #4

    TD

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    [tex]V = \pi \int_{0}^{\ln2} e^{2x} + 2e^{x} +1 dx[/tex]

    This is still correct, the next step isn't. Show the primitive function first, then substitute the boundaries.
     
  6. Feb 24, 2006 #5

    Hootenanny

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    I dear, I see that I've forgot to intergrate. That was a bit stupid wasn't it?
     
  7. Feb 24, 2006 #6

    TD

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    I think that happens at least once to everyone :smile:
     
  8. Feb 24, 2006 #7

    Hootenanny

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    Have I intergrated correctly? I get:
    [tex]V = \pi \left[ \frac{1}{2} e^{2x} + 2e^{x} + x \right]^{\ln2}_{0}[/tex]
     
  9. Feb 24, 2006 #8

    TD

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    Looks good, see if you can come to the given answer.
     
  10. Feb 24, 2006 #9

    Hootenanny

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    The subsituting the values ln2 and 0 in gives:
    [tex] V = \pi \left[ \left( \frac{1}{2}\cdot4 + 4 + \ln2 \right) - \left( \frac{1}{2} + 2 \right) \right] [/tex]
    Simplifying:
    [tex] \pi \left[ \frac{7}{2} + \ln2 \right] [/tex]
    I can't see where I go now. I know I'm close.
     
  11. Feb 24, 2006 #10

    TD

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    You are close indeed :smile:

    Factor out 1/2 so you get the [itex]\pi/2[/itex] which is there in the given answer, then think of a property of logarithms...
     
  12. Feb 24, 2006 #11

    Hootenanny

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    O yes, I forgot [itex] 2\cdot \ln2 = ln4[/itex]. My brains obviously not in gear today. Thanks very much.
     
  13. Feb 24, 2006 #12

    TD

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    Well done :smile:
     
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