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Volumes last week of calc

  1. May 28, 2008 #1
    [SOLVED] volumes... last week of calc

    i was absent when we went over the volume section.

    y= 1/x, x= 1, x= 2, y= 0,; about the x-axis

    [tex]
    \int^{1}_{2} \pi \frac{1}{x} dx
    [/tex]

    i don't know whats next. can anyone inform me, please
     
  2. jcsd
  3. May 28, 2008 #2

    Dick

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    You could actually do the integration, but that would be wrong because you haven't set it up right either. You integrate pi*r^2 where r is the radius of the disk over the volume. I'd suggest checking a few examples in your textbook.
     
  4. May 28, 2008 #3
    [tex]\int^{1}_{2} \pi [\frac{1}{x}]^2 dx[/tex]

    if you help me with this problem, i am sure i will get the rest of the problems.
     
  5. May 28, 2008 #4

    Dick

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    It's a power law integral. Like x^n. What's n in this case? What's the antiderivative?
     
  6. May 28, 2008 #5
    [tex] \pi \int^{2}_{1} \frac{1}{x^2} dx[/tex]
     
    Last edited: May 28, 2008
  7. May 28, 2008 #6

    Dick

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    Fine start. Now what's the antiderivative of 1/x^2?
     
  8. May 28, 2008 #7
    [tex] \pi [ \frac{x^{-3}}{-3}] [/tex]
     
  9. May 28, 2008 #8

    Dick

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    Beep. Wrong. The antiderivative of x^n is x^(n+1)/(n+1). What's n in this case? Unless that's a careless error because you are paying more attention to texing than thinking, you may have missed more than 'volumes'.
     
    Last edited: May 28, 2008
  10. May 28, 2008 #9
    since it's divided by 1, isn't it negative n
    =x^-2
     
  11. May 28, 2008 #10

    Dick

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    Yes, it is. Can you fix your antiderivative?
     
  12. May 28, 2008 #11
    (-1/3)x^-3
     
  13. May 28, 2008 #12

    Dick

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    Beep. Beep. Beep. You goofed it again. What's -2+1? Think this time.
     
  14. May 28, 2008 #13
    righttt

    pi [(1/-1)x^-1]
     
  15. May 28, 2008 #14

    Dick

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    Much better! You've got it now, right? And you promised that meant you could get all the others.
     
  16. May 28, 2008 #15
    thanks buddy

    your awesome
     
  17. May 28, 2008 #16

    Dick

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    It's "you're awesome". I'm correcting grammar tonight. Thanks.
     
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