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QRAwarrior

How would you draw a cylinder for the shell method for the region that is enclosed by x = pi/2, x = -pi/2 and y = cosx? This is rotated about y = -1.

  • one year ago
  • one year ago

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  1. QRAwarrior
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    @FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?

    • one year ago
  2. QRAwarrior
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    @jim_thompson5910

    • one year ago
  3. akamushi
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    the definite integrand of f(x)=cosx from -pi/2 to pi/2 sinx from -pi/2 to pi/2 = 1-(-1)=2

    • one year ago
  4. QRAwarrior
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    I do not want the answer. @akamushi

    • one year ago
  5. QRAwarrior
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    @FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?

    • one year ago
  6. QRAwarrior
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    @FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?

    • one year ago
  7. lgbasallote
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    @lgbasallote lol jk...idk shell using trig functions

    • one year ago
  8. PROSS
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    |dw:1337557715702:dw|

    • one year ago
  9. QRAwarrior
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    The height would be 2cosx, and the radius would be cosx - (-1)?

    • one year ago
  10. QRAwarrior
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    ie. h = 2y r = y - (-1) ?

    • one year ago
  11. QRAwarrior
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    |dw:1337558085638:dw|

    • one year ago
  12. QRAwarrior
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    The height (which are the horizontal lines of the cylinder) would be given by h = 2cosx? And the radius would be given by r = cos(x) - (-1)?

    • one year ago
  13. QRAwarrior
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    Wait, I think the height would be: h = 2x, due to symmetry

    • one year ago
  14. QRAwarrior
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    |dw:1337558202025:dw|

    • one year ago
  15. QRAwarrior
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    So then, radius is r = cos(x) - (-1) for sure, right?

    • one year ago
  16. QRAwarrior
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    And I can restate that cos(x) as y So then, r = y +1, h = 2x = 2arccos(y), I believe.

    • one year ago
  17. PROSS
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    now you are on the right track.

    • one year ago
  18. PROSS
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    if you are familiar with winplot, you can make a nice 3d representation.

    • one year ago
  19. QRAwarrior
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    @shivam_bhalla! Yes you finally came.

    • one year ago
  20. QRAwarrior
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    |dw:1337558589994:dw|

    • one year ago
  21. QRAwarrior
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    I have one question for either of you: I drew a vertical line from the x-axis to the function cos(x). Can that generically be represented as y?

    • one year ago
  22. shivam_bhalla
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    @QRAwarrior , try this http://openstudy.com/updates/4fb44286e4b05565342a2a03 You will find it a lot more helpful :) The answer to your question you asked before is yes

    • one year ago
  23. amistre64
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    x = pi/2, x = -pi/2 and y = cosx|dw:1337561871090:dw|

    • one year ago
  24. amistre64
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    |dw:1337562804283:dw|

    • one year ago
  25. QRAwarrior
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    Ok so it is: h = x y = cosx - (-1)

    • one year ago
  26. amistre64
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    we only need half of this, then double the results height is a measure of arccos(y) radius is from from 1 to 2 by perspective \[2*2pi\int_{0}^{1}(y+1)*\cos^{-1}(y)\ dy\]

    • one year ago
  27. QRAwarrior
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    I think I came up with this a few posts back

    • one year ago
  28. amistre64
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    height is a function of x; y=cos(x) arccos(y) = x = height

    • one year ago
  29. QRAwarrior
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    -------------------------------------------------------- Ok consider this question: Find the length of y^2 = x^3 from (0,0) to (4,8) I know the formula for arclength, but I end up getting this: |dw:1337563218324:dw|

    • one year ago
  30. QRAwarrior
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    |dw:1337563244465:dw|

    • one year ago
  31. QRAwarrior
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    How the heck can you resolve the radical when you have: |dw:1337563293983:dw| As your argument?

    • one year ago
  32. amistre64
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    have you tried looking at it form both forms? \[\sqrt{1+(x')^2}\ or\ \sqrt{1+(y')^2}\]

    • one year ago
  33. QRAwarrior
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    No. I think I will try the other form, BUT the problem is that I am not sure whether to pick the POSITIVE root or the NEGATIVE root..

    • one year ago
  34. amistre64
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    for the most part, i think you stick with the positive root

    • one year ago
  35. QRAwarrior
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    Why do you say "most part"

    • one year ago
  36. amistre64
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    distance tends to be a measure in absolute value, so the positive root seems most apt

    • one year ago
  37. QRAwarrior
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    WOW! I just realized how EASY it is doing it wrt x.

    • one year ago
  38. QRAwarrior
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    |dw:1337563549037:dw|

    • one year ago
  39. QRAwarrior
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    Easy U-sub. Thanks a lot amistre64. What level of education is your mathematics?

    • one year ago
  40. QRAwarrior
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    I was just curious

    • one year ago
  41. amistre64
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    i dunno what level it is, im guessing its at least on a 3rd grade level ;)

    • one year ago
  42. QRAwarrior
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    Ok anyways, thanks for your help

    • one year ago
  43. amistre64
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    youre welcome. Levelwise is difficult since im self taught for the most part of this stuff

    • one year ago
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