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Dido525 Group Title

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = -5. y=x^2 , x=y^2

  • one year ago
  • one year ago

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  1. Dido525 Group Title
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    Anyone want to help me set up the intergal? I can do the rest.

    • one year ago
  2. exraven Group Title
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    sketch the region first

    • one year ago
  3. Dido525 Group Title
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    |dw:1358400866711:dw|I did:

    • one year ago
  4. Dido525 Group Title
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    Something like that roughly.

    • one year ago
  5. exraven Group Title
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    can you determine the radius and the height?

    • one year ago
  6. Dido525 Group Title
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    No I can't actually lol.

    • one year ago
  7. Dido525 Group Title
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    Would the rdiube 5?

    • one year ago
  8. Dido525 Group Title
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    radius*

    • one year ago
  9. exraven Group Title
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    |dw:1358400896304:dw|

    • one year ago
  10. exraven Group Title
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    the radius and the height must be a function of y

    • one year ago
  11. Dido525 Group Title
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    Right.

    • one year ago
  12. Dido525 Group Title
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    So y^2-(-5)

    • one year ago
  13. exraven Group Title
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    yes, and the height?

    • one year ago
  14. Dido525 Group Title
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    1?

    • one year ago
  15. exraven Group Title
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    the height is the rightmost curve minus the leftmost curve right?

    • one year ago
  16. exraven Group Title
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    the height must also be a function of y

    • one year ago
  17. Dido525 Group Title
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    |dw:1358401360267:dw|

    • one year ago
  18. Dido525 Group Title
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    Ohh....I see....

    • one year ago
  19. exraven Group Title
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    actually its not like that, wait ill draw again

    • one year ago
  20. Dido525 Group Title
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    Allright. Thanks for thing the timet o help me :) . Really hard vizualzing all this stuff.

    • one year ago
  21. exraven Group Title
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    |dw:1358401483221:dw|

    • one year ago
  22. Dido525 Group Title
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    Yeah, that's better.

    • one year ago
  23. exraven Group Title
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    |dw:1358401575777:dw|

    • one year ago
  24. Dido525 Group Title
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    Right - left the I suppose right?

    • one year ago
  25. Dido525 Group Title
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    x^2-sqrt(x) .

    • one year ago
  26. exraven Group Title
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    yes, but it must be a function of y

    • one year ago
  27. Dido525 Group Title
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    Why not x?

    • one year ago
  28. Dido525 Group Title
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    Same function regardless right?

    • one year ago
  29. exraven Group Title
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    because we integrate it with respect to y later

    • one year ago
  30. Dido525 Group Title
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    Yeah I see it now haha.

    • one year ago
  31. Dido525 Group Title
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    so sqrt(y)-y^2?

    • one year ago
  32. exraven Group Title
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    yes

    • one year ago
  33. exraven Group Title
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    and here is the way to determine the radius|dw:1358402147166:dw| the radius is y + 5

    • one year ago
  34. Dido525 Group Title
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    the limits would be from 0 to 1 I believe or am I wrong?

    • one year ago
  35. Dido525 Group Title
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    The radius would simply be the parbola - the line.

    • one year ago
  36. Dido525 Group Title
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    + the line sorry.

    • one year ago
  37. exraven Group Title
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    \[\Delta V = 2 \pi R h \Delta y\]\[\Delta V = 2 \pi (y + 5)(\sqrt{y} - y^2) \Delta y\]\[V = \int\limits_{0}^{1} 2 \pi (y + 5)(\sqrt{y} - y^2) dy\]

    • one year ago
  38. Dido525 Group Title
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    1 sec,I need to process that.

    • one year ago
  39. Dido525 Group Title
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    Woudn't it b sqrt(y)+5?

    • one year ago
  40. Dido525 Group Title
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    be*

    • one year ago
  41. exraven Group Title
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    this is for the radius|dw:1358403009035:dw|this is for the height|dw:1358403125789:dw|

    • one year ago
  42. Dido525 Group Title
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    The answer I got was 2 which is wrong.

    • one year ago
  43. Dido525 Group Title
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    Nvm. It's right.

    • one year ago
  44. Dido525 Group Title
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    109pi/30

    • one year ago
  45. exraven Group Title
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    feel free to ask if its hard to follow

    • one year ago
  46. Dido525 Group Title
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    Why isn't the height sqrt(x)+5?

    • one year ago
  47. exraven Group Title
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    it must be a function of y, because we integrate it with respect to y, sqrt(x) + 5 = y + 5

    • one year ago
  48. exraven Group Title
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    and it is not the height, it is the radius

    • one year ago
  49. exraven Group Title
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    the radius is sqrt(y) - y^2

    • one year ago
  50. Dido525 Group Title
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    y^2 + 5 sorry.

    • one year ago
  51. Dido525 Group Title
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    Why not that?

    • one year ago
  52. exraven Group Title
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    |dw:1358403804913:dw|

    • one year ago
  53. Dido525 Group Title
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    Right. That's root x.

    • one year ago
  54. Dido525 Group Title
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    Ahh, that distance is y and add 5.

    • one year ago
  55. exraven Group Title
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    yup, its sqrt(x), because the curve is x = y^2 sqrt(x) = sqrt(y^2) = y

    • one year ago
  56. Dido525 Group Title
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    What's dy?

    • one year ago
  57. Dido525 Group Title
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    Thickness right?

    • one year ago
  58. exraven Group Title
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    yes, because the cylinder looks like this|dw:1358404151779:dw|we integrate it with respect to y, because we fill the cylinder downward

    • one year ago
  59. Dido525 Group Title
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    Thanks so much!

    • one year ago
  60. exraven Group Title
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    here is more details, I am sorry if I cannot explain further, because I teach myself all this stuff http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

    • one year ago
  61. Dido525 Group Title
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    Thanks :) .

    • one year ago
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