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Dido525

  • one year ago

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

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  1. Dido525
    • one year ago
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    Anyone want to help me set up the intergal? I can do the rest.

  2. exraven
    • one year ago
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    sketch the region first

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

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

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

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

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

  8. Dido525
    • one year ago
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    radius*

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

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

  11. Dido525
    • one year ago
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    Right.

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

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

  14. Dido525
    • one year ago
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    1?

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

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

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

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

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

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

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

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

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

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

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

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

  27. Dido525
    • one year ago
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    Why not x?

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

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

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

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

  32. exraven
    • one year ago
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    yes

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

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

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

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

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

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

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

  40. Dido525
    • one year ago
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    be*

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

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

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

  44. Dido525
    • one year ago
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    109pi/30

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

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

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

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

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

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

  51. Dido525
    • one year ago
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    Why not that?

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

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

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

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

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

  57. Dido525
    • one year ago
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    Thickness right?

  58. exraven
    • one year ago
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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

  59. Dido525
    • one year ago
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    Thanks so much!

  60. exraven
    • one year ago
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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

  61. Dido525
    • one year ago
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    Thanks :) .

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