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anonymous

  • 4 years ago

The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the y-axis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36. These are two separate rotation axis and answers.

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  1. amistre64
    • 4 years ago
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    which method?

  2. amistre64
    • 4 years ago
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    |dw:1327531365366:dw|

  3. amistre64
    • 4 years ago
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    we need to know the point of intersection here

  4. anonymous
    • 4 years ago
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    Hey sorry, keyboard died. The point of intersection is (4,16).

  5. amistre64
    • 4 years ago
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    really? i see it as (6,36) meself (y=6) = (y=sqrt(x)) 6 = sqrt(x) at x=36

  6. anonymous
    • 4 years ago
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    |dw:1327531523131:dw|

  7. amistre64
    • 4 years ago
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    your pic has way more parts then the question asks for

  8. anonymous
    • 4 years ago
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    It's a rotation of y=6, but the shape is bounded by y=4, so the intersection of y=4 and y=sqrt(x) would be (16,4).

  9. anonymous
    • 4 years ago
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    |dw:1327531686984:dw|

  10. anonymous
    • 4 years ago
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    So I'm thinking of using the cylindrical shell method, 2Pi(radius)(height)

  11. amistre64
    • 4 years ago
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    ok, since typos happen; lets review what youve posted: The region bounded above by the line, y = 6, below by the curve y=sqrt(x), and to the left by the y-axis. The solid is rotated along the following lines: C. The line y=6 D. The line x=36

  12. anonymous
    • 4 years ago
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    Yes, that is correct.

  13. amistre64
    • 4 years ago
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    |dw:1327531783399:dw|

  14. amistre64
    • 4 years ago
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    then this is our graph of that region

  15. amistre64
    • 4 years ago
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    shell is fine, but we need to establish that this is our representation

  16. anonymous
    • 4 years ago
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    so you this is a graph for rotation around x=36?

  17. amistre64
    • 4 years ago
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    yes, in fact its the same graph for rotate y=6 AND x=36

  18. anonymous
    • 4 years ago
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    Oh oh

  19. amistre64
    • 4 years ago
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    the inverse of y=sqrt(x) that we can use for this is x=y^2 that will give us the shell height for y=6 rotation

  20. anonymous
    • 4 years ago
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    so when rotating around a Y-axis, we should put the curve as Y too?

  21. amistre64
    • 4 years ago
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    first we should simply draw out the graphs; after we have the graphs down, then we worry about where to rotate at

  22. amistre64
    • 4 years ago
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    the graph itself has nothing to do with what we rotate around to begin with

  23. anonymous
    • 4 years ago
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    true

  24. anonymous
    • 4 years ago
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    so the graph I had illustrated was inccorect

  25. anonymous
    • 4 years ago
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    visualizing the graphs has been the toughest part for me!

  26. amistre64
    • 4 years ago
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    |dw:1327532250390:dw|this is our shell representing rotate about y=6

  27. amistre64
    • 4 years ago
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    the graph you drew was something else altogther :)

  28. anonymous
    • 4 years ago
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    haha, I'm sorry.

  29. anonymous
    • 4 years ago
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    I HAVE NO CLUE WHY I PUT Y = 4!!

  30. amistre64
    • 4 years ago
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    'sok :) so\[Area_{shell}=\int_{0}^{6}2pi\ y\ f(y)\ dy\] \[Area_{shell}=2pi\int_{0}^{6} y\ y^2\ dy\] \[Area_{shell}=2pi\int_{0}^{6} y^3\ dy\] we agree so far?

  31. anonymous
    • 4 years ago
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    |dw:1327532422099:dw|

  32. amistre64
    • 4 years ago
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    thats better lol

  33. amistre64
    • 4 years ago
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    i think i want to "move" our graph stuff to a better location casue I think im forgetting something; moving the graph changes no value, just positions

  34. anonymous
    • 4 years ago
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    Since we are rotating around y=6, wouldn't we do pi(6-y)^2

  35. amistre64
    • 4 years ago
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    |dw:1327532555854:dw|

  36. amistre64
    • 4 years ago
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    ugh, yes, your right; been awhile :) x=(y-6)^2

  37. anonymous
    • 4 years ago
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    so we could do the cross-section method

  38. amistre64
    • 4 years ago
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    \[A=2pi\int_{0}^{6}y(y-6)^2dy\] \[A=2pi\int_{0}^{6}y(y^2-12y+36)dy\] \[A=2pi\int_{0}^{6}(y^3-12y^2+36y)dy\] looks better to me for the shell method.

  39. anonymous
    • 4 years ago
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    Ahhh, it does..

  40. amistre64
    • 4 years ago
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    disc would be: \[A=pi\int_{0}^{36}(x^{1/2}-6)dx\]

  41. anonymous
    • 4 years ago
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    |dw:1327532793666:dw|

  42. anonymous
    • 4 years ago
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    what i posted above is for the cross-section, or it works for it..

  43. amistre64
    • 4 years ago
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    \[A=pi\int_{0}^{36}(x^{1/2}-6)^2dx\] forgot to r^2 the area :)

  44. amistre64
    • 4 years ago
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    the disc gives me: 216 pi the shell gives me: 216 pi

  45. anonymous
    • 4 years ago
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    That is right!

  46. amistre64
    • 4 years ago
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    yay!!

  47. anonymous
    • 4 years ago
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    ok so I'm still a little confused how we got there..

  48. amistre64
    • 4 years ago
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    lets turn this on its side so it looks like an upsidedown cone; the math doesnt care how it looks :)

  49. anonymous
    • 4 years ago
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    2pi(r)(h) 2pi\[2\pi \int\limits_{0}^{6} y(6-y)\]

  50. amistre64
    • 4 years ago
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    |dw:1327533114400:dw|

  51. amistre64
    • 4 years ago
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    good, r = y and h=f(y), which is the inverse of f(x) which we moved down by 6

  52. amistre64
    • 4 years ago
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    y=sqrt(x)-6 y+6 = sqrt(x) (y+6)^2 = x = f(y) I used the other side of the line so I got (y-6)^2 which makes no difference here

  53. amistre64
    • 4 years ago
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    well, it does according to wolfram but .... fer some reason i did it right to begin with lol

  54. anonymous
    • 4 years ago
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    hmm.. so what would be the equation we are trying to integrate and with what limits? 0 to 36?

  55. amistre64
    • 4 years ago
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    Volume, not Area; named it wrong .... it bites getting old\[V=2\pi\int_{0}^{-6}y(y+6)^2dy=216\pi\]

  56. amistre64
    • 4 years ago
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    the 0 to 36 is for the disc method for this thing

  57. amistre64
    • 4 years ago
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    i need to downlaod google chrome; this thing is just way too slow in IE

  58. anonymous
    • 4 years ago
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    haha it runs great in chrome

  59. anonymous
    • 4 years ago
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    so why is the (y+6) being squared?

  60. anonymous
    • 4 years ago
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    Thanks so much for your help, you are great!

  61. amistre64
    • 4 years ago
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    college computers here clear out and reboot so any downloads are erased

  62. amistre64
    • 4 years ago
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    w have to tell the integration to look at the the proper rendition of out equation for height

  63. anonymous
    • 4 years ago
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    ohhhhh I see!! I was looking at the limit from x=y instead of the axis of revolution.

  64. amistre64
    • 4 years ago
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    the direction we are looking at for the shell measuers up from the y axis, or over ; but we have to rewrite the height line in terms of y for it to make sence in out integration using shells

  65. amistre64
    • 4 years ago
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    since w is the inverse of a normal xy graph we inverse our y=sqrt(x) to match it

  66. amistre64
    • 4 years ago
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    since our view is the inverse ....

  67. anonymous
    • 4 years ago
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    do you mean height instead of width?

  68. anonymous
    • 4 years ago
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    I have the correct answer on my paper now, sweet!

  69. anonymous
    • 4 years ago
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    Could you help me do a couple more? :)

  70. amistre64
    • 4 years ago
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    i can, i got the chrome downlaoded now; doesnt help the typos, but its a bit quicker lol

  71. anonymous
    • 4 years ago
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    cool, should I start an entirely new question so I can give you another medal?

  72. amistre64
    • 4 years ago
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    not that the medals matter, but starting a new one might help my computer to keep up the pace :)

  73. amistre64
    • 4 years ago
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    we didnt get thru rotate about the x=36 yet tho, so lets do that one on another post

  74. anonymous
    • 4 years ago
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    Alright I'll do that!

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