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anonymous

  • 5 years ago

Consider the given curves to do the following. y=5sqrt(x) y=0 x=1 Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about x = -3.

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  1. amistre64
    • 5 years ago
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    wha!?!

  2. amistre64
    • 5 years ago
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    :) just kiddin' this ones easy too

  3. anonymous
    • 5 years ago
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    yup :( i went on this online tutor thing from the hw website i use, they got me set up but i think i have the boundaries wrong

  4. anonymous
    • 5 years ago
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    lol xD the integral they told me to use is

  5. anonymous
    • 5 years ago
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    \[2\pi \int\limits5x^(3/2) + 15x^(1/2)\]

  6. anonymous
    • 5 years ago
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    sorry it should be 5x^(3/2) and 15x^(1/2)

  7. anonymous
    • 5 years ago
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    use {} when using exponent

  8. anonymous
    • 5 years ago
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    kay i will

  9. anonymous
    • 5 years ago
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    i thought the boundaries for the integral would be 1 and -3, but when i plug in -3 to the integral itll give me trouble lol

  10. amistre64
    • 5 years ago
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  11. amistre64
    • 5 years ago
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    this is a rough idea of what you got and since we are spinning around an x; we integrate with respect to y.....lets move that x line; and I got the rotation axis in the wrong spot

  12. anonymous
    • 5 years ago
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    so if we integrate with respect to y, that means we have to get x = in terms of y ?

  13. amistre64
    • 5 years ago
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    like this

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  14. amistre64
    • 5 years ago
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    the shells have a radius along the x; so lets stick to that.... i had to draw the pic for clarity in me head :)

  15. amistre64
    • 5 years ago
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    lets +3 to all the 'x' values; to move this thing; NOT f(x) + 3; but f(x+3)

  16. anonymous
    • 5 years ago
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    how come were adding 3?

  17. amistre64
    • 5 years ago
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    y=5sqrt(x+3) y=0 = x axis x=1+3; you see that instruction to spin about x=-3? get that to zero dear lol

  18. anonymous
    • 5 years ago
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    LOL oh. haha thank youu

  19. amistre64
    • 5 years ago
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    our inteval is then ; [0+3,1+3], or simply; [3,4] right?

  20. anonymous
    • 5 years ago
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    yes

  21. amistre64
    • 5 years ago
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    Our farthest Raduis =4 and our inner radius = 3 so we integrate from r to R roght?

  22. anonymous
    • 5 years ago
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    so like lets say the instruction told us to spin about x = 1, would we subtract 1 from the x values?

  23. anonymous
    • 5 years ago
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    and correct

  24. amistre64
    • 5 years ago
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    and our height only changes as 5sqrt(x+3) changes now; so lets do this :)

  25. amistre64
    • 5 years ago
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    make that 5sqrt(x-3) ; you know why?

  26. anonymous
    • 5 years ago
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    mmm, im not too sure why

  27. amistre64
    • 5 years ago
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    we want 5sqrt(0) to equal 5sqrt(3+___) so remember that we moved it but we dont want to change its value; just its position

  28. amistre64
    • 5 years ago
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    the value at 3 away from center of rotation nedds to be the same regardless of where we put this thing right?

  29. anonymous
    • 5 years ago
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    correct

  30. amistre64
    • 5 years ago
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    so; lets integrate this thing like this: 2pi {S} x*5sqrt(x-3) dx ; [3,4]

  31. anonymous
    • 5 years ago
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    you see the 5sqrt(x-3)

  32. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=int%282pi*x*5sqrt%28x-3%29%29+from+3+to+4

  33. anonymous
    • 5 years ago
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    can we put it as 5(x-3)^1/2

  34. amistre64
    • 5 years ago
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    i see it :) we can pull out that 5 right? nothing is stopping that from not being a constant

  35. anonymous
    • 5 years ago
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    ooh that makes it easier than what i was about to do lol

  36. amistre64
    • 5 years ago
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    we can put that lonely x inside the sqrt

  37. amistre64
    • 5 years ago
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    x*sqrt(x-3) = sqrt(x^2(x-3)) (x^3 -3x^2)^(1/2) right?

  38. anonymous
    • 5 years ago
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    can i ask one thing though. can we use that integral of 2pi x f(x) dx for any cylindrical shell problem? because when i went to the hw tutor from the hw website, they told me to use the 2pi r h or something like that

  39. amistre64
    • 5 years ago
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    r = x or y depending on if you spin it up or sideways h = f(x) regardless

  40. amistre64
    • 5 years ago
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    well; maybe h can = f(y) too lol

  41. anonymous
    • 5 years ago
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    ohh so they're basically the same thing, just different way of putting it

  42. anonymous
    • 5 years ago
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    x*sqrt(x-3) = sqrt(x^2(x-3)) (x^3 -3x^2)^(1/2) right? and yes lol xD

  43. amistre64
    • 5 years ago
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    yes; its like calling your dad by his given name; hell respond to either right? the person is the same, the name can change

  44. anonymous
    • 5 years ago
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    haha nice comparison xD

  45. amistre64
    • 5 years ago
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    we could integrate this numercally instead; with like the simpsons rule or trap rule if trying to find a morph that works proves too difficult

  46. anonymous
    • 5 years ago
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    i havent learned those rules yet :(

  47. amistre64
    • 5 years ago
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    you have, you just havent learned how to apply them yet lol its like this shell methid; you already knew how to find the area of a sheet of paper; but dint realize how that fact could aid you

  48. anonymous
    • 5 years ago
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    lol oh xD im sorry, im still a little confused on why we added 3. i understand to get x=0 but im still confused on why we did it.

  49. amistre64
    • 5 years ago
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    becasue in analysis it is always easier to bring things to a standard working table that is right in front of you. If we dont move it, it is stuck up high on a shelf and that is no way to examine it right? so, we bring it to the table by moving it off the shelf

  50. amistre64
    • 5 years ago
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    we change its position ; but retain its value

  51. amistre64
    • 5 years ago
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    this is the basic of the trapazoid rule

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  52. anonymous
    • 5 years ago
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    you mean everything is the same, the value is the same, its just on a different point on the graph?

  53. amistre64
    • 5 years ago
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    exactly :)

  54. amistre64
    • 5 years ago
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    so the answer we receive from it will be the same; if you move your breakfast form the stove to your plate to the table; what has changed?

  55. anonymous
    • 5 years ago
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    the place of where its at lol xD i gotcha, thank you for clarifying that for me. i was so confused before lol :p

  56. amistre64
    • 5 years ago
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    lets see if we can int this function :) x*sqrt(x-3) then :) any tricks up your sleeve thatll be useful for us?

  57. amistre64
    • 5 years ago
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    ohohoh...maybe int by parts lol

  58. amistre64
    • 5 years ago
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    u = x and v = sqrt(x-3) ??

  59. anonymous
    • 5 years ago
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    are we using substitution right now?

  60. amistre64
    • 5 years ago
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    its a kind of subsitution based on the product rule for derivatives

  61. amistre64
    • 5 years ago
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    remember the product rule? Dx(uv) = uv' + vu' right?

  62. anonymous
    • 5 years ago
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    yesss i remember that :)

  63. anonymous
    • 5 years ago
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    so it would be equal to x/2 (x-3)^-1/2 + sqrt of x-3 ?

  64. amistre64
    • 5 years ago
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    {S} (uv)' = {S} u v' + {S} v u' right? uv = {S} u dv + {S} v du

  65. amistre64
    • 5 years ago
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    uv - {S} v du = {S} u dv

  66. amistre64
    • 5 years ago
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    we want to integrate " u = x sqrt(x-3)" and it has 2 parts; so we use this trick to find it lol

  67. anonymous
    • 5 years ago
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    how did you get from {S} (uv)' = {S} u v' + {S} v u' right? to uv = {S} u dv + {S} v du

  68. amistre64
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  69. amistre64
    • 5 years ago
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    the {S} of a derivate (y') = y; {S} y' = y; right?

  70. anonymous
    • 5 years ago
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    correct

  71. amistre64
    • 5 years ago
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    v' = dv; just different notations

  72. amistre64
    • 5 years ago
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    lets focus on that part of my "u" column... we start with: (x-3)^(1/2) ; and that ints up to: 2/3 (x-3)^(3/2) ; in tthat again... pull out the 2/3 and do: (x-3)^(3/2) goes to what?

  73. amistre64
    • 5 years ago
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    (2/5) (x-3)^(5/2) right?

  74. anonymous
    • 5 years ago
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    so we take the antiderivative twice for it?

  75. amistre64
    • 5 years ago
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    \[\int\limits_{}(x.\sqrt{x-3}).dx = [\frac{2x}{3}.\sqrt{(x-3)^3}] - [\frac{4}{15}.\sqrt{(x-3)^{5}}]\]

  76. amistre64
    • 5 years ago
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    right?

  77. amistre64
    • 5 years ago
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    we integrate it till we can stop pretty much

  78. amistre64
    • 5 years ago
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    if we pick a value for u that derives to 0 then it doesnt matter what v suits up to cause when it gets to u=0 its done

  79. anonymous
    • 5 years ago
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    okay i see what you mean now

  80. amistre64
    • 5 years ago
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    u =x u' = 1 u'' = 0 right?

  81. anonymous
    • 5 years ago
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    correct

  82. amistre64
    • 5 years ago
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    v = sqrt(...) {S} v = {S} sqrt(...) {S} [{S} v] = {S} [{S} sqrt(...)]

  83. amistre64
    • 5 years ago
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    +u * {S} v - u' * {S} [{S} v] + 0 * ...doesnt matter lol

  84. amistre64
    • 5 years ago
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    this is the method known as integration by parts; and it can get tricky but the key is to pick smart values for u and v to begin with and itll work itself out in the end :)

  85. amistre64
    • 5 years ago
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    int by parts is just using the product rule as a basis; and then using it again, and again, till you get to 0 lol

  86. anonymous
    • 5 years ago
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    seems so hard :( lol but i think i got a jist of what you mean. thank you for all your help and for staying and explaining all of it to me. :))

  87. amistre64
    • 5 years ago
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    youre welcome :) the more i explain it the better i understand it meself ;)

  88. amistre64
    • 5 years ago
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    its says your typing.... its like waiting for the waffle iron

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