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anonymous

  • 5 years ago

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the y-axis. y = x^2 y=0 x=1

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  1. amistre64
    • 5 years ago
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    HA!! shells, i knew they were comin

  2. amistre64
    • 5 years ago
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    since we is going around the y axis; its simpler right now

  3. anonymous
    • 5 years ago
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    yup :( i never used this method before ><

  4. amistre64
    • 5 years ago
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    the radius of each shell moves from x=0 to x=1 right?

  5. anonymous
    • 5 years ago
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    yup

  6. amistre64
    • 5 years ago
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    and the height of each shell is determined by: f(x)

  7. anonymous
    • 5 years ago
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    wait, i know why its x=1, but why is it x=0

  8. amistre64
    • 5 years ago
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    so the area of each shell, when flattened out is height* width...

  9. amistre64
    • 5 years ago
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    umm... y axis means x=0 they are the same thing

  10. anonymous
    • 5 years ago
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    ohh okay gotcha

  11. amistre64
    • 5 years ago
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    the height = f(x) the width = 2pi [x] the area of any given shell = 2pi x[f(x)] so integrate that from [a,b] = [0,1] in this case

  12. amistre64
    • 5 years ago
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    2pi [S] x(x^2) dx ; [0,1] 2pi x^4 ------- = (pi x^4)/4 = pi/4 4

  13. anonymous
    • 5 years ago
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    how did you get rid of the 4th power in the numerator?

  14. amistre64
    • 5 years ago
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  15. amistre64
    • 5 years ago
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    you dont... why would you?

  16. anonymous
    • 5 years ago
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    but you put (pi x^4)/4 = pi/4 where did the 4th power go?

  17. amistre64
    • 5 years ago
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    F(0) = 0 F(1) = pi/4 F(1) - F(0) = pi/4

  18. amistre64
    • 5 years ago
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    over 2 lol

  19. amistre64
    • 5 years ago
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    i forgot i changed 2/4 to 1/2.... doh!

  20. amistre64
    • 5 years ago
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    \[\frac{2\pi x^4}{4} = \frac{\pi x^4}{2}\]

  21. anonymous
    • 5 years ago
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    much clearer, lol thank you :D

  22. amistre64
    • 5 years ago
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    when im wrong; say im wrong...not ask me how im right lol. I just assume im right ;)

  23. anonymous
    • 5 years ago
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    lol will do xD

  24. amistre64
    • 5 years ago
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    you understand the mechanics of the shell method tho?

  25. anonymous
    • 5 years ago
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    all the methods like the washer, disk, etc all those methods seem so similiar to eachother and i get them all confused. because in the end youre always taking the antiderivative and plugging in the boundaries right?

  26. anonymous
    • 5 years ago
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    like i understand theres certain formulas you have to follow for each method, which i have to memerize lol all of them seem really similiar

  27. amistre64
    • 5 years ago
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  28. anonymous
    • 5 years ago
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    so for the shell method, basically youre taking half of it, and spreading it out which is why it turns to be a rectangle shape?

  29. amistre64
    • 5 years ago
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    when we flatten out the "shell" we get a flat rectangular piece that the area is eay to determine right?

  30. anonymous
    • 5 years ago
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    correct

  31. amistre64
    • 5 years ago
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    the width = 2pi x...can you tell me why?

  32. amistre64
    • 5 years ago
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    we aint using half the shell, but the whole thing

  33. anonymous
    • 5 years ago
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    i think it has to do with the radius, is that why the width is 2 pi x?

  34. anonymous
    • 5 years ago
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    ohh is it the circumference of the cylnder?

  35. amistre64
    • 5 years ago
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    it does, but i want to make sure you understand this :) its very basic and easy...exactly lol

  36. anonymous
    • 5 years ago
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    lol i took a careful look at the cylinder and realized haha xD

  37. amistre64
    • 5 years ago
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    you did good :)

  38. amistre64
    • 5 years ago
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    the shell method makes things easier for some problems

  39. anonymous
    • 5 years ago
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    thanks ;) so basically for the shell method we're plugging in f(x) into the formula and taking the antiderivative of it? then plug in the boundaries and solve?

  40. amistre64
    • 5 years ago
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    lets think that thruough; we want to add up all the areas...which is what intagrating does; each area = 2pi x[f(x)]

  41. amistre64
    • 5 years ago
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    {S} 2pi x[f(x)] dx is what we do right?

  42. amistre64
    • 5 years ago
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    if there is a constant we can pull it aside right?

  43. anonymous
    • 5 years ago
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    mhmm

  44. anonymous
    • 5 years ago
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    the 2pi

  45. amistre64
    • 5 years ago
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    2 pi goes out and what are we left with inside?

  46. anonymous
    • 5 years ago
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    x*f(x)

  47. amistre64
    • 5 years ago
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    excatly; so we integrate 'x*f(x)'

  48. amistre64
    • 5 years ago
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    do you se how that differes from your first statement of ; we int f(x)?

  49. anonymous
    • 5 years ago
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    yesss, thanks for clearing that up for me :) this method seems a bit easier than the others lol

  50. amistre64
    • 5 years ago
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    look at the disc method

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  51. anonymous
    • 5 years ago
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    oh, i have one question. when you solve for an integral, is it possible for it to be negative? or does it always have to be positive

  52. amistre64
    • 5 years ago
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    the disc method take all the areas of a given circle; with a radius of f(x) and adds them up

  53. amistre64
    • 5 years ago
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    volume is always a positive number

  54. amistre64
    • 5 years ago
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    what is the formula for the the area of a circle ?

  55. anonymous
    • 5 years ago
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    uh oh. darn i should of took the absolute value. i was so sure i did it right too. lol but thank you :)

  56. anonymous
    • 5 years ago
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    and area of circle is pi r^2

  57. amistre64
    • 5 years ago
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    so when we add up all the areas of the cicles thru integration we do: {S} pi [f(x)]^2 dx right? adding up the areas of circles

  58. anonymous
    • 5 years ago
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    so f(x) is the radius?

  59. amistre64
    • 5 years ago
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    look at the drawing i did in this one

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  60. amistre64
    • 5 years ago
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    and tell me what the radius of each given circle is?

  61. anonymous
    • 5 years ago
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    oh, f(x) xD haha

  62. amistre64
    • 5 years ago
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    that is all these volume of rotation problems amopunt to :)

  63. anonymous
    • 5 years ago
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    i remember there was this one method i think where you had to divide by two, is there a method that involves that? or am i making stuff up lol :p

  64. amistre64
    • 5 years ago
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    i dont think thats a volume of rotation one :) in some area propblems we are asked to find the area under a curve that hopes the y axis... and if the symmetry is the same from left to right side of the y axis; we get a skewed result

  65. amistre64
    • 5 years ago
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    lets try something simple and see; suppose we want to find the area of a square that is sitting halfway between the sides

  66. amistre64
    • 5 years ago
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    lets say 4 high and 4 wide; the area should be 16 right?

  67. anonymous
    • 5 years ago
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    correct

  68. amistre64
    • 5 years ago
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    if we integrate f(x) = y = 4; from -2 to 2 we know we should get 16 right?

  69. anonymous
    • 5 years ago
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    how come its from -2 and 2?

  70. amistre64
    • 5 years ago
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    {S} 4 dx -> 4x 4(2) - 4(-2) = 8 -- 8 = 16, so that one doesnt apply to the /2 thing you discussed lol

  71. amistre64
    • 5 years ago
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    the distance from -2 to 2 = 4 right?

  72. anonymous
    • 5 years ago
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    yupp

  73. amistre64
    • 5 years ago
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    and a square with sides = 4 would straddle the y axis would sit from -2 to 2 on the x axis right?

  74. anonymous
    • 5 years ago
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    correct. i gotcha now :) lol

  75. anonymous
    • 5 years ago
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    if the prob asked to rotate around the x-axis, its generally the same idea of solving it like we did with the y axis right?

  76. amistre64
    • 5 years ago
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    i pic is worth a thousand words :)

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  77. anonymous
    • 5 years ago
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    it really is. thank you for that :D

  78. amistre64
    • 5 years ago
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    yes, but we just need to make sure we are rotating it properly; make sure the numbers come out correctly

  79. amistre64
    • 5 years ago
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    once we can draw a picture, the rest is just intuitive

  80. anonymous
    • 5 years ago
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    gotcha. i shall try and attempt the rest of these shell method questions. thank you so so much for all your help. :D you make it possible for me to understand calculus lol xD thank you times infinity !

  81. amistre64
    • 5 years ago
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    youre welcome :)

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