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anonymous

  • 5 years ago

Find volume of revolution for region bounded by y = x^2, y = x , in first quadrant, revolve it about the line x = 6 using shell method

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  1. anonymous
    • 5 years ago
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    i think i got it , integral 2pi ( x - x^2) (6-x) dx from 0 to 1 ?

  2. anonymous
    • 5 years ago
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    the part i dont get is the 6 - x

  3. anonymous
    • 5 years ago
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    isnt there an inner radius and an outer radius ?

  4. anonymous
    • 5 years ago
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    like with the washer method , integral pi f(x)^2 - pi (g(x)) ^2 , where f(x) > g(x)

  5. anonymous
    • 5 years ago
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    where exactly does 6-x come from

  6. anonymous
    • 5 years ago
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    volume will be \[2\pi \int\limits_{0}^{1} (5+x)(x-x^2).dx -\pi 5^2 .1\]

  7. anonymous
    • 5 years ago
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    i believe thats wrong

  8. anonymous
    • 5 years ago
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    integral 2pi ( x - x^2) (6-x) dx from 0 to 1

  9. anonymous
    • 5 years ago
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    i'm giving you the graph...

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  10. anonymous
    • 5 years ago
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    thats not correct, what do you get as an answer

  11. watchmath
    • 5 years ago
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    I agree with cantor set. The 6-x coming from the distance from a slice to the axis of symmetry.

  12. anonymous
    • 5 years ago
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    first find the volume made by revolving only y=x, then subtract the volume made by y=x^2. and that also due to the cylinder of radius 5.

  13. anonymous
    • 5 years ago
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    , watchmath, what about this question Determine the volume of the solid obtained by rotating the region bounded by x^(1/3), x=8, y=0, about the x-axis.

  14. anonymous
    • 5 years ago
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    \[\pi \int\limits_{x=0}^{8}y^2 dx \] where y=x^1/3 \[\pi \int\limits_{0}^{8}x ^{2/3}dx =3\pi 8^{5/3}/2=48\pi\]

  15. anonymous
    • 5 years ago
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    thats wrong too

  16. anonymous
    • 5 years ago
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    the answer is 96 pi/5

  17. anonymous
    • 5 years ago
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    the question is here http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx

  18. anonymous
    • 5 years ago
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    sorry i did a mistake in doing the integration...\[\pi \int\limits_{0}^{8}x ^{2/3}dx=3\pi (x^{5/3}|_{0}^{8})/5=3\pi 8^{5/3}/5\]\[=3\pi 2^5 /5 =96\pi /5\]

  19. anonymous
    • 5 years ago
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    oh nevermind

  20. anonymous
    • 5 years ago
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    ok small error

  21. anonymous
    • 5 years ago
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    but i should have said, use shell method :)

  22. anonymous
    • 5 years ago
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    i make it 48pi

  23. anonymous
    • 5 years ago
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    shell method stumps me on this problem

  24. watchmath
    • 5 years ago
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    Shell method \(\int_0^2 2\pi y (8-y^3)\, dy\)

  25. anonymous
    • 5 years ago
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    \[2\pi \int\limits_{x=0}^{8}y (8-x)dy\]

  26. anonymous
    • 5 years ago
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    yeah, how come its 8-y^3

  27. anonymous
    • 5 years ago
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    use x=y^3

  28. anonymous
    • 5 years ago
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    watchman has already given it.....

  29. anonymous
    • 5 years ago
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    sorry - i made the same mistake as dipank

  30. anonymous
    • 5 years ago
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    how come normally we dont do 8 - y^3 , like shell method for region bounded by y = x^2 , x axis, y axis

  31. anonymous
    • 5 years ago
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    you need one more boundary...

  32. watchmath
    • 5 years ago
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    Because the axis of rotation now is the x-axis (before the rotation axis was parallel to the y-axis)

  33. anonymous
    • 5 years ago
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    is there a similiar problem to this revolving about y axis

  34. watchmath
    • 5 years ago
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    Assuming all the information are the same but now we rotate about the y-axis then the volume is given by \(2\pi\int_0^8 x\cdot x^{1/3}\, dx\)

  35. anonymous
    • 5 years ago
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    shell method for region y= x^2 bounded by y=0, x=2, revovling about y=-2

  36. watchmath
    • 5 years ago
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    \(\int_0^4 2\pi (y+2)(2-\sqrt{y})\,dy\)

  37. anonymous
    • 5 years ago
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    thx

  38. anonymous
    • 5 years ago
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    whats your trick for shell methods, i have trouble setting them up

  39. anonymous
    • 5 years ago
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    whats the shell method for region bounded by y = e^(-x^2) , x = 2, y = 0. revolved about x axis . only need to set up

  40. anonymous
    • 5 years ago
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    \[2\pi \int\limits_{0}^{2}x e^{-x^2/2}dx\]

  41. watchmath
    • 5 years ago
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    Well there is not trick. First make a slice that is parallel to the axis of symmetry. In your last problem we slice along the y-axis. Then imagine that you rotate that slice. You will have a cyllinder. Try to find the radius of the cyllinder. If the slice is of distance y from the x-axis, then it is of distance y+2 from the axis of rotation (y=-2). So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8-sqrt(y)).

  42. anonymous
    • 5 years ago
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    i think this will be the ans...

  43. watchmath
    • 5 years ago
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    Give it a try Cantorset for you very last problem.

  44. anonymous
    • 5 years ago
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    wait, which problem are you explaining , you said the height of the cylinder is bounded by x = 8, for which problem

  45. anonymous
    • 5 years ago
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    So the circumference of the base of the cyllinder is 2pi(y+2). Now the height of the cyllinder is bounded to the right by x=8 and to the left by x= sqrt(y). So the height is (8-sqrt(y)). for which problem was this?

  46. watchmath
    • 5 years ago
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    Oh sorry, the height should be 2- sqrt(y)

  47. anonymous
    • 5 years ago
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    dont you mean, width of the cylinder?

  48. anonymous
    • 5 years ago
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    oh height, if you look at it sideways

  49. watchmath
    • 5 years ago
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    Try this one. Find the volume of solid resulted from rotating the region bounded by the circle (x-2)^2+y^2=1 around the y-axis (the solid is a torus).

  50. anonymous
    • 5 years ago
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    ok one sec, before i do that one. what if i have a region bounded y = x^2 , y=4, x = 2, about x axis

  51. anonymous
    • 5 years ago
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    then it would be integral 2pi * y *sqrt y dy ?

  52. anonymous
    • 5 years ago
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    from 0 to 4

  53. anonymous
    • 5 years ago
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    for the second problem the height is same for all the shells and it is 2 and the radius is equal to y so volume will \[2\pi \int\limits_{y=0}^{1} y *2dy\]be

  54. anonymous
    • 5 years ago
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    please tell me is the integral correct for the problem y=e^(-x^2/2)

  55. anonymous
    • 5 years ago
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    no

  56. anonymous
    • 5 years ago
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    for which one? first solve for x

  57. watchmath
    • 5 years ago
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    Need to go ... I'll be back in 20 minute(ish)

  58. anonymous
    • 5 years ago
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    ok

  59. anonymous
    • 5 years ago
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    graph the curve y = e^(-x^2)

  60. anonymous
    • 5 years ago
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    and we are revolving about x axis

  61. anonymous
    • 5 years ago
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    for the volume bounded by the curve y=e^(-x^2/2), x=2,y=0, around x-axis..

  62. anonymous
    • 5 years ago
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  63. anonymous
    • 5 years ago
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    oh wait, i meant y = e^(-x^2)

  64. anonymous
    • 5 years ago
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    i have done a rough graph can you check it...

  65. anonymous
    • 5 years ago
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    not the right graph, should look like a bell shape

  66. anonymous
    • 5 years ago
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    ok i will try it.......

  67. anonymous
    • 5 years ago
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    here http://www.wolframalpha.com/input/?i=graph+e^%28-x^2%29

  68. anonymous
    • 5 years ago
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    thanx

  69. anonymous
    • 5 years ago
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    its easier to check with disc method, so lets try that

  70. anonymous
    • 5 years ago
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    integral pi *e^(-x^2) ^2 from 0 to 2

  71. anonymous
    • 5 years ago
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    918800

  72. watchmath
    • 5 years ago
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    If you want to use the cyllindrical shell to the e^(-x^2/2) problem, then remember we need to make a slice parallel to the x-axis. This slice is bounded to the right by x=2 and to the left by \(x=\sqrt{-\ln y}\).

  73. watchmath
    • 5 years ago
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    sorry I mean to the right by \(x=\sqrt{-2\ln y}\)

  74. watchmath
    • 5 years ago
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    to the left ... :)

  75. anonymous
    • 5 years ago
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    for the sake of graphing and checking i changed the problem

  76. anonymous
    • 5 years ago
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    y = 4 e^(-x^2/4) (which stretches it horizontally and vertically)

  77. anonymous
    • 5 years ago
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    bounded by x = 2 , x = 0 , y =0

  78. anonymous
    • 5 years ago
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    sorry i keep changing problem , im checking this with calculator

  79. anonymous
    • 5 years ago
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    using the disc method, just to make sure we are doing the shell method right (to check later)

  80. watchmath
    • 5 years ago
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    and we rotate about which axis?

  81. anonymous
    • 5 years ago
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    disc method is int pi ( 4 e^(-x^2/4) ) ^2 , from 0 to 2 , about x axis

  82. anonymous
    • 5 years ago
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    i get 60.13 as my volume

  83. anonymous
    • 5 years ago
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    ok shell method ... x= sqrt (- 4 ln (y/4) )

  84. watchmath
    • 5 years ago
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    It is not nice using the shell method here. You need to split into two integrals.

  85. anonymous
    • 5 years ago
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    so there are two pieces to the shell method int 2pi* y*sqrt (- 4 ln (y/4) ) from 2 to 4 +

  86. anonymous
    • 5 years ago
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    + int 2pi * y* 2 from 0 to 2

  87. watchmath
    • 5 years ago
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    0 to 4e^(-1) and from 4e^(-1) to 4

  88. anonymous
    • 5 years ago
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    oh

  89. anonymous
    • 5 years ago
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    int 2pi * y* 2 from 0 to 4e^-1 + int 2pi* y*sqrt (- 4 ln (y/4) ) from 4e^-1 to 4

  90. anonymous
    • 5 years ago
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    answer does not match disc method

  91. anonymous
    • 5 years ago
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    ok it does,

  92. anonymous
    • 5 years ago
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    wow, what a problem !!!

  93. anonymous
    • 5 years ago
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    int 2pi * y* 2 from 0 to 4e^-1 + int 2pi* y*sqrt (- 4 ln (y/4) ) from 4e^-1 to 4 = int pi (4*e^(-x^2/4)) ^2 from 0 to 2

  94. anonymous
    • 5 years ago
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    i guess i have problems with upper and lower limits, i confuse them , which ones we are doing

  95. anonymous
    • 5 years ago
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    im just doing these problems to be thorough,

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