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anonymous

  • 5 years ago

find the volume of the solid generated by revolving the line x=-4.The region bounded by the two parabolas x=(y-y^2),x=(y^2-3).

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  1. amistre64
    • 5 years ago
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    sounds like a torus...

  2. amistre64
    • 5 years ago
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    need to figure out the solution to the system of equations to know what our "bounds" are gonna be...

  3. amistre64
    • 5 years ago
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    i got an idea, but its just to early in the morning to get it right in me head.....

  4. amistre64
    • 5 years ago
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    the bounds for the integral are gonna be [-1,(3/2)]

  5. anonymous
    • 5 years ago
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    That is an odd region...is it the odd oval shape?

  6. amistre64
    • 5 years ago
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    (S) y-y^2 dy = (1/2)y^2 -(1/3) y^3 |-1-3/2

  7. amistre64
    • 5 years ago
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    its a funky shape fer sure :)

  8. anonymous
    • 5 years ago
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    use circular ring method..

  9. amistre64
    • 5 years ago
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    F(-1) = 5/6

  10. anonymous
    • 5 years ago
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    That shape isn't circular...

  11. anonymous
    • 5 years ago
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    i know the answer but i dont know the solution..

  12. amistre64
    • 5 years ago
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    F(3/2) = 9/8

  13. anonymous
    • 5 years ago
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    v=85.902 cubic units

  14. amistre64
    • 5 years ago
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    your one step ahead of me then :) I am just trying to find the area of the shape

  15. anonymous
    • 5 years ago
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    .plz help me..

  16. amistre64
    • 5 years ago
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    for F(x) + get an area of 7/24....which could be totally wrong, but ill see :)

  17. anonymous
    • 5 years ago
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    I don't understand exactly which region he is asking for...

  18. amistre64
    • 5 years ago
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    G(x) = (S) y^2-3 dy = (1/3)y^3 -3y | -1,(3/2)

  19. anonymous
    • 5 years ago
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    These are the graphs http://www.wolframalpha.com/input/?i=x+%3D+y-y^2%2C+x%3D+y^2-3%2C+x%3D-4

  20. anonymous
    • 5 years ago
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    Why does this site suck at showing links...

  21. amistre64
    • 5 years ago
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    the region for the area is between y=-1 and y=(3/2)

  22. anonymous
    • 5 years ago
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    So it's the odd oval shape in the center http://www.wolframalpha.com/input/?i=x+%3D+y-y^2%2C+x%3D+y^2-3

  23. amistre64
    • 5 years ago
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    i solved the system of equations and got y=-1 and y=(3/2) as solutions

  24. anonymous
    • 5 years ago
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    Yes i see those bounds

  25. amistre64
    • 5 years ago
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    If I did it right; the area of F(x) = 7/24. Now I need to subtract area of G(x) right?

  26. amistre64
    • 5 years ago
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    G(-1) = 8/3

  27. anonymous
    • 5 years ago
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    So where exactly does the line x = -4 come in then

  28. amistre64
    • 5 years ago
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    you spin the area around x=-4 to find the area of the torus shape

  29. anonymous
    • 5 years ago
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    Oh that's the center

  30. amistre64
    • 5 years ago
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    i hope its the center :)

  31. amistre64
    • 5 years ago
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    G(3/2) = -27/8 if Im doing it right :)

  32. anonymous
    • 5 years ago
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    where should i put the strip??is it horizontal or vertical?

  33. amistre64
    • 5 years ago
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    area of G(x) = -145/24 ? Horizontal I beleive

  34. anonymous
    • 5 years ago
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    well it can't be

  35. amistre64
    • 5 years ago
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    total area of shape is F(x) - G(x) = 152/24

  36. amistre64
    • 5 years ago
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    how do we spin it? multiply by circumference of the circle?

  37. anonymous
    • 5 years ago
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    integrate from 0 to 2*pi

  38. anonymous
    • 5 years ago
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    i thought

  39. amistre64
    • 5 years ago
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    I can find the area between curves alright, but after that I get lost :)

  40. amistre64
    • 5 years ago
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    Area = 38/6 = 19/3...thats as far as I can reduce it; if its right

  41. amistre64
    • 5 years ago
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    how do we integrate the spin? Do we need to find the domain of the area are anything?

  42. anonymous
    • 5 years ago
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    I'm not exactly sure

  43. amistre64
    • 5 years ago
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    lol.....comeon..your smarter than me for sure :)

  44. anonymous
    • 5 years ago
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    i thought it was the integral from 0 to 2*pi, but that wouldn't make much sense in this case

  45. anonymous
    • 5 years ago
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    What about using the ring method

  46. anonymous
    • 5 years ago
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    That would probably just be much more difficult

  47. amistre64
    • 5 years ago
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    not that versed in the ring method...

  48. amistre64
    • 5 years ago
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    sounds like we'd integrate with respect to the radius....

  49. anonymous
    • 5 years ago
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    yes the inner radius would be the maximum for the first equation, and the outer radius would be the maximum for the 2nd i believe

  50. amistre64
    • 5 years ago
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    gotta go back and find the bends then :)

  51. amistre64
    • 5 years ago
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    one of those bounds is x=1/4

  52. amistre64
    • 5 years ago
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    is that right? y-y^2 x'=1-2y=0 y=1/2. .5-.5^2 = .25 x=1/4 right?

  53. anonymous
    • 5 years ago
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    My brain hurts lol. Okay so you're deriving y-y^2, solving for y, and plugging it back in?

  54. amistre64
    • 5 years ago
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    y^2 -3 x'=2y=0 y=0 other bound is x=-3

  55. anonymous
    • 5 years ago
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    Yeah that seems right

  56. amistre64
    • 5 years ago
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    yep, only thing I can do :)

  57. amistre64
    • 5 years ago
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    sp we got our bends at x=-3 and x=1/4....now what?

  58. amistre64
    • 5 years ago
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    spin it at x=-4 somehow

  59. amistre64
    • 5 years ago
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    each slice is spin from -3 to 1/4 in dx increments and then add them all up

  60. anonymous
    • 5 years ago
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    Im confused on the area

  61. amistre64
    • 5 years ago
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    make it easier and spin add 4 to x so that we have the middle at 0 and our limits at 1 and the other at 4.25

  62. anonymous
    • 5 years ago
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    Yeah that's what i did actually :)

  63. amistre64
    • 5 years ago
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    which area :) the area between the curves?

  64. anonymous
    • 5 years ago
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    yes, the integral of x = y-y^2 would give the area under the parabola, but how would you remove the sections on the sides to get the torus

  65. anonymous
    • 5 years ago
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    Or i should say...to the right of the parabola

  66. amistre64
    • 5 years ago
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    the two equations meet at the points -1 and 3/2; so these are the bounds we use. to find the area right?

  67. anonymous
    • 5 years ago
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    Ok i see

  68. amistre64
    • 5 years ago
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    back in algebra here :) when y-y^2 = y^2-3 we get solutions to the equations :)

  69. amistre64
    • 5 years ago
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    2y^2 -y -3 =0 (y-...)(y+....)

  70. amistre64
    • 5 years ago
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    I use that to find the area of F(x) and G(x) then: F(x)-G(x) gives us the area we need

  71. amistre64
    • 5 years ago
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    now, since we are talking about spinning a circle; do we (S) 2pi(x)(area)? dx

  72. amistre64
    • 5 years ago
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    it should be the same concept as finding the area between curves; use the lower and upper bounds of x and subtract them?

  73. amistre64
    • 5 years ago
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    cant hurt to try: (S) 2pi(x)(19/3) dx = 19pi(x^2)/3 | 1,4.25

  74. amistre64
    • 5 years ago
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    114.39583333(pi) - 6.3333333333(pi)

  75. amistre64
    • 5 years ago
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    I get 108.0625 pi....but am I right?

  76. amistre64
    • 5 years ago
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    ill need alot more practice for this I guess :)

  77. amistre64
    • 5 years ago
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    v=85.902 cubic units is what the person said to begin with; and that doesnt match mine

  78. anonymous
    • 5 years ago
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    Yeah it doesn't, my brain is too fried for this right now, I've got diff eq in 5 hours, best of luck though

  79. amistre64
    • 5 years ago
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    the key lies in the y axis. distance from the y axis bounds i think..not the x-axis

  80. anonymous
    • 5 years ago
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    what is the integral of 1/((x+2)√(x+3))

  81. amistre64
    • 5 years ago
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    im wrong somewhere :)

  82. amistre64
    • 5 years ago
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    i need more practice on integrals... .

  83. amistre64
    • 5 years ago
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    have you intergrated by parts?

  84. anonymous
    • 5 years ago
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    yeah...but dunno if its correct

  85. amistre64
    • 5 years ago
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    did you do it more than once?

  86. amistre64
    • 5 years ago
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    (S) 1/x+2 dx = ln(x+2) that much works out

  87. amistre64
    • 5 years ago
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    i gotta get some sleep :)....chat with ya later think...

  88. anonymous
    • 5 years ago
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    okay amistre...btw u have taken u=1/x+2?

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