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anonymous

  • 5 years ago

Find the surface area that results when the curve is rotated around the x-axis: x = t ^(2) + 1 , y = t ; 0 ≤ t ≤ 2 .

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  1. anonymous
    • 5 years ago
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    ????

  2. myininaya
    • 5 years ago
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    use int(2*pi*y*sqrt((x')^2+(y')^2),t)

  3. myininaya
    • 5 years ago
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    int(2*pi*t*sqrt(4t^2+1),t)

  4. myininaya
    • 5 years ago
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    let u=4t^2+1 so du=8tdt so we have int(2/8*u^(1/2),u) that should help you

  5. anonymous
    • 5 years ago
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    ok i'm tryin it now

  6. amistre64
    • 5 years ago
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    would this be equivalent to: x = y^2 +1 ?

  7. anonymous
    • 5 years ago
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    im still workin on it ha

  8. amistre64
    • 5 years ago
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    i spose the line integral dont care if its parametric or not eh :)

  9. anonymous
    • 5 years ago
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    what im confused ha

  10. amistre64
    • 5 years ago
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    the line integral;\[\int\limits_{} \sqrt{f'(x) + g'(x)} dx\]

  11. anonymous
    • 5 years ago
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    i dont know what exactly you're asking but we're doing parametric stuff now?

  12. anonymous
    • 5 years ago
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    haha i'm so confused with this stuff right now sorry

  13. amistre64
    • 5 years ago
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    the line integral is primed for parametric equations; you insert the derivatives of your parameter equations and run with it :)

  14. anonymous
    • 5 years ago
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    so i should find the derivative of both and plug them into the integral you put earlier instead of what i just did?

  15. amistre64
    • 5 years ago
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    dunno what you just did :) but yeah. you find the length of the line, and spin it to find the surface area...

  16. amistre64
    • 5 years ago
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    \[2\pi \int\limits_{0}^{2}\sqrt{f'(t) + g'(t)} dt\]

  17. amistre64
    • 5 years ago
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    if im wrong, im sure someone will let me know :)

  18. amistre64
    • 5 years ago
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    x = t ^(2) + 1 , y = t x' = 2t ; y' = 1 fill em in

  19. amistre64
    • 5 years ago
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    and I was wrong...those whoud be "squared" lol

  20. amistre64
    • 5 years ago
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    \[2\pi \int\limits_{0}^{2}\sqrt{[f'(x)]^{2} + [g'(x)]^2}\]

  21. anonymous
    • 5 years ago
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    ok so i ended up with pi/4(2u^(3/2)/3) with the limits from 1 to 17 bc i plugged in 0 and 2 into u = 4t^2 +1

  22. anonymous
    • 5 years ago
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    and just going to plug in those limits to find out the answer is that right?

  23. amistre64
    • 5 years ago
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    [S] (4t^2 + 1)^(1/2) dt how did you go about integrating the square root? did you use a trig substitutiion? or a "u" substitution?

  24. anonymous
    • 5 years ago
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    u=4t^2 +1

  25. anonymous
    • 5 years ago
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    so du = 8tdt

  26. amistre64
    • 5 years ago
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    and how does that integrate?...good, and how did you get rid of the "t" from that?

  27. anonymous
    • 5 years ago
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    ohhhhh i guess i messed that up

  28. anonymous
    • 5 years ago
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    ahh now im confused again

  29. amistre64
    • 5 years ago
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    \[\int\limits_{}8t (\sqrt{u}) du =?\]

  30. amistre64
    • 5 years ago
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    itd be in the denominator, but still....its there lol

  31. amistre64
    • 5 years ago
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    trig identities are what you can use for substitution in this

  32. amistre64
    • 5 years ago
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    \[\sqrt{4x^2 + 1}\] tan^2 + 1 = sec^2 right?

  33. amistre64
    • 5 years ago
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    say u = 2x tan(x) u^2 = 4x^2 tan^2....plays a role in this someplace :)

  34. anonymous
    • 5 years ago
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    im not gettin it ha

  35. amistre64
    • 5 years ago
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    you pretty much want to make it looke like this:\[\sqrt{\tan^2(x) + 1} = \sec(x)\]

  36. amistre64
    • 5 years ago
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    so that tan(x) = 2x

  37. amistre64
    • 5 years ago
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    then you integrate sec(x) dx and turn the substitution around

  38. anonymous
    • 5 years ago
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    so would it be (2tan^2(x) +1)^1/2

  39. anonymous
    • 5 years ago
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    when i start the trig sub or what?

  40. amistre64
    • 5 years ago
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    no, then you cant "factor" out that "2" in front of the tan... just my thoughts :)

  41. anonymous
    • 5 years ago
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    ahhh ive been workin at this too long i cant think straight now im forgetting how to do trig sub stuff

  42. amistre64
    • 5 years ago
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    4x^2 needs to equal tan^2 2x = tan then

  43. amistre64
    • 5 years ago
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    we draw a triangle with the corresponding parts so that one leg = 2x; the other leg = 1 and the hypotenuse = whatever it equals :)

  44. amistre64
    • 5 years ago
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    that way we can convert between our stuff in the end right?

  45. anonymous
    • 5 years ago
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    ya i mean if we did the triangle it would be sqrt of 2x^2 + 1^1

  46. anonymous
    • 5 years ago
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    ok so heres where im at

  47. amistre64
    • 5 years ago
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    does this help?

  48. amistre64
    • 5 years ago
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  49. anonymous
    • 5 years ago
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    2 pi (ln(sec(t) + tan(t)) and whatever the limits are

  50. anonymous
    • 5 years ago
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    does that seem correct?

  51. amistre64
    • 5 years ago
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    that looks good, just need to replace sec and tan with their appropriate values

  52. anonymous
    • 5 years ago
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    ok so tan is 2x right?

  53. amistre64
    • 5 years ago
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    correct.... or 2t...whatever you wanna call it :)

  54. anonymous
    • 5 years ago
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    and sec is adj/opp? so 1/2x

  55. anonymous
    • 5 years ago
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    ?

  56. amistre64
    • 5 years ago
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    sec = hyp/adj

  57. amistre64
    • 5 years ago
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    its cos upside down

  58. anonymous
    • 5 years ago
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    ya thats right

  59. amistre64
    • 5 years ago
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    \[1/\sqrt{4t^2 +1}\]

  60. anonymous
    • 5 years ago
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    if its hyp/adj shouldnt it just be \[\sqrt{4t^2+1}\]

  61. amistre64
    • 5 years ago
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    .... i spose if you want to be absolutley technical lol

  62. amistre64
    • 5 years ago
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    [ln(sqrt(17) + 4)] - [ln(1)]

  63. amistre64
    • 5 years ago
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    ummm....4(4) = 16 + 1 = 17 right?

  64. anonymous
    • 5 years ago
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    ya

  65. amistre64
    • 5 years ago
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    and the other side is sqrt(4(0) + 1) + 2(0) = ln(1)

  66. amistre64
    • 5 years ago
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    i hope you got an answer key, cause id love to know if were right on this :)

  67. amistre64
    • 5 years ago
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    2.09 is what I get....

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