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anonymous

  • 5 years ago

Does anyone know how to do this one? Evaluate the line integral ∫ F dr C about the simple closed curve C, oriented in the positive direction, where C is the circle x2 + y2 = 16 and F = (-4y)(sin2 x) i + (2x + sin 2x) j

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  1. anonymous
    • 5 years ago
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    hold on i was just finished with line integrals =D

  2. anonymous
    • 5 years ago
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    use Green's theorem...

  3. anonymous
    • 5 years ago
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    ...wait, if and when im done with this one, can you help me w/ a parametrization?

  4. anonymous
    • 5 years ago
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    I can try

  5. anonymous
    • 5 years ago
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    so green's theorem says: doubleintegral(dQ/dx-dP/dy)dA

  6. anonymous
    • 5 years ago
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    so, dQ/dx=4x*cos(2x)+2sin(2x) dP/dy=-4sin(2x)

  7. anonymous
    • 5 years ago
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    doubleintegral(4x*cos(2x)-2sin(2x))dA

  8. anonymous
    • 5 years ago
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    ahh, wait you need to use a "power reducer"

  9. anonymous
    • 5 years ago
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    hold on...

  10. anonymous
    • 5 years ago
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    cos(2x)=1-2sin^2(x) sin(2x)=2sin(x)cos(x) ...ugly

  11. anonymous
    • 5 years ago
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    can someone help me with my problem

  12. anonymous
    • 5 years ago
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    doubleintegral(4x-8x*sin^2(x)-4sin(x)cos(x))dA

  13. anonymous
    • 5 years ago
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    you with me?

  14. anonymous
    • 5 years ago
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    yeah i got you. But shouldnt we paramatize the x and y values

  15. anonymous
    • 5 years ago
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    wait, that's not going to work... hold on

  16. anonymous
    • 5 years ago
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    i guess you know more about this than i do, we didn't parametrize anything when we were doing these. yes, i figure you can do that

  17. anonymous
    • 5 years ago
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    I was thinking that b/c this was a circle, we'll do a change of variables into polar coordinates

  18. anonymous
    • 5 years ago
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    so..x=4cos(t) y=4sin(t) ?

  19. anonymous
    • 5 years ago
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    yeah, but right now, as far as I am aware, you can't integrate cos(cos(t))

  20. anonymous
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    then t would be between 0 and 2pi..

  21. anonymous
    • 5 years ago
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    we have to work it out somehow so that the sines and cosines can disappear

  22. anonymous
    • 5 years ago
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    then we can do the change of variables

  23. anonymous
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    so right now we are at doubleintegral(4x*cos(2x)-2sin(2x))dA

  24. anonymous
    • 5 years ago
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    oh wait I see.. using greens theorem M or P=-4(sin^2(x)) N or Q=2x+sin2x then, we derive with respect to x on N and derive with respect to y on M

  25. anonymous
    • 5 years ago
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    that's what we did

  26. anonymous
    • 5 years ago
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    the problem is the double integral

  27. anonymous
    • 5 years ago
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    seems that i made a mistake when i did dQ/dx it's supposed to be 2cos(2x)+2

  28. anonymous
    • 5 years ago
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    what would be the bounds of integration then?

  29. anonymous
    • 5 years ago
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    right now, we haven't "parametrized" in polar coordinates, the integral would be from 0 to 2pi, 0 to 4

  30. anonymous
    • 5 years ago
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    ok..i see

  31. anonymous
    • 5 years ago
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    it's actually doubleintegral(4x*cos(2x)+2sin(2x)+2)dA

  32. anonymous
    • 5 years ago
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    i mean... doubleintegral(2cos(2x)+4sin(2x)+2)dA oops, can't keep track

  33. anonymous
    • 5 years ago
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    know of any property that can reduce this so that the trig disappears?

  34. anonymous
    • 5 years ago
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    hmm..well is there an identity for cos(2x) or sin(2x) ?

  35. anonymous
    • 5 years ago
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    yes, though i don't think it's going to help.

  36. anonymous
    • 5 years ago
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    you might try a classic integration, with y from -sqrt(4-x^2) to sqrt(4-x^2) and x from -4 to 4

  37. anonymous
    • 5 years ago
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    yes.

  38. anonymous
    • 5 years ago
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    in that case you need a calculator to help you with arcsin and what not

  39. anonymous
    • 5 years ago
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    im doing that now, but the calc's going terribly slowly

  40. anonymous
    • 5 years ago
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    this is not going to work, is it?

  41. anonymous
    • 5 years ago
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    i dont know I got a 32sqrt(4-x^2) using wolfram mathmatics

  42. anonymous
    • 5 years ago
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    see, now if we did the problem without using green's theorem, it won't work anyway, since F(r(t)) gets us another cosine within a cosine.

  43. anonymous
    • 5 years ago
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    remember, it's a double integral, x should be eliminated

  44. anonymous
    • 5 years ago
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    wait a minute, that sin2x you wrote, it that a "sine squared x" or a "sine double x"???

  45. anonymous
    • 5 years ago
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    sin squared

  46. anonymous
    • 5 years ago
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    darn, you caused me soooo much hell.

  47. anonymous
    • 5 years ago
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    this is much easier than i thought

  48. anonymous
    • 5 years ago
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    I mean no harm. I come in peace.

  49. anonymous
    • 5 years ago
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    jk =D

  50. anonymous
    • 5 years ago
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    can you rewrite the question a bit neater so i know what exactly im integrating?

  51. anonymous
    • 5 years ago
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    "sine squared x" should be (sin(x))^2

  52. anonymous
    • 5 years ago
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    ok..(-4y)(sin(x))^2 i + (2x + sin 2x) j

  53. anonymous
    • 5 years ago
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    hmmm...

  54. anonymous
    • 5 years ago
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    and 2x + sin2x i presume is "sine double x"?

  55. anonymous
    • 5 years ago
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    yes.

  56. anonymous
    • 5 years ago
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    much better

  57. anonymous
    • 5 years ago
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    lets start over

  58. anonymous
    • 5 years ago
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    dQ/dx=2cos(2x)+2 dP/dy=-4(sin(x))^2

  59. anonymous
    • 5 years ago
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    follow?

  60. anonymous
    • 5 years ago
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    did i get the derivative wrong?

  61. anonymous
    • 5 years ago
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    hold on

  62. anonymous
    • 5 years ago
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    perfect works out very nicely

  63. anonymous
    • 5 years ago
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    so doubleintegral(dQ/dx-dP/dy)dA is doubleintegral(2cos(2x)+2+4(sin(x))^2)dA

  64. anonymous
    • 5 years ago
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    ok

  65. anonymous
    • 5 years ago
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    using cos(2x)=2(cos(x))^2-1 we get doubleintegral(2(2cos(x))^2+2+4(sin(x))^2)dA

  66. anonymous
    • 5 years ago
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    which is doubleintegral(4(cos(x))^2-2+2+4(sin(x))^2)dA which is doubleintegral(4(cos(x))^2+4(sin(x))^2)dA

  67. anonymous
    • 5 years ago
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    follow?

  68. anonymous
    • 5 years ago
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    yesm

  69. anonymous
    • 5 years ago
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    ok, oops, typo in doubleintegral(2(2cos(x))^2+2+4(sin(x))^2)dA: should be doubleintegral(2(2cos(x)-1)^2+2+4(sin(x))^2)dA

  70. anonymous
    • 5 years ago
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    arrgh, i mean doubleintegral(2((2cos(x))^2-1)+2+4(sin(x))^2)dA

  71. anonymous
    • 5 years ago
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    ok..

  72. anonymous
    • 5 years ago
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    you understand, though? didn't mean to have all those typos in the way.

  73. anonymous
    • 5 years ago
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    that s what got us into this mess ;)

  74. anonymous
    • 5 years ago
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    ok so now we have doubleintegral(4(cos(x))^2+4(sin(x))^2)dA easy cake: (cos(x))^2+(sin(x))^2=1, so, it is just doubleintegral(4)dA

  75. anonymous
    • 5 years ago
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    now, you can do a change of variables!

  76. anonymous
    • 5 years ago
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    doubleintegral[theta,0,2pi][r,0,4](4r)dr dtheta evaluate that, and you're done

  77. anonymous
    • 5 years ago
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    do you get 32pi when you integrate r from 0 to 4 and theta from 0 to to pi

  78. anonymous
    • 5 years ago
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    64pi

  79. anonymous
    • 5 years ago
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    im pretty sure

  80. anonymous
    • 5 years ago
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    well, no r integrated is 16 pi, the full solution i worked out to be 64 pi

  81. anonymous
    • 5 years ago
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    you were supposed to integrate r from 0 to 4 and theta from 0 to 2pi, not pi.

  82. anonymous
    • 5 years ago
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    thats what i meant 2pi i still get 32pi

  83. anonymous
    • 5 years ago
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    integrate r, which is r^2/2. 4^2/2-0=8 integrate 1, which is theta 2pi-0=2pi multiply you get 16 pi

  84. anonymous
    • 5 years ago
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    oooooh....ok

  85. anonymous
    • 5 years ago
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    you are oh so smart

  86. anonymous
    • 5 years ago
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    nahhh, it was nothing i got problems of my own.

  87. anonymous
    • 5 years ago
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    i'll post my question (for the 4th time) =D

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