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Line integral with vector function on circular path.

  1. Jul 21, 2011 #1
    I'm not getting the answer from the back of the book for some reason. Is the book wrong or am I wrong?
    1. The problem statement, all variables and given/known data
    calculate [itex]\int[/itex]f · dr for the given vector field f(x, y) and curve C:

    f(x, y) = (x^2 + y^2) i; C : x = 2 + cos t, y = sin t, 0 ≤ t ≤ 2π


    2. Relevant equations
    itex]\int[/itex]f · dr
    r(t) = <x(t), y(t)>
    dr = r'(t) dt

    3. The attempt at a solution

    [itex]\int[/itex][itex]_{C}[/itex][itex]\widehat{F}[/itex]*dr

    F = <x^2+y^2, 0>
    C: x=2+cost, y = sint, 0<=t<=2pi

    r(t) = <2+cost, sint>
    r'(t) = <-sint, cost>

    F(r(t)) * r'(t) = <(2+cost)^2 + sin^2(t), 0> * <-sint, cost> =
    -sint(4 + 4cos(t) + cos^2(t) + sin^2(t)) = -sint(5+4cost)

    Letting u = 5+ 4cost
    du/4 = -sintdt
    [itex]\int[/itex] -sint(5+4cost)dt= 1/4 [itex]\int[/itex] udu = 1/8 (5+4cost)^2 |t = 0..2pi =
    81/8 -81/8 = 0

    But the answer in the back of the book says 4pi, what did I do wrong?

    Since the path taken is a closed curve, I also tried green's theorem to verify if it's right or not, but I got -2 2/3 using greens theorem. Green's theorem does apply in this case, does it not?
     
    Last edited: Jul 21, 2011
  2. jcsd
  3. Jul 21, 2011 #2
    Very strange problem to me. :\ Maybe someone here might know what I did wrong.
     
  4. Jul 21, 2011 #3

    ehild

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    What are x2 and y2?

    ehild
     
  5. Jul 21, 2011 #4
    Oops, sorry it is supposed to be F(x,y) = <(x^2 + y^2), 0>. Fixed it. XD
     
  6. Jul 21, 2011 #5

    ehild

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    Your work looks good.

    ehild
     
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