1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    Your work looks good.

    ehild
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...