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Need help in Stoke's Theorem

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data

    [tex]\oint_C{(x^2 + 2y + sin x^2)dx + (x + y + cos y^2)dy}[/tex]

    the contour C formed by 3 curves:
    [tex]C(x,y) = \begin{cases}x=0, \quad from (0,0) to (0,5)\\y = 5 - x^2,\quad from(0,5) to (2,1) \\ 4y = x^2, \quad from(2,1) to (0,0)\end{cases}[/tex]

    and the Stoke Theorem:
    [tex]\oint_C \vec F \, d\vec r = \iint_S \, curl \, \vec F \, d\vec S \\= \iint_S \, curl \, \vec F \cdot \vec n \, dS [/tex]

    3. The attempt at a solution

    So in this problem, can I say the vector field, F as follow?
    [tex]\vec F = <x^2 + 2y + sin x^2 ,\,\, x + y + cos y^2, \,\, 0>[/tex]
    then
    [tex]\oint_C \vec F \, d\vec r = \iint_S \, curl \, \vec F \, d\vec S \\ = \int_0^5 \int_\frac{x^2}{4}^{5-x^2} -1\, dy\, dx \\ = 27.0833[/tex]

    but probably there is something wrong in there, because the answer is 20/3
     
    Last edited: Jun 16, 2015
  2. jcsd
  3. Jun 16, 2015 #2

    RUber

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    Your x only ranges from 0 to 2, not 0 to 5.
     
  4. Jun 16, 2015 #3
    Oh yeah, what a silly mistake, thanks.
     
  5. Jun 16, 2015 #4

    Ray Vickson

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    Your notation is ambiguous and confusing: does ##sin x^2## mean ##\sin(x^2)## or ##\sin^2 x = (\sin x)^2##?
     
  6. Jun 22, 2015 #5
    it is [itex]sin(x^2)[/itex], i will make it clear next time
     
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