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Stoke's Theorem

  1. Aug 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Let S be the portion of the paraboloid ##z = 4 - x^2 - y^2 ## that lies above the plane ##z = 0## and let ##\vec F = < z-y, x+z, -e^{ xyz }cos y >##. Use Stoke's Theorem to find the surface integral ##\iint_S (\nabla × \vec F) ⋅ \vec n \,dS##.
    2. Relevant equations
    ##\iint_S (\nabla × \vec F) ⋅ \, d\vec S = \int_C \vec F ⋅\,d\vec r ##

    3. The attempt at a solution
    Parameterize C as ##\vec r(t) = 2<\cos t, \sin t, 0> , 0≤t≤2π ##

    ##\int_C \vec F ⋅\,d\vec r = \int_0^{2π} \vec F (\vec r(t))⋅\vec r'(t) \,dt##

    ## \vec r' (t) = 2<\sin t, -\cos t, 0> ##

    ## \vec F (\vec r (t)) = <-\sin t , \cos t, -\cos (\sin t)> ##

    ## 4\int_0^{2π} <-\sin t , \cos t, -\cos (\sin t)>⋅<\sin t, -\cos t, 0> \, dt = 4\int_0^{2π} (-\sin ^2 t -\cos ^2 t) \, dt = -4 \int_0^{2π} \, dt = -8π ##

    Just looking for someone to verify this was done correctly, it's been almost a year since I last took calculus and I don't remember if there's any way to check my work.
     
  2. jcsd
  3. Aug 26, 2016 #2

    vela

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    Close. You made a sign error when calculating ##\vec{r}'##, and your expression for ##\vec{F}(\vec{r}(t))## is missing factors of 2 here and there.
     
  4. Aug 26, 2016 #3
    that's embarrassing, thank you.
    ## \vec r' (t) = <-2\sin t, 2\cos t, 0> ##

    ## \vec F (\vec r (t)) = <-2\sin t , 2\cos t, -\cos (2\sin t)> ##

    ## \int_0^{2π} <-2\sin t , 2\cos t, -\cos (2\sin t)>⋅<-2\sin t, 2\cos t, 0> \, dt = \int_0^{2π} (4\sin ^2 t +4\cos ^2 t) \, dt = 4 \int_0^{2π} \, dt = 8π ##
     
  5. Aug 26, 2016 #4

    vela

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    Another way you can check your result is to do the surface integral over a simpler surface with the same boundary. In this case, try integrating over the circle of radius 2 with its center at the origin and lies in the xy plane. Since the normal points in the z direction, you only need to calculate the z-component of the curl.
     
  6. Aug 26, 2016 #5
    ## \iint_S (∇×\vec F)⋅\vec n \,dS = \iint_S <P, Q,\frac ∂ {∂x} (x+z) - \frac ∂ {∂y} (z-y) >⋅<0, 0, 1> \, dS = \iint_S 2\,dS = 8π ##

    That makes a lot of sense, thanks for the tip.
     
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