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Calc 3: Strokes Therom Orientatio Question

  1. Dec 17, 2006 #1
    1. The problem statement, all variables and given/known data

    [tex] F(x,y,z) = (x^2y^3z)i +(sin(xyz))j +(x^2yz)k [/tex]
    S is the part of the cone [tex] y^2=x^2+z^2 [/tex] that lies between the planes y = 0 and y = 3, oriented in the direction of the postive y-axis

    My question deals with orientation (see below sec 3)

    2. Relevant equations
    [tex] \int_c F dr = \int\int_S F dS [/tex]

    3. The attempt at a solution

    Alright so the boundary curve C is the circle [tex] x^2 + y^2 = 9 [/tex] [tex] y=3[/tex]

    My question is why does r(t) become
    [tex] r(t) = 3sin(t) i+ 3j + 3cos(t)k [/tex]
    instead of
    [tex] r(t) = 3cos(t) i+ 3j + 3sin(t)k [/tex]

    I am assuming it has something to do with the positive orientation towards postive y axis

    When I did the problem I got the exact negitive of what the answer should be

    i got [tex]\frac{-2187}{4}\pi[/tex] instead of [tex]\frac{2187}{4}\pi[/tex]

    so I was also going to ask if I mess up the cos and sin like i did on this one should I always get the opposite sign of the correct answer or was this just random?

    Also should all Strokes theorms problems be positive answers?
  2. jcsd
  3. Dec 18, 2006 #2


    User Avatar
    Homework Helper

    If you swap sin and cos in your parameterization of the curve C, you will get the negative answer (provided there aren't any other curves to be integrated along) since this effectively reverses the orientation of the curve from C to -C (C traversed backward), this gives a negative answer because

    [tex]\int_{-C}\vec{F}\cdot d\vec{r} = -\int_{C}\vec{F}\cdot d\vec{r}[/tex].

    No, absolutely not.
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