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Stokes theorem

  1. Aug 9, 2005 #1
    I'm trying to solve this problem:

    Compute [tex]\oint_c(y+z)dx + (z-x)dy + (x-y)dz [/tex] using Stoke's theorem, where c is the ellipse [tex]x(t) = asin^2t, \ y(t) = 2asintcost, z(t) = acos^2t, 0\leq t \leq \pi [/tex]

    The version of stoke's theorem I learned is:
    \int_c \overrightarrow{F} \cdot d\overrightarrow{r}
    = \int_s curl \overrightarrow{F} \cdot d\overrightarrow{S}
    =\iint_s curl \overrightarrow{F}\cdot \overrightarrow{n} \cdot dS

    where S is the elliptical surface bounded by the curve c, F is a vector field and n is the unit vector pointing out at that point.

    In this case, [tex]F = <y+z, z-x, x-y>[/tex], and I calculated curl F to be [tex]<-2, 0, -2>[/tex].

    So we have to find

    [tex]\iint_s <-2, 0, -2> \cdot \overrightarrow{n} \cdot dS [/tex]

    How would I find [tex]\overrightarrow {n} [/tex] and dS, and also the bounds of integration for the double integral?
  2. jcsd
  3. Aug 9, 2005 #2


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    Science Advisor

    Fortunately, since the curl is a constant vector, you don't have to do any integration! Just find [tex]curl \overrightarrow{F}\cdot \overrightarrow{n}[/tex] which will be a constant, of course, and multiply by the area of the ellipse.

    I don't know that this is the easiest way but here's what I did off the top of my head: When t= 0, a point on the ellipse is (0, 0, a). When t= [tex]\frac{\pi}{4}[/tex], a point on the ellipse is (a/2, a, a/2). When t= [tex]\frac{\pi}{2}[/tex], a point on the ellipse is (a, 0, 0). Since the interior of the ellipse is a plane, the vectors from one of those points to the other 2 lie in that plane and so perpendicular to any normal to the plane: Take the cross product of the two vectors to find a perpendicular (and then, of course, divide by its length to get a unit perpendicular).
    Take the dot product of that unit perpendicular with the curl you have already calculated and "integrate" that over the ellipse. Since, as I said before, the integrand is a constant, that is just the constant times the area of the ellipse. It should be easy to see, by looking at the points for t= 0, t= [tex]\frac{\pi}{4}[/tex], t= [tex]\frac{\pi}{2}[/tex] , and t= [tex]\frac{3\pi}{4}[/tex] that the ellipse has semi-axes of length a and a/2. You can calculate the area of the ellipse from that.
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