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Conservative vector fields

  1. Mar 18, 2008 #1
    Q: Let F= (-y/(x2+y2), x/(x2+y2), z) be a vector field and let U be the interior of the torus obtained by rotating the circle (x-2)2 + z2 = 1, y=0 about the z-axis.
    a) Show that curl F=0 but
    F . dx = 2pi where dx=(dx,dy,dz)
    and C (contained in U) is the circle x2+y2=4, z=0. Therefore F is NOT conservative.
    b) Determine the possible values of
    F . dx on a path in U.

    I am OK with part a. But part b is tricky!!

    Solution to part b:
    f(x,y,z)=arctan(y/x) + z2/2 is a function such that grad f = F
    => f(x,y,z)=arctan(rsin[tex]\theta[/tex]/rcos[tex]\theta[/tex]) (since z=0)
    => f=[tex]\theta[/tex]

    F . dx = f([tex]\theta[/tex]=pi/2) - f([tex]\theta[/tex]=0) = pi/2 by F.T.C. for line integrals

    Final Answer:
    F . dx = pi/2 + (2n)pi

    Now, I don't understand the parts in red.
    Why is z=0? This is NOT true for everywhere in U
    Why + (2n)pi ? What if the path is not a circle? This is possible because the torus has some "width" and "height"

    Can someone please explain the tough part (part b)? I would really appreciate it!
    Last edited: Mar 18, 2008
  2. jcsd
  3. Mar 18, 2008 #2


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    If the path is NOT a circle, then it can be changed to one: You can always replace a part of the path with a circle: draw lines from the center of the torus to two points on your path. draw a circular arc with center the center of the torus and ends on the two lines. Since the closed path- circular arc to one line to your path to other line to circular arc- does not enclose the center of the torus, and the field is conservative, the integral over that path is 0. Imagine doing that around the entire path, with the straight lines infintesmally close together so their "contribution" to the integral cancels. The integral around the path and the opposite way on the circle totals 0 so the two integrals cancel. The integral around the path and the integral on the circle, in the same direction, are equal.

    The reason for the 2n[itex]\pi[/itex] is that the arctan is an multi-valued function. tan(0)= tan(2[itex]pi[/itex]) etc.
  4. Mar 20, 2008 #3
    But they have assumed that z=0 in order to get f=[tex]\theta[/tex] where grad f= F.
    But clearly, in U it's not the case the everywhere z=0, there are many places in U in which z is nonzero. Why can they assume it's always the case that z=0?

    Also, from part a, we get that the line integral along a closed curve in U is nonzero, and thus we've concluded that F is NOT conservative, so the path taken from (2,0,0) to (0,2,0) matters, and there are infinitely many paths possible, why can't
    ∫ F . dx = pi/2 + 1 for some really complicated path from (2,0,0) to (0,2,0)?
    The above is on the xy-plane. The pink and yellow paths are different paths from (2,0,0) to (0,2,0), how can they give the same value pi/2 for the line integral while the vector field F is NOT conservative? Shouldn't they have different values?

    Also, I believe that arctan is defined to be a one-to-one function in calculus, so it can't be multi-valued...

    Thanks for explaining!!
  5. Mar 20, 2008 #4
    i believe they assume z=0 because it is given that the path integral is done with end points at z=0. Another point to note is that, the path contains the singular line x=y=0. This is the reason why the path integral have non-zero values. No specific value of z is singular, so the value of z along the path has no effect on the resulting integral value, since the path cannot contain extra singularity during integration.

    sorry i cannot be more specific since i dont remember the idea well...
  6. Mar 21, 2008 #5
    But in the path somewhere in between the starting and ending points, z may not be zero. It can go "up" and "down".

    But the line x=y=0, i.e. z-axis isn't even in the set U.

    I am still finding myself unable to understand the solution part b...totally confused :(
  7. Mar 24, 2008 #6
    Still wondering......
  8. Mar 24, 2008 #7


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    The given differential is conservative (I would say "exact") as long as you do NOT go around the z-axis. Any integration over a closed path that does NOT go around the z axis. My point before was that you can always deform parts of the path to go in a circle, with z= 0, say, without changing the integral.
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