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Green's theorem and regular region

  1. Mar 1, 2008 #1
    1) A vector form of Green's theorem states that under certain conditions,
    [​IMG]
    where n is the unit outward normal to the curve C and D is the region enclosed by C

    [Now, my question is: must n be a unit vector? Why or why not?]


    2) A "regular region" is a compact set S in Rn that is the closure of its interior. Equivalently, a compact set S in Rn is a regular region if every neighborhood of every point on the boundary of S contains points in the interior of S

    [I don't understand at all why these are equivalent. Can somebody please explain?]

    Thanks a lot!
     
    Last edited: Mar 1, 2008
  2. jcsd
  3. Mar 1, 2008 #2

    HallsofIvy

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    That's a strange question! Do you think the length of the vectors is not important in this integral? Of course the normal must be of unit length.

    Obviously the only difference is that one says "is the closure of its interior" and the other says "every neighborhood of every point on the boundary of S contains points in the interior of S".

    A consists precisely of its interior and its boundary. If there were points on the boundary of S having a neighborhood that did NOT "contain points in the interior of S", then that point would not be on the boundary of the interior of S and so would not be in the closure of the interior of S. Conversely, it there were points of S that were not in the closure of the interior of S, then that point could not have every neighbothood containing points of the interior.
    Under what conditions would a set, A, NOT be "the closure of its interior"? Certainly the closure of the interior of A must be at least a subset of A. Only if A has "isolated points"
     
  4. Mar 1, 2008 #3
    Thanks!

    1) Let N be any normal
    Let n be unit normal
    n=N/||N||
    dS=||N||dt
    The ||N|| just cancels, right? So why does the length matters?


    2) "Conversely, it there were points of S that were not in the closure of the interior of S, then that point could not have every neighbothood containing points of the interior."
    I get your first part, but I am still having trouble understanding this...
     
  5. Mar 2, 2008 #4

    HallsofIvy

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    1) That's a slightly different question than what you originally asked. Originally, you asked whether the unit vector [itex]\vec{n}[/itex] was necessary in
    [tex]\int \vec{f}\cdot\vec{n} dS[/itex]
    and I responded that it obviously was necesary.

    But, clearly, no reasonably intelligent person is going to calculate [itex]||\vec{N}||[/itex], in order to find [itex]\vec{n}= \vec{N}/||\vec{N}||[/itex], then calculate it again in order to find [itex]ds= ||\vec{N}||dt[/itex] and watch in amazement when they cancel. Not more than once, anyway!

    That's why I prefer to use the notatation [itex]d\vec{S}= \vec{N} dS[/itex] and write
    [tex]\int \vec{f}\cdot d\vec{S}[/tex]
    rather than
    [tex]\int \vec{f}\cdot\vec{n} dS[/itex]

    (Surely you understand that dS is not [itex]||\vec{S}||[/itex]dt for just any normal? If [itex]\vec{N}[/itex] is the vector formed by the cross product of [itex]d\vec{r}/du[/itex] and [itex]d\vec{r}/dv[/itex] ([itex]\vec{r}(u, v)[/itex] is the "position vector" of a point on the surface in terms of the parameters u and v) then [itex]dS= ||\vec{N}||du dv[/itex].)

    2) If p is a point that is not in "the closure of the interior of S", then, since closure of a set includes point in the set and boundary points of the set, p cannot be a boundary point of the interior of S. That, in turn, means that there is some neighborhood of p that contains no interior points of S.
     
  6. Mar 2, 2008 #5
    1) I have a related question about surface integrals of vector fields.
    [​IMG]

    ru x rv is simply a (not necessarily unit) normal to the surface

    The norms ||...|| always cancel, so can I directly jump to the last line and just substitute ANY normal N (in the right direction of course, but not unit length) for ru x rv on the last line?

    For example,
    z = f(x,y) is a surface
    => g(x,y,z) = z - f(x,y)=0
    => grad g = N = (a not necessarily unit) normal vector
    Can I just simply substitute this grad g for ru x rv directly on the last line?

    So to evaluate surface integrals of vector fields, all we need is a normal in the right direction, and not necessarily a unit normal, right?
     
  7. Mar 3, 2008 #6

    HallsofIvy

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    I think this is now the third time you have asked that question! YES, if you form
    [itex]d\vec{S}[/itex] as [itex](\vec{r}_u\times \vec{r}_v)dudv[/itex] then [itex]\vec{f}\cdot\vec{n}dS= \vec{f}\cdot d\vec{S}[/itex] and you do not need to calculate the normal vector separately.
     
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