Green's theorem and regular region

[kingwinner]

1) A vector form of Green's theorem states that under certain conditions,

where n is the unit[/color] 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[/color], 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!

 

[HallsofIvy]

1) A vector form of Green's theorem states that under certain conditions,

where n is the unit[/color] 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?]

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.

2) A "regular region" is a compact set S in Rⁿ that is the closure of its interior. Equivalently[/color], 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!

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"

 

[kingwinner]

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...

 

[HallsofIvy]

1) That's a slightly different question than what you originally asked. Originally, you asked whether the unit vector $\vec{n}$ was necessary in
[tex]\int \vec{f}\cdot\vec{n} dS[/itex]<br /> and I responded that it obviously was necesary.<br /> <br /> But, clearly, no reasonably intelligent person is going to calculate $||\vec{N}||$, in order to find $\vec{n}= \vec{N}/||\vec{N}||$, then calculate it <b>again</b> in order to find $ds= ||\vec{N}||dt$ and watch in amazement when they cancel. Not more than once, anyway!<br /> <br /> That's why I prefer to use the notatation $d\vec{S}= \vec{N} dS$ and write<br /> $$\int \vec{f}\cdot d\vec{S}$$<br /> rather than<br /> [tex]\int \vec{f}\cdot\vec{n} dS[/itex]<br /> <br /> (Surely you understand that dS is <b>not</b> $||\vec{S}||$dt for just <b>any</b> normal? If $\vec{N}$ is the vector formed by the cross product of $d\vec{r}/du$ and $d\vec{r}/dv$ ($\vec{r}(u, v)$ is the "position vector" of a point on the surface in terms of the parameters u and v) then $dS= ||\vec{N}||du dv$.)<br /> <br /> 2) If p is a point that is <b>not</b> 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.[/tex][/tex]

 

[kingwinner]

1) I have a related question about surface integrals of vector fields.

r_u x r_v 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 r_u x r_v 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 r_u x r_v 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?

 

[HallsofIvy]

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