# Stokes Theorem Boundry

1. Dec 19, 2008

### swraman

1. The problem statement, all variables and given/known data
This is a question about stokes theorem in general, not about a specific problem.

Directly from lecture:
"If S has no boundry (eg. if S is the boundry of a solid region) then $$\int\int_{S}Curl(\stackrel{\rightarrow}{F})\bullet ds = 0$$ "

because apparently "no boundry C exists"

However in the next example he gives S is a hemisphere or radius 1 above the XY plane; and he uses the projection of the hemesphere on the XY plane as the bondry curve C and it works out fine.
Isnt the hemisphere a boundry of a solid region? So shoudn't the integral be zero? Why can you use the boundry of the projection of S on the XY plane as the curve C?

This is hard to explain in words, if someone is so kind as to lok into this, you can see exactly what I am seeing here:
http://webcast.berkeley.edu/course_details_new.php?seriesid=2008-D-54472&semesterid=2008-D
Lecture 42, 4:00-5:00 is where he says the integral is zero, then at 9:40 he does the example with the hemesphere and it is not zero.

2. Relevant equations

$$\int\int_{S}Curl(\stackrel{\rightarrow}{F})\bullet ds = \int_{C}\stackrel{\rightarrow}{F}\bullet d\vec{r}$$

Thanks anyone willing to look/help :)

2. Dec 19, 2008

### tiny-tim

the boundary of a boundary is zero

Hi swraman!

(I haven't watched the webcast lecture you linked to but:)

The boundary of a boundary is zero (empty).

The boundary of a solid hemisphere is a hemipherical shell plus the "base".

In your example, the surface S is only the hemipherical shell.

So S is not the boundary of a solid object, and S's boundary is therefore not zero.

3. Dec 19, 2008

### swraman

Re: the boundary of a boundary is zero

Thanks for hte reply.

Im still a little confused; so is the shell of a full sphere considered to be a "boundry"? But a hemisphere's shell isnt?

4. Dec 19, 2008

### tiny-tim

The surface surrounding any solid object is a boundary …

a spherical shell does surround a sphere, so it is the boundary of the sphere …

but a hemi-spherical shell does not surround a hemi-sphere: the hemi-sphere is surrounded by the hemi-spherical shell and the base

5. Dec 19, 2008

### swraman

so...if the surface can be expressed as a single function it is a boundry?

6. Dec 19, 2008

### tiny-tim

Nooo … if the surface divides space into two, ie if it surrounds a solid volume, then it is a boundary.

This is even if it's in two parts (i assume that's what you would call "two functions"), like a hemispherical shell and a disc.

7. Dec 19, 2008

### swraman

doesnt the hemisphere shell surround a solid region?

8. Dec 20, 2008

### tiny-tim

i'm not sure we're talking about the same thing …

a solid hemisphere is surrounded by a curved piece and a flat piece …

i'm calling the curved piece the "hemisphere shell", and the flat piece the "base" or the "disc" …

so the hemisphere shell on its own is not the boundary of a solid region, and so its boundary C is not the boundary of a boundary.