Understanding Stokes Theorem: Solving Boundary Curve Dilemmas in Vector Calculus

In summary, the conversation discusses Stokes' theorem and the concept of boundaries in relation to solid objects. It is explained that the boundary of a boundary is zero and that the surface surrounding a solid object is considered a boundary. The example of a solid hemisphere and its boundary is used to illustrate this concept. Lastly, it is clarified that a hemisphere's shell is not considered a boundary on its own, but rather the combination of the hemisphere shell and the base of the hemisphere form its boundary.
  • #1
swraman
167
0

Homework Statement


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 [tex]\int\int_{S}Curl(\stackrel{\rightarrow}{F})\bullet ds = 0[/tex] "

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.


Homework Equations



[tex]\int\int_{S}Curl(\stackrel{\rightarrow}{F})\bullet ds = \int_{C}\stackrel{\rightarrow}{F}\bullet d\vec{r}[/tex]

Thanks anyone willing to look/help :)
 
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  • #2
the boundary of a boundary is zero

swraman said:
"If S has no boundry (eg. if S is the boundry of a solid region) then [tex]\int\int_{S}Curl(\stackrel{\rightarrow}{F})\bullet ds = 0[/tex] "

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?

Hi swraman! :smile:

(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. :wink:
 
  • #3


tiny-tim said:
Hi swraman! :smile:

(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. :wink:

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

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 :smile:
 
  • #5
so...if the surface can be expressed as a single function it is a boundry?
 
  • #6
swraman said:
so...if the surface can be expressed as a single function it is a boundry?

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. :smile:
 
  • #7
doesnt the hemisphere shell surround a solid region?
 
  • #8
swraman said:
doesnt the hemisphere shell surround a solid region?

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. :smile:
 

FAQ: Understanding Stokes Theorem: Solving Boundary Curve Dilemmas in Vector Calculus

1. What is Stokes' Theorem Boundary?

Stokes' Theorem Boundary, also known as Stokes' Theorem, is a mathematical formula that relates the surface integral of a vector field to the line integral of its curl along the boundary of the surface. It is named after the mathematician George Gabriel Stokes.

2. What is the significance of Stokes' Theorem Boundary?

Stokes' Theorem Boundary is significant because it allows us to evaluate a surface integral by instead computing a line integral over the boundary of the surface. This is often easier to calculate and can be applied in a variety of mathematical and scientific fields, such as physics, engineering, and fluid dynamics.

3. How is Stokes' Theorem Boundary used in physics?

In physics, Stokes' Theorem Boundary is used to calculate the circulation of a vector field around a closed surface. This is important in understanding the behavior of fluids and electromagnetic fields. It is also used in the study of fluid dynamics, where it helps determine the flow of fluids around objects.

4. What are the conditions for Stokes' Theorem Boundary to be applicable?

The conditions for Stokes' Theorem Boundary to be applicable include having a smooth surface with a well-defined boundary, a continuously differentiable vector field, and a closed boundary curve. The surface must also be oriented in a consistent manner, and the boundary curve must be traversed in a counterclockwise direction.

5. Can Stokes' Theorem Boundary be applied to non-orientable surfaces?

No, Stokes' Theorem Boundary can only be applied to orientable surfaces, meaning surfaces that have two distinct sides. Non-orientable surfaces, such as a Möbius strip, do not have a consistent orientation and therefore cannot be used with Stokes' Theorem.

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