Finding positive orientation for a surface

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Homework Help Overview

The discussion revolves around determining the positive orientation of a surface defined by a parameterization using a vector function. Participants explore the implications of the normal vector derived from the cross product of the parameterization's partial derivatives.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of finding the normal vector through the cross product of the parameterization's derivatives and question how to establish which direction constitutes a positive orientation. There is also a consideration of the arbitrary nature of orientation and the need for additional context to define "positive" versus "negative" orientation.

Discussion Status

The discussion is ongoing, with participants raising questions about the definitions and implications of surface orientation. Some have provided insights into the arbitrary nature of orientation, while others seek clarification on specific examples and the relationship between parameterization and orientation.

Contextual Notes

There is mention of needing specific wording or conditions to determine orientation, such as references to positive components or outward normals in closed surfaces. Participants also express uncertainty regarding the correct approach to visualizing the orientation towards the origin based on their parameterization.

Miike012
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If I parameterized a surface by the vector function r(a,b) I would then proceed to find the normal to the surface by crossing ra and rb. But how would I determine which normal raXrb or rbXra has the positive orientation?

Im assuming you would first need to know if the concave side of the surface is facing away or towards the origin? But from there I don't know.
 
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I don't see how it could be related to second order derivatives. Consider e.g. just moving around in the XY plane. Let's say the cross product is defined with a right-hand corkscrew rule, so (1,0,0)x(0,1,0) = (0,0,1). If increasing a moves r in the +ve x-direction and increasing b moves r in the +ve y-direction then raxrb is in the +ve z direction. But if you swap the sign of ra it will be in the negative z-direction.
 
The distinction between "positive" and "negative" orientation for a surface is purely arbitrary. A surface can have either of two "orientations", the normal vector pointing in either of two directions out of the surface. But to designate one of those as the "positive orientation" and the other as "negative orientation" requires some other information.

Could you give us the exact wording of an example. Usually such "surface orientation" probles will say "oriented by positive z component" or in the case of a closed surface oriented by "outward" or "inward" normals.
 
HallsofIvy said:
The distinction between "positive" and "negative" orientation for a surface is purely arbitrary. A surface can have either of two "orientations", the normal vector pointing in either of two directions out of the surface. But to designate one of those as the "positive orientation" and the other as "negative orientation" requires some other information.

Could you give us the exact wording of an example. Usually such "surface orientation" probles will say "oriented by positive z component" or in the case of a closed surface oriented by "outward" or "inward" normals.


I parameterized the surface
x = x
y = x^2+ z^2
z=z

Wrong picture look at the picture in post # 5
Now to find the orientation towards the origin do I do rxXrz or rzXrx?
 

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Correct problem
 

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