Problem in the parametrisation of a surface

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Homework Help Overview

The discussion revolves around the parametrization of a surface, specifically focusing on the orientation of loops and the implications of negative signs in the parametrization equations. Participants are examining the geometric interpretation of these concepts in relation to a cone and its loops.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the geometric reasoning behind the orientation of the loops, questioning why the inner loop's direction differs from the outer loop. There is also curiosity about the negative sign in the parametrization of the second loop.

Discussion Status

The discussion is active, with participants providing insights into the geometric reasoning and the right-hand rule. Some guidance has been offered regarding the traversal direction of the loops, and there is an ongoing exploration of the implications of reversing parameters in the context of the parametrization.

Contextual Notes

Participants are navigating the complexities of surface parametrization and its geometric interpretations, with specific attention to the assumptions made about the orientation and traversal directions of the loops.

Amaelle
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Homework Statement
Let Σ = [(x, y, z) ∈ R3 : z=-1/4+√(x^2+y^2) 0 <= z<=1/2]
be a surface oriented
so that the normal versor of Σ forms an obtuse angle with the fundamental versor of the z–axis.
Compute the flux of the curl of the vector field
F(x, y, z) = 2yz *exponontiel(x^2+z^2),3z*exponontiel(x^2+y^2),(0,5x+z)exponontiel(x^2+y^2)
Relevant Equations
parametrisation of surface
1598622105655.png
Good day my question is the following:
How did they determine the the orientation of the first and last loop?
and why in the parametrisation of the second loop we have: (3cost;-3 sint;1/2) ( the negative sign puzzles me) thanks

this is the solution of the book

1598622210202.png
 
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It's a bit tricky to see geometrically. But imagine that cone was much shallower, almost flat. In fact, suppose in the extreme it was flat, and you are looking down on two concentric circles, with the normal pointing away from you (downward), By the right hand rule, the outer circle would be traversed clockwise. Traverse the inner circle you must go the opposite direction. See the attached picture where we imagine a 0 width cut:
circuit.jpg
The integrals along the cut cancel so are ignored. The rest circulates around the area by the right hand rule. Remember the normal vector is pointing away from you.
 
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thanks a lot I understand now why the outer circle points clockwise (n outward) by wht the inner circle points differently shouldn't be the same?
 
Imagine the area being actually cut and pulled slightly apart at the cut. Start to go around it. There's no way to walk around it without going the opposite direction on the inner circle. Just follow the arrows.
 
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Thanks a million! and why in the second loop we have: (3cost;-3 sint;1/2) ( the minus sign puzzles me?)
 
##(\cos t, \sin t )## traverses the unit circle in the xy plane counterclockwise. If you want to go the other way, reverse the parameter, ##t \rightarrow -t##. What does that do to the expression?
 
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Amazing! thanks a million you made my day!
 

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