Problem in the parametrisation of a surface

  • Thread starter Amaelle
  • Start date
  • Tags
    Surface
In summary, the orientation of the first and last loop is determined by the right hand rule, with the outer circle being traversed clockwise and the inner circle in the opposite direction. The second loop's parametrisation, (3cost;-3 sint;1/2), has a negative sign because it reverses the parameter to go in the opposite direction. This concept can be better understood by imagining the area being cut and pulled apart at the cut.
  • #1
Amaelle
310
54
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
 
Physics news on Phys.org
  • #2
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.
 
  • Love
Likes Amaelle
  • #3
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?
 
  • #4
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.
 
  • Like
Likes Amaelle
  • #5
Thanks a million! and why in the second loop we have: (3cost;-3 sint;1/2) ( the minus sign puzzles me?)
 
  • #6
##(\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?
 
  • Love
Likes Amaelle
  • #7
Amazing! thanks a million you made my day!
 

Related to Problem in the parametrisation of a surface

1. What is parametrisation of a surface?

Parametrisation of a surface is a mathematical process of representing a surface in three-dimensional space using a set of parameters. It allows for the description of a surface in terms of its coordinates, which can be used to calculate various properties and perform geometric operations.

2. Why is there a problem in the parametrisation of a surface?

The problem in the parametrisation of a surface arises when the surface cannot be represented accurately using a set of parameters. This can occur when the surface is too complex or has irregularities that cannot be captured by a simple mathematical equation.

3. How does the problem in the parametrisation of a surface affect scientific research?

The problem in the parametrisation of a surface can affect scientific research in various ways. It can limit the accuracy of calculations and lead to incorrect results. It can also make it challenging to model and analyze complex surfaces, which are common in many scientific fields.

4. What are some techniques used to address the problem in the parametrisation of a surface?

There are several techniques used to address the problem in the parametrisation of a surface, such as using higher-order polynomials, piecewise parametrisation, or approximating the surface using simpler geometric shapes. Another approach is to use computer algorithms to generate a more accurate parametrisation based on data points.

5. Can the problem in the parametrisation of a surface be completely solved?

It is unlikely that the problem in the parametrisation of a surface can be completely solved. As surfaces become more complex, it becomes increasingly challenging to find a perfect parametrisation that accurately represents all aspects of the surface. However, ongoing research and advancements in computer technology may help improve the accuracy and efficiency of parametrisation techniques.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
864
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
978
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
834
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
3K
Replies
4
Views
200
Back
Top