Undergrad How do I parametrize a conical/cylindrical surface in cylindrical unit vectors?

Click For Summary
To parametrize a cylindrical surface, the equation is given as (r cos(θ)) i + (r sin(θ)) j + t k, with 0<θ<2π and 0<t<h. For a conical surface with base radius ρ and height h, the parametrization involves z=kr where k=h/ρ, leading to the expression (tρcos(θ)/h) i + (tρsin(θ)/h) j + t k. The discussion highlights the need to convert Cartesian basis vectors into cylindrical unit vectors, emphasizing that cylindrical coordinates are dependent on the coordinate system. A transformation matrix can be utilized to express the cylindrical basis in terms of Cartesian coordinates. Understanding these relationships is crucial for accurate parametrization of surfaces in cylindrical coordinates.
Trollfaz
Messages
143
Reaction score
14
Suppose I have a cylindrical shell of radius r, height h. I can easily express the surface as
$$(r cos(\theta)) i + (r sin(\theta)) j + t k$$
$$0<\theta<2π , 0<t<h$$
For a conical surface of base rad ρ and height h,
$$z=kr -> z=k, r=ρ$$
$$k=\frac{h}{ρ}$$
Then the surface is
$$ \frac {tρcos(\theta)}{h} i + \frac {tρsin(\theta)}{h} j + t k$$
$$0<\theta<2π , 0<t<h$$
But how do I parametrize the surfaces in ## s θ z## usual cylindrical unit vectors
 
Physics news on Phys.org
How are the cylinder coordinate unit vectors related to the usual Cartesian ones?
 
They are very different as cylindrical or spherical unit vectors are coordinate dependent
 
Trollfaz said:
They are very different as cylindrical or spherical unit vectors are coordinate dependent
Well, yes. But that does not really answer the question about how they are related to the Cartesian basis vectors, ie, how can you express the cylinder base vectors in terms of the Cartesian ones?
 
So you suggest doing in Cartesian vectors first then use the matrix to convert them to cylindrical vectors
 
You already have an expression that uses the Cartesian basis. All you need to do is to reexpress it in the cylinder basis.
 
The position vector in cylindrical polars is <br /> \mathbf{x}(r,\theta,z) = r \mathbf{e}_{r}(\theta) + z\mathbf{e}_z. You already have z = kr. or conversely r = z/k.
 
Orodruin said:
You already have an expression that uses the Cartesian basis. All you need to do is to reexpress it in the cylinder basis.
So you mean use that transformation matrix
 

Similar threads

Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
828
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K