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

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SUMMARY

This discussion focuses on parametrizing conical and cylindrical surfaces using cylindrical unit vectors. The cylindrical shell is expressed as (r cos(θ)) i + (r sin(θ)) j + t k for 0<θ<2π and 0. For a conical surface with base radius ρ and height h, the parametrization is given by (tρcos(θ)/h) i + (tρsin(θ)/h) j + t k. The relationship between cylindrical and Cartesian unit vectors is clarified, emphasizing the need to express cylindrical coordinates in terms of Cartesian coordinates using transformation matrices.

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  • Cylindrical coordinate systems
  • Parametrization of surfaces
  • Transformation matrices
  • Vector calculus
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This discussion is beneficial for mathematicians, physicists, and engineers who work with geometric modeling and need to understand the relationship between different coordinate systems, particularly in the context of cylindrical and conical surfaces.

Trollfaz
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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
 
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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
 

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