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

  • #1
Trollfaz
135
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
 
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  • #2
How are the cylinder coordinate unit vectors related to the usual Cartesian ones?
 
  • #3
They are very different as cylindrical or spherical unit vectors are coordinate dependent
 
  • #4
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?
 
  • #5
So you suggest doing in Cartesian vectors first then use the matrix to convert them to cylindrical vectors
 
  • #6
You already have an expression that uses the Cartesian basis. All you need to do is to reexpress it in the cylinder basis.
 
  • #7
The position vector in cylindrical polars is [tex]
\mathbf{x}(r,\theta,z) = r \mathbf{e}_{r}(\theta) + z\mathbf{e}_z.[/tex] You already have [itex]z = kr[/itex]. or conversely [itex]r = z/k[/itex].
 
  • #8
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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