Proving a form ##z=f(r)## to be a surface of revolution

[toforfiltum]

Homework Statement

Suppose that a surface has an equation in cylindrical coordinates of the form ##z=f(r)##. Explain why it must be a surface of revolution.

Homework Equations

The Attempt at a Solution

I consider ##z=f(r)## in terms of spherical coordinates.

## p cosφ = f \sqrt{(p sinφcosθ)^2 + (p sinφsinθ)^2} ##

## p cosφ= f\sqrt{(p sinφ)^2} ##

## p cosφ=f(p sinφ)##

##cosφ= f (sinφ)##

##∴φ= \cos^{-1} f(sinφ)##

Since equation is independent of ##\theta##, it describes a surface of revolution about the ##z## axis.

Is my prove right or acceptable?

 

[Ray Vickson]

Homework Statement

Suppose that a surface has an equation in cylindrical coordinates of the form ##z=f(r)##. Explain why it must be a surface of revolution.

Homework Equations

The Attempt at a Solution

I consider ##z=f(r)## in terms of spherical coordinates.

## p cosφ = f \sqrt{(p sinφcosθ)^2 + (p sinφsinθ)^2} ##

## p cosφ= f\sqrt{(p sinφ)^2} ##

## p cosφ=f(p sinφ)##

##cosφ= f (sinφ)##

##∴φ= \cos^{-1} f(sinφ)##

Since equation is independent of ##\theta##, it describes a surface of revolution about the ##z## axis.

Is my prove right or acceptable?

You are using spherical, not cylindreical coordinates. Also: in LaTeX, put a "\" before sin, cos, etc. Without it, the results are ugly and hard to read, like ##sin \phi cos \theta##; with it, they look good, as in ##\sin \phi \cos \theta##.

 

[toforfiltum]

You are using spherical, not cylindreical coordinates. Also: in LaTeX, put a "\" before sin, cos, etc. Without it, the results are ugly and hard to read, like ##sin \phi cos \theta##; with it, they look good, as in ##\sin \phi \cos \theta##.

Okay! Thanks for the advice! Just started using it, so sorry for the ugly text. I don't have any idea of starting the proof using cylindrical coordinates, that's why I converted it to spherical ones. But I will give it a try now.

The cylindrical coordinates are in the form ##(r,\theta, z)##. I assume that ##r \geq 0##
So ##r= \sqrt{(x^2 + y^2)}##

and ##z=f(\sqrt{(x^2 + y^2)})##, which gives a unique value. These gives a set of points that form a line in ##3D## space.

Since equation is independent of ##\theta##, line is the same for any value of ##\theta##. These similar set of lines form a surface of revolution.

Is it right in any way at all? I'm guessing here.

 

[LCKurtz]

Homework Statement

Suppose that a surface has an equation in cylindrical coordinates of the form ##z=f(r)##. Explain why it must be a surface of revolution.

Homework Equations

The Attempt at a Solution

I consider ##z=f(r)## in terms of spherical coordinates.

Why introduce spherical coordinates, which have nothing to do with this question?
If you have a surface of revolution revolved about the ##z## axis, that would mean ##z(r,\theta_1) = z(r,\theta_2)## for any ##\theta_1## and ##\theta_2## wouldn't it? Is that true in your case?

 

[toforfiltum]

Why introduce spherical coordinates, which have nothing to do with this question?
If you have a surface of revolution revolved about the ##z## axis, that would mean ##z(r,\theta_1) = z(r,\theta_2)## for any ##\theta_1## and ##\theta_2## wouldn't it? Is that true in your case?

Yes. So, is what I'm saying above right? But I'm not sure if having a unique set of points will form a line, though.
It depends on ##f##, right?

 

[toforfiltum]

Oh, I think I see now why ##z=f(r)## represents a surface of revolution. Using the explanation on ##zr## planes given by @LCKurtz , each value of ##r## gives a value of ##z##, and the set of values of ##r## gives its respective values of ##z##. Since equation is independent of ##\theta##, these set of points are the same for all values of ##\theta##, and this is the reason why it forms a surface of revolution.

Am I right?