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?