Prove that normals for surface of revolution $z=f(\sqrt{x^2+y^2})$ cross axis of revolution

Question

$z=f(\sqrt{x^2+y^2})$, $f'\not = 0$
My attempt: considering we have $F(x,y,z)=f(\sqrt{x^2+y^2}) - z$
And using normal equation: $$\frac{x-x_0}{\frac{\partial F}{\partial x}}=\frac{y-y_0}{\frac{\partial F}{\partial y}}=\frac{z-z_0}{\frac{\partial F}{\partial z}}$$ $\frac{\partial F}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}\cdot F'$
$\frac{\partial F}{\partial x} = \frac{y}{\sqrt{x^2+y^2}}\cdot F'$
$\frac{\partial F}{\partial z} = -1$
I got $\overline{n} = \{\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}},-1\}$ or $\overline{n} = \{x, y,-\sqrt{x^2+y^2}\}$ and I got stuck 'cause I don't know how to connect that with axis of revolution

Answer 1

The normal is not correctly stated. There are several errors. Correcting them,

$u=\sqrt{x^2+y^2}$

$\dfrac{\partial F}{\partial x}=\dfrac{\partial}{\partial x}(f(\sqrt{x^2+y^2})-z)=\dfrac{\partial}{\partial x}f(\sqrt{x^2+y^2})=\dfrac{df(u)}{du}\dfrac{\partial u}{\partial x}=\dfrac{x}{\sqrt{x^2+y^2}}\cdot f'(u)$

$\dfrac{\partial F}{\partial y}=\dfrac{\partial}{\partial y}(f(\sqrt{x^2+y^2})-z)=\dfrac{\partial}{\partial y}f(\sqrt{x^2+y^2})= \dfrac{df(u)}{du}\dfrac{\partial u}{\partial y}=\dfrac{y}{\sqrt{x^2+y^2}}\cdot f'(u)$

$\overline{n} = \{xf', yf',-\sqrt{x^2+y^2}\}$

Now, construct the line passing by the point in the surface $P(x,y,f(\sqrt{x^2+y^2}))$ and having as direction vector $\overline n$, the normal vector at this point $P$.

l:$\begin{cases} \tilde x=x+txf'\\ \tilde y=y+tyf'\\ \tilde z=f(\sqrt{x^2+y^2})-t\sqrt{x^2+y^2} \end{cases}$

The axis of revolution is the $z$ axis because $f$ is constant for points with constant distance to this axis ($d((x,y,z),(0,0,z))=\sqrt{x^2+y^2}$). Clearly, with the hypothesis of $f'\neq0$, $l$ passes thru the $z$ axis as $\tilde x=\tilde y=0$ for $t=-1/f'$