Fundamental vector product calculation

Question

Given the surface $S\in \mathbb{R}^3$; $y=x^2+z^2$ compute the fundamental outward normal. I seem to get two different answers depending on my parametrisation. Using $$\widetilde{r}_1(x,z)=\begin{pmatrix} x \\ x^2+z^2 \\ z \end{pmatrix}$$ We get the normal as $$\textbf{N}(x,z)=\begin{pmatrix} -f_x \\ 1 \\ -f_z \end{pmatrix}=\begin{pmatrix} -2x \\ 1 \\ -2z \end{pmatrix}$$ However using the parametrisation $$\widetilde{r}_2(r,\theta)=\begin{pmatrix} r\cos{\theta} \\ r^2 \\ r\sin{\theta} \end{pmatrix}$$ I get the normal as $$\begin{pmatrix} \cos{\theta} \\ 2r \\ \sin{\theta} \end{pmatrix}\times \begin{pmatrix} -r\sin{\theta} \\ 0 \\ r\cos{\theta} \end{pmatrix}= \begin{pmatrix} 2r^2\cos{\theta}\\ -r \\ 2r^2\sin{\theta} \end{pmatrix}$$ These clearly don't coincide. I seem to be factor of $r$ out for some reason. Are they both correct? Or have I made a mistake?

Answer 1

Since $r = \sqrt{x^2 + z^2}$ is distance from the $y$-axis, you are just off (in magnitude) by a scalar multiple of $r$. If you multiply the normal in your second parametrization by $\frac{1}{r}$, the two normals will agree in magnitude. Additionally note that the two normals are not pointing in the same direction. The normal from your first parametrization is pointing in the positive $y$-direction, whereas the normal from your second parametrization is pointing in the negative $y$ direction.

In general, different parametrizations will give rise to different normal vectors that point in different directions depending on whether the parametrization preserves or reverses the natural orienation on the parameter space. Assuming your surface is oriented (which is always the case locally), then to make the normal vectors identical, you need to specify an orientation (in this case "outwards") and "normalize" them to a common length.