Good morning,
I am posting this again because I am having trouble to understand this post (How do I check if the normal vector is pointing inside or outside?).
Here is my question :
I have the parametrization $$\sigma(r,\theta) = \left\langle r \cos \theta, r\sin \theta, \frac{r^2}{3} \right\rangle$$
I computed the cross product between $\sigma_r$ and $\sigma_\theta$ and got : $(-\frac{2}{3}r^2 \cos \theta,-\frac{2}{3}r^2 \sin \theta, r)$ with $r$ ranging from $[0, \sqrt(3)]$ and $\theta$ from $0$ to $2 \pi$.\
I really do not understand how to get that the normal is pointing inwards or outwards, I can visualize that the normal should be pointing outward by my cross product I am unsure where it points. Thank you very much for your help !
Just checking: $$ \sigma_r\times\sigma_\theta=\begin{pmatrix}\cos\theta\\\sin\theta\\\frac{2r}{3}\end{pmatrix}\times\begin{pmatrix}-r\sin\theta\\r\cos\theta\\0\end{pmatrix}=\begin{pmatrix}-\frac{2r^2}{3}\cos\theta\\-\frac{2r^2}{3}\sin\theta\\r\end{pmatrix}\,. $$ There is no way of telling what we mean by "inward" or "outward". The solid that is bounded by this surface could be above or below it. Also: who told you that the normal is $\sigma_r\times\sigma_\theta$ and not $\sigma_\theta\times\sigma_r\,?$ All we can say is that $\sigma_r\times\sigma_\theta$ points into the $z$-direction (obviously) since its $z$-component $r$ is nonnegative.