How to parametrize this surface $x^2 +2y^2+3z +4 = 0$?

Question

How do I parametrize this paraboloid S: $x^2 +2y^2+3z +4 =0$? I first isolated the z-component to make it a function of $f(x,y)$, howevere that gets me $z= (-x^2-2y^2-4)/3$ which leaves me stuck on how to parametrize this into $\vec{r}(r,\theta)$

Answer 1

You can work with a graph parametrization, solving for $z$. Namely, the parametrization $$(x,y) \mapsto \left(x,y, \frac{-x^2-2y^2-4}{3}\right)$$does the job. This might not be very useful, since $x^2$ and $2y^2$ together beg for some sort of modified polar coordinates. Then you can use $$(r,\theta)\mapsto \left(\sqrt{2}r\cos\theta, r\sin \theta, \frac{-2r^2-4}{3}\right).$$