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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 $\vec{r}(r,\theta)$

0000005
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  • Must you parametrize into cylindrical coordinates or is any parametrization acceptable? – Eric Towers Oct 23 '19 at 17:01
  • @EricTowers any – 0000005 Oct 23 '19 at 17:04
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    $(x,y)\mapsto (x,y,\frac{-4-x^2-2y^2}{3})$. If you need cylindrical coordinates then compose with the parametrization $(r,\theta)\mapsto (r\cos(\theta),r\sin(\theta))$ to get $(r,\theta)\mapsto (r\cos(\theta),r\sin(\theta), \frac{-4-r^2\cos^2(\theta)-2r^2\sin^2(\theta)}{3})$. – conditionalMethod Oct 23 '19 at 17:09
  • For a fixed value of $z$, you have an equation of an axis-aligned ellipse. Do you know any parameterizations for that? – amd Oct 23 '19 at 19:37

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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 $$(r,\theta)\mapsto \left(\sqrt{2}r\cos\theta, r\sin \theta, \frac{-2r^2-4}{3}\right).$$

Ivo Terek
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