I want to parametrize the surface of a given Two-Sheeted Hyperboloid expression:$$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1 $$I have tried with the parametrization that Wolfram|Mathworld give for the two-sheeted hyperboloid (with hiperbolic functions) but I would like to use the simplest way to do it. Maybe I just need a correct parametrization for the ellipse on the xy plane.
[Math] Simplest way of parameterize the surface of a two-sheeted hyperboloid
algebra-precalculuscalculusmultivariable-calculus
Related Solutions
We consider at first equalities. There are two surfaces of revolution $(a=\sqrt{18},b=4 )$
$$ y^2+z^2-x^2=a^2 $$
$$ y^2+z^2=b^2 $$
which are hyperboloid of one sheet and a circular cylinder respectively.
Intersection of surfaces is the boundary of region for further purposes of integration etc.
Eliminate $y$ we get for $z$
$$ z = \sqrt{ 2x^2+a^2-b^2 } \tag1 $$
Circle of radius $b$ in projection is parameterized as $ (x,y)= b(\cos \theta, \sin \theta) $ and for $z$ we obtain by plugging in from above (1)
$$ z=\pm \sqrt{ 2 b^2 \cos^2 \theta +(a^2-b^2)}$$
The connected regions and the surfaces that created them along with (white) lines of intersection are plotted in Mathematica below .
Note that when two conicoids intersect we can generally have
Null real intersection ( disjunct surfaces)
One real closed loop intersection
Two real intersections
depending on distance of centers of central conicoids. In each case we can refer to the patches so created arbitrarily connecting them as internal or external to intersection contour made or defined by the equality. Each sign defines its own boundary of demarcated intersection, which enables determining regions where inequalities apply.
To find demarcation and connection a sketch is often helpful in guiding the region continuity of the patch in reference.
So the parameterizations are same but for $\pm$ sign where each sign represents a region.
The situation is no different if you consider intersection of a big cylinder and a smaller cylinder regions so created!
You want to rotate the surface so that the direction $(1,0,0)$ becomes $\dfrac1{\sqrt3}(1,1,1)$. A possible rotation matrix is
$$\begin{pmatrix}\frac1{\sqrt3}&0&-\frac2{\sqrt6} \\\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6} \\\frac1{\sqrt3}&-\frac1{\sqrt2}&\frac1{\sqrt6}\end{pmatrix}$$
Now your equation is
$$\frac{(u+v+w)^2}3-\frac{(v-w)^2}2-\frac{(-2u+v+w)^2}6=1$$
or
$$u^2+v^2+w^2-4uv-4vw-4wu+3=0.$$
Caution: sign and permutations errors are guaranteed, I didn't check. But the final result must not be far from correct.
https://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%2Bz%5E2-4xy-4yz-4zx%2B3%3D0
Best Answer
We can combine the identities $\cos^2\phi+\sin^2\phi=1$ and $\sec^2\theta-\tan^2\theta=1$ to get:
$$x=a\tan \theta\cdot\cos \phi$$ $$y=b\tan \theta\cdot\sin \phi$$ $$z=c\sec \theta $$
This directly gives $$\frac{z^2}{c^2}-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=1$$
There are, of course, many other ways. You may consider this way simplest, compared to using hyperbolic functions.
I believe the parameter intervals are:
$$0\le\theta<\frac{\pi}2,\qquad \frac{\pi}2<\theta\le\pi$$ $$0\le\phi<2\pi$$
The two intervals for $\theta$ correspond to the two sheets.
And here is a parametrization for the ellipse on the $xy$ plane:
$$x=a\cos\phi$$ $$y=b\sin\phi$$
For $0\le\phi<2\pi$, this gives the Cartesian equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$