Parametrization for the ellipsoids Can someone help to describe some possible parametrizations for the ellipsoid:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1?$$
I am thinking polar coordinates, but there may be the concept of steographic projecting (not sure how to apply it here), and not sure how many ways I can provide such possible parametrizations for the ellipsoid.
Thanks
 A: I think in this way:
$x = a\sin(\varphi)\cos(\theta),\quad y = b\sin(\varphi)\sin(\theta),\quad z = c\cos(\varphi)$
A: I know this is old but wanted to offer an answer.
If you want to use spherical coordinates, you need to parameterize $r$ as a function of $\theta$ and $\phi$.
Start with
$x = r\cos(\phi)\sin(\theta)$
$y = r\sin(\phi)\sin(\theta)$
$z = r\cos(\theta)$  
Plug into the equation for an ellipsoid and get
$$r = \frac{1}{\sqrt{\left ((\cos(\phi)/a)^2 + (\sin(\phi)/b)^2)\sin(\theta)^2 + (\cos(\theta)/c)^2 \right )}}$$
Given an angle pair $(\theta, \phi)$ the above equation will give you the distance from the center of the ellipsoid to a point on the ellipsoid corresponding to $(\theta, \phi)$.  This may be a little more work than some of the other parameterizations but has its uses.
A: You mention polar coordinates, which allow you to parameterise a sphere.
Let $X = \frac{x}{a}$, $Y = \frac{y}{b}$, and $Z = \frac{z}{c}$. Then the equation becomes
$$X^2 + Y^2 + Z^2 = 1.$$
This is the equation of a sphere of radius $1$, so you can parameterise it using polar coordinates. Once you have done that, use the fact that $x = aX$, $y = bY$, and $z = cZ$ to obtain a parameterisation of the ellipsoid.
A: The answer to your question will depend on what you want to do with the
ellipsoid.  The Wikipedia page on geodesics on ellipsoids gives
three possible parametrizations of the surface: (1) geographic latitude
and longitude (useful if you're determining your position by
astronomical observations); (2) parametric coordinates, probably the
simplest to deal with computationally; (3) ellipsoidal coordinates, nice
in that they are lines of curvature and hence orthogonal.
