Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know how to change 2D cartesian equations into polar equations, however I'm having some difficulty with a 3D equation. I am trying to take the cartesian equation x^2+(.75y+4)^2+(z+3)^2=20 and turn it into a polar equation. Is that possible with z in the mix? I know that x=rcosθ and y=rsinθ, but is there also one for z? Because r=4 in polar 3D plots a sphere with radius 4, which is the same as y^2+x^2+z^2=16. So I figure it must work somehow. Does anyone know how I could plot the equation x^2+(.75y+4)^2+(z+3)^2=20 in polar?

share|cite|improve this question
what you want are either spherical coordinates or cylindrical coordinates. Both of them are generalizations of the polar coords for 3 dimension. – example Apr 3 '12 at 20:35
up vote 1 down vote accepted

There's no "polar coordinates" as such in 3 dimensions. You have two standard choices: cylindrical coordinates (where $z$ is left as-is), or spherical coordinates where you have two angles $\theta$ and $\phi$ and one radial coordinate $\rho$. Neither of those will be very nice for your ellipsoid. If you're interested in plotting, you might use the parametrization $$ \eqalign{x &= \sqrt{20} \cos(u) \sin(v)\cr y &= (4/3) \sqrt{20} \sin(u) \sin(v) - 16/3\cr z &= \sqrt{20} \cos(v) - 3\cr & 0 \le u \le 2 \pi,\ 0 \le v \le \pi \cr}$$

enter image description here

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.