# How do I change a 3D cartesian equation into a polar equation?

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?

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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

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}