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 following cartesian equation and turn it into a polar equation. Is that possible with $z$ in the mix?
$$ x^2+(0.75\,y+4)^2+(z+3)^2=20 $$
I know that $x=r\,\cosθ$ and $y=r\,\sinθ$, 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 $x^2+(0.75\,y+4)^2+(z+3)^2=20$ in polar?
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 $$ \eqalign{x &= \sqrt{20} \cos(u) \sin(v)\cr y &= (4/3) \sqrt{20} \sin(u) \sin(v) - 16/3\cr z &= \sqrt{20} \cos(v) - 3\cr & 0 \le u \le 2 \pi,\ 0 \le v \le \pi \cr}$$