Converting Spherical Equations to Cylindrical and Rectangular

[DeadxBunny]

Question:

(Note: p=rho and o=phi)
Convert p(1-2cos^2(o))=-psin^2(o) into cylindrical and rectangular coordinates and describe or sketch the surface.

The part that I don't know how to do is converting the spherical equation into cylindrical or rectangular coordinates. I know all the equations like x=psin(o)cos(theta) and y=psin(o)sin(theta) but I don't see how I can manipulate the given equation so that I could use those equations. Any help would be greatly appreciated!

 

[cepheid]

Don't try manipulating them directly. Look at the geometry.Yeah...I'm working w/ spherical coord systems as well a lot in Electromagnetism right now.

Think about what it means for something to be radially outward...sweeps out a sphere...every point equidistant (\rho) no matter the direction:

\rho = \sqrt{x^2 + y^2 + z^2}

For some reason our convention in physics class is opposite what we did in math (our thetas and phis are reversed from yours. No matter...I'll convert)

See the projection of \rho onto the xy plane? It represents a line "radially" outward in that plane i.e.:

\rho\sin\phi = \sqrt{x^2 + y^2}

This projection into the plane forms a right triangle with z, the hypotenuese of which is \rho.

The angle between z and \rho is just \phi, so from the geometry of the right triangle:

\phi = \tan^{-1}\left(\frac{\sqrt{x^2 + y^2 }}{z}\right)

It shouldn't be too hard to see that the azimuthal angle (\theta in your case) is given by:

\theta = \tan^{-1}\left(\frac{y}{x}\right)

After all that, cylindrical coords should be easy

 

[HallsofIvy]

You know x= ρcos(θ)sin(&phi), y= ρsin(θ)sin(φ), z= ρcos(φ) but you need to know them the other way:


\rho= \sqrt{x^2+ y^2+ z^2}
\theta= arctan(\frac{y}{x})
\phi= arccos(\frac{z}{\sqrt{x^2+y^2+z^2})

Replace each occurance in your equation by the corresponding formula.