# Converting Spherical Equations to Cylindrical and Rectangular

1. Oct 20, 2004

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!

2. Oct 20, 2004

### cepheid

Staff Emeritus
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

3. Oct 21, 2004

### HallsofIvy

Staff Emeritus
You know x= &rho;cos(&theta;)sin(&phi), y= &rho;sin(&theta;)sin(&phi;), z= &rho;cos(&phi;) 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.