# Sphere Radius = 1 centered at origin

1. Nov 4, 2008

### Philosophaie

r^2 = x^2 + y^2 + z^2

I would like to know what would be the equations be for:

A sphere of radius = r in Rectangular Coordinates
fx(x,y,x)*x +fy(x,y,x)*y + fz(x,y,x)*z
fx(x,y,x)=?
fy(x,y,x)=?
fz(x,y,x)=?

A sphere of radius = r in Rectangular Coordinates with spherical members
fx(r,Ɵ,Ø)*x + fy(r,Ɵ,Ø)*y + fz(r,Ɵ,Ø)*z
fx(r,Ɵ,Ø)=?
fy(r,Ɵ,Ø)=?
fz(r,Ɵ,Ø)=?

A sphere of radius = r in Spherical Coordinates
fr(r,Ɵ,Ø)*r + fƟ(r,Ɵ,Ø)*Ɵ + fØ(r,Ɵ,Ø)*Ø
fr(r,Ɵ,Ø)=?
fƟ(r,Ɵ,Ø)=?
fØ(r,Ɵ,Ø)=?

2. Nov 4, 2008

### HallsofIvy

Staff Emeritus
The parametric equations for a point on a sphere of radius r are just the spherical coordinates with variable $\rho$ replaced by the constant r:
$x= r cos(\theta) sin(\phi)$
$y= r sin(\theta) sin(\phi)$
$z= r cos(\phi)$

The vector equation would be
$(r cos(\theta) sin(\phi)\vec{i}+ r sin(\theta) sin(\phi)\vec{j}+ r cos(\phi)\vec{k}$

3. Nov 4, 2008

### Philosophaie

If you have an equation:

Br*r + BƟ*Ɵ

Br=μ0/(4*pi)*2*m*cosƟ/r^3
BƟ = μ0/(4*pi)*2*m*cosƟ/r^3

How do you get the magnitude of B in x-y-z reference frame?

4. Nov 5, 2008

### HallsofIvy

Staff Emeritus
It's impossible to understant what you have written. First you don't have an equation. Is that first expression supposed to be equal to the position vector of a point?

Second what does " / /" mean? Are those two divisions? If so isn't the first one just $(\mu_0 r^3)(8m\pi cos(\theta))$? Or do you mean the product of two fractions: $(\mu_0/(4\pi))(2m cos(\theta))/r^3)$?

Assuming it is the latter,
$$\frac{\mu_0}{4\pi}\frac{2 m cos(\theta)}{r^3}= \frac{\mu_0}{2\pi}\frac{m rcos(\theta)}{r^4}$$
$$= \frac{\mu_0}{2\pi}\frac{m r cos(\theta)}{(r^2)^2}= \frac{\mu_0}{2\pi}\frac{m x}{(x^2+ y^2)^2}$$