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Sphere Radius = 1 centered at origin

  1. Nov 4, 2008 #1
    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

    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

    A sphere of radius = r in Spherical Coordinates
    fr(r,Ɵ,Ø)*r + fƟ(r,Ɵ,Ø)*Ɵ + fØ(r,Ɵ,Ø)*Ø
  2. jcsd
  3. Nov 4, 2008 #2


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

    The vector equation would be
    [itex](r cos(\theta) sin(\phi)\vec{i}+ r sin(\theta) sin(\phi)\vec{j}+ r cos(\phi)\vec{k}[/itex]
  4. Nov 4, 2008 #3
    If you have an equation:

    Br*r + BƟ*Ɵ

    BƟ = μ0/(4*pi)*2*m*cosƟ/r^3

    How do you get the magnitude of B in x-y-z reference frame?
  5. Nov 5, 2008 #4


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    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 [itex](\mu_0 r^3)(8m\pi cos(\theta))[/itex]? Or do you mean the product of two fractions: [itex](\mu_0/(4\pi))(2m cos(\theta))/r^3)[/itex]?

    Assuming it is the latter,
    [tex]\frac{\mu_0}{4\pi}\frac{2 m cos(\theta)}{r^3}= \frac{\mu_0}{2\pi}\frac{m rcos(\theta)}{r^4}[/tex]
    [tex]= \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}[/tex]
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