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### Parametric Sphere Formula

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Date: 7/7/96 at 20:31:40
From: Anonymous
Subject: Parametric Sphere Formula

How are circles generally described in 3D space instead of on the
plane? In 2D a parametric form is

x=rsin(theta), y=rcos(theta)

What is the equivalent in 3D? I know a sphere centered at the origin
can be written as:

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

Ideally I want a parametric form of a circle in 3 space that passes
through a particular 3 points.
```

```
Date: 7/7/96 at 23:25:30
From: Doctor Pete
Subject: Re: Parametric Sphere Formula

As you may already know, the parameterization of a surface in 3-space
requires two parameters.  We can derive such a parameterization for
the form of a circle in 3 space that passes here from the one for the
circle.  Consider

x = v cos(u)
y = v sin(u)     u = [0, 2*Pi) ,
z = 0

which describes a circle of radius v in the xy-plane.  Now, let's vary
v from 0 to some constant radius r.  We obtain a series of concentric
circles in the xy-plane.  These circles correspond to what would
happen if we sliced our desired sphere with radius r perpendicular to
the z-axis.  So visualize what this looks like in the xz-plane.  The
hemisphere above the xy-plane now looks like a semicircle or radius r
in the xz-plane.  For some v, we see that the sphere has z-coordinate
sqrt(r^2-v^2), where we take the positive square root.  It follows
that

x = v cos(u)
y = v sin(u)       u = [0, 2*Pi)
z = sqrt(r^2-v^2)  v = [0, r]

parameterizes a hemisphere above and including the xy-plane.  To get
the whole sphere, we could mirror it in the xy-plane and let the
square root take on both positive and negative values, but this is not
very elegant. Rather, consider replacing v with r cos(v).  As this new
v goes from 0 to Pi/2, r cos(v) goes from r to 0.  So let this v go
from -Pi/2 to Pi/2, and z = sqrt(r^2-r^2 cos^2(v)) = r sin(v).  Hence

x = r cos(v) cos(u)
y = r cos(v) sin(u)     u = [0, 2*Pi)
z = r sin(v)            v = [-Pi/2, Pi/2]

parameterizes the whole sphere of radius r in 3-space (verify this by
substituting this into the Cartesian formula and simplifying).

The point here is that there generally exists more than one
parameterization for a surface, just as in the 1-parameter case.

As for your second question, this is not so simple.  If I wanted to
find a parameterization of a circle that passes through three
arbitrary (non-collinear) points in 3-space, I would first translate
the coordinate system so that one of the points is at the origin, then
apply a few rotations in 3-space to get all 3 points in the xy-plane,
and finally, calculate the parameterization that gives the circle
passing through these three points.  Then one would apply the inverse
transformations on this parameterization to obtain the desired result.

To obtain these transformations, it might help to find the equation of
the plane that passes through the three points.  This is done by
taking the general equation ax+by+cz = 1, plugging in the coordinates,
obtaining a system of three equations in the unknowns a,b,c, and
solving for these. Needless to say, this is a task much better suited
for a computer.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Higher-Dimensional Geometry

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