petertheta said:
I'm lost. could use parameterization to check something like the volume or surface area to confirm it was a sphere?!?
There are many volumes that are enclosed by surfaces that have the same volume and surface area as a given sphere, so that would not identify a surface as a sphere.
The usual way that we identify a collection of points as a sphere is that every point is the same distance from some central point.
In 3-dimensional Euclidean space, the distance between two arbitrary points is defined by the Pythagorean theorem applied to the Cartesian coordinates of the points. That is the distance d between the two points with Cartesian coordinates (x, y, z) and (x', y', z') is given by the equation d^2 = (x - x')^2 + (y - y')^2 + (z - z')^2.
Treating (x, y, z) as three variables now, if a sphere is the collection of points that are the same distance from a central point (x', y', z'), then every point on the sphere must satisfy the equation above for a constant distance d, which we now call the radius of the sphere. Therefore, if we can rearrange an equation to look exactly like the one above, where x, y, and z are variables and d, x', y' and z' are constants, those points (x, y, z) must lie on a sphere of radius d around the point (x', y', z').
For example, the set of points (x, y, z) that satisfy the equation x^2 + y^2 - 2y + z^2 + 4z = 2 lie on the surface of a sphere of radius \sqrt{7} around the point (0, 1, -2). The reason we know this is because we can rearrange this equation algebraically to see that the same points must satisfy the equation (x - 0)^2 + (y - 1)^2 + (z - (-2))^2 = (\sqrt{7})^2. We compare this to the standard equation of the sphere above to find out the details.