Surfaces in Spherical Coordinates

Since x = rsin( f) cos( q) , y = rsin( f) sin( q) , and z = rcos( f), the position vector of a point in space is

r( f,q) = á rsin( f) cos( q) ,rsin( f) sin( q) ,rcos( f) ñ = r er(f,q)

If we define e_r(f,q) = á sin( f) cos(q) , sin(f) sin(q), cos(f) ñ, then can be written more compactly as

r( f,q) =  r er(f,q)

Thus, graph of a function r = f( f,q) forms a surface in R³ which is parameterized by

r( f,q) = f( f,q) á sin( f) cos( q) ,sin( f) sin( q) ,cos( f) ñ

(4)

or equivalently, r( f,q) = f( f,q) e_r(f,q). 

In particular, if we substitute x = rsin( f) cos( q) , y = rsin( f) sin( q) , and z = rcos( f) into the equation of a level surface and solve for ρ, then (4) parameterizes at least a section of that level surface.      

EXAMPLE 5    What is the spherical parameterization of the hyperboloid in two sheets

z2-x2-y2 = 1

Solution: Substituting from (2) and simplifying yields

r2 cos2( f) -r2sin2( f)cos2( q) -r2sin2( f) sin2( q)  = 1 r2cos2( f) -r2sin2( f)[ cos2( q) +sin2( q)]  = 1 r2cos2( f) -r2sin2( f)  = 1

However, cos²( f) -sin²( f) = cos( 2f) , so that r²cos( 2f) = 1 and

r2 = sec( 2f)

(5)

Thus, the upper sheet of the hyperboloid is parameterized by r(f, q) = [ sec( 2f) ]^1/2 u, which yields

r( f,q) =  sec( 2f)    á sin( f) cos( q) ,sin( f) sin( q) ,cos( f) ñ

       

If 0 £ a £ p and if f( f) ³ 0 on [ 0,a] does not depend on q, then the surface r = f( f) on [ 0,a] is said to be radially symmetric and has a parameterization of

r( f,q) = f( f) er( f,q)

 If f( f) is positive and continuous on [ 0,p] , then r = f( f) is called a radially symmetric deformation of the sphere.

The intersection of the surface r = f( f) with the xz-plane results in the curve parameterized by

r( f,0) = á f( f) sin( f) ,0,f( f) cos( f) ñ , f in [ 0,a]

The surface itself is obtained by revolving this curve about the z-axis.        

EXAMPLE 6    Discuss the graph of the radially-symmetric surface

r =  5 + 0.1fsin(7f) ,  f in [ 0,p]

Solution:  The parameterization is given by

r( f,q) = ( 5 + 0.1fsin( 7f) ) er( f,q)

which by definition of e_r( f,q) leads to

x = ( 5+0.1fsin( 7f) ) sin( f) cos( q) ,    y = ( 5+0.1fsin(7f) ) sin( f) sin( q)

z = ( 5+0.1fsin( 7f) ) cos( f)

The surface is the revolution of the curve

r( f,0) = á ( 5+0.1fsin(7f) ) sin( f) ,0,( 5+0.1fsin( 7f) ) cos( f) ñ

about the z-axis. The curve is shown below:
The resulting surface of revolution then follows:

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