Undergraduate

To this end, we examine again the surface of revolution ( u , v ) = ( f ( u ) cos v , f

To this end, we examine again the surface of revolution
(u, v)=(f(u)cos v, f(u)sin v, g(u))
obtained by rotating the unit-speed curve u
(f(u),0,g(u)) in the xz-plane around the z-axis. We found in Example 8.1.4 that its Gaussian curvature is
f f
K =.(8.6)
Suppose first that K =0 everywhere. Then, Eq.8.6 gives f =0, so f(u)= au + b for some constants a and b.Since f 2+g2=1, we get g =1 a2
(so we must have |a|1) and hence g(u)=1 a2u+c,where c is another
constant. By applying a translation along the z-axis we can assume that c =0, and by applying a rotation by about the x-axis, if necessary, we can assume that the sign is +. This gives the ruled surface
(u, v)=(b cos v, b sin v,0)+ u(a cos v, a sin v,1 a2).
If a =0 this is a circular cylinder; if |a|=1 it is the xy-plane; and if 0<|a|<1 it is a circular cone (to see this, put u = au + b). Now suppose that K>0,say K =1/R2,where R>0 is a constant. Then,
Eq.8.6 becomes f + R2=0, f
which has the general solution f(u)= a cos
u R + b
,
where a and b are constants. We can assume that b =0 by performing a reparametrization u = u + Rb,v = v. Then, up to a change of sign and adding a constant,
g(u)=1
a2 R2 sin2 u
R, g(u)= R sin u R du. The integral in the formula for g(u) can be evaluated in terms of elementary
functions only when a =0 or R.The case a =0 does not give a surface, and if a = R then f(u)= Rcos u
R, and we have a sphere of radius
R (the case a =R can be reduced to this by rotating the surface by around the z-axis). Suppose finally that K<0. We can restrict ourselves to the case K =1,
as the general case can be obtained from this by applying a dilation of R3(see Exercise 8.1.5). In view of the preceding case, we can think of a surface with K =1 as a sphere of imaginary radius
1, or a pseudosphere.

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