Surfaces of revolution and special examples

Try

Rotate the profile r(s)=sin s, z(s)=cos s, 0<s<π, around the z-axis. It produces the unit sphere away from its poles. The profile has unit speed, so its revolution metric is ds²+sin²s dθ².

A surface of revolution reduces many calculations to a planar profile, but a convenient parameter must still be identified. If the profile is not parametrized by arc length, the coefficient of ds² is not automatically one.

Predict

Predict which coordinate degenerates when r=0. The angular derivative vanishes, so this revolution chart loses rank. The resulting surface could be smooth, as at a sphere pole, or singular, as at a cone vertex.

Predict whether every parallel circle is a geodesic. The metric and connection will later show that only parallels at stationary profile radius have that property, except in special cases.

Experiment

For a unit-speed profile (r(s),z(s)) with r>0, compare exact coefficients.

Quantity	Formula
Metric	ds²+r²dθ²
Chosen unit normal	(−z′cos θ,−z′sin θ,r′)
Meridian curvature	r′z″−z′r″
Parallel curvature	z′/r

Insert the sphere profile: both principal values are −1 with this outward normal. Insert a vertical line profile r=R, z=s: the chosen normal points inward and the values are zero and 1/R. The sign difference is an orientation choice, not a contradiction.

Understand

Differentiate X(s,θ)=(r cos θ,r sin θ,z). The coordinate vectors have norms one and r and are orthogonal. Their ordered cross product divided by r gives the displayed N.

The second derivatives give b_ss=r′z″−z′r″, b_sθ=0 and b_θθ=rz′. Dividing by the diagonal metric coefficients yields the two principal values. Their product simplifies to K=−r″/r. Indeed differentiating r′²+z′²=1 gives r′r″+z′z″=0, and substitution into (r′z″−z′r″)z′ eliminates z″. The identity remains valid at points where z′=0 by using the undivided relation.

For a nonplanar minimal example use the catenoid X(u,v)=(cosh v cos u,cosh v sin u,v). Here E=G=cosh²v and F=0. With N from X_u×X_v, the second coefficients are e=−1, f=0, g=1. Thus the principal values are −1/cosh²v and 1/cosh²v, proving H=0 and K=−1/cosh⁴v.

Apply

Repair “The catenoid is minimal because its profile looks like a hanging chain.” A physical analogy does not verify the surface equation. The two fundamental forms and their trace supply an exact proof. Moreover, global area-minimizing behavior depends on the selected boundary and allowable competitors; stationarity alone is insufficient.

For optional depth, the torus of revolution has both positive and negative K regions because its profile radius has changing second derivative. Its total curvature will later be constrained by topology. A single revolution chart also has periodic seams, so integration requires correct coverage accounting.

At a profile pole, return to the chart definition rather than canceling r in formulas that assumed r>0.

Continue

Use the profile’s parameter, orientation and regular region explicitly in every revolution calculation. Continue to ruled and developable surfaces to compare another geometric construction that simplifies one second derivative.