We calculate the equation of a surface of revolution with constant mean curvature H. Without loss of generality, we suppose that the profile curve is a planar curve in the xz-plane and that the z-axis is the rotational axis. Also, suppose that the curve is a graph on the z-axis, that is, a parametrization of the profile curve is (f(z),0,z), z∈I. Then a parametrization of the rotational surface is X(t,s)=(f(t)cos(s),f(t)sin(s),t), t∈I,s∈[0,2π]. Thus H satisfies
−f″(1+f′2)3/2+1f√1+f′2=2H.
The key of this equation is that because H is constant, it is possible to obtain a first integral of this equation (which is of second order). Indeed, multiplying by ff′ we have −ff′f″(1+f′2)3/2+f′√1+f′2=2Hff′,which can be written as
(f√1+f′2)′=(Hf2)′. Therefore there exists c∈R such that
f(z)√1+f′(z)2=Hf(z)2+c.
For example, the sphere and the cylinder can be obtained from (*). For the sphere, take c=0. Then we have
f′=1H√1f2−H2,
or
f′√1f2−H2=1H. By integrating, we have
1H2√1−H2f2=1Hx.
Definitively, f(z)=√1H2−z2 which is a circle of radius 1/|H|, and the surface is a sphere of radius 1/|H|.
For the cylinder, we have to come back to the initial equation for H. If f(z)=r, then H=1/(2r).
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