Surfaces of revolution with constant mean curvature (I)

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 $(f(z),0,z)$, $z\in I$. Then a parametrization of the rotational surface is $$X(t,s)=(f(t)\cos(s),f(t)\sin(s),t),\ t\in I,s\in [0,2\pi].$$ Thus $H$ satisfies 

$$\frac{-f''}{(1+f'^2)^{3/2}}+\frac{1}{f\sqrt{1+f'^2}}=2H.$$

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 $$\frac{-ff'f''}{(1+f'^2)^{3/2}}+\frac{f'}{\sqrt{1+f'^2}}=2Hff',$$which can be written as 

$$\left(\frac{f}{\sqrt{1+f'^2}}\right)'=(Hf^2)'.$$ Therefore there exists $c\in {\mathbb R}$ such that 

$$\frac{f(z)}{\sqrt{1+f'(z)^2}}=Hf(z)^2+c.$$

For example, the sphere and the cylinder can be obtained from (*). For the sphere, take $c=0$. Then we have 

$$f'=\frac{1}{H}\sqrt{\frac{1}{f^2}-H^2},$$

or

$$\frac{f'}{\sqrt{\frac{1}{f^2}-H^2}}=\frac{1}{H}.$$ By integrating, we have

$$\frac{1}{H^2}\sqrt{1-H^2 f^2}=\frac{1}{H}{x}.$$

Definitively, $$f(z)=\sqrt{\frac{1}{H^2}-z^2}$$ 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  $H=1/(2r)$.