Surface of revolution

A surface of revolution is invariant under rotations around a fixed axis and is locally generated by rotating a profile curve.

A solid of revolution is the three-dimensional region swept out by rotating a planar region about an axis in its plane. Perpendicular slices are disks or annuli, making volume, center-of-mass, and moment-of-inertia integrals one-dimensional.

Forthe coordinate tangent vectors are orthogonal and the first fundamental form is

Put $q=(f^{\prime 2}+g^{\prime 2}{)}^{1/2}$ for the profile parametrizationWith unit normal $N=(-g^{\prime }\cos u,-g^{\prime }\sin u,f^{\prime })/q$, the principal curvatures areConsequently $H=(k_1+k_2)/2$ and $K=k_1{k}_2$.

A parallel of a surface of revolution is the circle obtained by fixing the profile parameter and varying the rotation angle. Its radius is the distance $f(v)$ from the rotation axis.

For the arc-length meridian metric $d{s}^2+f(s{)}^{2}d{\theta }^2$, a unit-speed parallel has $\dot{s}=0$ and $\dot{\theta }=\pm 1/f(s_0)$. Its radial geodesic equation is satisfied exactly when $f\prime (s_0)=0$. Thus extrema of the radius, and any other stationary radii, give geodesic circles. If every parallel is geodesic, the surface is a circular cylinder.

A meridian is the intersection of a surface of revolution with a plane containing the rotation axis. It is formed from a profile curve and its half-turn about the axis.

Suppose a smooth profile $(f(v),g(v))$ has $f(a)=0$, $f>0$ just to the right of $a$, and $f^{\prime }(a)\ne 0$. Use $r=f(v)$ as a local parameter and write $g(v)=h(r)$. The rotated surface is smooth at the pole if and only ifnear zero for a smooth function $H$. It is then locally the smooth graph $z=h(0)+H(x^2+y^2)$. Equivalently, the even reflection $h(|r|)$ is smooth; in particular all odd derivatives of $h$ at zero vanish and the profile meets the rotation axis orthogonally. The analogous condition holds at the other endpoint.

A catenoid is the surface of revolution obtained by rotating a catenary about its directrix. One conformal parametrization is

For the standard catenoid, the two principal curvatures are opposite, so its mean curvature vanishes. Its Gaussian curvature is $K=-{\mathrm{sech}}^{4}u<0$, which also shows that no neighbourhood is locally isometric to a plane, cylinder, or regular part of a cone, all of which have zero Gaussian curvature.

For the catenoid profile $y(x)=E\cosh (x/E)$ spanning fixed-endpoint rings, the second variation of $F[y]=2\pi \int y\sqrt{1+y^{\prime 2}}dx$ isThis refers to axisymmetric graph variations, not all possible changes in surface topology. Strict positivity for every nonzero admissible variation gives a strict local minimum in the $C^1$ topology: the positive lowest Dirichlet eigenvalue of the regular Sturm-Liouville problem gives coercivity, and sufficiently small coefficient perturbations preserve it. The zero variation must be excluded from a strict-positivity hypothesis.

Forthe area density after integrating over $\theta$ is $2\pi \varphi \sqrt{1+{\varphi }^{\prime 2}}$. If every height interval has area $2\pi r$ times its length, continuity forcesWhere $0<\varphi <r$, the sign of ${\varphi }^{\prime }$ is constant and $\sqrt{r^2-{\varphi }^2}$ has derivative $\pm 1$. Hence the profile lies on a circle of radius $r$.

For the surfaces generated by $(e^u,0,u)$ and $(\cosh s,0,s)$,The diffeomorphism $u=\log \sinh s$ matches these curvatures, although comparison of the first fundamental forms shows that no local isometry exists.

For $X(u,v)=(Y(u)\cos v,Y(u)\sin v,Z(u))$ with $Y^{\prime 2}+Z^{\prime 2}=1$, the induced metric is

With unit normal $(-Z^{\prime }\cos v,-Z^{\prime }\sin v,Y^{\prime })$, the principal-form coefficients give

For an arc-length profile with radius $Y(u)>0$, the equation $K=c$ isThus $Y=A\cos u$ gives $K=1$ locally, while $Y=A\cosh u$ gives $K=-1$ wherever $|Y^{\prime }|<1$. At $u=0$ their mean curvatures are respectivelyso different choices of $A$ give locally noncongruent surfaces with the same constant Gaussian curvature.

For $0<A<1$, rotate the unit-speed profilenear $s=0$. Since $K=-Y^{\prime \prime }/Y$, the resulting regular surface of revolution has $K=1$. At $s=0$ its principal curvatures are $A$ and $A^{-1}$, so the point is not umbilical when $A\ne 1$ and no neighbourhood of it lies on a sphere.

For the parametrizationthe Gaussian curvature and area element are

For $a\le x\le b$, the curvature formula is an exact derivative and gives

A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm 1/R$, according to orientation. Hence $K=0$ and $|H|=1/(2R)$.

A circular cylinder of radius $\rho$ has induced metric locally equal to $d{u}^2+d{z}^2$ with $u=\rho \phi$. Its Riemann curvature tensor therefore vanishes, despite its nonzero extrinsic curvature. With outward unit normal and convention $K_{ij}=e_i^a{e}_j^b{\nabla }_a{n}_b$, its trace is $1/\rho$ and its principal curvatures are $1/\rho$ and zero. Its averaged mean curvature is $1/(2\rho )$. The angular identification changes global topology, not local flatness.

For a geodesic in the metric $d{u}^2+Y(u{)}^{2}d{v}^2$, the cyclic coordinate $v$ givesEquivalently, the geodesic meets parallels according to Clairaut's relation.

Let a compact smooth surface of revolution have exactly two poles, and let $c=f(v{)}^2\dot{u}=f(v)\cos \theta$ be the Clairaut first integral for a surface of revolution of a complete unit-speed geodesic. If $c\ne 0$, then $f(v)\ge |c|$, so the geodesic avoids open neighbourhoods of both poles. If $c=0$, then $u$ is constant away from the poles and the geodesic lies in a meridian of a surface of revolution, whose complement contains a nonempty open set. Thus no complete geodesic has dense image.