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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.

Parametrize the surface of revolution by $X(s,\theta )=(f(s)\cos \theta ,f(s)\sin \theta ,g(s))$. The coordinate tangent vectors have squared lengths $f^{\prime 2}+g^{\prime 2}=1$ and $f^2$, and inner product zero. Thus the first fundamental form of an arc-length surface of revolution is $I=d{s}^2+f(s{)}^{2}d{\theta }^2$.

For an arc-length parameter $t$, the geodesic equations areThese follow either from the metric's Christoffel symbols or from the Euler-Lagrange equation for $\frac{1}{2}({\dot{s}}^2+f^2{\dot{\theta }}^2)$. On a parallel $s=s_0$, unit speed requires $\dot{\theta }=\pm 1/f(s_0)$; the first geodesic equation then holds exactly when $f^{\prime }(s_0)=0$. Thus geodesic parallels of a surface of revolution are precisely the critical points of the radius function.

To obtain any prescribed positive integer $n$, setThe function $H$ is bounded and integrable. Choose $\epsilon >0$ so $\epsilon \sup |H|<1$, and choose $R>\epsilon {\int }_{\mathbb{R}}|H|$. Then $f>0$, $g$ is smooth, and $f^{\prime 2}+g^{\prime 2}=1$. The derivative $f^{\prime }=\epsilon H$ vanishes at exactly $1,\dots ,n$. Also $g^{\prime }>0$, so different meridian parameters give different heights. The resulting surface has exactly $n$ geodesic parallels.

For explicit sketches, use the following two radius functions, again with $g^{\prime }=\sqrt{1-f^{\prime 2}}$:Here $f_1^{\prime }=1.2s{e}^{-s^2}$ has only the zero $0$, while $f_2^{\prime }=0.6(s^2-1)e^{-s^2}$ has exactly the zeros $-1,1$. Both radii stay positive and both derivatives have absolute value below one. The highlighted rings are the geodesic parallels; the surfaces continue beyond the displayed window.

If every parallel is a geodesic, then $f^{\prime }=0$ everywhere, so $f=R$ is a positive constant. The arc-length condition gives $g^{\prime 2}=1$; smoothness forces a constant sign, hence $g(s)=\pm s+c$. The surface is a circular cylinder of radius $R$. It is a circular cylinder because the meridian is a straight line parallel to the axis.

In cylindrical coordinates $r=\sqrt{x^2+y^2}$, the bounded solid enclosed by the two conical sheets and the circular cylinder isIts meridional section consists of two triangular regions; revolving them about the $z$-axis gives the solid. The two conical sheets meet at the origin and meet the circular cylinder in its circles at $z=\pm 1$.

Since $\mathbf{F}=(x,y,z)$, its divergence is $3$. Integrate directly with the cylindrical coordinates volume element $rdzdrd\theta$:The same calculation gives $\mathrm{Vol}(V)=4\pi /3$.

Close the truncated two-sheet double circular cone by the lateral circular cylinder $r=1$, $-1\le z\le 1$. Together these form the boundary of the solid in part (c); no planar caps are required because the cone meets the circular cylinder along both end circles. On the circular cylinder the outward unit normal is $(\cos \theta ,\sin \theta ,0)$, so $\mathbf{F}\cdot \mathbf{n}=1$, and $dS=d\theta dz$. Its flux integral isThe divergence theorem now gives ${\int }_S\mathbf{F}\cdot d\mathbf{S}+4\pi =4\pi$, using the volume integral from part (c). Henceagreeing with tangency. The isolated vertex can be handled by excising a ball of radius $\epsilon$ and passing to the limit: since $|\mathbf{F}|=O(\epsilon )$ there and the added area is $O({\epsilon }^2)$, its extra flux is $O({\epsilon }^3)$ and tends to zero.

Lift the angular coordinate continuously to a real coordinate and write $\Delta z=z_B-z_A$ and $\Delta {\varphi }_m={\varphi }_B-{\varphi }_A+2\pi m$ for a prescribed winding number $m\in \mathbb{Z}$. Assume $z_A<z_B$ for the calculation; reversing the endpoints changes no length. The arc length functional on the circular cylinder isThe Euler-Lagrange equation says $R^2{\varphi }^{\prime }/\sqrt{1+R^2{\varphi }^{\prime 2}}$ is constant. This expression is strictly increasing in ${\varphi }^{\prime }$, so ${\varphi }^{\prime }$ is constant. The candidate is the helixIt is a global minimum within the chosen winding class: the integrand is strictly convex, and Jensen inequality bounds the functional below by $(z_B-z_A)\sqrt{1+R^2(\Delta {\varphi }_m/\Delta z{)}^2}$, with equality only for constant slope. Equivalently, unrolling the cylinder makes the minimizing path a straight segment.

The signed pitch of a helix, measured for an increase of $2\pi$ in angle, isIts geometric magnitude is $|P_m|$. If only the physical endpoints are fixed, choose the integer $m$ minimizing $|\Delta {\varphi }_m|$; opposite points may give two equally short helices. If the lifted endpoint angles are prescribed, use that winding class. A zero angular increment gives a straight generator, the limiting infinite-pitch case.

Let $c>0$ denote the stated unit cost of cylindrical wall, and assume $V>0$. The circular cylinder and hemisphere give the volume constraint and total costOnly the walls and cap have specified costs, so no separate base cost is included. Eliminating $h$ giveswith $0<r\le (3V/(2\pi ){)}^{1/3}$ from $h\ge 0$. Now $C^{\prime }(r)=-2cV/r^2+(16/3)\pi cr$ vanishes at $r^3=3V/(8\pi )$. Moreover $C^{\prime \prime }(r)=4cV/r^3+(16/3)\pi c>0$ and the stationary point is feasible, so it is the unique global minimum. The optimal dimensions are

Take the circular cylinderIts first fundamental form is $R^{2}d{\theta }^2+d{z}^2$. The universal covering parametrization ${\mathbb{R}}^2\to S$ is therefore a local isometry from the Euclidean plane, after scaling the first coordinate. By geodesic preservation by a local isometry, every geodesic on the cylinder lifts to a straight line defined for all $t\in \mathbb{R}$. Thus the cylinder is geodesically complete. It is noncompact because $z$ is unbounded, and it is not a plane because one of its principal curvatures is $1/R$.

A regular surface is a minimal surface when its mean curvature vanishes identically, equivalently when its two principal curvatures sum to zero everywhere.

The catenoidis an example. In part (a), take $f(v)=\cosh v$ and $g(v)=v$, so $q=\cosh v$. The formula there givesA plane and a circular cylinder have zero Gaussian curvature, as does a flat cone away from its vertex. Because a local isometry preserves Gaussian curvature, the catenoid is minimal and is not locally isometric to any of those surfaces; this is the catenoid is a minimal surface.

A plane has both principal curvatures zero, hence $K=H=0$. A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, henceBecause the absolute value of mean curvature is preserved even if one reverses the chosen orientation, no Euclidean motion can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv 0$ does not characterize planar pieces.

There are analogous examples for both signs of constant Gaussian curvature. Start with the arc-length surface of revolutionOn a sufficiently small interval about $u=0$, this is a regular surface and the curvatures of an arc-length surface of revolution giveAt $u=0$ its mean curvature isFor $A=1$ this is a piece of the unit sphere and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, useagain on a sufficiently small interval. ThenThe choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The constant Gaussian curvature surfaces of revolution therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.