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For a surface of revolution whose induced metric is $g=E(\rho )d{\rho }^2+{\rho }^{2}d{\varphi }^2$, with $\rho >0$ and $E>0$ smooth on a coordinate interval, define $\sigma (\rho )={\int }_{{\rho }_{\ast }}^{\rho }\sqrt{E(r)}dr/r$. Then ${\sigma }^{\prime }>0$ and the conformal factor ${\rho }^{-1}$ gives ${\rho }^{-2}g=d{\sigma }^2+d{\varphi }^2$. This proves the rescaled metric is locally the Euclidean metric without solving a curvature equation. Poles and places where $\rho$ is not a coordinate must be treated in other charts; the angular coordinate is also understood locally.

The lateral surface obtained by rotating a straight segment in a plane containing the rotation axis has endpoint radii $a,c$ and slant length $\ell$. Its surface area is $\pi (a+c)\ell$: the radius is linear in the segment's arc length, so the surface area of a surface of revolution is $2\pi {\int }_0^{\ell }\rho (s)ds=2\pi \ell (a+c)/2$. This formula excludes the end disks and assumes that the segment's radial direction is fixed. A straight segment whose endpoint horizontal directions differ generally sweeps a different surface of revolution.

For a sphere of radius $R>0$, the spherical cap above $z=R-h$, with $0\le h\le 2R$, has curved surface area $2\pi Rh$. On the surface use $r(z)=\sqrt{R^2-z^2}$. The surface of revolution area element is $2\pi r\sqrt{1+(dr/dz{)}^2}dz=2\pi Rdz$. Integrating from $R-h$ to $R$ proves the formula. It includes only the spherical surface; if the planar closing disk is required, add $\pi (2Rh-h^2)$. The coordinate expression is interpreted by a limit at a pole.

An elliptic paraboloid is a smooth bowl-shaped quadratic surface with elliptical horizontal sections. In the circular case $A=B$, it is a surface of revolution. For $z=x^2+y^2$, polar coordinates give the parametrization $(r\cos \theta ,r\sin \theta ,r^2)$ and upward vector area element $(-2r^2\cos \theta ,-2r^2\sin \theta ,r)drd\theta$. A band between two positive heights has two boundary circles and no cap.

The area of a surface of revolution about the horizontal axis is $2\pi \int |y|\sqrt{1+y^{\prime 2}}dx$. For a curve below the axis with endpoints $(\pm a,0)$ and fixed arc length, its stationarity is equivalent to stationarity of $\int (y+\lambda )\sqrt{1+y^{\prime 2}}dx$. The Beltrami identity gives $(y+\lambda )/\sqrt{1+y^{\prime 2}}=C$. Choosing $C=1/k$, imposing the endpoints and integrating the length gives the displayed curve. The function $\sinh s/s$ increases strictly from $1$ to infinity on $s>0$, so every $l>2a$ determines exactly one positive $k$. Stationarity alone does not establish a global minimum.

The surface of revolutionis geodesically incomplete: a meridian reaches the missing origin in finite length. Its Gaussian curvature, computed using the curvatures of a parametrized surface of revolution, iswhich tends to infinity as $r\downarrow 0$. Every finite-time endpoint of a constant-speed geodesic must approach the origin, the only finite point in the ambient closure missing from $S$, so this surface satisfies the curvature-blowup criterion.

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 surface of revolution of the circle with center $(a,R)$ and radius $r<R$ in the upper half-plane, use $X(u,v)=(a+r\cos u,(R+r\sin u)\cos v,(R+r\sin u)\sin v)$ and unit normal $n=(\cos u,\sin u\cos v,\sin u\sin v)$. Exactly four surface normals pass through the displayed $P$. The equation $P-X=tn$ forces $(t+r)\cos u=0$. The case $t=-r$ would put $P$ at radial distance $R$, so $\cos u=0$; its nonzero radial coordinate then forces $v=0$ or $\pi$. All four combinations work. By contrast, infinitely many normals pass through $Q$: both complete circles with $u=\pi /2$ or $3\pi /2$ have radial normals through that point. This is a concrete illustration of degenerate distance-critical sets versus isolated critical points.

For the surface of revolution, the arc-length condition on its profile gives $E=1$, $F=0$, $G=f(u{)}^2$. Thus the geodesic equations reduce toConsequentlythe Clairaut first integral for a surface of revolution.

On a meridian, take $v=b$ and $u=\pm s+u_0$. Both equations hold, so every meridian is a geodesic. On a cross-section $u=c$, arc length requires $v^{\prime }=\pm 1/f(c)$. The first equation then holds exactly when $f^{\prime }(c)=0$, while the second holds automatically. Hence

Work on a regular patch with $f(u)>0$. Differentiating the surface of revolution givesThe arc-length condition gives $E=1$, $F=0$, $G=f^2$. Also $\Vert r_u\times r_v\Vert =f$, so takeThe second derivatives yieldThus the first fundamental form and second fundamental form areThe opposite normal reverses the second displayed form. At a smooth pole the revolution coordinates degenerate, so curvature is understood by a regular local chart or by continuity.

Choose $f(u)=\sin u$ and $h(u)=\cos u$ for $0\le u\le \pi$. Adding the smooth poles at the endpoints gives the unit sphere. Its meridian is parametrized by arc length, and $-f^{\prime \prime }/f=1$ away from the poles, hence everywhere by continuity. Thus the unit sphere has constant Gaussian curvature one.

There is no compact smooth surface of revolution without boundary of constant Gaussian curvature minus one in this setting. On a compact such surface, the distance $f$ from the rotation axis attains a positive maximum. The maximum lies on a regular revolution patch, where the meridian arc length coordinate gives $f^{\prime \prime }\le 0$. Hence $K=-f^{\prime \prime }/f\ge 0$ there, contradicting $K=-1$. Equivalently, constant negative curvature would require $f^{\prime \prime }=f>0$, excluding any positive radial maximum. Compact patches with boundary would not be excluded by this argument, but they are not closed surfaces.

For the surface of revolution,The mixed inner product vanishes, givingTaking $f(u)=\cosh u$ makes both coefficients ${\cosh }^{2}u$, as used in the parent calculation.

Using the two first fundamental forms calculated below, chooseThe surface of revolution is then a catenoid, with metric ${\cosh }^{2}s(d{s}^2+d{v}^2)$. On the helicoid, take $u=\sinh s/a$, assuming $a\ne 0$ so that its parametrization is regular. Substitution givesThus the coordinate mapis a local isometry on sufficiently small patches. The constant $a$ only changes the radial parametrization of the given helicoid, whose pitch is fixed at one. For $a=0$ the displayed helicoid parametrization is not a regular surface.

Use the cylindrical coordinate system basis ${\mathbf{e}}_r,{\mathbf{e}}_{\phi },{\mathbf{e}}_z$. The meridional unit tangent vector and outward unit normal vector consistent with the given geometry areAn surface of revolution has $\theta$ independent of $\phi$. ConsequentlyApply the product rule to the vector force balance:Equating the three independent components to zero, and using $r_s=\cos \theta$, givesThus, away from the rotation axis where $r>0$, the scalar mechanical equilibrium equations areThe two geometric factors ${\theta }_s$ and $\sin \theta /r$ are the signed meridional and azimuthal principal curvatures in this convention. The normal equation is the Young–Laplace equation generalized to unequal principal membrane tensions.

The radius of the surface of revolution is $|y(x)|$, so its area and the curve's arc length areFor the proposed curve, $y<0$ on the interior. On this branch $A=-2\pi S$ with $S=\int y\sqrt{1+y^{\prime 2}}dx$. Extremizing $S$ subject to $L=l$ is therefore equivalent to making the actual area stationary. Introduce a Lagrange multiplier $\lambda$ and the augmented integrand $F=(y+\lambda )\sqrt{1+y^{\prime 2}}$. The Beltrami identity givesFor the fixed-length stationary surface of revolution, takeIndeed $y^{\prime }=\sinh (kx)$ and $\sqrt{1+y^{\prime 2}}=\cosh (kx)$, so the Beltrami equation holds identically. It also follows directly that $F_y=\cosh (kx)$ and $d(F_{y^{\prime }})/dx=\cosh (kx)$, verifying the Euler-Lagrange equation. The endpoints vanish, and the length condition becomesFor endpoint-fixed variations preserving this length, the augmented first variation is zero; subtracting the zero length variation proves that the area's first variation is zero. Its value, if desired, is $A=\pi [\sinh (2ka)-2ka]/k^2$.

To prove uniqueness, put $s=ka>0$. The equation is $\sinh s/s=l/(2a)>1$. Its derivative is $(s\cosh s-\sinh s)/s^2$. The numerator is zero at $s=0$ and has derivative $s\sinh s>0$, so the ratio increases strictly from one to infinity. The intermediate value theorem gives exactly one positive $k$. The stated upper bound on $l$ is not needed for this uniqueness argument. Stationarity has been proved, without making an unsupported global-minimum claim.

For a regular parametrized surface $\sigma (u,v)$, the first fundamental form is the induced metricChoose the unit normal $N=({\sigma }_u\times {\sigma }_v)/|{\sigma }_u\times {\sigma }_v|$. With the convention $I{I}_{ij}=N\cdot {\sigma }_{ij}$, the second fundamental form isThe Gaussian curvature, the product of the principal curvatures, isFor a compact smooth embedded surface without boundary, the Gauss-Bonnet theorem giveswhere $e(S)$ is the Euler characteristic. The no-boundary condition is needed for this form; a smooth embedded surface with boundary has the additional boundary geodesic curvature integral.

For the surface of revolution, useThe parameters are periodic, with local open parameter rectangles providing the coordinate charts. Put $R=r+a\sin u$ and $D=\sqrt{a^2{\cos }^{2}u+b^2{\sin }^{2}u}$. Since $r>a>0$ and $b>0$, both $R$ and $D$ are positive. The tangent vectors areTheir inner products are $E=D^2$, $F=0$, and $G=R^2$. ThusTheir cross product is $R(b\sin u\cos v,b\sin u\sin v,a\cos u)$, so its normalization gives the outward unit normalNext,Taking their inner products with $N$ yields $e=-ab/D$, $f=0$, and $g=-bR\sin u/D$. Hence the fundamental forms of an elliptic ring torus giveChoosing the inward unit normal changes both displayed coefficients of $II$ to their negatives and leaves $I$ and $K$ unchanged. As a consistency check, $K=a{b}^2\sin u/(R{D}^4)$; its integral over the torus is zero, since $KdA=a{b}^2\sin ududv/D^3$ cancels under $u\mapsto u+\pi$. This agrees with $e(S)=0$ in the Gauss-Bonnet theorem.

The assertion is false without further global hypotheses. For a concrete non-spherical surface of constant Gaussian curvature one, fix $a=1/2$ and setHere $f>0$, $g^{\prime }>0$ and $f^{\prime 2}+g^{\prime 2}=1$, so this is a connected regular embedded surface of revolution with unit-speed meridian. Its metric is $d{u}^2+f(u{)}^{2}d{v}^2$, whose Gaussian curvature is $-f^{\prime \prime }/f=1$. This formula also follows by multiplying its principal curvatures $-f^{\prime \prime }/g^{\prime }$ and $g^{\prime }/f$.

At $u=0$ these principal curvatures are $a=1/2$ and $1/a=2$. Every point of a radius-one sphere is umbilical with both principal curvature magnitudes one. The displayed surface therefore cannot lie in any such sphere, despite having $K=1$ everywhere.

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.

Along a sufficiently smooth stationary curve, differentiation and the Euler-Lagrange equation give the Beltrami identity:The cancellation uses the absence of explicit $x$ dependence.

For an axisymmetric surface of revolution, its area is $I[y]=2\pi {\int }_{-l}^{l}y\sqrt{1+y^{\prime 2}}dx$. A smooth positive-radius minimizer is stationary, so the Beltrami identity, with the constant factor omitted, givesEquivalently the Euler-Lagrange equation gives $y{y}^{\prime \prime }=1+y^{\prime 2}$, whence $y^{\prime \prime }=y/k^2$. The first integral selectsEqual radii at $\pm l$ imply $x_{\ast }=0$ (subtract the two cosh values), givingThis is a catenoid. The conclusion is a necessary form for a smooth connected minimizer; it does not declare every stationary catenoid a minimum.

Put $a=l/k>0$. The boundary equation is $r/l=\cosh a/a$. Its derivative is $(a\sinh a-\cosh a)/a^2$, so its unique critical point solves $a\tanh a=1$. The left side increases strictly from zero to infinity. Thus $\cosh a/a$ decreases to a unique minimum and then increases, diverging as $a\downarrow 0$ and $a\to \infty$. There are exactly two positive values of $k$ whenAt equality there is one value, and below it there is none.

To match the printed second variation convention, write $I[y+\epsilon \eta ]=I[y]+\epsilon \delta I+{\epsilon }^2{\delta }^{2}I+o({\epsilon }^2)$; its ${\delta }^{2}I$ is the quadratic coefficient, half the second derivative with respect to $\epsilon$. Expanding the area integrand givesOn the catenoid these coefficients are $k{\mathrm{sech}}^2(x/k)$ and $2\tanh (x/k)$. Integrate the mixed term by parts; the boundary term is zero because $\eta (\pm l)=0$. Since $(\tanh (x/k){)}^{\prime }={\mathrm{sech}}^2(x/k)/k$, this yieldsThis is the axisymmetric second variation of a catenoid, with the explicitly stated quadratic-coefficient convention.

Stokes theorem equates the circulation around the induced boundary orientation of an oriented surface to the normal flux of curl:The usual version requires a piecewise smooth oriented surface and a vector field continuously differentiable on a neighbourhood of it. We use the upward orientation wherever the surface is a graph; each outer circle is counterclockwise viewed from above, and an inner circle is clockwise.

Write $\rho =(x^2+y^2{)}^{1/2}$. The surface of revolution has $z=\sqrt{{\rho }^2+1-\lambda }$ with $0\le z\le 1$. For $0<\lambda <1$, it is a disk-shaped cap of a two-sheeted hyperboloid, running from $\rho =0,z=\sqrt{1-\lambda }$ to $\rho =\sqrt{\lambda },z=1$. For $\lambda =1$, it is a right circular cone, $z=\rho$, including its apex. For $\lambda >1$, it is an annular strip of a one-sheet hyperboloid, between the circles $\rho =\sqrt{\lambda -1},z=0$ and $\rho =\sqrt{\lambda },z=1$.

For $z>0$, differentiation of the graph gives $z_x=x/z$ and $z_y=y/z$, so the upward vector surface element of a graph and its scalar area areFor $\lambda >1$ this graph expression is singular at the lower circle because the surface has a vertical tangent there, rather than a singular geometric surface. A regular parametrization isIt proves that the graph surface integral is interpreted by its integrable endpoint limit. The same parametrization works away from the pole or apex in the other regimes.

Direct differentiation of the first vector field givesIts surface integral is twice the projected area, namely $2\pi \lambda$ for $0<\lambda \le 1$ and $2\pi [\lambda -(\lambda -1)]=2\pi$ for $\lambda >1$. On an oriented circle of radius $R$, $\mathbf{F}\cdot d\mathbf{x}=R^{2}d\varphi$. The outer circulation is $2\pi \lambda$; an inner circle, when present, contributes $-2\pi (\lambda -1)$. ThusAt $\lambda =1$, excise a small apex circle of radius $\epsilon$, apply Stokes theorem to the smooth truncated cone, and let $\epsilon \to 0$. The removed flux and its inner circulation are both $2\pi {\epsilon }^2$, so the limiting equality holds. This avoids silently applying a smooth-surface theorem at the cone's nonsmooth apex.

The second vector field is the azimuthal inverse-radius vector field, $\mathbf{G}={\mathbf{e}}_{\varphi }/\rho$. Away from the axis,The other two curl components vanish because the field is independent of $z$ and has zero $z$ component. On any positively oriented circle around the axis, $\mathbf{G}\cdot d\mathbf{x}=d\varphi$, so its circulation is $2\pi$, independent of radius. ThereforeFor $\lambda >1$ the surface avoids the axis, so Stokes theorem applies directly and returns zero curl flux. For $0<\lambda <1$ the disk cap meets the axis, and for $\lambda =1$ the apex lies on it: $\mathbf{G}$ is undefined there, so the neighbourhood hypothesis of Stokes theorem fails. Removing a small circle gives cancelling inner and outer circulations; the inner circulation remains $-2\pi$ as its radius tends to zero. Hence it cannot be discarded as it could for $\mathbf{F}$. A nonzero circulation here is compatible with zero curl away from the excluded axis.