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Constant Gaussian curvature one gives $f^{\prime \prime }+f=0$, so, after a shift of the arc-length parameter, $f=C\cos u$ with $C>0$ on an interval contained in $(-\pi /2,\pi /2)$. To close the surface by adding only two points, the limiting parallels at both ends must collapse to points on the axis. If an end had limiting radius $f>0$, it would contribute a whole missing circle, which two added points cannot supply. Thus the parameter interval is the full interval between consecutive zeros, and its two ends are $u=\pm \pi /2$.

At a smooth axis point the tangent plane of a surface of revolution is invariant under all rotations about the axis, so it must be horizontal. The unit meridian tangent must consequently have $g^{\prime }\to 0$ and $|f^{\prime }|\to 1$ there. But $|f^{\prime }(\pm \pi /2)|=C$, so smoothness forces $C=1$. The arc-length identity also illustrates the alternatives: $C>1$ prevents reaching either zero with real $g^{\prime }$, while $0<C<1$ gives a nonhorizontal limiting meridian and a conical singularity.

Now $g^{\prime 2}=1-f^{\prime 2}={\cos }^{2}u$. On the connected interior, $g^{\prime }$ has one fixed sign because $\cos u>0$. Therefore $g=g_0\pm \sin u$, andThe revolved surface is the unit sphere centered at $(0,0,g_0)$, with only its two poles $(0,0,g_0\pm 1)$ omitted. Those points are antipodal. This proves the claim using smooth endpoint criterion for a surface of revolution, without integrating a general square root.

No. The missing point is not assumed to complete $S$ as a smooth surface, so the hypotheses of the global Gauss-Bonnet theorem do not apply to $S\cup \{p\}$.

For a concrete counterexample, rotate a smooth simple arc $(r(s),z(s))$, parametrized by arc length for $0\le s<L$, about the $z$-axis. Arrange that it starts smoothly on the axis,stays in $r>0$ for $0<s<L$, and converges to another point of the axis as $s\uparrow L$, withThe endpoint is omitted. The resulting surface of revolution is diffeomorphic to ${\mathbb{S}}^2$ minus one point, and adjoining its limiting endpoint makes the set compact.

For an arc-length surface of revolution,Therefore the total Gaussian curvature of a punctured surface with a singular compactification point iswhich can take values different from $4\pi$. Compactness of the one-point closure alone therefore does not determine the integral.

No. Fix $0<A<1$ and, near $s=0$, letThen $f^{\prime 2}+g^{\prime 2}=1$, so this is a regular unit-speed profile with $f>0$ on a sufficiently small interval. Rotate it around the $z$-axis. For an arc-length surface of revolution, $K=-f^{\prime \prime }/f$, and thereforeAt $s=0$, the formulas from part (a) give the principal curvaturesThey are unequal because $A\ne 1$, so the point is not an umbilical point. Every point of a sphere is umbilical. Consequently no neighbourhood of this point can lie on a sphere, giving a non-spherical surface of constant Gaussian curvature one.

The curve is symmetric in both coordinate axes and under interchange of $x$ and $z$. Its shape, including its eight inflection points, is:

Writing $r^2=x^2+y^2$, the surface of revolution has equationLetIts gradient isIf this vanished, then $r^2\in \{0,1\}$ and $z^2\in \{0,1\}$, but at such a point $(r^2-1{)}^2+(z^2-1{)}^2\le 2$, not five. Hence zero is a regular value, and the regular level set theorem shows that $S=F^{-1}(0)$ is a smooth embedded surface.

For a surface of revolution, the two principal curvatures are the curvature of the meridian and the normal curvature of a parallel. Along $r=1$,so the normal is vertical in the meridian plane and the parallel principal curvature vanishes. Therefore the Gaussian curvature, the product of the principal curvatures, is zero there.

These are not the only zero-curvature points. The meridian itself has inflection points, at which its principal curvature vanishes. Setting $u=r^2$ and $v=z^2$, their positive squared coordinates satisfyBesides interchanging $u$ and $v$, the solution is approximatelyEach corresponding meridian inflection point sweeps out another circle on $S$, so $r=1$ does not exhaust the zero set of the Gaussian curvature.

Withthe coordinate tangent vectors satisfyThus the first fundamental form isGeometrically, $S$ is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolutionIt has the same first fundamental form, so $(u,v)\mapsto Y(u,v)$ defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives

The planemeets the one-sheet hyperboloid in the two disjoint curvesThey are the two opposite meridians of the surface of revolution. Parametrize the generating profile by arc length. Its acceleration lies in the meridian plane and is orthogonal to the profile tangent; it is also orthogonal to the azimuthal tangent. It is therefore normal to the surface, so both meridians are geodesics. This is the disjoint case in plane-section geodesics of the unit one-sheet hyperboloid.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

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 $K=-1$.

The endpoint radii are $5$ and $13$, but substituting those two numbers in the finite conical frustum formula is invalid. A rotation about the $z$-axis preserves the horizontal angle between the endpoint position vectors; here $(3,4)$ and $(12,5)$ are not parallel. The radius along the generating segment is not linear in height.

Parametrize the actual segment byIts rotated circle at height $1+6t$ has radiusThus the rotated image is the surface of revolution $\mathbf{R}(t,\theta )=(\rho (t)\cos \theta ,\rho (t)\sin \theta ,1+6t)$. Applying the conical frustum formula to infinitesimal pieces of this radial profile, or directly using the cross product as in ii, gives the surface area of a surface of revolutionTo evaluate it explicitly, put $u=82t+31$. Then $9676t^2+7316t+1861=(59u^2+19602)/41$. DefineThe chain rule gives $G^{\prime }(u)=2\sqrt{59u^2+19602}$, so the answer isThe positive logarithm arguments make this a real-valued expression. Rotating the segment's endpoints independently into one vertical plane before joining them would instead create a different surface, with surface area $180\pi \simeq 565.4867$.

Interpolate the radius and height byA parametrized surface for the conical frustum isThe strict positivity of $a,c$ gives $\rho (t)>0$. Since $d>b$, the height identifies $t$ uniquely; the angle then identifies a point of each circle uniquely modulo $2\pi$. Hence this covers the lateral surface of revolution once. If $a=c$, it is the lateral surface of a circular cylinder.

On the surface of revolution, a normal vector is proportional to $-H^{\prime }(r){\mathbf{e}}_r+{\mathbf{e}}_z$. Thus the normal force and gravity have no ${\mathbf{e}}_{\theta }$ component. The ${\mathbf{e}}_{\theta }$ component of Newton's second law, using the given linear drag force, isMultiplication by $r$ therefore givesThe printed target equation omits the factor $1/m$. The matching original PDF, page 5, gives both ${\mathbf{F}}_D=-\alpha \dot{\mathbf{r}}$ and $\dot{h}=-\alpha h$, so this is a source inconsistency. If its $\alpha$ in the target equation is intended to mean the drag coefficient per unit mass, put $\lambda =\alpha /m$ and write $\dot{h}=-\lambda h$. Equivalently, if the intended force was ${\mathbf{F}}_D=-m\alpha \dot{\mathbf{r}}$, the requested convention givesThese formulas determine $\dot{\theta }$ once $r(t)$ is known; they do not assume a constant radius.

Use the parametric curveHere $|y(t)|\le \sqrt{5}<3$, so the ellipse stays strictly above the axis of rotation $y=-3$. Rotating in three dimensions gives a surface of revolution with parametrizationThe two parameter tangent vectors are orthogonal, with Euclidean norms $\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$ and $y(t)+3$. Hence the area element of a surface isThe generating ellipse is simple and has positive distance from the axis of rotation; its rotation sweeps the surface once, apart from the identified parameter endpoints. AlsoWriting $w(t)=\sqrt{8{\sin }^{2}t+2{\cos }^{2}t}$, the surface area is therefore $2\pi {\int }_0^{2\pi }(3+y(t))w(t)dt$. Since $y(t+\pi )=-y(t)$ and $w(t+\pi )=w(t)$,ConsequentlyThe cancellation is an instance of the surface area of revolution of a centrally symmetric curve: the surface area is the generating curve’s arc length times the distance $6\pi$ traveled by its center.

Parametrize the surface of revolution byAn oriented vector surface area element isUse the vector field $\mathbf{F}=(x/2,y/2,0)$, whose divergence is one. If this parametrization has the outward orientation on an enclosing surface, the divergence theorem givesThe second equality is integration by parts, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi /2$.

The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the flux formula is an algebraic signed volume. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric volume is insteadThis explicitly handles a reversal or change in the order of the two branches.

The rotation axis is $L=\{(x,1,0):x\in \mathbb{R}\}$. The distance of the generating function graph from this axis is $\sqrt{x}$, so a parametrization isThe surface of revolution therefore has equation $(y-1{)}^2+z^2=x$: it is a circular elliptic paraboloid opening along the positive $x$ direction, with its vertex and its rim at $x=1$ excluded. Translating the axis does not change the surface area formula, soThe cancellation makes the improper integral finite at the vertex. Neither the missing vertex nor the missing boundary circle changes the surface area, and no cap is included.

A parametrization of the surface of revolution isIts tangent vectors are orthogonal, with norms $\sqrt{1+f^{\prime }(x{)}^2}$ and $|f(x)|$. ThusThe absolute value is necessary if the function graph passes below the axis. The formula gives the curved surface area; it adds no end discs. Zeros of $f$ can be treated by splitting the integral into regular portions and taking limits.

Let $\mathbf{N}$ be the normal reaction of the stationary smooth surface of revolution. Its normal lies in the span of ${\mathbf{e}}_r,{\mathbf{e}}_z$, so it has no ${\mathbf{e}}_{\theta }$ component. The transverse component of Newton's second law, $m\ddot{\mathbf{r}}=\mathbf{N}-mg{\mathbf{e}}_z$, is thereforewhich gives $\boxed{h=r^2\dot{\theta }=\text{constant}}$.

The velocity is tangent to the stationary surface, so $\mathbf{N}\cdot \dot{\mathbf{r}}=0$. Take the dot product of Newton's second law with $\dot{\mathbf{r}}$:Using $z=f(r)$ proves conservation of mechanical energy per unit mass:The ideal constraint does no work. These equations describe motion while the particle remains constrained to the surface.

Take $a>0$ and the intended horizontal launch around a parallel of the surface of revolution: $\dot{r}(0)=\dot{z}(0)=0$, $a\dot{\theta }(0)=v$. Then $h=av$ and $E=v^2/2+gf(a)$. Since $\dot{z}=f^{\prime }(r)\dot{r}$, the conserved energy becomesHere $U$ is the radial effective potential on a surface of revolution. Rearrangement givesFor a nonconstant bounded radial motion with $v>0$, there must be another finite radial bound $b\ne a$ satisfying $D(b)=0$, equivalentlyThe interval between the radial bounds must be dynamically accessible: $D\ge 0$ inside, and a simple endpoint must have a forbidden region $D<0$ beyond it. Equivalently the motion must lie in a bounded part of $\{r>0:U(r)\le E\}$ separated from escape by a potential barrier. When $f$ stays finite near zero and $v>0$, the centrifugal barrier already excludes $r=0$; an outer barrier is the additional requirement. An arbitrary second algebraic root on a disconnected allowed branch would not establish boundedness.

The initial radial acceleration, obtained by projecting Newton's second law along ${\mathbf{e}}_r+f^{\prime }(a){\mathbf{e}}_z$, isThus $v^2/g>a{f}^{\prime }(a)$ starts the particle outward, and $v^2/g<a{f}^{\prime }(a)$ starts it inward. Equality gives the exact circular orbit $r=a$ and needs no second distinct radial bound. A double root at a different radius can instead be approached asymptotically: this separatrix is bounded, although the radial motion is not periodic. If $v=0$, the centrifugal barrier disappears and the particle may pass through the axis, still with bounded $r$.

The source leaves a direction ambiguity at points where $f^{\prime }(a)=0$: every horizontal tangent direction is then possible, so “projected horizontally” alone does not imply $h=av$. The displayed inequality uses the azimuthal launch specified above; at $f^{\prime }(a)\ne 0$ that direction is forced by horizontality and tangency.

For a regular isoperimetric constraint, introduce a constant Lagrange multiplier $\lambda$ and set $L=F-\lambda G$. With fixed endpoint values, the Euler-Lagrange equation isIf $F$ and $G$ have no explicit $x$ dependence, the Beltrami identity gives the first integralThe regularity assumption excludes the exceptional case in which the constraint itself has zero first variation for every admissible perturbation.

Parametrize the surface of revolution by $r(z,\theta )=(f(z)\cos \theta ,f(z)\sin \theta ,z)$. The magnitude of $r_z\times r_{\theta }$ is $f\sqrt{1+f^{\prime 2}}$. Integrating the resulting surface area element over $\theta$, and integrating the circular cross-sectional areas for the volume, givesHere $f\ge 0$ because it is a radius. For a regular connected closed surface, take $f>0$ in the interior; the closing endpoints can have unbounded $f^{\prime }$ even when the surface itself is smooth.

After dividing $A-\lambda V$ by $\pi$, its integrand is $L=2f\sqrt{1+f^{\prime 2}}-\lambda f^2$. The Euler-Lagrange equation and Beltrami identity becomeAs $f$ tends to zero at an endpoint, $0\le 2f/\sqrt{1+f^{\prime 2}}\le 2f$ also tends to zero, so $C=0$. On the positive interior this implies $\lambda >0$ and, writing $R=2/\lambda$,The Euler-Lagrange equation also yields $f^{\prime \prime }=-R^2/f^3$. Thereforeincluding at the maximum where $f^{\prime }=0$. Integrating and using the first integral givesThus the regular extremal is a sphere. The two zeros fix its center and radius:Equivalently, the prescribed volume fixes $R=(3V/(4\pi ){)}^{1/3}$. If $a,b$ are fixed in advance, their separation and the prescribed volume must be compatible with these equations; there is no positive regular stationary profile of the stated type for an incompatible volume.

For minimality, the three-dimensional isoperimetric inequality states that every sufficiently regular closed surface enclosing volume $V$ haswith equality precisely for a sphere. Our extremal has $A=4\pi R^2$ and $V=4\pi R^3/3$, so it attains equality and is a global minimum of area at the fixed volume. This also makes clear the connected-surface assumption behind uniqueness: allowing several disconnected components introduces stationary collections of spheres, rather than the single positive generating profile used here.

For the surface of revolution $r(\eta ,\zeta )=(\eta ,f(\eta )\cos \zeta ,f(\eta )\sin \zeta )$, the coordinate tangent vectors have squared lengths $1+f^{\prime 2}$ and $f^2$, and their dot product is zero. Hence its first fundamental form isUse the energy Lagrangian $L=A{\dot{\eta }}^2+f^2{\dot{\zeta }}^2$. Its Euler-Lagrange equations areFor a meridian $\zeta ={\zeta }_0$, choose arc length as parameter, so $\dot{\eta }=1/\sqrt{A}$. Differentiating gives $2A\ddot{\eta }+A^{\prime }{\dot{\eta }}^2=0$, while the second equation is automatic. Part (b) makes this path an extremal of length. Thus every meridian is a geodesic.

For a nonconstant parallel $\eta ={\eta }_0$, choose constant nonzero $\dot{\zeta }$. The second equation holds, and the first reduces to $f({\eta }_0)f^{\prime }({\eta }_0){\dot{\zeta }}^2=0$. Since $f>0$, this is equivalent toConversely that condition makes both equations hold at constant speed, hence supplies a length extremal. Geometrically, a parallel is geodesic exactly at a stationary radius of the surface.

Parametrize the surface of revolution byThe magnitude of the cross product ${\mathbf{r}}_x\times {\mathbf{r}}_{\theta }$ is $y\sqrt{1+y^{\prime 2}}$. Integrating the surface area element over $\theta$ givesPositivity of $y$ removes an otherwise necessary absolute value in the area element.