Geodesics in a surface of revolution

Consider a surface of revolution with respect to $z$-axis and suppose that the generating curve $s\mapsto (f(t),0,g(t))$ is parametrized by the arc-length. Let $\alpha(s)=X(u(s),v(s))$ be a geodesic where $X$ is the parametrization of the surface given by

$$X(u,v)=(f(u)\cos(v),f(u)\sin(v),g(u)).$$Suppose that $\alpha$ is a geodesic. Then we know that the functions $u(s)$, and $v(s)$ satisfy  two equations involving the Christoffel symbols. We pay attention in the second one, which is $$v''+\Gamma_{11}^2 u'^2+2\Gamma_{12}^2u'v'+\Gamma_{22}^2v'^2=0.$$

The computation of these symbols are: $\Gamma_{11}^2=\Gamma_{22}^2=0$ and $\Gamma_{12}^2=f'/f$. Thus the equation writes as

$$v''+\frac{f'}{2f}u'v'=0.$$ We deduce that $$(f^2v')'=2ff'v'+f^2v''=0.$$

If we compute the angle $\theta(s)$ that makes $\alpha$ with each parallel that meets, we have

$$\cos\theta=\frac{\langle u'X_u+v' X_v,X_v\rangle}{1}=v'$$ because $\alpha$ is parametrized by the arc-lengt since it is a geodesic. As a conclusion

Theorem: In a surface of revolution, we have that the angle that makes a geodesic with every parallel that intersects satisfies

$$f(u(s))^2\cos\theta(s)=\mbox{constant}.\quad (*)$$

Application. A sailor what to find/keep a specific path along a trip on the Earth. He needs to know at every time what is its position. The knowledge of the parallel is easy measuring with respect to the Polar star and this gives the value of $f$ by the radius of the Earth. Thus if the sailor want to follow a way doing a given angle $\theta$ with the parallel, he only has to maintain the value of the constant at (*): the value of $v(s)$ indicates which is the meridian where he is positioned.

Surfaces of revolution with constant mean curvature (II)

Following the above entry, the case $H=0$ is known: the plane and the catenoid are the only rotational minimal surfaces. If $H=0$, then we have 

$$\frac{f(z)}{\sqrt{1+f'(z)^2}}=c,\ c>0,$$ that is, $$\frac{f'}{\sqrt{f^2-c^2}}=\frac{1}{c}.$$ Then the solution is $$f(z)=c\cosh(\frac{1}{c}z+d),\ d\in{\mathbb R} (*).$$

Minimal surfaces are models of soap films. In this particular case of the catenoid, the surface is the soap film formed by two coaxial circles, that is, two circles $C_1\cup C_2$ in parallel planes and the straight-line joining their centers is orthogonal to the planes containing the circles. It is natural to ask if there exists a soap film joining two given circles in parallel planes. In order to simplify the arguments, we suppose 

1. the radii of the circles are identical, namely, $r>0$. 
2. the circles $C_1$ and $C_2$  are contained in the planes of equation $z=-h$ and $z=h>0$, respectively. 

Then we pose the next:

Problem. Under what conditions on $r$ and $h$ does exist a catenoid $S$ joining $C_1$ and $C_2$? In such a case, how many catenoids do exist?

By the symmetry of the hyperbolic cosine, and since $f(h)=f(-h)$, we conclude $d=0$ in (*). Thus, the problem reduces to find $c>0$ such that $$c\cosh(\frac{1}{c}h)=r (**).$$

It is natural to think that if the circles lie very close, then there do exists a catenoid, that is, if $h$ is small, then there exists a solution of (**). 

We propose the problem in the next direction. We suppose that the circles are given (the radius $r$, which we suppose $r=1$). If they are close, there exists a catenoid, but if we separate far then the catenoid is destroyed, that is, there do no exist a catenoid between both circles.

In order to simplify the problem, we do a homothety of the ambient space from the origin and we suppose that the value $r$ of the radius is $r=1$. Consider the function $$g(c)=c\cosh(\frac{1}{c}h).$$ Our idea is using the mean value theorem. It is not difficult to see that $$\lim_{c\rightarrow 0}g(c)= \lim_{c\rightarrow \infty}g(c)=\infty,$$

so we have to study carefully the monotonicity intervals of $g$. We calculate the critical points of $g$. We have

$$g'(c)=\cosh(h/c)-\frac{h}{c}\sinh(h/c).$$ By letting $y=h/c$, this is equivalent to find $y>0$ such that $$\frac{1}{y}=\tanh(y).$$ The function $1/y$ is decreasing from $-\infty$ to $0$ and $\tanh(y)$ is increasing from $0$ to $1$, s   there is only one critical point $c_0$ (with $h>c_0$). See the next figure:

Because the limits are $\infty$, then this critical point is a minimum, $c=c_0$. This the graphic of $g$ when $h=0,5$, $h=1$ and $h=4$, and the graphic of $y=1$.

Let us observe that the minimum increases with $h$! so we have to find that it is possible to choose $h$ so the value of this minimum is $1$ at more.

We compute the value of the minimum, that is, $g(c_0)$. We know that $c_0/h=\tanh(h/c_0)$, and numerically we obtain, $h/c_0=1.19968$. Then 

$$g(c_0)=\frac{hc_0}{\sqrt{h^2-c_0^2}}=\frac{h}{\sqrt{(h/c_0)^2-1}}=1,50888 h.$$

When $h$ is close to $0$, $g(c_0)<1$, thus the graphic of $g$ has points under the line $y=1$, proving that there exists two catenoids spanning $C_1\cup C_2$. For a certain height $h=h_0$, this minimum is exactly $1$, so there exists only one catenoid and when $h>h_0$ there do not exist a catenoid joining $C_1$ and $C_2$. The value of $h_0$ is

$$h_0=\frac{1}{1,50888}=0,6627.$$
Thus, and after a homothety, we obtain:

Theorem. Let $d$ be the distance $d$ between two coaxial circles of radii $r>0$.

1. If $d<1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
2. If $d=1,3254 r$, there exists exactly two catenoids spanning $C_1\cup C_2$.
3. If $d>1,3254 r$, there do not  exist a catenoid spanning $C_1\cup C_2$.

Now we give one example of two circles that bound two catenoids. Take $r=1$ and we choose $h=0.5$. The solutions of $g(c)=1$ are: $c_1=0,235095$ and $c_2=0,848338$. The picture of the two catenoids is

Finally a remark: when one dips two coaxial circles in a soapy water container, only one catenoid is formed. In the above case, it would be the blue catenoid. Among the two catenoids, the physical systems chooses that catenoid with minimum area (minimum energy) and in this case, is the 'exterior' catenoid.

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  $H=1/(2r)$.  

Surfaces of revolution with positive constant Gauss curvature

We know that when we write $K=c$ in the family of rotational surfaces, then this equation is an ordinary differential equation, so there is a unique solution for each initial conditions. We show this phenomenon when $K=1$. Suppose that the profile curve is locally a graph on the rotation axis, that is, $z\mapsto (f(z),0,z)$ for $z\in I$, $f(z)>0$. The parametrization of the surface is $X(z,s)=(f(z)\cos(s),f(z)\sin(s),z)$. Equation $K=1$ writes as 

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

Thus we have $f''+f(1+f'^2)^2=0$. This is differential equation is not possible to integrate, up to special cases. We think that sphere should easily solve. The initial conditions are put on $z=0$, that is $f(0)=xo$ and $f'(0)=0$. With this last condition, we are imposing that the tangent line at $z=0$ is vertical. Moreover, by this condition, we can suppose that the solution is symmetric with respect to $z=0$. 

We use Mathematica to solve numerically the initial value problem $$(*) \left\{\begin{array}{l} f''+f(1+f'^2)^2=0\\ f(0)=xo\\ f'(0)=0\end{array}\right.$$ We study the solutions depending on the initial value $xo$, that is, the intersection point of the profile curve with the $x$-axis.

When $xo=1$, we know that the solution is the sphere, exactly, $f(z)=\sqrt{1-z^2}$ is a solution of (*). 

In order to study with Mathematica (*) we write here the sentences: 

profile =  NDSolve[{F''[z] + F[z] (1 + F'[z]^2)^2 == 0, F[0] == xo, F'[0] == 0}, F[z], {z, -Zo, Zo}]

f[z_] := F[z] /. profile[[1]]

ParametricPlot[{{z, 0}, {f[z], z}}, {z, -Zo, Zo}, PlotRange -> All]

The first line numerically solves the ODE with initial conditions as we have presented. Here $Zo$ is the width of the interval when the solution $f$ is defined. The second line `takes' the numerical value f in order to manage in the next line, where we plot the profile curve. In fact, the last line indicates that we also draw the $x$-line. I write this because Mathematica `reduces' the picture to the interval where is defined the solution and we want to compare the profile curve with its position with respect to the rotation axis. Finally, we use

ParametricPlot3D[{f[z] Cos[s], f[z] Sin[s], z}, {s, 0, 2 Pi}, {z, -Zo, Zo}]

for drawing the surface.

We begin with the study and sphere is our starting point: sphere appears when $xo=1$ and the domain of $f$ is for $Zo=1$. We now increase $xo$, for example $xo=1.5$. If we put $Zo=1$, Mathematica says that the solution is not defined in the interval $(-Zo,Zo)$ because appear errors. In fact, Mathematica says what is the maximum interval. In this example, the output is

NDSolve::ndsz: At z == -0.559099, step size is effectively zero; singularity or stiff system suspected.

This means that we have to take $Zo=0.5590$, obtaining the profile curve in its maximum domain, namely:

If we increase $xo$, that is, we move far the point $(f(xo),0,0)$, the profile moves far from the rotation axis: let us observe that the profile curve does not meet the rotation axis. In the figure, it indicates that the tangent plane at the boundary circles is horizontal, and the surface one `hole'.

Now we let $xo\rightarrow 0$. If $xo=0.7$, and for $Zo=2$ we see that the profile curve meets the $z$-axis, which is not possible.

Then, and after some trials, we see that for $Zo=1.35$, the profile meets exactly the $z$-axis. The figures are:

Now the surface presents two `singularities' exactly in the intersection points with the $z$-axis.

On surfaces of revolution (II)

The idea about the concept of a surface of revolution is as 'something that rotates'. However there is a characterization of this class of surfaces in terms of the tangent planes. It is not difficult to see that in a surface of revolution the normal lines through any point meets the rotation axis. Now, and it is here the surprise, this property characterizes a surface of revolution. Thus the result is the following.

Theorem. If all normal lines in a surface meet a given straight-line $L$, then the surface is included in a surface of revolution and $L$ is the rotation axis.

The proof consists into prove that the intersection of any orthogonal plane to $L$ with the surface $S$ is a circle centered at $L$. Then it suffices to finish the result: the surface is formed by the union of (arcs of)  circles centered at $L$ and this is just the definition of a surface of revolution. Denote by $N$ the unit normal vector to $S$.

First step. Let $P$ be a orthogonal plane to $L$ that meets $S$. In particular, for any $p\in S\cap P$, $P\not= T_pS$: on the contrary, the normal line is parallel to $L$ so it does not meet $L$. Thus  $S$ and $P$ meet transversally and $S\cap P$ can parametrized as a regular curve, namely, $\alpha=\alpha(s)$. 

Second step. The normal line of $\alpha$, as curve of ${\mathbb R}^3$, meets $L$. First, recall that the normal line is included in $P$. Furthermore, the normal vector $n(s)$ of $\alpha$ at $s$ is orthogonal to $\alpha'(s)$, which lies in $P$. But the orthogonal projection $\pi(N(\alpha(s)))$ of $N(\alpha(s))$ on $P$ is also a vector orthogonal to $\alpha'(s)$. Thus $n(s)$ and $N(\alpha(s))$ are collinear. Since the normal line through $\alpha(s)$ meets $L$, the same occurs for the line through $\alpha(s)$ and with direction $\pi(N(\alpha(s)))$. 

Third step. The only planar curve whose normal lines meet at one point $p_0$ is a circle centered at $p_0$. Indeed, for each $s\in I$, there exists $\lambda(s)$ such that $p_0=\alpha(s)+\lambda n(s)$, where we are assuming that $\alpha$ is parametrized by the length-arc. If we differentiate with respect to $s$ and using the Frenet equations, we obtain, $$0=\alpha'(s)+\lambda'(s)n(s)-\lambda(s)\kappa(s)\alpha'(s).$$ This proves that $\lambda'=0$ on $I$, that is, $\lambda$ is a non-zero constant and $1-\lambda(s)\kappa(s)=0$, so $\kappa(s)=1/\lambda$, that is, $\alpha$ is included in a circle. 

On surfaces of revolution

There are two ways to define a surface of revolution in Euclidean space ${\mathbb R}^3$.

1. A surface $S$ is a surface of revolution with respect to the line $L$ is $\phi(S)=S$ for any rotation of axis $L$ (type I)

2. A surface of revolution is a surface constructed as follows. Fix $L$ a straight-line and let $P$ be a plane containing $L$. Consider a curve $C$ contained in $P$. Then the surface of revolution generated by $C$ is the set of points obtained when we rotate $C$ about the axis $L$ (type II). We denote this surface as $S(C)$

It is clear that any surface of type II is a surface of type I by the definition given in 2. It is less clear if a surface of type I is of type II, that is, if $S$ satisfies I, is there exists a curve $C$ in a plane $P$ containing $L$ such that $S=S(C)$? After a rigid motion of ${\mathbb R}^3$, we suppose that $L$ is the $z$-axis. 

Take $P$ a plane containing $L$. Then at any point   $p\in S\cap P$, the surfaces $S$ and $P$ are transversal, that is, $T_pS\not=T_pP$. Indeed, if $p=(x,y,z)\in S$ with $x^2+y^2\not=0$ (we are assuming that $S$ does not intersect the axis $L$), the rotation about the $z$ axis is the curve $$\alpha(\theta)=  \left(\cos\theta x-\sin\theta y,\sin\theta x+\cos\theta y,z \right).$$ Since $\alpha(0)=p$, then $\alpha'(0)\in T_pS$, that is, $(-y,x,0)$. The plane $P$ containing $p$ and the $z$ axis is the plane orthogonal to $(-y,x,0)$ through $p$. This proves $T_pS\not=T_pP$.

As a consequence $S\cap P$ defines a regular curve $C_p$ around $p$. Since $C_p\subset S$, then $\phi_\theta(C_p)\subset S$ for any rotation $\phi_\theta$ about the $z$-axis. In particular, the surface of revolution $S(C_p)\subset S$. 

Therefore we have prove that if $C=S\cap P$ is the intersection curve (with possible many components), then $S(C)\subset S$.

For the other inclusion, if $(x,y,z)\in S$, it is immediate that the circle $\alpha(\theta)$ defined previously intersects $P$. For example, if $P$ is the $xz$-plane, we are asking if there exists $\theta$ such that  $$\sin\theta x+\cos\theta y=0.$$ It suffices by taking $\theta$ such that $\tan\theta=-y/x$ if $x\not=0$ and $\theta=\pi/2$ if $x=0$ (it is not possible $x=y=0$). If $q=\alpha(\theta)$, then it is immediate that a suitable rotation of $q$ (exactly that rotation with angle $-\theta$) gives $p$.

Surfaces constructed from curves (II)

Surfaces of revolutions are other type of surfaces constructed by curves. In the previous entry, a cylinder is noting a planar curve $\alpha$ moved along a fix direction $a$, that is, we translate $\alpha$ along a direction. If $a$ is the given direction, a translation in this direction is $T_t(x)=x+t a$, $x\in{\mathbb R}^3$ and $t\in{\mathbb R}$. Then the cylinder on basis $\alpha$ is $$\cup_{t\in {\mathbb R}}T_t(\alpha(s)):s\in I\}.$$

In order to  define a surface of revolution, we consider a   curve $\alpha$ contained in a plane $P$ and we rotate $\alpha$ about a line $L$ contained in the plane $P$. We know that the parametrization is $X(s,\theta)=(f(t)\cos\theta,f(t)\sin\theta,g(t))$, where $\alpha(t)=(f(t),0,g(t))$. Again, the difficulties appear when we prove that $X$ is an embedding. For this reason we assume again that $\alpha$ is an embedding or a simple closed curve. 

With the curve $\alpha(t)=( \sin(t),0,1+\cos(t)\cos(2t))$, with $t\in (0.5,2.5)$, we observe that there is a self-intersection, so it does not define a surface. 

Other example is the torus generated by the circle $\alpha(t)=(1+2\cos(t),0,2\sin(t))$ because it intersects the $z$-axis.

Definitively, we impose that the curve $\alpha$ is an embedding or a simple closed curve. In the next picture, the surface is generated by the simple closed curve  $\alpha(t)=(2+\sin(t),0,\cos(t)+\cos(2t))$.

Parametrizations of a surface of revolution

Consider $S$ the surface of revolution obtained by rotating the planar curve $\alpha(t)=(f(t),0,g(t))$, $t\in I$ which it is contained in the halfplane $y=0, x\geq 0$. We know that $\alpha$ is regular and we have two possible types of curves, i) $\alpha$ is an embedding or ii) $\alpha$ is a simple closed curve. The surface $S$ is $X(I\times{\mathbb R})$, where $$X(t,\theta)=(f(t)\cos\theta,f(t)\sin\theta,g(t)).$$ In order to prove that $S$ is a surface, it appears the problem about how many of parametrizations are needed and how to prove that they are homeomorphisms.

First consider that $\alpha$ is an embedding. We compute the inverse function of $X$ so we will discover what is the right domain of $X$. By the injectivity, it is necessary that $\theta$ moves in an interval of length $2\pi$ at most. Thus a possibility is $X:U_1:=I\times (0,2\pi)\rightarrow X(U_1)$. The set $X(U_1)$ is an open set of $S$ because $$X(U_1)=S-\alpha(I)=S-S\cap(\{y=0,x\geq 0\}).$$ Here we use that $f(t)>0$. Because we need to cover the curve $\alpha(I)$, then the other parametrization is $Y:U_2:=I\times (-\pi,\pi)\rightarrow X(U_2)$. For $Y$ we have to remove `the other side' of $S$, that is $\Phi_\pi(\alpha(I))$, where $\Phi_\theta$ is the rotation of angle $\theta$. In other words, $Y(U_2)=S-(S\cap(\{y=0,x\leq 0\})$.

We compute the inverse of $X$ (or $Y$). We have to write $t$ and $\theta$ in terms of $x,y,z$ from the next equations $$\left\{\begin{array}{l}x=f(t)\cos\theta\\ y=f(t)\sin\theta\\ z=g(t)\end{array}\right.$$ From $x^2+y^2=f(t)^2$, we obtain $f(t)=\sqrt{x^2+y^2}$, and using that $\alpha$ is an embedding, then $t=\alpha^{-1}(\sqrt{x^2+y^2},0,z)$. For $\theta$ we have two possibilities:

1. If we want to write `something' with the arc tangent, then it would be $\theta=\mbox{arc tan}(y/x)$. But in such a case, $\theta$ is defined in $(-\pi/2,\pi/2)$. Because the initial interval is $(0,2\pi)$, we have to change the domain of $X$. We write now $X:U_1:=I\times(-\pi/2,\pi/2)\rightarrow X(U_1)$. The only difference is that we have to prove that $X(U_1)$ is an open set of $S$. But by the picture, $X(U_1)=S-(S\cap \{x\leq 0\})$. Other parametrization is with $I\times (\pi/2,3\pi/2)$ where it is possible to define the inverse of the tangent. And what about the points with $\theta=\pi/2$ or $\theta=3\pi/2$? Here, the $x$-coordinate of the point $(x,y,z)\in S$ vanishes. Now we consider that inverse of the cotangent and taking $\theta={\mbox arc cot}(x/y)$. We need two more domains, namely, $I\times (0,\pi)$ and $I\times (\pi,2\pi)$. Finally, we observe that, using the inverse functions of the tangent of the cotangent function, we need 4 parametrizations: all them write 'very similar', but  the domain goes changing.
2.  If we do not want to use trigonometric functions, we can do the following. If one looks the picture of a surface of revolution, it is clear that the parametrization $X$ can be defined in $I\times (0,2\pi)$, because the angle $\theta$ is well defined. The problem appeared in how to catch the variable $\theta$. The trigonometric functions have added a bit of confusion. Other way is the following.                                                                                                                        Consider $\beta:(0,2\pi)\rightarrow {\mathbb S}^1-\{(0,0\})$ a parametrization of the circle minus one point. The key is the following: the map $\beta$ is a homeomorphism! It is clear that $\beta$ is one-to-one and continuous. One could do the following argument. It is well known that  ${\mathbb S}^1$ minus one point is homeomorphic to the real line ${\mathbb R}$, which it is homeomorphic to the interval $(0,2\pi)$. But the problem is if $\beta$ is, indeed, a homeomorphism. The only trouble is about the inverse. But $\beta$ is an open map because the image of an interval of $(0,2\pi)$   is an open set of ${\mathbb S}^1$.             Once proved that $\beta$ is a homeomorphism, from $x=f(t)\cos\theta$ and $y=f(t)\sin\theta$ we conclude $$\theta=\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$ Then $$X^{-1}(x,y,z)=\left(\alpha^{-1}(\sqrt{x^2+y^2}),\beta^{-1}\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)\right).$$ Thus we need 2 parametrizations.

If $\alpha$ is a simple closed curve, for covering the curve $\alpha$ by embeddings, we need two parametrizations, namely, $\alpha:(0,T)\rightarrow \alpha(0,T)$ and $\alpha:(T/2,3T/2)\rightarrow \alpha(T/2,3T/2)$. Then for the surface, we need 4 parametrizations.

There is another elegant argument. We observe that when one has proved that $X$ is a parametrization, then we think $Y$ as a `rotation' of $X$. Recall that if $S$ is a parametrization and $\psi$ is a diffeomorphism of ${\mathbb R}^3$, then $\psi(S)$ is a surface. But the parametrizations of $\psi(S)$ are of type  $\psi\circ X$, where $X$ is a parametrization of $S$. With this idea in mind, consider now $\Phi_\theta$ the rotation about the $z$-axis of angle $\theta$. By the definition of $S$, we have $S=\Phi_\theta(S)$. Suppose that we have proved that $X:U_1:=I\times (-\pi/2,\pi/2)\rightarrow {\mathbb R}^3$ is a parametrization of $S$: it is only of a part of $S$, exactly, $V_1=X(U_1)$. Now we take $\theta\in{\mathbb R}$. Because $\Phi_\theta$ is a homeomorphism, it is clear that $\Phi_\theta\circ X$ satisfies the properties of a parametrization, where now the coordinate open is $\Phi_\theta(V_1)$. We point out that $\Phi_\theta(V_1)$ is an open set of $S$. Therefore, if we are going taking many $\theta$, we cover all the surface $S$ by coordinate opens of type $\Phi_\theta(U_1)$. In fact, it is enough only one rotation!, namely, $\Phi_\pi$. With this argument, the effort lies only in the first parametrization $X$: the other ones are obtained `by rotating' $X$.

Finally, we conclude that it is clear that the parametrizations $X$ hold to prove that $S$ is a surface of revolution, for example, taking very small domains, or rotations of $X$. However, the proof is a bit tedious if one wants to write precise arguments.