Surface with only one parabolic point

We have proved that at an elliptic point, the surface lies in one side of the tangent plane at that point. Exactly, if $K(p)>0$, then there exists a neighborhood $V\subset S$ of $p$ such that $V\cap T_pS=\{p\}$ and $V-\{p\}$ lies in one of the two open halfspaces determined by $T_pS$. 

Here we show a surface with the same property but $K(p)=0$. The surface is obtained by rotating the curve $z=x^4$ around the $z$-axis. At the point $p=(0,0,0)$, $K(p)=0$, the tangent plane $T_pS$ is the $xy$-plane and $S-\{p\}$ lies in the halfspace $z>0$. Exactly, with the usual parametrization $X(x,s)=(x\cos(s), x\sin(s), x^4)$, we have 

$$K(x,s)=\frac{36 x^4}{(1+16x^6)^2},$$

so $K(p)=0$ and the rest of points are elliptic, that is, $p$ is the only parabolic point of $S$.

Metrics and Gauss curvature

The first and second fundamental form in a surface are symmetric bilinear forms, in particular they are metrics. Of course, the first fundamental form is positive definite because is the Euclidean metric in the tangent plane. We consider the second fundamental form $$\sigma_p:T_pS\times T_pS\rightarrow{\mathbb R}$$ $$\sigma_p(v,v)=\langle -dN_p(v),v\rangle.$$

If $X=X(u,v)$ is a parametrization of the surface, the matrix of $\sigma_p$ with respect to the basis $\{X_u,X_v\}$ is $$\left(\begin{array}{cc}e&f\\ f&g\end{array}\right).$$

We study the type of $\sigma_p$. However, it is better to choose a more suitable basis of $T_pS$, indeed, a basis of principal directions because in such a case, $$\sigma_p\rightarrow \left(\begin{array}{cc}\kappa_1(p)&0\\ 0&\kappa_2(p)\end{array}\right).$$ Because this basis diagonalizes the metric $\sigma_p$, we conclude:

1.  The point is elliptic ($K(p)>0$) is equivalent to $\kappa_i(p)>0$ for $i=1,2$, or $\kappa_i(p)<0$ for $i=1,2$ and this means that   $\sigma_p$ is  definite. 
2. The point is hyperbolic ($K(p)<0$) is equivalent to $\kappa_1(p)<0<\kappa_2(p)$. Then $\sigma_p$ is a non-degenerate indefinite metric with signature $(1,1)$.
3. $K(p)=0$, that is, some principal curvature is $0$. We have two subcases. 
• If the point is parabolic, then the non-zero principal curvature is positive or negative. This is equivalent to say that $\sigma_p$ is positive semidefinite or negative semidefinite and also, the metric is degenerate. Here the radical of $\sigma_p$ is the vector subspace spanned by the principal direction of the zero principal curvature.
• If the point is flat, then the metric is null.  

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.

Elliptic and parabolic points.

Two days ago, we have shown a surface where all its points are hyperbolic. Now, we consider the elliptic paraboloid $z=x^2+y^2$. Now $$K=\frac{4}{(1+4x^2+4y^2)^2},$$ proving that all its points are elliptic. 

On the other hand, the parabolic paraboloid $z=x^2$ has as Gauss curvature $K=0$ so all its points are parabolic or flat. In order to distinguish, we have to compute the principal curvatures. Since $K=0$, the principal curvatures are $\kappa_1=H$ and $\kappa_2=0$ by the relation $$\kappa_i=H\pm\sqrt{H^2-K}.$$ The mean curvature is given by $$H=\frac12\frac{(1+f_y^2)f_{xx}-2f_xf_yf_{xy}+(1+f_x^2)f_{yy}}{(1+f_x^2+f_y^2)^{3/2}}=\frac{1}{(1+4x^2)^{3/2}}\not=0$$ for any $x$. This proves that all its points are parabolic.

In the next pictures we have a elliptic paraboloid (left) and a parabolic paraboloid (right).

Surfaces with constant Gauss curvature

Consider surfaces where the Gauss curvature $K$ is a 'simplest' function, namely, $K$ is constant on the surface. For example, in a plane $K=0$ and in a sphere of radius $r$, $K=1/r^2$. Other surface with constant Gauss curvature is the cylinder, where $K=0$ again. If one want to obtain more examples, one may consider this problem in the family of surfaces of revolution. In such a case, the equation $K=c$ is an ordinary differential equation, and by the general theory, for each initial conditions, there is a solution. Thus this provides us a huge family of examples. 

If the profile curve is a graph on the $x$-line, that is, $\alpha(x)=(x,0,f(x))$, then 

$$K(X(x,\theta))=\frac{f'f''}{x(1+f'^2)^2}.$$

We know that when $K=0$, then $f''=0$, obtaining planes, circular cylinders and cones. On the other hand, the equation $$\frac{f'f''}{x(1+f'^2)^2}=c$$ is not possible to integrate completely and only some particular cases can solved: for example, the sphere and the pseudosphere. 

However, if we assume that the surface is compact, then panorama changes drastically, because we have

Theorem (Hilbert): Sphere is the only compact surface with constant Gauss curvature.

We observe that the topological assumption on compactness is essential in the result. It is known that in a compact surface there exists elliptic points, so the value of the constant $K$ must be positive. Then the key is based in the following result:

Lemma (Hilbert): Let $S$ be a surface and $p\in S$ an elliptic point. Consider $\kappa_1\geq \kappa_2$ the principal curvatures on $S$. If $p$ is a global maximum for $\kappa_1$ and it is a global minimum for $\kappa_2$, then $p$ is an umbilical point.

The proof of the Hilbert's theorem is then as follows. Since $K=\kappa_1\kappa_2>0$, the sign of the principal curvature is the same. After a change of orientation, we suppose $\kappa_i>0$. Let $p\in S$ a point where the (continuous) function $\kappa_2$ attains a global minimum. Since $\kappa_1=\frac{c}{\kappa_2}$, then $p$ is a global maximum for $\kappa_1$. By the lemma, $\kappa_1(p)=\kappa_2(p)$. Then we have for any $x\in  S$, $$\kappa_2(p)\leq\kappa_2(x)\leq\kappa_1(x)\leq\kappa_1(p)$$ and thus, $\kappa_1=\kappa_2$ on $S$. Then $S$ is an open of a plane or a sphere. Since $S$ is closed, by connectedness, $S$ is a plane or it is a sphere. But sphere is the only one that is compact.

A surface whose all points are hyperbolic

Consider the hyperbolic paraboloid, that is,  the surface $S$ given by the graph of $z=f(x,y)=x^2-y^2$. By the formula of the Gauss curvature for a surface which is a graph, we have $$K=\frac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}=-\frac{4}{(1+4x^2+4y^2)^2}<0.$$  

Then all points are negative. Consider the point $p=(0,0,0)\in S$, whose tangent plane is the plane $z=0$. Let $c>0$. Then the intersection of $S$ with planes parallel to $T_pS$ is $$S\cap \{z=c\}=\{(x,y,c): x^2-y^2=c\}=\{(x,y,c):\left(\frac{x}{\sqrt{c}}\right)^2-\left(\frac{y}{\sqrt{c}}\right)^2=1\},$$

that is, an ellipse, for each height $c>0$. Similarly, for $c<0$.

Comparison surfaces (IV): elliptic points

We prove the analogous result that was showed for curves about the position of the surface with respect to the tangent plane in terms of its Gauss curvature.

Since the result is local, we suppose that the surface at the point $p$ is tangent to the plane $z=0$ and writes as $z=f(x,y)$ with $p=(q,0)=(0,0,0)$. In such a case, we know that $$K(p)=(f_{xx}f_{yy}-f_{xy}^2)(q).$$

If $K(p)>0$, then the determinant of the Hessian is positive. Since $f_{xx}(q)\not=0$ (on the contrary, $K(p)\leq 0$), then $f_{xx}(q)$ is positive or negative, that is, the Hessian is positive definite or negative definite, respectively. This proves that $q$ is a local minimum or a local maximum, respectively, proving:

Theorem. If $K(p)>0$, then the surface lies in one side of $T_pS$ around $p$.

Corollary. If in any neighbourhood around $p$, $S$ has points in both sides of $T_pS$, then $K(p)\leq 0$.

Comparison of curves by curvatures (III)

By the theorem of the previous entry, we prove:
Theorem. Any compact surface has points with positive Gauss curvature.

Proof. Take $p_0\in S$ the fairest point of $S$ from the origin of ${\mathbb R}^3$: this point exists because $S$ is compact and the distance function to a fixed point is a continuous function. We do the next steps.

Take ${\mathbb S}^2(r)$ the sphere centered at the origin and radius $r=|p_0|$: this number is positive because on the contrary is only one point. 

The surfaces $S$ and ${\mathbb S}^2(r)$ are tangent at $p_0$. For ${\mathbb S}^2(r)$ we know that the tangent plane is orthogonal to the position vector $p_0$. For $S$, consider the function $f(p)=|p|^2$. Because $p_0$ is a maximum, it is a critical point, so $df_{p_0}=0$. But it is is immediate that $df_{p_0}(v)=2\langle p_0,v\rangle$ for any $v\in T_{p_0}S$. Thus $T_{p_0}S$ is orthogonal to $p_0$. 

We orient ${\mathbb S}^2(r)$ according the orientation pointing inside, so the normal curvature for any tangent vector is $1/r$. Consider the orientation on $S$ so $N(p_0)=-p_0/|p_0|$, that is, the same than ${\mathbb S}^2(r)$. Moreover, $S$ lies above ${\mathbb S}^2(r)$ around $p_0$.

By the Theorem in the previous day, $\kappa_n(v)\geq 1/r$, in particular, in along the principal directions, $\kappa_i(p_0)\geq 1/r$, so $K(p_0)\geq 1/r^2$.

In particular, we have an estimate of the Gauss curvature at the fairest point  $p_0$ from the origin: $$K(p_0)\geq\frac{1}{|p_0|^2}.$$

As a consequence of the inequality $H^2\geq K$, we have:

Corollary. There are no compact minimal surfaces.

Comparison of curves by curvatures (II)

The definition of the mean curvature is the average of the principal curvatures. But it is also the average of the normal curvatures in two orthogonal directions, that is, 

$$H(p)=\frac12\left(\kappa_n(\theta)+\kappa_n(\theta+\frac{\pi}{2})\right),\ \ (*)$$ where $\kappa_n(\theta)$ is the normal curvature in the direction that makes an angle $\theta$ with a fix direction of the tangent plane.

Suppose again  $S_1$ and $S_2$  two surfaces with $p\in S_1\cap S_2$, $T_pS_1=T_pS_2$ and $N_1(p)=N_2(p)$, where $N_i$ are the orientation in each surface $S_i$. Take all normal sections through $p$, that is, the intersection of the planes formed by $N_i(p)$ and the tangent vectors. The set of these planes in both surfaces is the same because $N_1(p)=N_2(p)$. Fix $v\in T_p S_i$ and $\Pi_v$ the corresponding normal section. This plane meets $S_i$ in two curves $\alpha_i^v$. We are computing the normal curvature with respect to the normal (of the curve) coincides with $N_i(p)$ at $p$ and this normal curvature is the curvature of the curve

Suppose that $S_1$ lies above $S_2$ around $p$. Then for each $v\in T_pS_i$, the curve $\alpha_1^v$ lies above $\alpha_2^v$ at $p$. But we know by the theory of curves that the curvature of $\alpha_1^v$ is greater or equal to the one of $\alpha_2^v$ at $p$. Thus, all normal curvatures of $S_1$ at $p$ are greater or equal to the ones of $S_2$ at $p$. We state this remarkable result:

Theorem. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then for avery $v\in T_pS_i$, we have the inequality
$$\kappa_n^1(v)\geq\kappa_n^2(v).$$

As a consequence,  it is immediate from (*) 

Corollary. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then $H_1(p)\geq H_2(p)$.

Comparison of surfaces by curvatures (I)

We address the following problem: suppose two surfaces tangent at a point $p$ and a surface lies above the other one around $p$. Could we say something about the curvatures of the surfaces at $p$?

Since the sign of the principal curvatures depend on the orientation, suppose that $S_1$ and $S_2$ are two surfaces with $p\in S_1\cap S_2$, $T_pS_1=T_pS_2$ and $N_1(p)=N_2(p)$, where $N_i$ are the orientation in each surface $S_i$. After a rigid motion, we suppose that $p=(0,0,0)$,  $T_pS_i$ is the horizontal plane of equation $z=0$ and $N_i(p)=(0,0,1)$. By this condition, $S_i$ is locally the graph of a function on the tangent plane $T_pS_i$ (exercise!). Let $S_1$ and $S_2$ be the graph of $u$ and $v$ respectively.

Because the tangent plane is $z=0$, then with the usual parametrization $X^i(x,y)$ of the surface, 

$X^1_x(0,0)=(1,0,u_x(0,0))$, $X^1_y(0,0)=(0,1,u_y(0,0))$. Since they belong to the $xy$-plane then $u_x(0,0)=u_y(0,0)=0$. Similarly, $v_x(0,0)=v_y(0,0)=0$. By the expression of the mean curvature $H$, we obtain

$$H_1(p)=\frac12(u_{xx}+u_{yy})(0,0),\ H_2(p)=\frac12(v_{xx}(0,0)+v_{yy}(0,0)\ \ (*).$$

Suppose that $S_1$ lies above $S_2$ around $p$ and with respect to the direction that indicates $N_i(p)$. This is equivalent to $u\geq v$ around $q=(0,0)$. Thus the function $u-v$ attains a minimum at $q$. In particular, the Hessian of $(u-v)$ at $q$ is positive semidefinite. But the Hessian is

$$\left(\begin{array}{cc} (u-v)_{xx}&(u-v)_{xy}\\ (u-v)_{xy} &(u-v)_{yy}\end{array}\right)(q).$$

Thus $$(u-v)_{xx}(q)\geq 0,\ \ (u-v)_{yy}(q)\geq 0$$

and (*) implies $H_1(p)\geq H_2(p)$. Summarizing, the result is:

Theorem. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then $H_1(p)\geq H_2(p)$.

Translation surfaces with constant zero curvature

A translation surface is a surface that is the sum of two planar curves contained in orthogonal planes. Thus a parametrization of the surfaces is $$X(u,v)=(x,0,f(x))+(0,y,g(y))$$ where $f$ and $g$ are smooth functions defined in some intervals of ${\mathbb R}$. We are interesting in the translation surfaces with $H=0$ or $K=0$ on the whole surface. 

For the mean curvature, it is immediate that $H=0$ is equivalent to $$\frac{f''(x)}{1+f'(x)^2}+\frac{g''(y)}{1+g'(y)^2}=0.$$ Then necessarily we have that $$\frac{f''(x)}{1+f'(x)^2}=-\frac{g''(y)}{1+g'(y)^2}=c$$ for some real number $c$. If $c=0$, then $f''=g''=0$, obtaining $f(x)=ax+b$, $g(y)=cy+d$ and $z=ax+cy+b+d$, that is, the surface is a plane. If $c\not=0$, integrating $f$ and $g$ we obtain $$f(x)=-\frac{1}{c}\log\cos(cx+m),\ g(y)=\frac{1}{c}\log\cos(cy+n),\ m,n\in{\mathbb R}.$$ Thus we write $$z=\frac{1}{c}\log\left(\frac{\cos(cy+n)}{\cos(cx+m)}\right).$$ This surface is called the Scherk's surface. In order to study the domain of the $z(x,y)$, we take $c=1$ and $m=n=0$. Then $$z=\log\left|\frac{\cos(y)}{\cos(x)}\right|.$$ Then the domain is 

$$(x,y)\in(-\frac{\pi}{2},\frac{\pi}{2})\times (-\frac{\pi}{2},\frac{\pi}{2}).$$

It is clear that in the sides of this square, the function $z=z(x,y)$ takes $\infty$ or $-\infty$ values, as it is shown in the next picture.

If we now study translation surfaces with $K=0$, then this identity is equivalent to 

$$f''g''=0.$$

Then $f''=0$ identically or $g''=0$ identically. Without loss of generality, we suppose $f''=0$, that is, $f(x)=ax+b$ for some numbers $a,b$. Then the surface writes as $z=ax+g(y)+b$ or in terms of $X$, $$X(x,y)=x(1,0,a)+(0,y,g(y)).$$ This surface is a ruled surface whose base curve is any curve as $\alpha(y)=(0,y,g(y))$ and the rulings as parallel to the direction $(1,0,a)$. In the picture we consider $a=0$ and $g(y)=\sin(y)$.

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 $p$.

Starshaped surfaces

Given a positive differentiable function $f:{\mathbb S}^2\rightarrow{\mathbb R}$, define $$S(f)=\{f(p)p:p\in {\mathbb S}^2\}.$$ This surface is called a starshaped surface because the half-line starting from the origin of coordinates meets only at one point of $S(f)$. For example, if $f=2$, then $S(2)$ is the sphere cetered at the origin of radius $2$. In order to give more explicit examples, consider the parametrization of the sphere $X(t,\theta)=(\cos(t)\cos\theta,\cos(t)\sin\theta,\sin(t))$. Then define two functions $f$ by $f(X(t,\theta))=1+\cos(t)^2$ and $f(t)=1+\cos(s)^2$. The pictures are: 

We observe in the second surface that there appears 'strange point'. Exactly, points where the tangent plane is not well-defined. This occurs because $X$ is not a parametrization that cover the whole sphere, but only a part. Exactly, $s\in (0,2\pi)$, that is, except a meridian, exactly the points where appear the problem.

We give some properties of these surfaces.

1. The set $S(f)$ is, indeed, a surface. The map $$\phi:{\mathbb S}^2\rightarrow {\mathbb R}^3,\ \phi(p)=f(p)p$$ is differentiable and $d\phi_p$ is one-to-one. The proof is as follows. If $v\in T_p{\mathbb S}^2$, then if $0=d\phi_p(v)=(df_p(v))p+f(p)v$, we have a linear combination of $p$ and $v$. We know that $T_p{\mathbb S}^2=<p>^\bot$. If $v\not=0$, then $f(p)=0$, a contradiction. This proves that $\phi({\mathbb S}^2)=S(f)$ is a surface.

2. The surface $S(f)$ is compact and connected because  $S(f)=\phi({\mathbb S}^2)$.

3. The map $\phi$ is one-to-one. If $\phi(p)=\phi(q)$, then $f(p)p=f(q)q$. Taking the modulus, we have $f(p)=f(q)$, so $p=q$. 

4. The map $\phi:{\mathbb S}^2\rightarrow S(f)$ is a diffeomorphism because the inverse is   is $\phi^{-1}(p)=p/|p|$, so it is differentiable. 

Connected components in a surface

A connected component of a topological space is a closed set because the closure of a connected set is connected again. Consider now $S_i$ a connected component of a surface $S$. Then we prove that $S$ is an open set, in particular, $S_i$ is a surface. This result holds because any surface is locally connected. In order to avoid the use of this terminology, take $p\in S_i$ and we prove that $p$ is an interior point of $S_i$. Consider $X:U\rightarrow V\subset S$ a parametrization around $p$, where $V\subset S$ is an open set around $p$. Since $U$ is an open set of ${\mathbb R}^2$, there exists a ball $B$ centered at $q=X^{-1}(p)$. In particular $B$ is connected and consequently, $X(B)$ is a connected open set around $p$. Since $S_i$ is the connected component of $p$, then $X(B)\subset S_i$, q.e.d.

A consequence: any component of a compact surface is a compact surface. Indeed, if $S_i$ is the component, then $S_i$ is closed and clearly bounded, so $S_i$ is a compact. And by the above paragraph, $S_i$ is an open set of $S$, so it is a surface.

A compact surface

Consider the set $$S^\{(x,y,z)\in{\mathbb R}^3: e^{x^2}+e^{y^2}+e^{z^2}=4\}.$$

We prove that $S$ is a compact surface diffeomorphic to the unit sphere ${\mathbb S}^2$.

Here it is the picture of $S$

First  $S$ is a surface because $S=f^{-1}(\{4\})$ and $4$ is a regular value of the function $f:{\mathbb R}^3\rightarrow{\mathbb R}$ given by $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$. To prove that $S$ is compact, the same fact $S=f^{-1}(\{0\}$ proves that $S$ is closed. On the other hand, $S$ is bounded because for any $(x,y,z)\in S$, we have

$e^{x^2}\leq 4$ so $x^2\leq\log(4)$. This shows that $|(x,y,z)|^2\leq 3\log(4)$.

Finally we prove that $\phi:S\rightarrow {\mathbb S}^2$, $\phi(p)=p/|p|$ is a diffeomorphism. We point out that $|p|\not=0$ for any $p\in S$. It is immediate that $\phi$ is differentiable.

We prove that $d\phi_p$ is an isomorphism. Since

$$d\phi_p(v)=\frac{v}{|p|}-\frac{\langle p,v\rangle}{|p|^3}p$$

if $d\phi_p(v)=0$, then $v$ is proportional to $p$, exactly $v=\langle p,v\rangle p/|p|^2$. If $\langle v,p\rangle=0$, then $v=0$. On the contrary, $p=\lambda v$ with $\lambda\not=0$. In particular, $p$ is a tangent vector. Because the tangent plane $T_pS$ is orthogonal to $\nabla f(p)=2(x e^{x^2},y e^{y^2},z e^{z^2})$,  $p=(x,y,z)$, then  $\langle p,\nabla f (p)\rangle=2(x^2 e^{x^2}+y^2 e^{y^2}+z^2 e^{z^2})=0$, which is not possible. Then the inverse function theorem proves that $\phi$ is a local diffeomorphism.

We prove that $\phi$ is tsurjective. Given $p\in {\mathbb S}^2$, we have to find $q\in S$ such that $q/|q|=p$. Then we take the half-straightline starting from the origin across $p$ until that we intersect with $S$ at one point. Then it is clear that $q$ is the desired point. Thus we have to find $\lambda\in {\mathbb R}$ such that if $(x,y,z)\in {\mathbb S}^2$, $$e^{\lambda^2 x^2}+e^{\lambda^2 y^2}+e^{\lambda^2 z^2}=4.$$ If we see the left hand-side as continuous function on $\lambda$, namely, $g(\lambda)$, we use the intermediate value theorem: if $\lambda\rightarrow 0$, $g(\lambda)\rightarrow 3$ and if $\lambda\rightarrow\infty$, then $g(\lambda)\rightarrow\infty$, obtaining the result.

Finally, and because $S$ is compact, it suffices to see that $\phi$ is injective. Suppose $\phi(p)=\phi(q)$. Then $p$ and $q$ are proportional, that is, $q=m p$, $m>0$. If $p=(x,y,z)$, then

$$e^{x^2}+e^{y^2}+e^{z^2}= e^{m^2 x^2}+e^{ m^2y^2}+e^{m^2 z^2}.$$ But it is clear that the function $g(m)=e^{m^2 x^2}+e^{ m^2y^2}+e^{m^2 z^2}$ is one-to-one because $g'(m)\not=0$ for $m>0$. This proves that in (*) $m$ is necessarily $1$.