Rudiments of Surface Geometry in $\mathbb{R}^3$

 Structures of $\mathbb{R}^3$ Regular parametrized curves and surfaces Various curvatures of regular parametrized surfaces Exercise 2.1 Exercise 2.2 Exercise 2.3 Example 2.1 Example 2.2 Exercise 2.4 Remark 2.2A Definition 2.2B Exercise 2.5 Example 2.3 Exercise 2.6 Exercise 2.7 Remark 2.4 Definition 2.5 Remark 2.6 Exercise 2.8 Example 2.7 Example 2.8 Example 2.9 Example 2.10 Exercise 2.9 Definition 2.11 Definition 2.12 Example 2.13 Example 2.14 Definition 2.14A Remark 2.14B Definition 2.14C Remark 2.14D Exercise 2.10 Theorem 2.15 Definition 2.15A Theorem 2.16 Definition 2.16A Theorem 2.17 Example 2.18 Example 2.19 Example 2.20

$\mathbb{R}^3$ is the Euclidean 3-dimensional space consisting of points represented by triples of real numbers $(x, y, z)$. The distinguished point $(0, 0, 0)$ is called the origin. In fact, $\mathbb{R}^3$ can also be regarded as a set of vectors, which are quantities that have both length and direction. In this way, $(x, y, z)$ can be thought of as an arrow emanating from the origin, with the tip at the point $(x, y, z)$, or any other arrow having the same length and direction.

There are various operations in $\mathbb{R}^3$. Addition of two vectors is done componentwise:

$(x_1, y_1, z_1)+(x_2, y_2, z_2)=(x_1+x_2, y_1+y_2, z_1+z_2)$
Similarly for subtraction:
$(x_1, y_1, z_1)-(x_2, y_2, z_2)=(x_1-x_2, y_1-y_2, z_1-z_2)$
There are three kinds of multiplications
• scalar multiplication involves a real number and a vector, and the product is a vector.
$\lambda\cdot(x, y, z)=(\lambda x, \lambda y, \lambda z)$
• Inner product of two vectors is a real number(a scalar).
$\langle(x_1, y_1, z_1), (x_2, y_2, z_2)\rangle= x_1x_2+y_1y_2+z_1z_2$
By Pythagoras' theorem, the length of $(x, y, z)$ is given by
$\sqrt{x^2+y^2+z^2}= \sqrt{\langle (x, y, z), (x, y, z)\rangle}$
We denote the length of a vector $\vec{u}$ by $\|\vec{u}\|$. So $\|\vec{u}\|^2=\langle u, u\rangle$. In fact, for two vectors $\vec{u}, \vec{v}\in\mathbb{R}^3$,
$\langle \vec{u}, \vec{v}\rangle= \|\vec{u}\|\cdot\|\vec{v}\|\cos\theta$
where $\theta$ is the included angle between $\vec{u}$ and $\vec{v}$.
Exercise 2.1 If $\vec{u}, \vec{v}\neq 0$ and $\langle \vec{u}, \vec{v}\rangle=0$, what can you say about $\vec{u}$ and $\vec{v}$?
Solution
• Vector product of two vectors is another vector which is perpendicular to that two vectors, and points in a direction determined by the right-hand rule: the direction given by the thumb of your right hand when the fingers curl from the direction of the first vector to that of the second vector. In coordinates,
$(x_1, y_1, z_1)\times (x_2, y_2, z_2)= (y_1z_2-y_2z_1, x_2z_1-x_1z_2, x_1y_2-x_2y_1)$
Exercise 2.2 Prove that $\|\vec{u}\times\vec{v}\|=\|u\|\|v\|\sin\theta$, where $\theta$ is the included angle between $\vec{u}$ and $\vec{v}$. Hint: prove that $\|u\times v\|^2+\langle u, v\rangle^2=\|u\|^2\|v\|^2$.
Solution
Exercise 2.3 Show that $\|\vec{u}\times\vec{v}\|$ is the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$.

Note: In this subsection, we will deal with some differential calculus of functions of two variables. Given $z=f(x, y)$, the partial differential of $z$ with respect to $x$, denoted by $\frac{\partial z}{\partial x}$, can be obtained by differentiating with respect to $x$ as if the variable $y$ were constant. Similarly for $\frac{\partial z}{\partial y}$. For instance, if $z=\cos xy+x^2(1+y)$, then $\frac{\partial z}{\partial x}=-y\sin xy+2x(1+y)$, $\frac{\partial z}{\partial y}=-x\sin xy+x^2$

Let $\alpha: I\to\mathbb{R}^3$ be given by

$\alpha(t)=(x(t), y(t), z(t))$
where $x, y, z$ are continuous functions mapping from an interval $I$ to $\mathbb{R}$. If $x, y, z$ can be differentiated infinitely many times, then we call $\alpha$ a smooth parametrized curve. If furthermore not all $x', y', z'$ are 0 for all $t\in I$, then we call $\alpha$ a regular parametrized curve. Define $\alpha'(t)=(x'(t), y'(t), z'(t))$.

Example 2.1 Let $\alpha$ be given by

$\alpha(t)=(R\cos t, R\sin t, 0), t\in[0, 2\pi]$
$\alpha$ is the circle of radius $R$ centered at the origin in the $xy$-plane. Since
$\alpha'(t)=(-R\sin t, R\cos t, 0)\neq\vec{0}, t\in[0, 2\pi]$
$\alpha$ is a regular parametrized curve.

Example 2.2 Let $\alpha$ be given by

$\alpha(t)=(R\cos t, R\sin t, bt), t\in(-\infty, +\infty)$
Then $\alpha$ is a helix winding along the surface of an infinite cylinder of radius $R$ with pitch $b$. Since
$\alpha'(t)=(-R\sin t, R\cos t, b)\neq\vec{0}, t\in(-\infty, +\infty)$
$\alpha$ is a regular parametrized curve.
The helix

Exercise 2.4 Show that, if $\alpha$ and $\beta$ are smooth parametrized curves,

$\frac{d}{dt}\langle\alpha(t), \beta(t)\rangle=\langle\alpha'(t), \beta(t)\rangle+\langle\alpha(t), \beta'(t)\rangle$
Deduce that, if $\alpha$ is a smooth parametrized curve lying on a sphere centered at the origin in $\mathbb{R}^3$, then $\langle\alpha'(t), \alpha(t)\rangle=0$.
Solution

Remark 2.2A In fact, $\alpha'(t_0)$ is the tangent vector of the curve $\alpha$ at $t=t_0$--it is the instantaneous direction at $t=t_0$ of a particle moving along the curve in such a way that it is at $\alpha(t)$ at time $t$. Put in another way, the assertion of the above exercise just says that any tangent vector of a smooth parametrized curve at a point is perpendicular to the position vector of that point, if the curve lies on a sphere centered at the origin.

Definition 2.2B Define the curvature $\kappa(t)$ of a regular parametrized curve $\alpha(t)$ by

$\kappa(t)=\frac{\|\alpha'(t)\times\alpha''(t)\|} {\|\alpha'(t)\|^3}$

Exercise 2.5 If $\|\alpha'(t)\|=1$ for all $t\in I$, then $\kappa(t)= \|\alpha''(t)\|$.
Solution

Example 2.3 The curvature of the circle of radius $R$, as in Example 2.1, is

\begin{align*} \kappa(t)&=\frac{\|(-R\sin t, R\cos t, 0)\times(-R\cos t, -R\sin t, 0)\|}{\|(-R\sin t, R\cos t, 0)\|^3}\\ &=\frac{\|(0, 0, R^2)\|}{R^3}\\ &=\frac{1}{R} \end{align*}

Exercise 2.6 Prove that the curvature of the helix as in Example 2.2 is $\frac{R}{R^2+b^2}$.
Solution

Exercise 2.7 Consider the smooth parametrized curve $\alpha(t)=(t^3, t^6, t^9)$, $t\in(-\infty, +\infty)$. Is it a regular parametrized curve? Can you compute the curvature at $t=0$?
Solution

Remark 2.4 Curvature is a measure of how curved a curve is. Example 2.3 shows that the greater the radius is, the smaller the curvature of the circle becomes. This coincides with our geometric intuition.

Let $f: I_1\times I_2\to\mathbb{R}^3$ be given by

$\vec{x}(u, v)=(x(u, v), y(u, v), z(u, v))$
where $I_1$ and $I_2$ are intervals and $x$, $y$ and $z$ are functions in two variables $(u, v)$. Suppose that $x$, $y$ and $z$ can be differentiated with respect to $u$ and $v$ infinitely many times. If we fix $v=v_0$ and let $u$ vary in $I_1$, we get a curve $\vec{x}(u, v_0)$, called a $u$-coordinate curve. Similarly, $\vec{x}(u_0, v)$ is a $v$-coordinate curve, with $u_0$ a constant. Put
$\vec{x}_u=\left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u}\right)$
$\vec{x}_v=\left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v}\right)$
They are the tangents of the $u$- and $v$-coordinate curves respectively.

Definition 2.5 If $\vec{x}_u$ and $\vec{x}_v$ are not multiple of each other for all $(u, v)\in I_1\times I_2$, then $\vec{x}$ is a regular parametrized surface.

Remark 2.6 The above definition is different from the usual definition found in the literature, which requires $\vec{x}$ to be a homeomorphism, i.e. $I_1\times I_2$ can be morphed into the image of $\vec{x}$ without cutting and pasting. For the ease of presentation we omit this requirement in our definition.

Exercise 2.8
1. Show that $\vec{u}$ and $\vec{v}$ are not multiple of each other iff $\vec{u}\times\vec{v}\neq \vec{0}$.
Solution
2. $\vec{u}$ and $\vec{v}$ are not multiple of each other iff there is a unique plane containing the two vectors passing through any particular point.

The above exercise shows that there exists a unique plane containing vectors $\vec{x}_u(u_0, v_0)$ and $\vec{x}_v(u_0, v_0)$ passing through $\vec{x}(u_0, v_0)$. We call this plane the tangent plane of $\vec{x}$ at $\vec{x}(u_0, v_0)$, and use $T_pS$ to denote the tangent plane of a surface $S$ at $p$. Moreover we know that $\vec{x}_u(u_0, v_0)\times \vec{x}_v(u_0, v_0)\neq \vec{0}$ for all $(u_0, v_0)\in I_1\times I_2$. Define

$N=\frac{\vec{x}_u\times \vec{x}_v} {\|\vec{x}_u\times\vec{x}_v\|}$
the unit normal vector of $\vec{x}$. By the definition of vector product, $N(u_0, v_0)$ is perpendicular to the tangent plane at $\vec{x}(u_0, v_0)$.

Example 2.7 The unit sphere Let $\vec{x}(u, v)=(\sin u\cos v, \sin u\sin v, \cos u)$, $(u, v)\in (0, \pi)\times[0, 2\pi)$. Then $\vec{x}$ is the unit sphere minus the two poles. The $u$-coordinate curves are the longitudes and the $v$-coordinate curves are the latitudes. Note that

\begin{align*} \vec{x}_u&=(\cos u\cos v, \cos u\sin v, -\sin u)\\ \vec{x}_v&=(-\sin u\sin v, \sin u\cos v, 0)\\ \|\vec{x}_u\times\vec{x}_v\|&=\|(\sin^2u\cos v, \sin^2u\sin v, \sin u\cos u)\|\\ &=\sin u\\ &\neq 0 \end{align*}
So $\vec{x}$ is a regular parametrized surface.
The unit sphere

Example 2.8 Torus Let $\vec{x}(u, v)=((R+r\cos u)\cos v, (R+r\cos u)\sin v, r\sin u)$, $R>r$, $(u, v)\in[0, 2\pi)\times[0, 2\pi)$. $\vec{x}$ is a torus obtained by revolving a circle of radius $r$ centered at $(R, 0, 0)$ around the $z$-axis. The $u$-coordinate curves are vertical circles on the torus, whereas the $v$-coordinate curves are horizontal circles on the torus. We have

\begin{align*} \vec{x}_u&=(-r\sin u\cos v, -r\sin u\sin v, r\cos u)\\ \vec{x}_v&=(-(R+r\cos u)\sin v, (R+r\cos u)\cos v, 0)\\ \vec{x}_u\times\vec{x}_v&=(-r(R+r\cos u)\cos u\cos v, -r(R+r\cos u)\cos u\sin v, -r(R+r\cos u)\sin u)\\ \|\vec{x}_u\times\vec{x}_v\|&=r(R+r\cos u)\neq 0\text{ since }R>r \end{align*}
So $\vec{x}$ is a regular parametrized surface.
The torus

Example 2.9 Surface of revolution In general, if $\alpha(u)=(p(u), q(u))$ is a regular parametrized curve in $\mathbb{R}^2$ with $p(u)\neq 0$, then the equation of the surface of revolution obtained by rotating $\alpha$ around the $z$-axis is

$\vec{x}(u, v)=(p(u)\cos v, p(u)\sin v, q(u))$
We have
\begin{align*} \vec{x}_u&=(p'\cos v, p'\sin v, q')\\ \vec{x}_v&=(-p\sin v, p\cos v, 0)\\ \vec{x}_u\times\vec{x}_v&=(-pq'\cos v, -pq'\sin v, pp')\\ \|\vec{x}_u\times\vec{x}_v\|&=p\sqrt{p'^2+q'^2}\neq 0\text{ since }p\neq 0\text{ and }\alpha\text{ is regular}. \end{align*}
So $\vec{x}$ is a regular parametrized surface.

Example 2.10 The graph of an infinitely differentiable function of two variables, $z=f(x, y)$, is the parametrized surface

$\vec{x}(u, v)=(u, v, f(u, v))$
We have
\begin{align*} \vec{x}_u&=(1, 0, \frac{\partial f}{\partial u})\\ \vec{x}_v&=(0, 1, \frac{\partial f}{\partial v})\\ \vec{x}_u\times\vec{x}_v&=(-\frac{\partial f}{\partial u}, \frac{\partial f}{\partial v}, 1)\neq\vec{0} \end{align*}
So the graph of $z$ is a regular parametrized surface.

Exercise 2.9 Show that the following are regular parametrized surfaces, and compute their normal vectors.

1. The helicoid. $\vec{x}(u, v)=(av\cos u, av\sin u, bu), (u, v)\in \mathbb{R}^2, a>0, b\neq 0$. Note: One can obtain a helicoid by dipping a wire in the shape of a helix into a soap solution and taking it out, as shown on the Homepage.
Solution can be found in Example 3.4
The helicoid
2. The catenoid. $\vec{x}(u, v)=(\cosh u\cos v, \cosh u\sin v, u)$, $(u, v)\in\mathbb{R}^2$, where $\cosh u=\frac{e^u+e^{-u}}{2}$. This is a special function which can be found in your calculator. Note: The curve which is revolved around the $z$-axis to get the catenoid is called the catenary. Its equation is $y=\cosh x$. It is in the shape of a freely hanging chain with each end attached to the top of a vertical pole. The catenoid can be obtained by dipping two identical circular wires into a soap solution and slowly drawing them apart.
Solution can be found here.
The catenoid

There are several notions of curvatures of surfaces, all of which are related to a special map called the Gauss map.

Definition 2.11 The Gauss map $N: S\to\mathbb{S}^2$ maps $p\in S$ to the unit normal vector of $S$ at $p$.

As you may see, the Gauss map measures how rough' a surface is. The image of the Gauss map of a plane consists of only one vector, whereas the Gauss map of a surface with lots of sharp bumps and troughs has a large' image which may be thought of as a large area on the unit sphere swept out by the unit normal vectors. Moreover, the local behaviour of the Gauss map gives a measure of how curved a surface is around a certain point.

Definition 2.12 The Gauss curvature $K$ at $p$ of a surface $S$ is the limit of the following quotient

$\pm\frac{\text{Area of }N(\mathcal{O}_p)\text{ on }\mathbb{S}^2} {\text{Area of a neighbourhood }\mathcal{O}_p\text{ around }p\text{ on }S}$
as $\mathcal{O}_p$ shrinks to $p$. The sign is determined as follows: if $\alpha(t)$ is a small closed curve around $p$ for $0\leq t\leq 1$, $\alpha(0)=\alpha(1)$ and traced out counterclockwise as $t$ ranges from 0 to 1 when you view the surface from the side to which $N(p)$ is pointing, and so is $N(\alpha(t))$, then the sign should be positive. Otherwise, negative sign should be attached.

Example 2.13 The Gauss map for the unit sphere $\mathbb{S}^2$ is the identity map. So the Gauss curvature is 1. On the other hand, the Gauss curvature of any plane is 0 because the image of the Gauss curvature consists of only one vector.

Example 2.14 The Gauss map for the cylinder $\vec{x}(u, v)=(\cos v, \sin v, u)$ has the equator of $\mathbb{S}^2$ as its image, which has no area. So the Gauss curvature of the cylinder is 0.

Though intuitive, the definition above will not be made used of to compute the Gauss curvature of other surfaces. Instead, we will introduce the first and second fundamental forms.

Definition 2.14A The first fundamental form $I_p$ at $p\in S$ maps a pair of tangent vectors of $S$ at $p$ to their inner product:

$I_p(\vec{v}, \vec{w})= \langle \vec{v}, \vec{w}\rangle, \vec{v}, \vec{w}\in T_pS$

Remark 2.14B The first fundamental form reflects the intrinsic property of surfaces, meaning that it gives the local information of length and area on the surface. For a regular parametrized surface $\vec{x}(u, v)$, put

$E=I(\vec{x}_u, \vec{x}_u), F=I(\vec{x}_u, \vec{x}_v), G=I(\vec{x}_v, \vec{x}_v)$

Definition 2.14C The second fundamental form $II_p$ at $p\in S$ is defined to be

$II_p(\vec{v}, \vec{w})= \left\langle-\frac{d}{dt}N(\alpha(t)), \vec{w}\right\rangle$
where $\vec{v}$ and $\vec{w}$ are tangent vectors at $p$, and $\alpha(t)$ is any curve on $S$ such that $\alpha(0)=p$ and $\alpha'(0)=\vec{v}$.

Remark 2.14D The second fundamental form reflects the extrinsic property of surfaces, meaning that it shows how surfaces are embedded in $\mathbb{R}^3$.

For a regular parametrized surface $\vec{x}(u, v)$, put

$e=II(\vec{x}_u, \vec{x}_u), f=II(\vec{x}_u, \vec{x}_v), g=II(\vec{x}_v, \vec{x}_v)$
In fact, one can show that
$e=-\left\langle \frac{\partial N}{\partial u}, \vec{x}_u\right\rangle, f=-\left\langle \frac{\partial N} {\partial u}, \vec{x}_v\right\rangle, g=-\left\langle \frac{\partial N} {\partial v}, \vec{x}_v\right\rangle$

Exercise 2.10 Show that $e=\langle N, \vec{x}_{uu}\rangle$, $f=\langle N, \vec{x}_{uv}\rangle$, $g=\langle N, \vec{x}_{vv}\rangle$, where $N=N(u, v)$, the normal vector at $\vec{x}(u, v)$, $\vec{x}_{uu}=\frac{\partial}{\partial u} \frac{\partial \vec{x}}{\partial u}$ and $\vec{x}_{uv}=\frac{\partial}{\partial u} \frac{\partial\vec{x}}{\partial v}$, etc.
Solution

In fact, for $\vec{x}(u, v)$ which can be differentiated infinitely many times, the order of taking partial differentiations is immaterial, i.e. $\vec{x}_{uv}=\vec{x}_{vu}$. By the above exercise, $f=II(\vec{x}_u, \vec{x}_v)= II(\vec{x}_v, \vec{x}_u)$. If $\vec{v}$ and $\vec{w}$ are tangent vectors at $\vec{x}(u_0, v_0)$, then $\vec{v}=a\vec{x}_u(u_0, v_0)+ b\vec{x}_v(u_0, v_0)$ and $\vec{w}=a'\vec{x}_u(u_0, v_0)+ b'\vec{x}_v(u_0, v_0)$ for some $a, b, a', b'\in\mathbb{R}$. It can be shown that

$II(\vec{v}, \vec{w})=aa'e+(ab'+a'b)f+bb'g=II(\vec{w}, \vec{v})$

Theorem 2.15 The Gauss curvature is given by

$K=\frac{eg-f^2}{EG-F^2}$

Another way to compute the Gauss curvature, which may be more tedious but relevant to the next section, is to use principal curvatures $k_1$ and $k_2$, which are defined as in

$k_1(p)=\underset{\|v\|=1, v\in T_pS}{\max}II_p(v, v)$
$k_2(p)=\underset{\|v\|=1, v\in T_pS}{\min}II_p(v, v)$

$\[K=k_1k_2$

Definition 2.16A Define the mean curvature $H=\frac{k_1+k_2}{2}$.

$H=\frac{Eg+Ge-2Ff}{2(EG-F^2)}$

Now with the various tools in our hands, we are able to compute $K$ and $H$ for all the regular parametrized surfaces in the previous examples.

Example 2.18 The torus.

$E=r^2, F=0, G=(R+r\cos u)^2$
\begin{align*} N&=(-\cos u\cos v, -\cos u\sin v, -\sin u)\\ \vec{x}_{uu}&=(-r\cos u\cos v, -r\cos u\sin v, -r\sin u)\\ \vec{x}_{uv}&=(r\sin u\sin v, -r\sin u\cos v, 0)\\ \vec{x}_{vv}&=(-(R+r\cos u)\cos v, -(R+r\cos u)\sin v, 0) \end{align*}
$e=r, f=0, g=\cos u(R+r\cos u)$
\begin{align*} K&=\frac{eg-f^2}{EG-F^2}=\frac{\cos u}{r(R+r\cos u)}\\ H&=\frac{Eg+Ge-2Ff}{2(EG-F^2)}=\frac{R+2r\cos u}{2r(R+r\cos u)} \end{align*}
Let us also verify the Gauss and mean curvatures by computing the principal curvatures. Let
\begin{align*} \vec{u}(u, v)&=\frac{\vec{x}_u(u, v)}{\|\vec{x}_u(u, v)\|}=(-\sin u\cos v, \sin u\sin v, \cos u)\\ \vec{v}(u, v)&=\frac{\vec{x}_v(u, v)}{\|\vec{x}_v(u, v)\|}=(-\sin v, \cos v, 0) \end{align*}
They are unit tangent vectors which are perpendicular to each other. It is clear that any unit tangent vector can be written as $a\vec{u}+b\vec{v}$ with $a^2+b^2=1$. Hence
\begin{align*} II(a\vec{u}+b\vec{v}, a\vec{u}+b\vec{v})&=a^2II(\vec{u}, \vec{u})+2abII(\vec{u}, \vec{v})+b^2II(\vec{v}, \vec{v})\\ &=\frac{a^2e}{r^2}+\frac{b^2g}{(R+r\cos u)^2}\\ &=\frac{a^2}{r}+\frac{b^2\cos u}{R+r\cos u} \end{align*}
It can be shown that the above expression attains maximum when $a=1, b=0$, and minimum when $a=0, b=1$. So
$k_1=\frac{1}{r}, k_2=\frac{\cos u}{R+r\cos u}$
Indeed, $K=k_1k_2$, $H=\frac{k_1+k_2}{2}$.

Example 2.19 Surface of revolution

$E=p'^2+q'^2, F=0, G=p^2$
\begin{align*} N&=\frac{1}{\sqrt{p'^2+q'^2}}(-q'\cos v, -q'\sin v, p')\\ \vec{x}_{uu}&=(p''\cos v, p''\sin v, q'')\\ \vec{x}_{uv}&=(-p'\sin v, p'\cos v, 0)\\ \vec{x}_{vv}&=(-p\cos v, -p\sin v, 0) \end{align*}
$e=\frac{p'q''-p''q'}{\sqrt{p'^2+q'^2}}, f=0, g=\frac{pq'}{\sqrt{p'^2+q'^2}}$
\begin{align*} K&=\frac{eg-f^2}{EG-F^2}=\frac{(p'q''-p''q')q'}{p(p'^2+q'^2)}\\ H&=\frac{(p'^2+q'^2)q'+p(p'q''-p''q')}{p(p'^2+q'^2)^{\frac{3}{2}}} \end{align*}

Example 2.20 The graph of an infinitely differentiable function $z=f(x, y)$

$E=1+\left(\frac{\partial f} {\partial u}\right)^2, F=\frac{\partial f}{\partial u}\frac{\partial f}{\partial v}, G=1+\left(\frac{\partial f}{\partial v}\right)^2$
\begin{align*} N&=\frac{1}{\sqrt{1+\left(\frac{\partial f}{\partial u}\right)^2+\left(\frac{\partial f}{\partial v}\right)^2}}\left(-\frac{\partial f}{\partial u}, -\frac{\partial f}{\partial v}, 1\right)\\ \vec{x}_{uu}&=\left(0, 0, \frac{\partial^2 f}{\partial u^2}\right)\\ \vec{x}_{uv}&=\left(0, 0, \frac{\partial^2 f}{\partial u\partial v}\right)\\ \vec{x}_{vv}&=\left(0, 0, \frac{\partial^2 f}{\partial v^2}\right) \end{align*}
$e=\frac{\frac{\partial^2f} {\partial u^2}}{\sqrt{1+\left(\frac{\partial f}{\partial u}\right)^2+\left(\frac{\partial f} {\partial v}\right)^2}}, f=\frac{\frac{\partial^2f}{\partial u\partial v}} {\sqrt{1+\left(\frac{\partial f}{\partial u}\right)^2+\left(\frac{\partial f}{\partial v} \right)^2}}, g=\frac{\frac{\partial^2f}{\partial v^2}}{\sqrt{1+\left(\frac{\partial f} {\partial u}\right)^2+\left(\frac{\partial f}{\partial v}\right)^2}}$
$K=\frac{\frac{\partial^2 f}{\partial u^2}\frac{\partial^2 f}{\partial v^2}- \left(\frac{\partial^2 f}{\partial u\partial v}\right)^2}{\left(1+\left(\frac{\partial f} {\partial u}\right)^2+\left(\frac{\partial f}{\partial v}\right)^2\right) \left(\left(1+\left(\frac{\partial f}{\partial u}\right)^2\right) \left(1+\left(\frac{\partial f}{\partial v}\right)^2\right)-\left(\frac{\partial f} {\partial u}\frac{\partial f}{\partial v}\right)^2\right)}$
$H=\frac{\left(1+ \left(\frac{\partial f}{\partial u} \right)^2\right)\frac{\partial^2 f}{\partial v^2}+\left(1+\left(\frac{\partial f} {\partial v}\right)^2\right)\frac{\partial^2 f}{\partial u^2}-2\frac{\partial f} {\partial u}\frac{\partial f}{\partial v}\frac{\partial^2 f}{\partial u\partial v}} {2\sqrt{1+\left(\frac{\partial f}{\partial u}\right)^2+\left(\frac{\partial f} {\partial v}\right)^2}\left(\left(1+\left(\frac{\partial f}{\partial u}\right)^2\right) \left(1+\left(\frac{\partial f}{\partial v}\right)^2\right)-\left(\frac{\partial f} {\partial u}\frac{\partial f}{\partial v}\right)^2\right)}$