Mathematics / Differential Geometry Curves, Surfaces, Fundamental Forms & Curvatures 100% Free Open Access
Chapter 4 โ€ข Theory & Derivations

Unit 4: Parametric Surfaces & The First Fundamental Form

Foundations of two-dimensional surface theory in Euclidean 3-space: smooth coordinate patches, regularity conditions, tangent planes and surface normals, the First Fundamental Form (surface metric) $I = E du^2 + 2F dudv + G dv^2$, positive definiteness, arc-length of surface curves, angles between tangent vectors, orthogonal coordinates, and the intrinsic surface area element.

ยง4.1 Parametric Surfaces, Coordinate Patches & Regular Points

1. Vector Parametrization of Surfaces in $\mathbb{R}^3$

Just as a curve is described by a single scalar parameter, a surface in $\mathbb{R}^3$ is locally parametrized by two independent real parameters $(u, v)$.

Definition 4.1 (Parametrized Surface Patch): Let $U \subseteq \mathbb{R}^2$ be an open connected domain in the $uv$-plane. A parametrized surface (or local coordinate patch) is a smooth vector-valued function:

$$\mathbf{r}: U \to \mathbb{R}^3, \quad (u, v) \mapsto \mathbf{r}(u, v) = \begin{pmatrix} x(u, v) \\ y(u, v) \\ z(u, v) \end{pmatrix}$$

The set of points $S = \mathbf{r}(U) \subset \mathbb{R}^3$ is the trace or image of the surface patch.


2. Partial Derivatives and Coordinate Curves

For a fixed $v = v_0$, the mapping $u \mapsto \mathbf{r}(u, v_0)$ traces a curve on $S$ called the $u$-coordinate curve (or $u$-parameter curve). Similarly, for a fixed $u = u_0$, $v \mapsto \mathbf{r}(u_0, v)$ traces a $v$-coordinate curve.

The partial derivative vectors are tangent to these coordinate curves:

$$\mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = \begin{pmatrix} \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial u} \\ \frac{\partial z}{\partial u} \end{pmatrix}, \qquad \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = \begin{pmatrix} \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial v} \\ \frac{\partial z}{\partial v} \end{pmatrix}$$

3. Regularity and the Tangent Plane

Definition 4.2 (Regular Point of a Surface): A point $p = \mathbf{r}(u_0, v_0)$ is called a regular point of the surface patch if the partial derivative vectors $\mathbf{r}_u$ and $\mathbf{r}_v$ are linearly independent at $(u_0, v_0)$:

$$\mathbf{r}_u \times \mathbf{r}_v \ne \mathbf{0}$$

Equivalently, the Jacobian matrix $J = \begin{pmatrix} x_u & x_v \\ y_u & y_v \\ z_u & z_v \end{pmatrix}$ has maximal rank $2$. A surface patch is regular if every point in $U$ is regular.

Definition 4.3 (Tangent Plane $T_p S$ and Unit Normal $\mathbf{n}$): At any regular point $p \in S$, the vectors $\mathbf{r}_u$ and $\mathbf{r}_v$ span a two-dimensional vector subspace of $\mathbb{R}^3$ called the tangent plane to $S$ at $p$, denoted $T_p S$:

$$T_p S = \operatorname{span}\{\mathbf{r}_u, \mathbf{r}_v\} = \{\lambda \mathbf{r}_u + \mu \mathbf{r}_v : \lambda, \mu \in \mathbb{R}\}$$

The unit normal vector $\mathbf{n}(u, v)$ to the surface at $p$ is defined by:

$$\mathbf{n}(u, v) = \frac{\mathbf{r}_u \times \mathbf{r}_v}{\|\mathbf{r}_u \times \mathbf{r}_v\|}$$

By construction, $\mathbf{n} \cdot \mathbf{r}_u = 0$, $\mathbf{n} \cdot \mathbf{r}_v = 0$, and $\|\mathbf{n}\| = 1$.

The Cartesian equation of the tangent plane passing through $\mathbf{r}(u_0, v_0)$ is:

$$\mathbf{n}(u_0, v_0) \cdot (\mathbf{R} - \mathbf{r}(u_0, v_0)) = 0 \iff (\mathbf{R} - \mathbf{r}) \cdot (\mathbf{r}_u \times \mathbf{r}_v) = 0$$

where $\mathbf{R} = (X, Y, Z)^T$ is an arbitrary point on the plane.

ยง4.2 The First Fundamental Form: Metric Coefficients E, F, G & Positive Definiteness

1. Differential of the Position Vector

Let $\mathbf{r}: U \to \mathbb{R}^3$ be a regular surface. Consider an infinitesimal displacement on the parameter domain $d\mathbf{u} = (du, dv)^T$. The corresponding infinitesimal displacement vector on the surface in $\mathbb{R}^3$ is given by the differential:

$$d\mathbf{r} = \mathbf{r}_u \, du + \mathbf{r}_v \, dv \in T_p S$$

The square of the infinitesimal Euclidean distance between $\mathbf{r}(u, v)$ and $\mathbf{r}(u + du, v + dv)$ is:

$$ds^2 = \|d\mathbf{r}\|^2 = d\mathbf{r} \cdot d\mathbf{r} = (\mathbf{r}_u \, du + \mathbf{r}_v \, dv) \cdot (\mathbf{r}_u \, du + \mathbf{r}_v \, dv)$$

Expanding the dot product bilinearly:

$$ds^2 = (\mathbf{r}_u \cdot \mathbf{r}_u) \, du^2 + 2(\mathbf{r}_u \cdot \mathbf{r}_v) \, du \, dv + (\mathbf{r}_v \cdot \mathbf{r}_v) \, dv^2$$

2. Definition of the First Fundamental Form

Definition 4.4 (First Fundamental Form): The First Fundamental Form of a surface $S$, denoted by $I$ or $I_p$, is the quadratic form on the tangent space $T_p S$ induced by the Euclidean metric of $\mathbb{R}^3$:

$$I(du, dv) = E \, du^2 + 2F \, du \, dv + G \, dv^2$$

where the metric coefficients (Gauss coefficients) are:

$$E = \mathbf{r}_u \cdot \mathbf{r}_u = \|\mathbf{r}_u\|^2$$
$$F = \mathbf{r}_u \cdot \mathbf{r}_v$$
$$G = \mathbf{r}_v \cdot \mathbf{r}_v = \|\mathbf{r}_v\|^2$$

In matrix notation, for a tangent vector $\mathbf{w} = \lambda \mathbf{r}_u + \mu \mathbf{r}_v \in T_p S$, represented in coordinates by $\mathbf{\xi} = \begin{pmatrix} \lambda \\ \mu \end{pmatrix}$:

$$I(\mathbf{w}) = \mathbf{\xi}^T g \, \mathbf{\xi} = \begin{pmatrix} \lambda & \mu \end{pmatrix} \begin{pmatrix} E & F \\ F & G \end{pmatrix} \begin{pmatrix} \lambda \\ \mu \end{pmatrix}$$

where $g = \begin{pmatrix} E & F \\ F & G \end{pmatrix}$ is the metric tensor matrix.


3. Positive Definiteness and the Metric Determinant

Theorem 4.1 (Positive Definiteness of the First Fundamental Form): At every regular point of a surface, the First Fundamental Form is strictly positive definite:

$$I(du, dv) > 0 \quad \text{for all } (du, dv) \ne (0, 0)$$

Moreover, the determinant of the metric tensor satisfies:

$$g = \det \begin{pmatrix} E & F \\ F & G \end{pmatrix} = EG - F^2 = \|\mathbf{r}_u \times \mathbf{r}_v\|^2 > 0$$

Proof: By Lagrange's vector identity for any two vectors $\mathbf{a}, \mathbf{b} \in \mathbb{R}^3$:

$$\|\mathbf{a} \times \mathbf{b}\|^2 = \|\mathbf{a}\|^2 \|\mathbf{b}\|^2 - (\mathbf{a} \cdot \mathbf{b})^2$$

Setting $\mathbf{a} = \mathbf{r}_u$ and $\mathbf{b} = \mathbf{r}_v$:

$$\|\mathbf{r}_u \times \mathbf{r}_v\|^2 = (\mathbf{r}_u \cdot \mathbf{r}_u)(\mathbf{r}_v \cdot \mathbf{r}_v) - (\mathbf{r}_u \cdot \mathbf{r}_v)^2 = EG - F^2$$

Because the point is regular, $\mathbf{r}_u \times \mathbf{r}_v \ne \mathbf{0}$, which implies:

$$EG - F^2 = \|\mathbf{r}_u \times \mathbf{r}_v\|^2 > 0$$

Next, consider $I(du, dv) = E du^2 + 2F dudv + G dv^2$. Since $\mathbf{r}_u \ne \mathbf{0}$, $E = \|\mathbf{r}_u\|^2 > 0$. We complete the square:

$$I(du, dv) = E \left[ du^2 + \frac{2F}{E} du dv + \frac{F^2}{E^2} dv^2 \right] + \left( G - \frac{F^2}{E} \right) dv^2$$
$$I(du, dv) = E \left( du + \frac{F}{E} dv \right)^2 + \frac{EG - F^2}{E} dv^2$$

Both coefficients $E > 0$ and $\frac{EG - F^2}{E} > 0$ are strictly positive. Thus $I(du, dv) \ge 0$, with equality holding if and only if $dv = 0$ and $du + \frac{F}{E} dv = 0 \implies du = 0$. Hence, $I$ is strictly positive definite. $\blacksquare$

ยง4.3 Arc-Length of Surface Curves, Isometries & Conformal Mappings

1. Arc-Length of Curves Lying on a Surface

Let $C$ be a smooth curve lying entirely on the surface $S$, defined parametrically by $u = u(t), v = v(t)$ for $t \in [a, b]$. The position vector of the curve in $\mathbb{R}^3$ is:

$$\mathbf{c}(t) = \mathbf{r}(u(t), v(t))$$

Applying the chain rule, the velocity vector is:

$$\mathbf{c}'(t) = \mathbf{r}_u \, u'(t) + \mathbf{r}_v \, v'(t)$$

The speed of the curve is the norm of the velocity vector:

$$\|\mathbf{c}'(t)\| = \sqrt{\mathbf{c}'(t) \cdot \mathbf{c}'(t)} = \sqrt{E (u')^2 + 2F u' v' + G (v')^2} = \sqrt{I(u', v')}$$

Theorem 4.2 (Arc-Length Formula on Surfaces): The arc-length $L$ of the curve $C$ from $t = a$ to $t = b$ is given intrinsically by:

$$L = \int_a^b \|\mathbf{c}'(t)\| \, dt = \int_a^b \sqrt{E \left(\frac{du}{dt}\right)^2 + 2F \left(\frac{du}{dt}\right)\left(\frac{dv}{dt}\right) + G \left(\frac{dv}{dt}\right)^2} \, dt$$

This fundamental result demonstrates that distance along curves on a surface can be computed solely from the metric coefficients $E(u, v), F(u, v), G(u, v)$ without knowing how the surface is embedded in 3D space!


2. Isometries and Intrinsic Geometry

Definition 4.5 (Local Isometry): A diffeomorphism $\phi: S_1 \to S_2$ between two surfaces is called a local isometry if it preserves the lengths of all curves. Equivalently, for any point $p \in S_1$ and tangent vectors $\mathbf{w}_1, \mathbf{w}_2 \in T_p S_1$:

$$I_p(\mathbf{w}_1, \mathbf{w}_2) = I_{\phi(p)}(d\phi(\mathbf{w}_1), d\phi(\mathbf{w}_2))$$

If $S_1$ and $S_2$ are parametrized by coordinates $(u, v)$ such that their metric coefficients satisfy:

$$E_1(u, v) = E_2(u, v), \quad F_1(u, v) = F_2(u, v), \quad G_1(u, v) = G_2(u, v)$$

then the mapping is an isometry.

Surfaces related by an isometry share all intrinsic geometric properties (such as arc-length, angles, and Gaussian curvature), even though their spatial embeddings may look completely different (e.g., a plane sheet rolling into a cylinder).


3. Conformal Mappings and Orthogonal Coordinates

Definition 4.6 (Conformal Mapping): A diffeomorphism $\phi: S_1 \to S_2$ is conformal (angle-preserving) if there exists a smooth positive function $\lambda(u, v) > 0$ (the conformal factor) such that:

$$I_2 = \lambda^2(u, v) I_1$$

Definition 4.7 (Orthogonal Coordinate Systems): A coordinate patch $(u, v)$ is called orthogonal if the coordinate curves intersect at right angles everywhere on $U$:

$$\mathbf{r}_u \cdot \mathbf{r}_v = 0 \iff F(u, v) \equiv 0$$

When $F \equiv 0$, the First Fundamental Form simplifies to:

$$I = E(u, v) \, du^2 + G(u, v) \, dv^2$$

In addition, if $E = G = \lambda^2(u, v)$ and $F = 0$, the coordinates are called isothermal (or conformal):

$$I = \lambda^2(u, v)(du^2 + dv^2)$$

ยง4.4 Angles Between Curves on Surfaces & Direction Fields

1. Angle Between Tangent Vectors on a Surface

Let $p \in S$ be a regular point, and let $\mathbf{w}_1, \mathbf{w}_2 \in T_p S$ be two non-zero tangent vectors. In local coordinates:

$$\mathbf{w}_1 = \mathbf{r}_u \, du + \mathbf{r}_v \, dv = d\mathbf{r}$$
$$\mathbf{w}_2 = \mathbf{r}_u \, \delta u + \mathbf{r}_v \, \delta v = \delta\mathbf{r}$$

The inner product of these tangent vectors is:

$$\mathbf{w}_1 \cdot \mathbf{w}_2 = (\mathbf{r}_u \, du + \mathbf{r}_v \, dv) \cdot (\mathbf{r}_u \, \delta u + \mathbf{r}_v \, \delta v)$$

Expanding:

$$\mathbf{w}_1 \cdot \mathbf{w}_2 = E \, du \, \delta u + F(du \, \delta v + dv \, \delta u) + G \, dv \, \delta v$$

Theorem 4.3 (Angle Between Directions on a Surface): The angle $\theta \in [0, \pi]$ between the directions $d\mathbf{r} = (du, dv)$ and $\delta\mathbf{r} = (\delta u, \delta v)$ is given by:

$$\cos \theta = \frac{\mathbf{w}_1 \cdot \mathbf{w}_2}{\|\mathbf{w}_1\| \|\mathbf{w}_2\|} = \frac{E \, du \, \delta u + F(du \, \delta v + dv \, \delta u) + G \, dv \, \delta v}{\sqrt{E \, du^2 + 2F \, du \, dv + G \, dv^2} \sqrt{E \, \delta u^2 + 2F \, \delta u \, \delta v + G \, \delta v^2}}$$

In particular, the two directions are orthogonal ($\theta = \pi/2$) if and only if:

$$E \, du \, \delta u + F(du \, \delta v + dv \, \delta u) + G \, dv \, \delta v = 0$$

2. Angle of a Curve with Coordinate Curves

For the $u$-coordinate curve ($dv = 0, du > 0$), the tangent vector is $\mathbf{r}_u$. The angle $\alpha$ between an arbitrary curve direction $(du, dv)$ and the $u$-coordinate curve satisfies:

$$\cos \alpha = \frac{\mathbf{r}_u \cdot (\mathbf{r}_u du + \mathbf{r}_v dv)}{\|\mathbf{r}_u\| \|d\mathbf{r}\|} = \frac{E du + F dv}{\sqrt{E} \sqrt{E du^2 + 2F dudv + G dv^2}}$$

For an orthogonal coordinate system ($F = 0$):

$$\cos \alpha = \frac{\sqrt{E} du}{\sqrt{E du^2 + G dv^2}} = \frac{\sqrt{E} du}{ds}, \qquad \sin \alpha = \frac{\sqrt{G} dv}{ds}$$

3. Orthogonal Trajectories of a Family of Curves

Suppose a family of curves on $S$ is defined by a differential equation:

$$P(u, v) \, du + Q(u, v) \, dv = 0 \implies \frac{dv}{du} = -\frac{P}{Q}$$

To find the family of orthogonal trajectories $(\delta u, \delta v)$, we apply the orthogonality condition:

$$E \, du \, \delta u + F(du \, \delta v + dv \, \delta u) + G \, dv \, \delta v = 0$$

Dividing by $du \, \delta u$:

$$E + F \left( \frac{\delta v}{\delta u} + \frac{dv}{du} \right) + G \left( \frac{dv}{du} \right)\left( \frac{\delta v}{\delta u} \right) = 0$$

Substituting $\frac{dv}{du} = -\frac{P}{Q}$:

$$E + F \left( \frac{\delta v}{\delta u} - \frac{P}{Q} \right) - G \frac{P}{Q} \frac{\delta v}{\delta u} = 0$$

Multiplying by $Q$:

$$(E Q - F P) \delta u + (F Q - G P) \delta v = 0$$

This first-order ODE governs the orthogonal trajectories across the surface patch.

ยง4.5 Surface Area Element, Jacobians & Integrals on Surfaces

1. Infinitesimal Area Element on a Surface

Consider an infinitesimal curvilinear parallelogram on the surface bounded by the vectors $\mathbf{r}_u \, du$ and $\mathbf{r}_v \, dv$. The area $dA$ of this infinitesimal parallelogram in $\mathbb{R}^3$ is the magnitude of their cross product:

$$dA = \|\mathbf{r}_u \, du \times \mathbf{r}_v \, dv\| = \|\mathbf{r}_u \times \mathbf{r}_v\| \, du \, dv$$

Using Lagrange's identity from Section 4.2:

$$\|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{EG - F^2}$$

Definition 4.8 (Surface Area Element): The intrinsic area element (or Riemannian volume element $d\sigma$) on a regular surface patch is:

$$dA = \sqrt{EG - F^2} \, du \, dv$$

The positive quantity $W = \sqrt{EG - F^2} > 0$ is the Gram determinant factor.


2. Surface Integral and Total Area

Definition 4.9 (Surface Area): Let $S = \mathbf{r}(U)$ be a regular surface patch where $U \subset \mathbb{R}^2$ is bounded. The surface area of $S$ is defined by the double integral:

$$\operatorname{Area}(S) = \iint_U dA = \iint_U \sqrt{EG - F^2} \, du \, dv$$

For a scalar function $f: S \to \mathbb{R}$, the surface integral of $f$ over $S$ is:

$$\iint_S f \, dA = \iint_U f(\mathbf{r}(u, v)) \sqrt{EG - F^2} \, du \, dv$$

3. Invariance Under Reparametrization

Theorem 4.4 (Invariance of Surface Area): The surface area $\operatorname{Area}(S)$ is invariant under orientation-preserving or orientation-reversing smooth reparametrizations.

Proof: Let $(\bar{u}, \bar{v})$ be an alternative coordinate system related to $(u, v)$ by a diffeomorphism $\Phi: \bar{U} \to U$, $(u, v) = \Phi(\bar{u}, \bar{v})$. By the multi-variable chain rule:

$$\mathbf{r}_{\bar{u}} = \mathbf{r}_u \frac{\partial u}{\partial \bar{u}} + \mathbf{r}_v \frac{\partial v}{\partial \bar{u}}, \qquad \mathbf{r}_{\bar{v}} = \mathbf{r}_u \frac{\partial u}{\partial \bar{v}} + \mathbf{r}_v \frac{\partial v}{\partial \bar{v}}$$

Taking the cross product:

$$\mathbf{r}_{\bar{u}} \times \mathbf{r}_{\bar{v}} = \left( \mathbf{r}_u \frac{\partial u}{\partial \bar{u}} + \mathbf{r}_v \frac{\partial v}{\partial \bar{u}} \right) \times \left( \mathbf{r}_u \frac{\partial u}{\partial \bar{v}} + \mathbf{r}_v \frac{\partial v}{\partial \bar{v}} \right)$$

Since $\mathbf{r}_u \times \mathbf{r}_u = \mathbf{0}$ and $\mathbf{r}_v \times \mathbf{r}_v = \mathbf{0}$, this simplifies to:

$$\mathbf{r}_{\bar{u}} \times \mathbf{r}_{\bar{v}} = \left( \frac{\partial u}{\partial \bar{u}}\frac{\partial v}{\partial \bar{v}} - \frac{\partial v}{\partial \bar{u}}\frac{\partial u}{\partial \bar{v}} \right) (\mathbf{r}_u \times \mathbf{r}_v) = \det(J_\Phi) (\mathbf{r}_u \times \mathbf{r}_v)$$

Taking norms on both sides:

$$\sqrt{\bar{E}\bar{G} - \bar{F}^2} = \|\mathbf{r}_{\bar{u}} \times \mathbf{r}_{\bar{v}}\| = |\det(J_\Phi)| \|\mathbf{r}_u \times \mathbf{r}_v\| = |\det(J_\Phi)| \sqrt{EG - F^2}$$

By the multivariable change-of-variables theorem for double integrals:

$$\iint_{\bar{U}} \sqrt{\bar{E}\bar{G} - \bar{F}^2} \, d\bar{u} \, d\bar{v} = \iint_{\bar{U}} \sqrt{EG - F^2} |\det(J_\Phi)| \, d\bar{u} \, d\bar{v} = \iint_U \sqrt{EG - F^2} \, du \, dv$$

Thus the surface area is completely independent of the choice of coordinate patch! $\blacksquare$

Tiered Solved Examination Problems & Rigorous Derivations

Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.

Foundational Example 4.1: First Fundamental Form and Total Surface Area of the 2-Sphere

Consider the 2-sphere of radius $R > 0$ parametrized by spherical angles $(\theta, \phi)$:

$$\mathbf{r}(\theta, \phi) = \begin{pmatrix} R \sin\theta \cos\phi \\ R \sin\theta \sin\phi \\ R \cos\theta \end{pmatrix}, \quad \theta \in (0, \pi), \; \phi \in (0, 2\pi)$$
  1. Compute the partial derivatives $\mathbf{r}_\theta$ and $\mathbf{r}_\phi$.
  2. Calculate the metric coefficients $E, F, G$ and write down the First Fundamental Form $I$.
  3. Compute the area element $dA = \sqrt{EG - F^2} \, d\theta \, d\phi$ and evaluate the total surface area of the sphere.

1. Partial Derivatives

Differentiating $\mathbf{r}(\theta, \phi)$ with respect to $\theta$:

$$\mathbf{r}_\theta = \begin{pmatrix} R \cos\theta \cos\phi \\ R \cos\theta \sin\phi \\ -R \sin\theta \end{pmatrix}$$

Differentiating with respect to $\phi$:

$$\mathbf{r}_\phi = \begin{pmatrix} -R \sin\theta \sin\phi \\ R \sin\theta \cos\phi \\ 0 \end{pmatrix}$$

2. Metric Coefficients and First Fundamental Form

Computing $E = \mathbf{r}_\theta \cdot \mathbf{r}_\theta$:

$$E = R^2 \cos^2\theta \cos^2\phi + R^2 \cos^2\theta \sin^2\phi + R^2 \sin^2\theta = R^2 \cos^2\theta (\cos^2\phi + \sin^2\phi) + R^2 \sin^2\theta = R^2 (\cos^2\theta + \sin^2\theta) = R^2$$

Computing $F = \mathbf{r}_\theta \cdot \mathbf{r}_\phi$:

$$F = (R \cos\theta \cos\phi)(-R \sin\theta \sin\phi) + (R \cos\theta \sin\phi)(R \sin\theta \cos\phi) + (-R \sin\theta)(0)$$
$$F = -R^2 \sin\theta \cos\theta \sin\phi \cos\phi + R^2 \sin\theta \cos\theta \sin\phi \cos\phi = 0$$

Since $F = 0$, the spherical coordinate lines are orthogonal everywhere!

Computing $G = \mathbf{r}_\phi \cdot \mathbf{r}_\phi$:

$$G = (-R \sin\theta \sin\phi)^2 + (R \sin\theta \cos\phi)^2 + 0^2 = R^2 \sin^2\theta (\sin^2\phi + \cos^2\phi) = R^2 \sin^2\theta$$

Thus, the First Fundamental Form of the sphere is:

$$I = R^2 \, d\theta^2 + R^2 \sin^2\theta \, d\phi^2 \quad \blacksquare$$

3. Surface Area Element and Total Area

The metric determinant is:

$$EG - F^2 = (R^2)(R^2 \sin^2\theta) - 0 = R^4 \sin^2\theta$$

Since $\theta \in (0, \pi)$, $\sin\theta > 0$, so:

$$\sqrt{EG - F^2} = R^2 \sin\theta$$

The area element is:

$$dA = R^2 \sin\theta \, d\theta \, d\phi$$

Evaluating the total surface area:

$$\operatorname{Area}(S^2) = \int_0^{2\pi} d\phi \int_0^\pi R^2 \sin\theta \, d\theta = 2\pi R^2 [-\cos\theta]_0^\pi = 2\pi R^2 (-(-1) - (-1)) = 2\pi R^2 (2) = 4\pi R^2 \quad \blacksquare$$
Advanced Example 4.2: Local Isometry Between the Catenoid and Helicoid

The Catenoid $S_{\text{cat}}$ and the Helicoid $S_{\text{hel}}$ are parametrized by:

$$\mathbf{r}_{\text{cat}}(u, v) = \begin{pmatrix} \cosh u \cos v \\ \cosh u \sin v \\ u \end{pmatrix}, \quad (u, v) \in \mathbb{R} \times (0, 2\pi)$$
$$\mathbf{r}_{\text{hel}}(\bar{u}, \bar{v}) = \begin{pmatrix} \bar{u} \cos \bar{v} \\ \bar{u} \sin \bar{v} \\ \bar{v} \end{pmatrix}, \quad (\bar{u}, \bar{v}) \in \mathbb{R} \times (0, 2\pi)$$
  1. Compute the metric coefficients $E, F, G$ of the Catenoid.
  2. Compute the metric coefficients $\bar{E}, \bar{F}, \bar{G}$ of the Helicoid.
  3. Show that under the coordinate transformation $\bar{u} = \sinh u$ and $\bar{v} = v$, the two First Fundamental Forms coincide, proving that the Catenoid and Helicoid are locally isometric.

1. Metric Coefficients of the Catenoid

For $\mathbf{r}_{\text{cat}}(u, v) = (\cosh u \cos v, \cosh u \sin v, u)^T$:

$$\mathbf{r}_u = \begin{pmatrix} \sinh u \cos v \\ \sinh u \sin v \\ 1 \end{pmatrix}, \qquad \mathbf{r}_v = \begin{pmatrix} -\cosh u \sin v \\ \cosh u \cos v \\ 0 \end{pmatrix}$$

Now compute the dot products:

$$E = \mathbf{r}_u \cdot \mathbf{r}_u = \sinh^2 u (\cos^2 v + \sin^2 v) + 1 = \sinh^2 u + 1 = \cosh^2 u$$
$$F = \mathbf{r}_u \cdot \mathbf{r}_v = -\sinh u \cosh u \sin v \cos v + \sinh u \cosh u \sin v \cos v + 0 = 0$$
$$G = \mathbf{r}_v \cdot \mathbf{r}_v = \cosh^2 u (\sin^2 v + \cos^2 v) + 0 = \cosh^2 u$$

Thus, the First Fundamental Form of the catenoid is:

$$I_{\text{cat}} = \cosh^2 u \, du^2 + \cosh^2 u \, dv^2 = \cosh^2 u (du^2 + dv^2) \quad \blacksquare$$

2. Metric Coefficients of the Helicoid

For $\mathbf{r}_{\text{hel}}(\bar{u}, \bar{v}) = (\bar{u} \cos \bar{v}, \bar{u} \sin \bar{v}, \bar{v})^T$:

$$\mathbf{r}_{\bar{u}} = \begin{pmatrix} \cos \bar{v} \\ \sin \bar{v} \\ 0 \end{pmatrix}, \qquad \mathbf{r}_{\bar{v}} = \begin{pmatrix} -\bar{u} \sin \bar{v} \\ \bar{u} \cos \bar{v} \\ 1 \end{pmatrix}$$

Now compute the metric coefficients:

$$\bar{E} = \mathbf{r}_{\bar{u}} \cdot \mathbf{r}_{\bar{u}} = \cos^2 \bar{v} + \sin^2 \bar{v} + 0 = 1$$
$$\bar{F} = \mathbf{r}_{\bar{u}} \cdot \mathbf{r}_{\bar{v}} = -\bar{u} \sin \bar{v} \cos \bar{v} + \bar{u} \sin \bar{v} \cos \bar{v} + 0 = 0$$
$$\bar{G} = \mathbf{r}_{\bar{v}} \cdot \mathbf{r}_{\bar{v}} = \bar{u}^2 (\sin^2 \bar{v} + \cos^2 \bar{v}) + 1 = \bar{u}^2 + 1$$

Thus, the First Fundamental Form of the helicoid is:

$$I_{\text{hel}} = d\bar{u}^2 + (\bar{u}^2 + 1) \, d\bar{v}^2 \quad \blacksquare$$

3. Coordinate Transformation and Local Isometry

Let $\bar{u} = \sinh u$ and $\bar{v} = v$. Then:

$$d\bar{u} = \cosh u \, du, \qquad d\bar{v} = dv$$

Substitute these into $I_{\text{hel}}$:

$$d\bar{u}^2 = (\cosh u \, du)^2 = \cosh^2 u \, du^2$$
$$\bar{u}^2 + 1 = \sinh^2 u + 1 = \cosh^2 u$$
$$(\bar{u}^2 + 1) \, d\bar{v}^2 = \cosh^2 u \, dv^2$$

Therefore:

$$I_{\text{hel}} = \cosh^2 u \, du^2 + \cosh^2 u \, dv^2 = I_{\text{cat}}$$

Since the First Fundamental Forms match identically under this smooth bijection, the Catenoid and Helicoid are locally isometric! $\blacksquare$

Honors / Proof Challenge Example 4.3: Angle Preservation of Isothermal Coordinates & Conformal Invariance

A regular coordinate system $(u, v)$ on a surface $S$ is called isothermal (or conformal) if the First Fundamental Form takes the form:

$$I = \lambda^2(u, v)(du^2 + dv^2)$$

where $\lambda(u, v) > 0$ is a smooth non-vanishing function.

  1. Let $\mathbf{w}_1 = \mathbf{r}_u \, du_1 + \mathbf{r}_v \, dv_1$ and $\mathbf{w}_2 = \mathbf{r}_u \, du_2 + \mathbf{rv} \, dv_2$ be two tangent vectors at $p = \mathbf{r}(u, v)$. Prove that the geometric angle $\theta \in [0, \pi]$ between $\mathbf{w}_1$ and $\mathbf{w}_2$ in $\mathbb{R}^3$ equals the Euclidean angle $\alpha$ between the parameter displacement vectors $\mathbf{v}_1 = (du_1, dv_1)^T$ and $\mathbf{v}_2 = (du_2, dv_2)^T$ in the parameter plane $\mathbb{R}^2$.
  2. Conclude that the parameter mapping $\mathbf{r}: U \subset \mathbb{R}^2 \to S \subset \mathbb{R}^3$ is a conformal map, preserving all angles and shapes of infinitesimal figures.

1. Inner Product and Norms Under Isothermal Coordinates

Assume the metric coefficients satisfy:

$$E = \lambda^2(u, v), \quad F = 0, \quad G = \lambda^2(u, v)$$

with $\lambda(u, v) > 0$. The inner product of the tangent vectors $\mathbf{w}_1, \mathbf{w}_2 \in T_p S$ is given by the bilinear form:

$$\mathbf{w}_1 \cdot \mathbf{w}_2 = E \, du_1 \, du_2 + F(du_1 \, dv_2 + dv_1 \, du_2) + G \, dv_1 \, dv_2$$

Substituting $E = G = \lambda^2$ and $F = 0$:

$$\mathbf{w}_1 \cdot \mathbf{w}_2 = \lambda^2 \, du_1 \, du_2 + 0 + \lambda^2 \, dv_1 \, dv_2 = \lambda^2 (du_1 \, du_2 + dv_1 \, dv_2)$$

Notice that:

$$du_1 \, du_2 + dv_1 \, dv_2 = \mathbf{v}_1 \cdot \mathbf{v}_2$$

where $\mathbf{v}_1 = (du_1, dv_1)^T$ and $\mathbf{v}_2 = (du_2, dv_2)^T$ are vectors in the flat parameter plane $\mathbb{R}^2$. Thus:

$$\mathbf{w}_1 \cdot \mathbf{w}_2 = \lambda^2 (\mathbf{v}_1 \cdot \mathbf{v}_2)$$

Next, compute the norms of $\mathbf{w}_1$ and $\mathbf{w}_2$:

$$\|\mathbf{w}_1\|^2 = I(du_1, dv_1) = \lambda^2 (du_1^2 + dv_1^2) = \lambda^2 \|\mathbf{v}_1\|^2 \implies \|\mathbf{w}_1\| = \lambda \|\mathbf{v}_1\|$$
$$\|\mathbf{w}_2\|^2 = I(du_2, dv_2) = \lambda^2 (du_2^2 + dv_2^2) = \lambda^2 \|\mathbf{v}_2\|^2 \implies \|\mathbf{w}_2\| = \lambda \|\mathbf{v}_2\|$$

2. Angle Equivalence

The cosine of the 3D angle $\theta$ between $\mathbf{w}_1$ and $\mathbf{w}_2$ on the surface is:

$$\cos \theta = \frac{\mathbf{w}_1 \cdot \mathbf{w}_2}{\|\mathbf{w}_1\| \|\mathbf{w}_2\|}$$

Substituting the expressions derived above:

$$\cos \theta = \frac{\lambda^2 (\mathbf{v}_1 \cdot \mathbf{v}_2)}{(\lambda \|\mathbf{v}_1\|) (\lambda \|\mathbf{v}_2\|)} = \frac{\lambda^2 (\mathbf{v}_1 \cdot \mathbf{v}_2)}{\lambda^2 \|\mathbf{v}_1\| \|\mathbf{v}_2\|} = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{\|\mathbf{v}_1\| \|\mathbf{v}_2\|}$$

The right-hand side is precisely the definition of $\cos \alpha$, where $\alpha$ is the standard Euclidean angle between $\mathbf{v}_1$ and $\mathbf{v}_2$ in $\mathbb{R}^2$:

$$\cos \theta = \cos \alpha$$

Since both $\theta, \alpha \in [0, \pi]$, we have:

$$\theta = \alpha \quad \blacksquare$$

3. Conclusion: Conformal Mapping

Because $\theta = \alpha$ for any two arbitrary non-zero tangent directions at every point $p \in S$:

  1. Every angle between curves on the surface is identical to the angle between their preimage curves in the $uv$-plane.
  2. The coordinate mapping $\mathbf{r}: U \to S$ is a conformal map (angle-preserving).
  3. Infinitesimal circles in the parameter plane are mapped to infinitesimal circles on the surface, scaled uniformly in all directions by the factor $\lambda(u, v)$ without angular shearing. $\blacksquare$