---
title: "Surfaces"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 19
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/19-surfaces
---

# Chapter 19 — Surfaces

After curves, surfaces: two-parameter objects in $\R^3$. The differential calculus of [Chapter 15](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#ch-b2-diffcalc) provides everything we need — partial derivatives give tangent vectors, the cross product gives the normal, [determinants](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) give [areas](#def-b2-surfaces-area). We define regular [parametrized surfaces](#def-b2-surfaces-param), their [tangent planes](#def-b2-surfaces-tangent), and the *[first fundamental form](#def-b2-surfaces-fff)*, which encodes all [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) and [area](#def-b2-surfaces-area) measurements on the surface. Surfaces also arise as level sets $f(x, y, z) = c$; the gradient then directs the normal.

## 19.1 Parametrized surfaces

**Definition 19.1 (Regular parametrized surface).**

Let $U \subseteq \R^2$ be [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology). A *parametrized surface* of class $\mathcal{C}^k$ ($k \geq 1$) is a map $\sigma \colon U \to
\R^3$, $(u, v) \mapsto \sigma(u, v)$, of class $\mathcal{C}^k$. A point is *regular* if the partial derivative vectors

$$
\sigma_u(u, v) = \frac{\partial\sigma}{\partial u}(u,v),
\qquad
\sigma_v(u, v) = \frac{\partial\sigma}{\partial v}(u,v)
$$

are linearly independent, i.e. $\sigma_u \wedge \sigma_v \neq 0$; the surface is *regular* if every point is.

**Example 19.2 (The three standard descriptions).**

1. *Graph:* $\sigma(u, v) = (u,\ v,\ f(u, v))$ for $f \in \mathcal{C}^1(U)$ . Always regular: $\sigma_u = (1, 0, f_u)$ and $\sigma_v = (0, 1, f_v)$ are independent.
2. *Sphere (spherical coordinates):* for the sphere of radius $R$, $$\sigma(\theta, \varphi) = (R\cos\theta\cos\varphi,\ R\sin\theta\cos\varphi,\ R\sin\varphi), \qquad (\theta, \varphi) \in \R \times \bigl(-\tfrac\pi2, \tfrac\pi2\bigr),$$ with $\theta$ the longitude and $\varphi$ the latitude. One checks $\norm{\sigma_\theta \wedge \sigma_\varphi} = R^2\cos\varphi > 0$: regular away from the poles (which this chart omits).
3. *Level set:* $S = \{(x,y,z) : f(x, y, z) = c\}$ where $f$ is $\mathcal{C}^1$ and $\nabla f \neq 0$ on $S$ . Near each point, one coordinate can be expressed as a function of the other two by the implicit function theorem ( [Chapter 15](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#ch-b2-diffcalc) ), so $S$ is locally a graph.

**Example 19.3 (From level set to graph).**

The implicit function theorem in item 3 deserves one explicit run. Take the sphere $x^2 + y^2 + z^2 = R^2$ near its north pole $(0, 0, R)$: there $\frac{\partial f}{\partial z} = 2z =
2R \neq 0$, and solving for $z$ gives the graph chart

$$
z = \sqrt{R^2 - x^2 - y^2},
\qquad x^2 + y^2 < R^2 ,
$$

regular everywhere on its ([open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology)) domain — including the pole that the spherical chart missed. Near an equator point like $(R, 0, 0)$ the same theorem solves for $x$ instead ($\frac{\partial f}{\partial x} = 2R \neq 0$). The rule of thumb: a level surface is a graph over the coordinate plane *orthogonal to the largest component of the gradient*, and by covering the sphere with six such graph charts one checks its smoothness everywhere with no trigonometry at all.

## 19.2 Tangent plane and normal

**Definition 19.4 (Tangent plane).**

Let $\sigma$ be regular at $(u_0, v_0)$, $M_0 = \sigma(u_0, v_0)$. The *tangent plane* $T_{M_0}S$ is the plane through $M_0$ directed by $\operatorname{Vect}(\sigma_u, \sigma_v)$ (partials at $(u_0, v_0)$). The *unit normal* is

$$
n(u_0, v_0)
= \frac{\sigma_u \wedge \sigma_v}
       {\norm{\sigma_u \wedge \sigma_v}} .
$$

**Proposition 19.5 (Tangent vectors are velocity vectors).**

The direction of $T_{M_0}S$ is exactly the set of vectors $\gamma'(0)$, where $\gamma = \sigma \circ c$ ranges over the $\mathcal{C}^1$ curves drawn on the surface through $M_0$ (i.e. $c \colon (-\varepsilon, \varepsilon) \to U$ is $\mathcal{C}^1$ with $c(0) = (u_0, v_0)$).

**Proof.** If $c(t) = (u(t), v(t))$, the chain rule ([Chapter 15](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#ch-b2-diffcalc)) gives

$$
\gamma'(0) = u'(0)\,\sigma_u + v'(0)\,\sigma_v
\in \operatorname{Vect}(\sigma_u, \sigma_v).
$$

Conversely, the vector $a\sigma_u + b\sigma_v$ is attained by the curve $c(t) = (u_0 + at,\ v_0 + bt)$, which stays in the [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) set $U$ for $\abs t$ small. ∎

**Example 19.6 (The helicoid’s tangent plane).**

For the helicoid $\sigma(u, v) = (v\cos u,\ v\sin u,\ au)$, at the point $\sigma(0, 1) = (1, 0, 0)$:

$$
\sigma_u = (0,\ 1,\ a), \qquad \sigma_v = (1,\ 0,\ 0),
\qquad
\sigma_u \wedge \sigma_v = (0,\ a,\ -1),
$$

so the [tangent plane](#def-b2-surfaces-tangent) is $a\,y = z$. It contains the whole horizontal ruling $\{(t, 0, 0)\}$ (direction $\sigma_v$): as for the cone of [Exercise 19.1](#exo-b2-surfaces-1), a surface ruled by straight lines has each ruling lying inside the [tangent plane](#def-b2-surfaces-tangent) along it. The other tangent direction $\sigma_u$ is the velocity of the helix $u \mapsto \sigma(u, 1)$: one chart, two drawn curves, and the whole [tangent plane](#def-b2-surfaces-tangent) is spanned — [Proposition 19.5](#prop-b2-surfaces-velocity) in action.

**Proposition 19.7 (Normal of a level surface).**

Let $S = \{f = c\}$ with $f$ of class $\mathcal{C}^1$ and $\nabla
f(M_0) \neq 0$. Then the [tangent plane](#def-b2-surfaces-tangent) of $S$ at $M_0$ is the plane through $M_0$ orthogonal to $\nabla f(M_0)$:

$$
T_{M_0}S :\quad
\langle \nabla f(M_0),\ M - M_0\rangle = 0 .
$$

**Proof.** For any curve $\gamma$ drawn on $S$ through $M_0$, $f(\gamma(t)) =
c$ identically, so the chain rule gives $\langle
\nabla f(M_0), \gamma'(0)\rangle = 0$: all velocity vectors are orthogonal to the gradient, so the tangent direction is contained in the plane $\nabla f(M_0)^\perp$. Both are $2$-dimensional subspaces — the tangent direction because $S$ is locally a regular graph ([Example 19.2](#ex-b2-surfaces-standard)), the orthogonal complement because $\nabla f(M_0) \neq 0$ — hence they are equal. ∎

**Example 19.8.**

For the sphere $x^2 + y^2 + z^2 = R^2$: $\nabla f = 2(x, y, z)$, so the [tangent plane](#def-b2-surfaces-tangent) at $M_0$ is orthogonal to the radius $\vect{OM_0}$ — the classical fact that radius and [tangent plane](#def-b2-surfaces-tangent) are perpendicular, with equation $\langle M_0, M\rangle = R^2$.

**Example 19.9 (Graph tangent plane).**

For $z = f(x, y)$ at $(x_0, y_0)$: applying [Proposition 19.7](#prop-b2-surfaces-gradient) to $F(x,y,z) = f(x,y) - z$,

$$
z = f(x_0, y_0)
+ f_x(x_0, y_0)(x - x_0)
+ f_y(x_0, y_0)(y - y_0) ,
$$

the [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) part of the first-order Taylor expansion — the [tangent plane](#def-b2-surfaces-tangent) is the graph of the [differential](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#def-b2-diffcalc-differential), as it must be.

**Example 19.10 (The closest point of a surface).**

Which point of the paraboloid $z = x^2 + y^2$ is nearest to $P = (0, 0, 1)$? Minimize the squared distance along the surface: with $\rho^2 = x^2 + y^2$,

$$
g(\rho^2) = \rho^2 + (\rho^2 - 1)^2,
\qquad
g'(\rho^2) = 1 + 2(\rho^2 - 1) = 0
\iff \rho^2 = \tfrac12 ,
$$

giving the circle of points at height $z = \frac12$ and distance $\sqrt{\tfrac12 + \tfrac14} = \frac{\sqrt3}2$. The geometric signature of minimality: at such a point $M$, the vector $\vect{MP}$ must be *normal* to the surface — otherwise sliding along a drawn curve with velocity having a component toward $P$ would decrease the distance. Check: $\nabla(z - x^2 - y^2) = (-2x, -2y, 1)$ at $M = (x, y,
\tfrac12)$, while $\vect{MP} = (-x, -y, \tfrac12) =
\tfrac12(-2x, -2y, 1)$: parallel, as predicted. The first-order condition “foot of the perpendicular” is the same one that will drive extrema on level sets in [Exercise 19.12](#exo-b2-surfaces-12).

**Example 19.11 (Tangent planes of quadrics: the polarization rule).**

Let $S : xy + yz + zx = 1$ and $M_0 = (1, 1, 0) \in S$. Here $\nabla f = (y + z,\ x + z,\ x + y)$, so $\nabla f(M_0) = (1,
1, 2)$ and the [tangent plane](#def-b2-surfaces-tangent) is

$$
(x - 1) + (y - 1) + 2z = 0,
\qquad\text{i.e.}\qquad x + y + 2z = 2 .
$$

The same answer comes from the *[polarization](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def) rule* that generalizes [Example 19.8](#ex-b2-surfaces-spheretangent) and [Exercise 19.2](#exo-b2-surfaces-2): in the [quadric](#pb-b2-surfaces-1)’s equation, replace $x^2$ by $x_0x$, and each product $xy$ by $\frac{x_0y +
y_0x}2$ (and cyclically):

$$
\frac{x_0y + y_0x}2 + \frac{y_0z + z_0y}2 + \frac{z_0x +
x_0z}2 = 1
\;\xrightarrow{\ M_0 = (1,1,0)\ }\;
\frac{x + y}2 + \frac z2 + \frac z2 = 1 ,
$$

which is $x + y + 2z = 2$ again. The rule works because $\nabla$ of a [quadratic form](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def) is the associated bilinear form evaluated against the base point — tangency to a [quadric](#pb-b2-surfaces-1) is [polarization](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def), one more face of [Chapter 12](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#ch-b2-quadratic).

## 19.3 The first fundamental form

**Definition 19.12 (First fundamental form).**

Let $\sigma \colon U \to \R^3$ be a regular $\mathcal{C}^1$ surface. Its *first fundamental form* at $(u,v)$ is the positive definite [quadratic form](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def) on $\R^2$

$$
I(h, k) = \norm{h\,\sigma_u + k\,\sigma_v}^2
= E\,h^2 + 2F\,hk + G\,k^2,
$$

where

$$
E = \norm{\sigma_u}^2,
\qquad
F = \langle \sigma_u, \sigma_v\rangle,
\qquad
G = \norm{\sigma_v}^2 .
$$

**Remark 19.13.**

$I$ is the restriction of the ambient Euclidean scalar product to the [tangent plane](#def-b2-surfaces-tangent), read in the basis $(\sigma_u, \sigma_v)$: it is positive definite precisely because $\sigma_u, \sigma_v$ are independent ([Chapter 12](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#ch-b2-quadratic)). Every metric quantity on the surface — [lengths](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) of drawn curves, angles between them, [areas](#def-b2-surfaces-area) — is computed from $E, F, G$ alone. Two surfaces with the same $E,
F, G$ in suitable parameters are *isometric* even if they sit differently in space: this is the starting point of intrinsic geometry.

**Example 19.14 (Angles between coordinate curves).**

The [first fundamental form](#def-b2-surfaces-fff) also measures angles: the coordinate curves $u \mapsto \sigma(u, v_0)$ and $v \mapsto
\sigma(u_0, v)$ meet at the angle $\theta$ with

$$
\cos\theta = \frac{\langle\sigma_u,
\sigma_v\rangle}{\norm{\sigma_u}\,\norm{\sigma_v}}
= \frac{F}{\sqrt{EG}} :
$$

the single coefficient $F$ decides orthogonality of the parameter net. For the sphere chart and the helicoid, $F =
0$: meridians cut parallels, and helices cut the horizontal rulings, at right angles — which is why their [area](#def-b2-surfaces-area) integrands collapsed to $\sqrt{EG}$. For a graph chart, $F = f_xf_y$ vanishes only where a partial derivative does: the coordinate net of a tilted graph is *not* orthogonal, even though the $(x, y)$-net downstairs is. When computations on a surface look heavy, the first move is to seek a chart with $F = 0$.

**Example 19.15 (The saddle chart).**

For the saddle $z = xy$ with chart $\sigma(u, v) = (u, v,
uv)$:

$$
\sigma_u = (1, 0, v), \qquad \sigma_v = (0, 1, u),
\qquad
E = 1 + v^2, \quad F = uv, \quad G = 1 + u^2 ,
$$

and $EG - F^2 = 1 + u^2 + v^2 > 0$: regular everywhere. The two coordinate curves through a point are *straight lines* of $\R^3$ (fix $u$ or fix $v$: the rulings of the doubly ruled saddle), yet $F \neq 0$ off the axes: rulings through a generic point are not orthogonal. Both rulings lie in the [tangent plane](#def-b2-surfaces-tangent), which they span — so the [tangent plane](#def-b2-surfaces-tangent) cuts the surface along two whole lines, the extreme opposite of the sphere, whose [tangent planes](#def-b2-surfaces-tangent) touch at one point only. The sign of the “second-order contact” between a surface and its [tangent planes](#def-b2-surfaces-tangent) is a [curvature](https://one-course.com/books/math/4/en/chapter/18-curves#thm-b2-curves-frenet2d) story, taken up in the Year 3 volume.

**Proposition 19.16 (Length of a curve drawn on a surface).**

If $\gamma(t) = \sigma(u(t), v(t))$, $t \in [a, b]$, is $\mathcal{C}^1$, then

$$
L(\gamma) = \int_a^b
\sqrt{E\,u'^2 + 2F\,u'v' + G\,v'^2}\;\dd t ,
$$

with $E, F, G$ evaluated at $(u(t), v(t))$.

**Proof.** $\gamma' = u'\sigma_u + v'\sigma_v$ by the chain rule, so $\norm{\gamma'}^2 = I(u', v') = Eu'^2 + 2Fu'v' + Gv'^2$; integrate $\norm{\gamma'}$ ([Definition 18.6](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length)). ∎

**Example 19.17 (Why airliners fly over the pole).**

Two airports sit at latitude $\varphi_0$ and opposite longitudes: $A = \sigma(0, \varphi_0)$ and $B = \sigma(\pi,
\varphi_0)$ on the sphere of radius $R$. Along the parallel ($\varphi \equiv \varphi_0$), the [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) is $\int_0^\pi R\cos\varphi_0\,\dd\theta = \pi R\cos\varphi_0$. Along the route over the pole (up the meridian $\theta = 0$, down the meridian $\theta = \pi$), it is $2R(\frac\pi2 -
\varphi_0)$. At latitude $\varphi_0 = \frac\pi3$ (sixty degrees): parallel route $\pi R/2 \approx 1.571\,R$, polar route $\pi R/3 \approx 1.047\,R$ — a third shorter. In fact $\pi\cos\varphi_0 \geq \pi - 2\varphi_0$ on $\intcc0{\pi/2}$ (the function $\pi\cos\varphi - \pi +
2\varphi$ vanishes at both ends and its derivative $2 -
\pi\sin\varphi$ changes sign once, so it is first increasing then decreasing, hence nonnegative): the polar route never loses. The [first fundamental form](#def-b2-surfaces-fff) turned a navigation question into two one-line integrals; [Exercise 19.6](#exo-b2-surfaces-6) pushes the idea to a genuine minimality proof for meridians.

**Lemma 19.18 (Lagrange identity).**

For all $a, b \in \R^3$: $\norm{a \wedge b}^2 = \norm a^2 \norm b^2 - \langle a, b\rangle^2$. In particular

$$
\norm{\sigma_u \wedge \sigma_v} = \sqrt{EG - F^2}.
$$

**Proof.** Both sides are unchanged if we replace $b$ by its component $b_\perp
= b - \frac{\langle a, b\rangle}{\norm a^2}a$ orthogonal to $a$ (for $a \neq 0$; the case $a = 0$ is trivial): the left side because $a \wedge a = 0$, the right side by expanding $\norm
{b_\perp}^2 = \norm b^2 - \frac{\langle a,b\rangle^2}{\norm a^2}$ and $\langle a, b_\perp\rangle = 0$. So it suffices to prove the identity for orthogonal $a, b$, where it reads $\norm{a \wedge
b} = \norm a \norm b$: true, since for orthogonal vectors the cross product has [norm](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) $\norm a\norm b\,\abs{\sin\frac\pi2}$. The displayed formula is the case $a = \sigma_u$, $b = \sigma_v$. ∎

**Remark 19.19.**

The Lagrange identity says $EG - F^2 = \det\left(\begin{smallmatrix}
E & F\\ F & G\end{smallmatrix}\right)$ is the *Gram [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det)* of $(\sigma_u, \sigma_v)$: the squared [area](#def-b2-surfaces-area) of the parallelogram they span. Regularity, positive definiteness of the [first fundamental form](#def-b2-surfaces-fff), and positivity of the Gram [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) are three phrasings of one condition — which is why the [area](#def-b2-surfaces-area) integrand below never vanishes on a regular chart.

**Definition 19.20 (Area).**

Let $\sigma \colon U \to \R^3$ be a regular injective $\mathcal{C}^1$ surface and $K \subseteq U$ a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) domain on which double integrals make sense ([Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint)). The *area* of the piece $\sigma(K)$ is

$$
\mathcal{A}
= \iint_K \norm{\sigma_u \wedge \sigma_v}\,\dd u\,\dd v
= \iint_K \sqrt{EG - F^2}\;\dd u\,\dd v .
$$

**Remark 19.21 (Why this formula).**

The rectangle $[u, u + \dd u] \times [v, v + \dd v]$ is mapped, to first order, onto the parallelogram spanned by $\sigma_u\,\dd u$ and $\sigma_v\,\dd v$, whose [area](#def-b2-surfaces-area) is $\norm{\sigma_u \wedge
\sigma_v}\,\dd u\,\dd v$: the definition integrates the local area-distortion factor, exactly as [arc length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) integrates the local speed. Consistency with [change of parameters](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-reparam) is [Exercise 19.8](#exo-b2-surfaces-8); consistency with the change-of-variables formula for double integrals is discussed in [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint).

**Example 19.22 (Two different graphs, one area).**

Over the unit disk, compare the bowl $z = \frac12(x^2 + y^2)$ and the saddle $z = xy$. Their [area](#def-b2-surfaces-area) integrands ([Exercise 19.5](#exo-b2-surfaces-5)) are

$$
\sqrt{1 + x^2 + y^2}
\qquad\text{and}\qquad
\sqrt{1 + y^2 + x^2} :
$$

identical. The two surfaces — one curving the same way in all directions, the other saddle-shaped — have exactly equal [areas](#def-b2-surfaces-area) over every domain, $\frac{2\pi}3(2\sqrt2 - 1)$ over the unit disk. The [area](#def-b2-surfaces-area) element only sees the *[length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length)* of the gradient, not the arrangement of the bending; telling the bowl from the saddle requires second-order data (the sign structure exhibited in [Figure 19.1](#fig-b2-surfaces-saddle)), which no amount of [area](#def-b2-surfaces-area) measurement detects. [First fundamental form](#def-b2-surfaces-fff): metric, blind to shape; the shape-seeing second form belongs to Year 3.

**Example 19.23 (Area of the sphere).**

For the spherical chart of [Example 19.2](#ex-b2-surfaces-standard):

$$
\sigma_\theta = R(-\sin\theta\cos\varphi,\
\cos\theta\cos\varphi,\ 0),
\qquad
\sigma_\varphi = R(-\cos\theta\sin\varphi,\
-\sin\theta\sin\varphi,\ \cos\varphi),
$$

so $E = R^2\cos^2\varphi$, $F = 0$, $G = R^2$ and $\sqrt{EG - F^2}
= R^2\cos\varphi$. Hence

$$
\mathcal{A}
= \int_{-\pi/2}^{\pi/2}\!\!\int_0^{2\pi}
R^2\cos\varphi\;\dd\theta\,\dd\varphi
= 2\pi R^2\,\bigl[\sin\varphi\bigr]_{-\pi/2}^{\pi/2}
= \boxed{4\pi R^2} .
$$

**Example 19.24 (The cone, checked against the school formula).**

For the cone $z = \sqrt{x^2 + y^2}$ over the annulus $a \leq
\rho \leq b$, the graph formula of [Exercise 19.5](#exo-b2-surfaces-5) gives $1 + f_x^2 + f_y^2 = 2$ (compute $f_x = x/\rho$, $f_y = y/\rho$), so

$$
\mathcal A = \sqrt2\,\pi\,(b^2 - a^2) .
$$

Consistency check with the slant-height formula $\pi\rho\ell$ of [Example 19.26](#ex-b2-surfaces-revolution): the full cones of base radii $b$ and $a$ have lateral [areas](#def-b2-surfaces-area) $\pi b\cdot b\sqrt2$ and $\pi a\cdot a\sqrt2$, whose difference is exactly $\sqrt2\pi(b^2 - a^2)$. Two charts, two formulas, one [area](#def-b2-surfaces-area) — the invariance proved in [Exercise 19.8](#exo-b2-surfaces-8), seen in the wild.

**Remark 19.25 (Sanity checks for areas).**

Three instant checks catch most errors in an [area](#def-b2-surfaces-area) computation. *Scaling*: dilating a surface by $\lambda$ multiplies $E, F, G$ by $\lambda^2$ and the [area](#def-b2-surfaces-area) by $\lambda^2$ — an answer whose dependence on $R$ is not quadratic (like $4\pi R^2$) is wrong. *Positivity of the element*: $\sqrt{EG - F^2}$ must be strictly positive on the chart’s interior; a vanishing value flags a chart degeneracy, to be excised as in the sphere’s poles. *Symmetry*: a computation over a [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) piece must be consistent with summing its congruent parts — the hemisphere had better give $2\pi R^2$.

**Example 19.26 (Surface of revolution).**

Rotate the curve $z \mapsto (r(z), 0, z)$, $r > 0$ of class $\mathcal{C}^1$, around the $z$-axis:

$$
\sigma(\theta, z) = (r(z)\cos\theta,\ r(z)\sin\theta,\ z).
$$

Then $E = r(z)^2$, $F = 0$, $G = 1 + r'(z)^2$, so

$$
\mathcal{A} = \int_a^b\!\!\int_0^{2\pi}
r(z)\sqrt{1 + r'(z)^2}\;\dd\theta\,\dd z
= 2\pi\int_a^b r(z)\sqrt{1 + r'(z)^2}\,\dd z ,
$$

the classical formula (circumference $2\pi r$ times the slant [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) element). For the cone $r(z) = kz$, $z \in [0, h]$: $\mathcal
A = 2\pi k\sqrt{1 + k^2}\,\frac{h^2}{2} = \pi \rho \ell$ with $\rho = kh$ the base radius and $\ell = h\sqrt{1 + k^2}$ the slant height — the school formula, now derived rather than admitted.

**Example 19.27 (The catenoid).**

Rotate the catenary $r(z) = \cosh z$, $z \in \intcc{-1}{1}$, around its axis: the resulting *catenoid* has, by the revolution formula and $1 + \sinh^2 z = \cosh^2 z$,

$$
\mathcal A = 2\pi\int_{-1}^1\cosh z\,\sqrt{1 + \sinh^2z}\,
\dd z = 2\pi\int_{-1}^1\cosh^2z\,\dd z
= \pi\bigl[z + \sinh z\cosh z\bigr]_{-1}^{1},
$$

that is

$$
\mathcal A = 2\pi + \pi\sinh 2 \approx 17.68 .
$$

The slant factor $\sqrt{1 + r'^2}$ merged with the radius into a perfect square — the same identity that made the catenary’s [arc length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) elementary in the curves chapter. This is no accident of algebra: among all surfaces of revolution spanning the two boundary circles, the catenoid minimizes [area](#def-b2-surfaces-area) (it is the shape of a soap film between two rings), and this variational property is precisely what singles out $\cosh$; the Year 3 volume proves it with the calculus of variations.

![The saddle z = x2 - y2 near the origin, with its coordinate curves (u-curves in blue, v-curves in green), the tangent plane at M_0 = (0,0,0) (dashed) and the unit normal n. The surface crosses its tangent plane — the two-dimensional analogue of an inflection.](https://one-course.com/images/onecourse/chapters/math-4/b2-surfaces/fig-b28608515b89.svg)

***Figure 19.1.** The saddle $z = x^2 - y^2$ near the origin, with its coordinate curves ($u$-curves in blue, $v$-curves in green), the [tangent plane](#def-b2-surfaces-tangent) at $M_0 = (0,0,0)$ (dashed) and the [unit normal](#def-b2-surfaces-tangent) $n$. The surface crosses its [tangent plane](#def-b2-surfaces-tangent) — the two-dimensional analogue of an inflection.*

**Remark 19.28 (Common pitfalls).**

(i) *Chart singularities are not surface singularities*: the spherical chart degenerates at the poles ($\cos\varphi = 0$), but the sphere is perfectly smooth there — another chart (exchange the roles of the axes) is regular at the poles. Before declaring a point singular, try a second parametrization. (ii) *Regularity of $\sigma$ concerns the parametrization, not the image*: $\sigma(u, v) = (u^3, v,
0)$ is non-regular along $u = 0$ although its image is a plane. (iii) The [area](#def-b2-surfaces-area) formula requires $\sigma$ *injective* on $K$: a chart covering a piece twice counts it twice ($\theta$ running over $\intcc0{4\pi}$ doubles the sphere’s [area](#def-b2-surfaces-area)). (iv) The [unit normal](#def-b2-surfaces-tangent) is defined up to sign by the surface but is *chosen* by the chart (order of $u, v$); statements involving orientation must fix that choice. (v) Finally, $EG - F^2 > 0$ is not an extra hypothesis: it is exactly regularity, by the Lagrange identity — if it vanishes somewhere, the problem is the chart, and no [area](#def-b2-surfaces-area) or tangent-plane formula applies there.

**Remark 19.29 (Perspectives within this volume).**

Forward links from here. The [area](#def-b2-surfaces-area) element $\sqrt{EG -
F^2}\,\dd u\,\dd v$ is a two-dimensional Jacobian in disguise, and [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint) makes the analogy exact with the change-of-variables theorem — the surface integrals there are this chapter’s [areas](#def-b2-surfaces-area) with an integrand on board. The [first fundamental form](#def-b2-surfaces-fff) is a field of positive [quadratic forms](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def), handled [pointwise](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) by the tools of [Chapter 12](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#ch-b2-quadratic), whose spectral theorem also powers this chapter’s weekend classification of [quadrics](#pb-b2-surfaces-1). And the normal line drives extremum problems on constraint sets ([Example 19.10](#ex-b2-surfaces-closest)), the geometric germ of the Lagrange multiplier method sketched with [Theorem 15.11](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#thm-b2-diffcalc-extrema).

![Four of the nine quadric surfaces classified in the weekend problem, sketched by their silhouettes and a level curve (red): the bounded ellipsoid, the doubly ruled one-sheet hyperboloid with its waist, the bowl of the elliptic paraboloid, and the saddle, whose two parabolic sections bend in opposite directions.](https://one-course.com/images/onecourse/chapters/math-4/b2-surfaces/fig-efda384d63fe.svg)

![Four of the nine quadric surfaces classified in the weekend problem, sketched by their silhouettes and a level curve (red): the bounded ellipsoid, the doubly ruled one-sheet hyperboloid with its waist, the bowl of the elliptic paraboloid, and the saddle, whose two parabolic sections bend in opposite directions.](https://one-course.com/images/onecourse/chapters/math-4/b2-surfaces/fig-3c9fa3d0b4ca.svg)

![Four of the nine quadric surfaces classified in the weekend problem, sketched by their silhouettes and a level curve (red): the bounded ellipsoid, the doubly ruled one-sheet hyperboloid with its waist, the bowl of the elliptic paraboloid, and the saddle, whose two parabolic sections bend in opposite directions.](https://one-course.com/images/onecourse/chapters/math-4/b2-surfaces/fig-f4ae063bae8a.svg)

![Four of the nine quadric surfaces classified in the weekend problem, sketched by their silhouettes and a level curve (red): the bounded ellipsoid, the doubly ruled one-sheet hyperboloid with its waist, the bowl of the elliptic paraboloid, and the saddle, whose two parabolic sections bend in opposite directions.](https://one-course.com/images/onecourse/chapters/math-4/b2-surfaces/fig-8a7851124f87.svg)

*Four of the nine [quadric](#pb-b2-surfaces-1) surfaces classified in the weekend problem, sketched by their silhouettes and a level curve (red): the bounded ellipsoid, the doubly ruled one-sheet hyperboloid with its waist, the bowl of the elliptic paraboloid, and the saddle, whose two parabolic sections bend in opposite directions.*

**Remark 19.30 (Where this is used).**

The [area](#def-b2-surfaces-area) element $\norm{\sigma_u\wedge\sigma_v}\,\dd u\,\dd v$ is the surface-integral measure of [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint), where it meets Green’s formula; the [first fundamental form](#def-b2-surfaces-fff) is the prototype of a field of [quadratic forms](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def), studied [pointwise](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) with the tools of [Chapter 12](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#ch-b2-quadratic); and this chapter’s weekend problem classifies all [quadric](#pb-b2-surfaces-1) surfaces with the spectral theorem. The Year 3 volume returns to surfaces with [differential](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#def-b2-diffcalc-differential) forms and the divergence theorem, and intrinsic [curvature](https://one-course.com/books/math/4/en/chapter/18-curves#thm-b2-curves-frenet2d) — what $E, F, G$ know about bending — is the gateway to differential geometry proper.

## 19.4 Exercises

**Exercise 19.1 ★.**

Show that the [tangent planes](#def-b2-surfaces-tangent) of the cone $z = \sqrt{x^2 + y^2}$ (minus its apex) all pass through the apex. *(Parametrize by $\sigma(\theta, r) = (r\cos\theta, r\sin\theta, r)$, $r > 0$.)*

**Solution of Exercise 19.1.**

With $\sigma(\theta, r) = (r\cos\theta,\ r\sin\theta,\ r)$:

$$
\sigma_\theta = (-r\sin\theta,\ r\cos\theta,\ 0),
\qquad
\sigma_r = (\cos\theta,\ \sin\theta,\ 1),
$$

independent for $r > 0$. The [tangent plane](#def-b2-surfaces-tangent) at $M_0 =
\sigma(\theta_0, r_0)$ passes through $M_0$ with directions $\sigma_\theta, \sigma_r$. Now $M_0 - O = r_0(\cos\theta_0,
\sin\theta_0, 1) = r_0\,\sigma_r(\theta_0, r_0)$ is itself a tangent direction: the apex $O$ lies on the [tangent plane](#def-b2-surfaces-tangent). (This is the general behaviour of cones: they are ruled by lines through the apex, and a [tangent plane](#def-b2-surfaces-tangent) contains the ruling line through the point of tangency.)

**Exercise 19.2 ★.**

Find the [tangent plane](#def-b2-surfaces-tangent) of the ellipsoid $\frac{x^2}{a^2} +
\frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ at a point $(x_0, y_0,
z_0)$ of the surface.

**Solution of Exercise 19.2.**

Apply [Proposition 19.7](#prop-b2-surfaces-gradient) to $f(x,y,z) =
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}$: $\nabla
f(x_0,y_0,z_0) = 2\bigl(\frac{x_0}{a^2}, \frac{y_0}{b^2},
\frac{z_0}{c^2}\bigr) \neq 0$ on the surface. The [tangent plane](#def-b2-surfaces-tangent) is

$$
\frac{x_0}{a^2}(x - x_0) + \frac{y_0}{b^2}(y - y_0)
+ \frac{z_0}{c^2}(z - z_0) = 0,
\qquad\text{i.e.}\qquad
\frac{x_0\,x}{a^2} + \frac{y_0\,y}{b^2} + \frac{z_0\,z}{c^2} = 1,
$$

using that $(x_0, y_0, z_0)$ satisfies the ellipsoid equation — the “split the squares” rule generalizing the sphere’s $\langle
M_0, M \rangle = R^2$.

**Exercise 19.3 ★.**

Compute $E, F, G$ for the helicoid $\sigma(u, v) = (v\cos u,\
v\sin u,\ au)$, $a > 0$, and the [area](#def-b2-surfaces-area) of the piece $0 \leq u \leq
2\pi$, $0 \leq v \leq 1$, as an integral (evaluate it using $\int\sqrt{v^2 + a^2}\,\dd v = \frac12\bigl(v\sqrt{v^2+a^2} +
a^2\ln(v + \sqrt{v^2 + a^2})\bigr) + C$).

**Solution of Exercise 19.3.**

$\sigma_u = (-v\sin u,\ v\cos u,\ a)$ and $\sigma_v = (\cos u,\
\sin u,\ 0)$, so

$$
E = v^2 + a^2, \qquad F = 0, \qquad G = 1,
\qquad
\sqrt{EG - F^2} = \sqrt{v^2 + a^2} > 0
$$

(the helicoid is regular everywhere, including on its axis $v =
0$). [Area](#def-b2-surfaces-area) of the piece:

$$
\mathcal{A} = \int_0^{2\pi}\!\!\int_0^1
\sqrt{v^2 + a^2}\;\dd v\,\dd u
= 2\pi\cdot\frac12\Bigl[v\sqrt{v^2 + a^2}
+ a^2\ln\bigl(v + \sqrt{v^2 + a^2}\bigr)\Bigr]_0^1 ,
$$

i.e. $\mathcal{A} = \pi\Bigl(\sqrt{1 + a^2} +
a^2\ln\frac{1 + \sqrt{1 + a^2}}{a}\Bigr)$.

**Exercise 19.4 ★★.**

(Torus) Parametrize the torus obtained by rotating the circle of center $(R, 0, 0)$ and radius $r < R$ in the $xz$-plane around the $z$-axis:

$$
\sigma(\theta, \psi) = \bigl((R + r\cos\psi)\cos\theta,\
(R + r\cos\psi)\sin\theta,\ r\sin\psi\bigr).
$$

Compute $E, F, G$, check regularity, and show that the [area](#def-b2-surfaces-area) is $4\pi^2 R r$ (Pappus: mean circumference $2\pi R$ times circle [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) $2\pi r$).

**Solution of Exercise 19.4.**

Derivatives:

$$
\sigma_\theta = \bigl(-(R + r\cos\psi)\sin\theta,\
(R + r\cos\psi)\cos\theta,\ 0\bigr),
\qquad
\sigma_\psi = \bigl(-r\sin\psi\cos\theta,\
-r\sin\psi\sin\theta,\ r\cos\psi\bigr).
$$

Then

$$
E = (R + r\cos\psi)^2,
\qquad
F = r\sin\psi\cos\psi\,(R + r\cos\psi)
\bigl(\sin\theta\cos\theta - \sin\theta\cos\theta\bigr) = 0,
\qquad
G = r^2 ,
$$

so $\sqrt{EG - F^2} = r(R + r\cos\psi) \geq r(R - r) > 0$: regular everywhere. [Area](#def-b2-surfaces-area):

$$
\mathcal{A}
= \int_0^{2\pi}\!\!\int_0^{2\pi} r(R + r\cos\psi)
\,\dd\theta\,\dd\psi
= 2\pi r \Bigl(2\pi R + r\int_0^{2\pi}\cos\psi\,\dd\psi\Bigr)
= 4\pi^2 R r ,
$$

the $\cos\psi$ term integrating to zero — Pappus’ theorem: [area](#def-b2-surfaces-area) $=$ ([length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) of the rotated circle) $\times$ (distance travelled by its center).

**Exercise 19.5 ★★.**

Show that the [area](#def-b2-surfaces-area) of the graph of $f \in \mathcal{C}^1(K)$ is $\iint_K \sqrt{1 + f_x^2 + f_y^2}\,\dd x\,\dd y$, and compute it for the paraboloid piece $z = \frac12(x^2 + y^2)$ over the disk $x^2 + y^2 \leq 1$ (polar coordinates, [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint)).

**Solution of Exercise 19.5.**

For $\sigma(x, y) = (x, y, f(x,y))$: $\sigma_x = (1, 0, f_x)$, $\sigma_y = (0, 1, f_y)$, so $E = 1 + f_x^2$, $F = f_xf_y$, $G = 1
+ f_y^2$ and

$$
EG - F^2 = (1 + f_x^2)(1 + f_y^2) - f_x^2f_y^2
= 1 + f_x^2 + f_y^2 ,
$$

giving the stated [area](#def-b2-surfaces-area) formula. For $f = \frac12(x^2 + y^2)$ on the unit disk: $1 + f_x^2 + f_y^2 = 1 + x^2 + y^2$, and in polar coordinates ($x = \rho\cos\alpha$, $y = \rho\sin\alpha$, Jacobian $\rho$, [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint)):

$$
\mathcal{A}
= \int_0^{2\pi}\!\!\int_0^1 \sqrt{1 + \rho^2}\;\rho
\,\dd\rho\,\dd\alpha
= 2\pi\Bigl[\tfrac13(1 + \rho^2)^{3/2}\Bigr]_0^1
= \frac{2\pi}{3}\bigl(2\sqrt2 - 1\bigr).
$$

**Exercise 19.6 ★★.**

A drawn curve $\gamma(t) = \sigma(u(t), v(t))$ on the sphere of radius $R$ (spherical chart) has $u = \theta(t)$, $v =
\varphi(t)$. Write its [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) as an integral in $\theta, \varphi$ and prove that among curves joining two points of the same meridian $\theta = \theta_0$, the meridian arc is the shortest. *(Bound the integrand below by $R\abs{\varphi'}$.)*

**Solution of Exercise 19.6.**

From [Example 19.23](#ex-b2-surfaces-spherearea)’s computation, $E =
R^2\cos^2\varphi$, $F = 0$, $G = R^2$, so by [Proposition 19.16](#prop-b2-surfaces-curvelength)

$$
L = \int_a^b R\sqrt{\cos^2\varphi(t)\,\theta'(t)^2
+ \varphi'(t)^2}\;\dd t .
$$

Let the endpoints be $(\theta_0, \varphi_1)$ and $(\theta_0,
\varphi_2)$, $\varphi_1 < \varphi_2$. For any joining curve,

$$
L \geq \int_a^b R\,\abs{\varphi'(t)}\,\dd t
\geq R\,\Bigl|\int_a^b \varphi'(t)\,\dd t\Bigr|
= R\,(\varphi_2 - \varphi_1),
$$

dropping the nonnegative $\cos^2\varphi\,\theta'^2$ term and using the triangle inequality for integrals. The meridian arc $\theta \equiv \theta_0$, $\varphi$ increasing from $\varphi_1$ to $\varphi_2$, has [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) exactly $R(\varphi_2 - \varphi_1)$: it is shortest. (Meridians are great circles; this is the first, elementary case of the fact that geodesics of the sphere are great circles.)

**Exercise 19.7 ★★★.**

(Normal lines of a sphere) Let $S$ be a regular level surface $\{f = c\}$, [connected](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-connected), all of whose normal lines pass through a fixed point $\Omega$. Show that $S$ is contained in a sphere centered at $\Omega$. *(Show that $\norm{M - \Omega}^2$ has zero derivative along every curve drawn on $S$.)*

**Solution of Exercise 19.7.**

Fix a curve $\gamma$ drawn on $S$ and let $g(t) = \norm{\gamma(t)
- \Omega}^2$. Then $g'(t) = 2\langle \gamma'(t),\ \gamma(t) -
\Omega\rangle$. The normal line at $M = \gamma(t)$ passes through $\Omega$ by hypothesis, so $\gamma(t) - \Omega$ is a *normal* vector, orthogonal to the [tangent plane](#def-b2-surfaces-tangent), in particular to the velocity $\gamma'(t)$ ([Proposition 19.5](#prop-b2-surfaces-velocity)): $g' =
0$, and $g$ is constant along every drawn curve.

Now the set $S_c = \{M \in S : \norm{M - \Omega}^2 = c\}$ is closed in $S$; it is also [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) in $S$: near any of its points, $S$ is a regular graph, so any nearby point of $S$ is joined to it by a drawn curve (a lifted segment), along which $g$ is constant. As $S$ is [connected](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-connected) and $S_c$ nonempty for the right $c$, $S = S_c
\subseteq$ the sphere of center $\Omega$ and radius $\sqrt c$ ([Chapter 4](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#ch-b2-metric): [connectedness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-connected) argument).

**Exercise 19.8 ★★★.**

([Area](#def-b2-surfaces-area) is geometric) Let $\Phi \colon U' \to U$ be a $\mathcal{C}^1$ diffeomorphism between [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) sets of $\R^2$ and $\tilde\sigma =
\sigma \circ \Phi$. Show that

$$
\norm{\tilde\sigma_{u'} \wedge \tilde\sigma_{v'}}
= \abs{\det J_\Phi}\,
\norm{(\sigma_u \wedge \sigma_v)\circ\Phi} ,
$$

and deduce, using the change-of-variables formula of [Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint), that the [area](#def-b2-surfaces-area) of [Definition 19.20](#def-b2-surfaces-area) does not depend on the chosen regular parametrization.

**Solution of Exercise 19.8.**

Write $\Phi(u', v') = (u, v)$. By the chain rule,

$$
\tilde\sigma_{u'} = \frac{\partial u}{\partial u'}\sigma_u
+ \frac{\partial v}{\partial u'}\sigma_v,
\qquad
\tilde\sigma_{v'} = \frac{\partial u}{\partial v'}\sigma_u
+ \frac{\partial v}{\partial v'}\sigma_v ,
$$

partials of $\sigma$ evaluated at $\Phi(u',v')$. Expanding the cross product bilinearly and using $\sigma_u \wedge \sigma_u =
\sigma_v \wedge \sigma_v = 0$, $\sigma_v \wedge \sigma_u =
-\sigma_u \wedge \sigma_v$:

$$
\tilde\sigma_{u'} \wedge \tilde\sigma_{v'}
= \Bigl(\frac{\partial u}{\partial u'}
\frac{\partial v}{\partial v'}
- \frac{\partial v}{\partial u'}
\frac{\partial u}{\partial v'}\Bigr)\,
\sigma_u \wedge \sigma_v
= \det J_\Phi \cdot (\sigma_u \wedge \sigma_v)\circ\Phi .
$$

Taking [norms](https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces#def-b2-nvs-norm) gives the identity. Then, by the change-of-variables formula ([Chapter 20](https://one-course.com/books/math/4/en/chapter/20-line-integrals-and-multiple-integrals#ch-b2-multint)) applied to the map $\Phi$ on $K' =
\Phi^{-1}(K)$:

$$
\iint_{K'} \norm{\tilde\sigma_{u'} \wedge \tilde\sigma_{v'}}
\,\dd u'\dd v'
= \iint_{K'} \norm{\sigma_u \wedge \sigma_v}\circ\Phi\;
\abs{\det J_\Phi}\,\dd u'\dd v'
= \iint_K \norm{\sigma_u \wedge \sigma_v}\,\dd u\,\dd v :
$$

the two parametrizations assign the same [area](#def-b2-surfaces-area) to the same piece of surface.

**Exercise 19.9 ★.**

(Archimedes’ hat-box theorem) On the sphere of radius $R$, the *zone* between the latitudes with $z_1 \leq z \leq z_2$ ($-R \leq z_1 < z_2 \leq R$) has [area](#def-b2-surfaces-area) $2\pi R\,(z_2 - z_1)$: prove it with the spherical chart, and conclude that a zone’s [area](#def-b2-surfaces-area) depends only on its height — slicing an orange into equal-thickness slices gives equal amounts of peel.

**Solution of Exercise 19.9.**

In the spherical chart, $z = R\sin\varphi$, and the zone corresponds to $\varphi_1 \leq \varphi \leq \varphi_2$ with $z_i = R\sin\varphi_i$. With the [area](#def-b2-surfaces-area) element $R^2\cos\varphi\,\dd\theta\,\dd\varphi$ ([Example 19.23](#ex-b2-surfaces-spherearea)):

$$
\mathcal A = \int_{\varphi_1}^{\varphi_2}\!\!\int_0^{2\pi}
R^2\cos\varphi\,\dd\theta\,\dd\varphi
= 2\pi R^2(\sin\varphi_2 - \sin\varphi_1)
= 2\pi R\,(z_2 - z_1) .
$$

The result depends only on the height $z_2 - z_1$: slices of equal thickness carry equal [areas](#def-b2-surfaces-area), whether cut at the equator or at the pole — Archimedes’ hat-box theorem, and the reason the lateral [area](#def-b2-surfaces-area) of the circumscribed cylinder ($2\pi R \cdot
2R = 4\pi R^2$) equals the [area](#def-b2-surfaces-area) of the sphere.

**Exercise 19.10 ★★.**

Show that every normal line of a surface of revolution $\sigma(\theta, z) = (r(z)\cos\theta,\ r(z)\sin\theta,\ z)$ ($r > 0$ of class $\mathcal C^1$) meets the axis of revolution, and locate the intersection point.

**Solution of Exercise 19.10.**

$\sigma_\theta = (-r\sin\theta,\ r\cos\theta,\ 0)$ and $\sigma_z = (r'\cos\theta,\ r'\sin\theta,\ 1)$, so

$$
\sigma_\theta \wedge \sigma_z
= (r\cos\theta,\ r\sin\theta,\ -r\,r') ,
$$

a normal vector at $M = (r\cos\theta, r\sin\theta, z)$. The normal line is

$$
t \mapsto \bigl(r(1 + t)\cos\theta,\ r(1 + t)\sin\theta,\
z - t\,r\,r'\bigr),
$$

which at $t = -1$ reaches $(0,\ 0,\ z + r(z)\,r'(z))$: every normal line meets the axis, at the height $z + rr'$. (This is the three-dimensional reason rotational symmetry survives in the normal field.)

**Exercise 19.11 ★★.**

(Unrolling the cylinder) The chart $\sigma(u, v) = (\cos u,\
\sin u,\ v)$ of the unit cylinder has $E = G = 1$, $F = 0$: verify this, and explain why every drawn curve $t \mapsto
\sigma(u(t), v(t))$ has the same [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) as the plane curve $t
\mapsto (u(t), v(t))$. Deduce that the helix from $(1, 0, 0)$ to $(1, 0, 2\pi c)$ making one turn has [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) $2\pi\sqrt{1 + c^2}$, and that no drawn curve with the same endpoints and one full turn is shorter.

**Solution of Exercise 19.11.**

$\sigma_u = (-\sin u, \cos u, 0)$, $\sigma_v = (0, 0, 1)$: $E
= 1$, $F = 0$, $G = 1$. By [Proposition 19.16](#prop-b2-surfaces-curvelength) the [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) of a drawn curve is $\int\sqrt{u'^2 + v'^2}\,\dd t$ — the [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) of its parameter shadow $(u(t), v(t))$ in the plane: the chart is a local isometry (the unrolling of the cylinder). The helix $t \mapsto (\cos t, \sin t, ct)$, $t \in \intcc0{2\pi}$, has shadow the segment from $(0,0)$ to $(2\pi, 2\pi c)$, of [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) $2\pi\sqrt{1 + c^2}$. Any drawn curve from $(1,0,0)$ to $(1,
0, 2\pi c)$ making one full turn has a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) shadow joining $(0, 0)$ to $(2\pi, 2\pi c)$, of plane [length](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) $\geq$ the straight segment; since [lengths](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length) agree, the helix is shortest.

**Exercise 19.12 ★★★.**

Let $S = \{f = c\}$ be a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) regular level surface and $M_0 \in S$ a point at maximal distance from the origin. Show that $\nabla f(M_0)$ is collinear with $\vect{OM_0}$ — the normal at the farthest point is radial. Apply to the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ ($a >
b > c > 0$): find all points where the normal is radial, and identify the farthest ones.

**Solution of Exercise 19.12.**

The function $g(M) = \norm{M}^2$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on the [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) $S$, so it attains its maximum at some $M_0$. For every curve $\gamma$ drawn on $S$ with $\gamma(0) = M_0$, the function $t
\mapsto \norm{\gamma(t)}^2$ has a maximum at $t = 0$, so its derivative $2\langle\gamma'(0), M_0\rangle$ vanishes: $M_0$ is orthogonal to every tangent vector, i.e. normal to $S$ at $M_0$. As $\nabla f(M_0) \neq 0$ also directs the normal line ([Proposition 19.7](#prop-b2-surfaces-gradient)), $\nabla f(M_0)$ and $\vect{OM_0}$ are collinear. For the ellipsoid, radiality means

$$
\Bigl(\frac{x_0}{a^2}, \frac{y_0}{b^2},
\frac{z_0}{c^2}\Bigr) = \mu\,(x_0, y_0, z_0) :
$$

each coordinate satisfies $x_0(\frac1{a^2} - \mu) = 0$, etc.; since $a^{-2}, b^{-2}, c^{-2}$ are distinct, at most one coordinate is nonzero, and the solutions on the surface are the six axis endpoints $(\pm a, 0, 0)$, $(0, \pm b, 0)$, $(0,
0, \pm c)$. The farthest points are $(\pm a, 0, 0)$, at distance $a = \max(a,b,c)$.

## 19.5 Problem: the classification of quadrics of $\R^3$

**Problem 19.1.**

Weekend problem — every quadric surface, sorted by the spectral theorem

A *quadric* is the zero set in $\R^3$ of a degree-two polynomial

$$
q(X) = X^{\mathsf T}\!AX + 2\,\langle b, X\rangle + c,
\qquad A \in \mathcal S_3(\R),\ A \neq 0,\ b \in \R^3,\
c \in \R .
$$

The surfaces of this chapter’s figures — spheres, ellipsoids, saddles, cones, cylinders — are all [quadrics](#pb-b2-surfaces-1). This problem classifies them *all*: the spectral theorem ([Theorem 12.13](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#thm-b2-quadratic-spectral)) straightens the quadratic part, [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) translations ([Chapter 17](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#ch-b2-affine)) absorb the linear part, and what remains is a short, complete list of normal forms.

**Part I — The reduction machine.**

1. Let $X = PY + t$ with $P \in O(3)$ and $t \in \R^3$ (a rigid change of coordinates). Show that $q(PY + t) =  Y^{\mathsf T}\!A'Y + 2\langle b', Y\rangle + c'$ with $$A' = P^{\mathsf T}\!AP, \qquad  b' = P^{\mathsf T}(At + b), \qquad  c' = q(t) .$$ Deduce that the [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $A$ (hence its rank and signature) is a rigid invariant of the equation, and explain why the equation of a given [quadric](#pb-b2-surfaces-1) is only determined up to a nonzero scalar factor.
2. Using the spectral theorem, show that after a rotation the equation becomes $\sum_i \lambda_i y_i^2 +  2\sum_i\beta_iy_i + c = 0$ with $\lambda_1, \lambda_2,  \lambda_3$ the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $A$ .
3. For every $i$ with $\lambda_i \neq 0$ , absorb $\beta_iy_i$ by a translation ( $y_i \mapsto y_i -  \beta_i/\lambda_i$ ). Write the reduced equation when $\operatorname{rank} A = r$ : $\sum_{i\leq r}\lambda_i  z_i^2 + 2\sum_{i > r}\beta_iz_i + c'' = 0$ .
4. A *center* of the [quadric](#pb-b2-surfaces-1) of equation $q = 0$ is a point $\Omega$ with $q(2\Omega - X) = q(X)$ for all $X$ : the point reflection in $\Omega$ preserves the equation, hence the surface. Show that $q(2\Omega - X)  - q(X) = -4\langle A\Omega + b, X\rangle + 4\langle  A\Omega + b, \Omega\rangle$ , and deduce: the centers are exactly the solutions of $A\Omega = -b$ ; they exist iff $b \in \operatorname{im}A$ , and the center is unique iff $A$ is invertible.

**Part II — Central [quadrics](#pb-b2-surfaces-1) ($\operatorname{rank}A = 3$).** Here the reduced equation is $\lambda_1z_1^2 + \lambda_2z_2^2 + \lambda_3z_3^2 = \delta$.

5. Multiplying by $-1$ if needed, assume at least two $\lambda_i > 0$ . Enumerate the possibilities: signature $(3, 0)$ with $\delta > 0$ , $= 0$ , $< 0$ , and signature $(2, 1)$ with $\delta > 0$ , $= 0$ , $<  0$ ; name the six resulting sets (ellipsoid, point, empty set, one-sheet hyperboloid, cone, two-sheet hyperboloid) and put each in its Euclidean normal form ( $\frac{x^2}{a^2} + \frac{y^2}{b^2} +  \frac{z^2}{c^2} = 1$ , etc.).
6. Classify $x^2 + y^2 + z^2 + 4xy + 4yz + 4zx = 1$ : show $A = 2J - I$ with $J$ the all-ones matrix, compute the [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\{5, -1, -1\}$ , and identify a two-sheet hyperboloid of revolution about the axis $\R(1,1,1)$ .
7. (Rulings) For the one-sheet hyperboloid $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2}  = 1$, factor $$\Bigl(\frac xa - \frac zc\Bigr)  \Bigl(\frac xa + \frac zc\Bigr)  = \Bigl(1 - \frac yb\Bigr)\Bigl(1 + \frac yb\Bigr)$$ and produce two one-parameter families of straight lines lying on the surface.
8. Show that through every point of the one-sheet hyperboloid passes exactly one line of each family: the surface is *doubly ruled* .
9. The *asymptotic cone* of the one-sheet hyperboloid is $C : \frac{x^2}{a^2} + \frac{y^2}{b^2}  - \frac{z^2}{c^2} = 0$ . With $\rho^2 = \frac{x^2}{a^2}  + \frac{y^2}{b^2}$ , show that any point of the hyperboloid is at distance at most $c\bigl(\rho -  \sqrt{\rho^2 - 1}\bigr) = \frac{c}{\rho +  \sqrt{\rho^2 - 1}}$ from $C$ , so the surface hugs its cone at infinity. What are the sections of the hyperboloid by the planes $x = \pm a$ ?

**Part III — Rank $2$ and rank $1$: paraboloids, cylinders, planes.**

10. Suppose $\operatorname{rank}A = 2$, say $\lambda_1,  \lambda_2 \neq 0 = \lambda_3$. Starting from question 3, split into two cases according to $\beta_3 \neq 0$ (no center, by question 4) or $\beta_3 = 0$ (a line of centers), and reduce to $$\lambda_1z_1^2 + \lambda_2z_2^2 + 2\beta_3z_3 = 0  \qquad\text{or}\qquad  \lambda_1z_1^2 + \lambda_2z_2^2 + c'' = 0 :$$ elliptic/hyperbolic paraboloids in the first case, cylinders over central conics (or pairs of intersecting planes, a line, the empty set) in the second.
11. Show that the saddle $z = xy$ is a hyperbolic paraboloid: rotate by $\pi/4$ in the $xy$ -plane to reach $z = \tfrac12(u^2 - v^2)$ , the surface of [Figure 19.1](#fig-b2-surfaces-saddle) up to scale.
12. Show that the saddle $z = xy$ carries the two line families $\{x = x_0,\ z = x_0y\}$ and $\{y = y_0,\ z  = xy_0\}$ , with exactly one line of each through every point: the second doubly ruled [quadric](#pb-b2-surfaces-1) .
13. Classify $x^2 + y^2 - 2x + 4y + 3 = 0$ in $\R^3$ (complete the squares; identify a right circular cylinder, and give its axis and radius).
14. Now let $\operatorname{rank}A = 1$, say $\lambda_1  \neq 0 = \lambda_2 = \lambda_3$. Rotating within the kernel plane and translating, reduce to $$\lambda_1z_1^2 + 2\beta z_2 = 0 \quad (\beta \neq 0)  \qquad\text{or}\qquad  \lambda_1z_1^2 + c'' = 0 :$$ a parabolic cylinder, or a pair of parallel planes, a double plane, or the empty set. Classify $(x + y)^2 =  z$ completely (normal form, axis of translation invariance).

**Part IV — The classification theorem.**

15. Assemble Parts I–III into a theorem: *every [quadric](#pb-b2-surfaces-1) of $\R^3$ is mapped by a rigid motion onto exactly one normal form* . List the seventeen [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) types (count the empty variants and degenerate sets), and single out the nine [quadric](#pb-b2-surfaces-1) *surfaces* : ellipsoid, one- and two-sheet hyperboloids, cone, elliptic and hyperbolic paraboloids, elliptic, hyperbolic and parabolic cylinders.
16. Write the classification *algorithm* : given $(A,  b, c)$ , which quantities do you compute, in which order, and which branch decides which type? Justify that each step is effective ( [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of a [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) $3\times3$ matrix, rank, solvability of $A\Omega = -b$ ).
17. Run the algorithm on $x^2 + y^2 + z^2 - 2xy - 2yz -  2zx = 1$ : show $A = 2I - J$ has [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\{2, 2,  -1\}$ and conclude: a one-sheet hyperboloid of revolution about $\R(1,1,1)$ .
18. Run it on $x^2 + y^2 - z^2 - 2x + 4y + 2z + 4 = 0$ : find the center and identify the [quadric](#pb-b2-surfaces-1) .
19. Run it on $x^2 + 4xy + y^2 = 2z$ : diagonalize the $xy$ -block ( $u = \frac{x+y}{\sqrt2}$ , $v =  \frac{x-y}{\sqrt2}$ ) and identify the [quadric](#pb-b2-surfaces-1) .
20. Euclidean versus [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) . Show that two central [quadrics](#pb-b2-surfaces-1) in normal form are *rigidly* equivalent iff they have the same coefficient lists (up to permutation and a common positive scalar on the equation), while *affinely* , only the signature data survives: every ellipsoid is an [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) image of the round sphere. Which theorem guarantees that the signature cannot change along the way ( [Theorem 12.6](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#thm-b2-quadratic-sylvester) )?

**Part V — Dividends.**

21. Show that every section of a [quadric](#pb-b2-surfaces-1) by an [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) plane is a conic (possibly degenerate) of that plane. Identify the sections of the saddle $z = xy$ by the planes $z = c$ ( $c \neq 0$ and $c = 0$ ).
22. Which [quadric](#pb-b2-surfaces-1) surfaces contain straight lines? Show that the ellipsoid, the two-sheet hyperboloid and the elliptic paraboloid contain none *(restrict $q$ to a line and use the Cauchy–Schwarz inequality for the two-sheet case)* ; that cone and cylinders are ruled by one family; and conclude that the doubly ruled [quadric](#pb-b2-surfaces-1) surfaces are exactly the one-sheet hyperboloid and the hyperbolic paraboloid.
23. When only the *[affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace)* type is wanted, Gauss’s reduction ( [Theorem 12.5](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#thm-b2-quadratic-gauss) ) is cheaper than diagonalizing. Redo question 17 with Gauss’s algorithm and check the signature $(2, 1)$ ; what Euclidean information does Gauss lose?
24. All [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of a [symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) matrix are real; show that consequently the *signs* of the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $A$ can be read from the [characteristic polynomial](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-charpoly) by Descartes’ rule of signs, and verify it on question 6: $\chi_A(\lambda) = \lambda^3 - 3\lambda^2  - 9\lambda - 5$ has exactly one sign change, hence signature $(1, 2)$ .
25. Synthesis. Summarize the algorithm in a few lines; state the exact role played by (i) the spectral theorem, (ii) the center equation $A\Omega = -b$ , (iii) Sylvester’s inertia theorem, (iv) Gauss’s reduction. What does the same machine yield in $\R^2$ , and what changes in $\R^n$ ?

**Solution of Problem 19.1.**

**1.** Expanding, and using the symmetry of $A$ ($t^{\mathsf T}\!APY = (At)^{\mathsf T}PY$):

$$
q(PY + t) = Y^{\mathsf T}P^{\mathsf T}\!APY +
2\,(At + b)^{\mathsf T}PY + \bigl(t^{\mathsf T}\!At +
2b^{\mathsf T}t + c\bigr),
$$

which is the displayed triple. $A' = P^{\mathsf T}\!AP =
P^{-1}AP$ is similar to $A$: same [characteristic polynomial](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-charpoly), [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen), rank, signature. Finally $\{q = 0\} = \{sq = 0\}$ for $s \neq 0$, so only the equation *up to a scalar* is attached to the set; scaling by $s$ multiplies all [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) by $s$.

**2.** The spectral theorem provides $P \in O(3)$ with $P^{\mathsf T}\!AP = \operatorname{diag}(\lambda_1, \lambda_2,
\lambda_3)$; question 1 with $t = 0$ turns the equation into $\sum\lambda_iy_i^2 + 2\sum\beta_iy_i + c = 0$, where $\beta =
P^{\mathsf T}b$.

**3.** For $\lambda_i \neq 0$: $\lambda_iy_i^2 +
2\beta_iy_i = \lambda_i\bigl(y_i +
\frac{\beta_i}{\lambda_i}\bigr)^2 -
\frac{\beta_i^2}{\lambda_i}$; the translation $z_i = y_i +
\beta_i/\lambda_i$ (and $z_i = y_i$ for $i > r$) gives

$$
\sum_{i \leq r}\lambda_iz_i^2 + 2\sum_{i > r}\beta_iz_i +
c'' = 0, \qquad
c'' = c - \sum_{i\leq r}\frac{\beta_i^2}{\lambda_i}.
$$

**4.** $q(2\Omega - X) = (2\Omega - X)^{\mathsf T}
A(2\Omega - X) + 2b^{\mathsf T}(2\Omega - X) + c$; expanding and subtracting $q(X)$, the quadratic terms cancel and

$$
q(2\Omega - X) - q(X) = -4\,\langle A\Omega + b,\ X\rangle +
4\,\langle A\Omega + b,\ \Omega\rangle .
$$

If $A\Omega + b = 0$ this vanishes identically: the point reflection preserves $q$, hence the [quadric](#pb-b2-surfaces-1). Conversely, “$q(2\Omega - X) = q(X)$ for all $X$” says that the [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) function above vanishes on all of $\R^3$, which forces its linear part $A\Omega + b$ to be zero. So centers $=$ solutions of $A\Omega = -b$: a nonempty set iff $b \in \operatorname{im}A$ (an [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) subspace directed by $\ker A$), and a single point iff $A$ is invertible.

**5.** Signature $(3,0)$ (all $\lambda_i > 0$): $\delta >
0$ gives $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}
= 1$ with $a = \sqrt{\delta/\lambda_1}$, etc. — an ellipsoid; $\delta = 0$: the single point $O$; $\delta < 0$: empty. Signature $(2,1)$ ($\lambda_1, \lambda_2 > 0 >
\lambda_3$): $\delta > 0$: $\frac{x^2}{a^2} + \frac{y^2}{b^2}
- \frac{z^2}{c^2} = 1$, the one-sheet hyperboloid; $\delta =
0$: the cone $\frac{x^2}{a^2} + \frac{y^2}{b^2} =
\frac{z^2}{c^2}$; $\delta < 0$: $\frac{z^2}{c^2} -
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the two-sheet hyperboloid ($\abs z \geq c$: two components).

**6.** The quadratic part has matrix $A$ with diagonal $1$ and off-diagonal $2$: $A = 2J - I$. Since $J$ has [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\{3, 0, 0\}$ ([eigenvector](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $(1,1,1)$ for $3$), $A$ has [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\{5, -1, -1\}$, the [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $5$ carried by $\R(1,1,1)$. In the rotated coordinates: $5u^2 - v^2 - w^2 =
1$, i.e. $\frac{u^2}{1/5} - v^2 - w^2 = 1$: a two-sheet hyperboloid, of revolution (equal [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $-1$) about the axis $\R(1,1,1)$.

**7.** The surface is $\bigl(\frac xa - \frac
zc\bigr)\bigl(\frac xa + \frac zc\bigr) = \bigl(1 - \frac
yb\bigr)\bigl(1 + \frac yb\bigr)$. For $(\lambda : \mu) \neq
(0:0)$ define the line

$$
D_{\lambda:\mu} :\quad
\lambda\Bigl(\frac xa - \frac zc\Bigr) = \mu\Bigl(1 - \frac
yb\Bigr), \qquad
\mu\Bigl(\frac xa + \frac zc\Bigr) = \lambda\Bigl(1 + \frac
yb\Bigr)
$$

(two independent [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) equations: a line). Multiplying the two equations shows every point of $D_{\lambda:\mu}$ lies on the surface when $\lambda\mu \neq 0$; the cases $\lambda = 0$ or $\mu = 0$ are checked directly (e.g. $\lambda = 0$: $y =
b$, $\frac xa = -\frac zc$, which satisfies the equation). The second family $D'_{\lambda:\mu}$ swaps the two right-hand factors.

**8.** Fix $M$ on the surface. The conditions for $M \in
D_{\lambda:\mu}$ form a $2\times2$ homogeneous linear system in $(\lambda, \mu)$ whose [determinant](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-det) is

$$
\Bigl(\frac{x^2}{a^2} - \frac{z^2}{c^2}\Bigr) - \Bigl(1 -
\frac{y^2}{b^2}\Bigr) = 0
$$

precisely because $M$ lies on the [quadric](#pb-b2-surfaces-1): a nontrivial solution $(\lambda : \mu)$ exists. The coefficient matrix is never zero (that would force $1 - \frac yb = 1 + \frac yb =
0$), so its rank is $1$ and the solution is unique up to scale: exactly one line of the family passes through $M$. The same holds for the second family, and the two lines are distinct (at $(a, 0, 0)$ they are $\{x = a,\ \frac zc = \frac
yb\}$ and $\{x = a,\ \frac zc = -\frac yb\}$): the one-sheet hyperboloid is doubly ruled.

**9.** Let $M = (x, y, z)$ on the hyperboloid, $\rho^2 =
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{z^2}{c^2} \geq
1$. The point $N = (x,\ y,\ \varepsilon c\rho)$ with $\varepsilon$ the sign of $z$ satisfies $\frac{x^2}{a^2} +
\frac{y^2}{b^2} - \frac{(c\rho)^2}{c^2} = 0$: $N \in C$, and

$$
d(M, C) \leq \abs{z - \varepsilon c\rho}
= c\bigl(\rho - \sqrt{\rho^2 - 1}\bigr)
= \frac{c}{\rho + \sqrt{\rho^2 - 1}} .
$$

If $\norm M \to \infty$ then $\rho \to \infty$ (all three coordinates are bounded by multiples of $\rho$), so $d(M, C)
\to 0$. The section $x = a$: $\frac{y^2}{b^2} -
\frac{z^2}{c^2} = 0$, the pair of crossing lines of question 8 — and likewise at $x = -a$.

**10.** With $\lambda_3 = 0$, question 3 leaves $\lambda_1z_1^2 + \lambda_2z_2^2 + 2\beta_3z_3 + c'' = 0$. In the eigenbasis, $\operatorname{im}A = \operatorname{Vect}(e_1,
e_2)$, so by question 4 centers exist iff $\beta_3 = 0$. If $\beta_3 \neq 0$: the translation $z_3 \mapsto z_3 -
c''/(2\beta_3)$ removes the constant, leaving $\lambda_1z_1^2
+ \lambda_2z_2^2 + 2\beta_3z_3 = 0$, i.e. $z_3 = px^2 +
qy^2$ after renaming: an *elliptic paraboloid* if $\lambda_1\lambda_2 > 0$, a *hyperbolic paraboloid* if $\lambda_1\lambda_2 < 0$ — and indeed no center. If $\beta_3 = 0$: the equation $\lambda_1z_1^2 + \lambda_2z_2^2 +
c'' = 0$ does not involve $z_3$: the [quadric](#pb-b2-surfaces-1) is a cylinder over the corresponding plane conic — elliptic cylinder, line, or empty set when $\lambda_1\lambda_2 > 0$; hyperbolic cylinder or pair of intersecting planes when $\lambda_1\lambda_2 < 0$ — with a whole line of centers $\{(z_1^*, z_2^*)\} \times \R$.

**11.** $xy - z = 0$. Substituting $x = \frac{u +
v}{\sqrt2}$, $y = \frac{u - v}{\sqrt2}$ (rotation by $\pi/4$): $xy = \frac{u^2 - v^2}2$, so the equation becomes $z =
\frac12(u^2 - v^2)$: a hyperbolic paraboloid — the saddle of the figure, up to the factor $\frac12$.

**12.** The line $\{x = x_0,\ z = x_0y\}$ (parametrized by $y$) clearly lies on $z = xy$, as does $\{y = y_0,\ z =
xy_0\}$; through $(x_0, y_0, x_0y_0)$ pass the two of them. Uniqueness: if $t \mapsto (x_0 + tv_1, y_0 + tv_2, z_0 +
tv_3)$ stays on the surface, the coefficient of $t^2$ in $(x_0
+ tv_1)(y_0 + tv_2) - z_0 - tv_3$ gives $v_1v_2 = 0$, so $v_1 = 0$ or $v_2 = 0$, landing in one of the two families: one line of each through each point — the second doubly ruled [quadric](#pb-b2-surfaces-1).

**13.** Completing squares: $(x - 1)^2 + (y + 2)^2 = 2$, with no condition on $z$: a right circular cylinder of radius $\sqrt2$ and axis the vertical line $\{(1, -2, z) : z \in
\R\}$ — a line of centers, as question 10 predicts.

**14.** With $\lambda_2 = \lambda_3 = 0$ the reduced equation is $\lambda_1z_1^2 + 2\beta_2z_2 + 2\beta_3z_3 + c''
= 0$. A rotation of the kernel plane $(z_2, z_3)$ aligns the linear form: $2\beta_2z_2 + 2\beta_3z_3 = 2\beta w$ with $\beta = \sqrt{\beta_2^2 + \beta_3^2}$. If $\beta \neq 0$, translate $w$ to absorb $c''$: $\lambda_1z_1^2 + 2\beta w =
0$, a *parabolic cylinder*; if $\beta = 0$: $\lambda_1z_1^2 = -c''$ gives two parallel planes ($c''
\lambda_1 < 0$), a double plane ($c'' = 0$), or the empty set. For $(x + y)^2 = z$: with $u = \frac{x + y}{\sqrt2}$ the equation reads $z = 2u^2$: a parabolic cylinder, invariant under translations along $(1, -1, 0)$.

**15.** Every [quadric](#pb-b2-surfaces-1) is carried by a rotation plus translations onto one of: (rank 3) ellipsoid, point, empty set, one-sheet hyperboloid, cone, two-sheet hyperboloid; (rank 2) elliptic paraboloid, hyperbolic paraboloid, elliptic cylinder, line, empty set, hyperbolic cylinder, pair of intersecting planes; (rank 1) parabolic cylinder, pair of parallel planes, double plane, empty set. Identifying the three empty variants as distinct *[affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) types* of equations, the count is seventeen; among them nine are honest surfaces: ellipsoid, the two hyperboloids, the cone, the two paraboloids, and the three cylinders.

**16.** Algorithm. (i) Read off $(A, b, c)$; compute the [characteristic polynomial](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-charpoly) of $A$, its [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) (real, by the spectral theorem) and $r = \operatorname{rank}A$. (ii) Solve $A\Omega = -b$ (Gaussian elimination): solvable or not — centers or not. (iii) If solvable, translate to a center: the equation becomes $\sum\lambda_iz_i^2 + c'' = 0$ with $c''
= c + \langle b, \Omega\rangle$; sort by $r$, the signature, and the sign of $c''$ using questions 5, 10, 14. (iv) If not solvable ($r \leq 2$), rotate and reduce as in questions 10 and 14: paraboloid ($r = 2$) or parabolic cylinder ($r = 1$), elliptic/hyperbolic according to the sign of $\lambda_1\lambda_2$. Each step is a finite computation: roots of a cubic with real roots, ranks, linear systems.

**17.** $A$ has diagonal $1$, off-diagonal $-1$: $A = 2I
- J$, [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\{2 - 3,\ 2,\ 2\} = \{-1, 2, 2\}$ with $-1$ on $\R(1,1,1)$. Signature $(2,1)$, $b = 0$, right side $\delta = 1 > 0$: $2u^2 + 2v^2 - w^2 = 1$, a one-sheet hyperboloid of revolution about the axis $\R(1,1,1)$.

**18.** Complete squares: $(x-1)^2 + (y+2)^2 - (z-1)^2 +
(-1 - 4 + 1 + 4) = 0$, i.e.

$$
(x-1)^2 + (y+2)^2 = (z-1)^2 :
$$

the constant vanished — a right circular *cone* with vertex (and unique center) $(1, -2, 1)$ and axis parallel to $Oz$.

**19.** The quadratic part $x^2 + 4xy + y^2$ has matrix $\left(\begin{smallmatrix}1 & 2\\ 2 & 1\end{smallmatrix}\right)$ (in the $xy$-plane), [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $3$ (on $(1,1)$) and $-1$ (on $(1,-1)$): with $u = \frac{x+y}{\sqrt2}$, $v =
\frac{x-y}{\sqrt2}$ it equals $3u^2 - v^2$, and the [quadric](#pb-b2-surfaces-1) is

$$
z = \tfrac32u^2 - \tfrac12v^2 :
$$

a hyperbolic paraboloid ($A$ has rank $2$ and $b$ has a component along $\ker A = \R e_z$: no center).

**20.** A rigid motion transforms the equation’s data by $A \mapsto P^{\mathsf T}\!AP$ (same [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen)) and the normal forms have no residual freedom except permuting coordinates and multiplying the whole equation by a scalar ($> 0$ to preserve the writing): two central normal forms coincide up to isometry iff the coefficient lists agree up to permutation and common positive factor — for the ellipsoid, iff the semiaxes $(a, b, c)$ agree. Affinely, one may also scale each coordinate separately ($z_i \mapsto z_i\sqrt{\abs
{\lambda_i}}$), which erases the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) and leaves only their signs: every ellipsoid becomes $u^2 + v^2 + w^2 = 1$, the sphere. Sylvester’s inertia theorem ([Theorem 12.6](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#thm-b2-quadratic-sylvester)) guarantees the signature survives any invertible linear change: the [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) types of question 15 are genuinely distinct.

**21.** Parametrize the plane affinely: $M = P + su +
tv$. Then $q(P + su + tv)$ is a polynomial of degree $\leq 2$ in $(s, t)$ (expand the [quadratic form](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def) bilinearly), so the section $\{q = 0\}$ is a conic of the plane, possibly degenerate. For $z = xy$ and the plane $z = c$: $xy = c$, a hyperbola for $c \neq 0$, and for $c = 0$ the two coordinate lines — the pair of rulings through the origin.

**22.** *Ellipsoid*: bounded, contains no line. *Two-sheet hyperboloid* $\frac{z^2}{c^2} -
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: restrict to $p + tv$; the $t^2$ coefficient $\frac{v_3^2}{c^2} - \frac{v_1^2}{a^2}
- \frac{v_2^2}{b^2}$ must vanish, so $v_3 \neq 0$ (else $v =
0$); the $t$ coefficient gives $\frac{p_3v_3}{c^2} =
\frac{p_1v_1}{a^2} + \frac{p_2v_2}{b^2}$, and Cauchy–Schwarz yields

$$
\frac{p_3^2}{c^2}
= \frac{c^2}{v_3^2}\Bigl(\frac{p_1v_1}{a^2} +
\frac{p_2v_2}{b^2}\Bigr)^2
\leq \frac{c^2}{v_3^2}\Bigl(\frac{p_1^2}{a^2} +
\frac{p_2^2}{b^2}\Bigr)\frac{v_3^2}{c^2}
= \frac{p_1^2}{a^2} + \frac{p_2^2}{b^2},
$$

so the constant term is $\leq 0 \neq 1$: no line. *Elliptic paraboloid* $z = \frac{x^2}{a^2} +
\frac{y^2}{b^2}$: the $t^2$ coefficient forces $v_1 = v_2 =
0$, then the equation is linear nonconstant in $t$: no line. *Cone* $x^2 + y^2 = z^2$: a line on it satisfies $q(p) =
q(v) = B(p, v) = 0$ for the Lorentz form; equality in the plane Cauchy–Schwarz $\abs{p_1v_1 + p_2v_2} = \abs{p_3v_3} =
\sqrt{(p_1^2 + p_2^2)(v_1^2 + v_2^2)}$ forces $(p_1, p_2)
\parallel (v_1, v_2)$ and then $p = kv$: all lines pass through the vertex — one family. *Cylinders*: for the elliptic and parabolic cylinders the $t^2$ coefficient forces $v_1 = v_2 = 0$ (rulings only); for the hyperbolic cylinder $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, $\frac{v_1}a = \pm
\frac{v_2}b$ with $v_2 \neq 0$ leads via the $t$ coefficient to $\frac{p_1^2}{a^2} = \frac{p_2^2}{b^2}$, contradicting the constant term $1$: again only the vertical rulings. So the doubly ruled [quadric](#pb-b2-surfaces-1) surfaces are exactly the one-sheet hyperboloid and the hyperbolic paraboloid.

**23.** Gauss: $x^2 - 2xy - 2zx = (x - y - z)^2 - y^2 -
z^2 - 2yz$, so

$$
q = (x - y - z)^2 - 4yz = (x - y - z)^2 + (y - z)^2 - (y +
z)^2 :
$$

three independent squares with signs $(+, +, -)$ — signature $(2, 1)$, matching question 17, with no [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) computation. Gauss loses the metric data: the new coordinates are not orthonormal, so the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) (the shape of the hyperboloid, its axes and their [lengths](https://one-course.com/books/math/4/en/chapter/18-curves#def-b2-curves-length)) are gone; only the [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) type remains.

**24.** Let $p$, $n$, $z$ be the numbers of positive, negative and zero [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen), $p + n + z = 3$. Descartes’ rule bounds $p$ by the number $V$ of sign changes of $\chi_A$, and $n$ by the number $V'$ of sign changes of $\chi_A(-\lambda)$; moreover each pair of consecutive nonzero coefficients produces a change in exactly one of the two polynomials, so $V + V' \leq 3 - z$ (zero roots are visible as vanishing trailing coefficients). Then $p + n = 3 - z \geq
V + V' \geq p + n$: equality, so $p = V$ exactly — the signs of the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) can be read off. For $\chi_A(\lambda)
= \lambda^3 - 3\lambda^2 - 9\lambda - 5$: signs $+,-,-,-$ give $V = 1$, and $\chi_A(-\lambda) = -\lambda^3 - 3\lambda^2
+ 9\lambda - 5$ has signs $-,-,+,-$: $V' = 2$. Signature $(1,
2)$ — consistent with the exact factorization $\chi_A =
(\lambda - 5)(\lambda + 1)^2$ of question 6.

**25.** Algorithm: diagonalize the quadratic part orthonormally (spectral theorem: this is the only [analytically](https://one-course.com/books/math/4/en/chapter/11-power-series#def-b2-powerseries-analytic) deep step, and it is what makes the classification *Euclidean*); solve $A\Omega = -b$ to decide central versus parabolic types and to translate away the linear part where possible ([affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) geometry); read the type from rank, signature and the constant (Sylvester guarantees these are invariants); when only the [affine](https://one-course.com/books/math/4/en/chapter/17-affine-spaces#def-b2-affine-subspace) type matters, Gauss’s reduction replaces the spectral theorem at the cost of the metric information. In $\R^2$ the same machine classifies conics: ellipse, hyperbola, parabola, plus pairs of lines, a line, a point, and empty sets. In $\R^n$ nothing changes but bookkeeping: the types are indexed by the signature of $A$, the position of $b$ relative to $\operatorname{im}A$, and one constant — with the $(n{+}2)$-dimensional bordered matrix of $(A, b, c)$ providing a [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) invariant.