Calculus Volume 3 · Chapter 2 · Section 2.6

Quadric Surfaces

Planes come from first-degree equations. Square the variables and the surfaces curve: cylinders, ellipsoids, hyperboloids, cones, paraboloids.

Idea 1 · a missing variableA missing variable makes a cylinder.The curve slides along that axis. Idea 2 · slice itTraces show the shape.Every slice of a quadric is a conic. Idea 3 · read the signsSigns and powers name the surface.The odd one out is the axis.

Colour key surface slicing plane trace key idea · 3D figures turn when you drag them · page links open the textbook

L1One equation, two meanings
L1Cylinders
L1Traces: slice it and look
L1Six quadric surfaces
L1One picture
L2Reading an equation
L2Examples 2.55–2.59
L2Lab 2 · flip a sign
L2Lab 3 · complete the square
L2The focus of a dish
L2Common mistakes
L2Practice and answers
L3Why every trace is a conic
L3Straight lines on curved surfaces
L3A cylinder need not stand straight
L3When a quadric falls apart

Level 1 · see itSurfaces you can recognisebefore any algebra

One equation, two meanings

In the plane, $x^2+y^2=9$ is a circle of radius 3. In space it says nothing about $z$ , so every height is allowed.

Stack the circle at every height and you get a tube: a cylinder. (Figure 2.75)

Planes came from first-degree equations (§2.5). This section is about second degree.

The red circle is the equation in the plane; the cylinder is the same equation in space.

Cylinders

Definition · p193

Lines parallel to a given line, through a given curve, form a cylinder. The lines are its rulings.

If a variable is missing from the equation, the rulings run along that axis. The curve need not be a circle.

Example 2.55

Red: the curve in a coordinate plane. Grey lines: rulings along the missing variable.

Traces: slice it and look

Definition · p195

The traces of a surface are its cross-sections with planes parallel to the coordinate planes: set $x$ , $y$ or $z$ to a constant.

Lab 1 · slice a surface

surface

slice by

height of the plane, k3

Try: the ellipsoid at k = 5 — the trace shrinks to a point. The cone at k = 0 — a point for z, two crossing lines for x. Two sheets between z = −2 and 2 — no trace at all. A quadric's traces are ellipses, hyperbolas, parabolas, or what is left when those collapse.

Six quadric surfaces

Figures 2.87–2.88. Drag any of them.

One picture

Checkpoint 2.53: $\dfrac{x^2}{9}+\dfrac{y^2}{4}-\dfrac{z^2}{25}=1$ .

$z=0$ : the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$
$y=0$ : the hyperbola $\frac{x^2}{9}-\frac{z^2}{25}=1$
$x=0$ : the hyperbola $\frac{y^2}{4}-\frac{z^2}{25}=1$

A quadric surface is the conics it is made of.

Hyperboloid of one sheet (Checkpoint 2.53): one ellipse, two hyperbolas.

Level 2 · compute itFrom an equation to a nameand three labs

Reading an equation

Definition · p196

Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Jz+K=0

standard form	name	traces	axis
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$	ellipsoid	ellipses	—
$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$	hyperboloid of one sheet	ellipses ⟂ axis, hyperbolas ∥	the one minus sign
$\frac{z^2}{c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$	hyperboloid of two sheets	ellipses or nothing ⟂, hyperbolas ∥	the one plus sign
$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0$	elliptic cone	ellipses ⟂; hyperbolas, crossing lines ∥	the odd sign
$z=\frac{x^2}{a^2}+\frac{y^2}{b^2}$	elliptic paraboloid	ellipses ⟂, parabolas ∥	the linear variable
$z=\frac{x^2}{a^2}-\frac{y^2}{b^2}$	hyperbolic paraboloid	hyperbolas ⟂, parabolas ∥	the linear variable

A variable to the first power? A paraboloid: same signs elliptic, opposite signs hyperbolic.
Right side 0? A cone. Right side 1? Count the minus signs: none, ellipsoid; one, one sheet; two, two sheets.
Linear terms beside a square? Complete the square first; the centre moves, the name does not.

The axis is the variable that is different from the other two.

Examples 2.55–2.59

Example	given	result
2.55 · p193	$x^2+z^2=25$ ; $z=2x^2-y$ ; $y=\sin x$	cylinder of radius 5 along $y$ ; a parabolic surface; sine wave slid along $z$
2.56 · p196	$\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{25}=1$	ellipsoid; intercepts $\pm2,\ \pm3,\ \pm5$ ; three elliptical traces
2.57 · p198	$x^2+\frac{y^2}{4}=\frac z5$	$z=0$ : the origin; $z=5$ : $x^2+\frac{y^2}{4}=1$ ; $y=0$ : $z=5x^2$
2.58 · p200	dish $\frac{x^2}{100}+\frac{y^2}{100}=\frac z4$	$x^2=25z=4pz$ : focus $(0,0,6.25)$
2.59 · p202	$16x^2+9y^2+16z^2=144$	$\frac{x^2}{9}+\frac{y^2}{16}+\frac{z^2}{9}=1$ : ellipsoid
2.59b · p203	$9x^2-18x+4y^2+16y-36z+25=0$	$\frac{(x-1)^2}{4}+\frac{(y+2)^2}{9}=z$ : elliptic paraboloid, vertex $(1,-2,0)$

Erratum. The book's solution to Example 2.59b says "centered at (1, 2, 0)". $(y+2)^2$ vanishes at $y=-2$ , so the vertex is $(1,-2,0)$ .

Lab 2 · flip a sign

right side

Try: one sheet → change the right side to 0 (a cone) → make z² positive (a point). Pick z linear with + x²/4 − y² for the saddle. A variable to the first power is always a paraboloid's axis.

Lab 3 · complete the square

equation

Each square you complete adds a number on one side; subtract it back. $(y+2)^2$ puts the centre at $y=-2$ .

The focus of a dish

Example 2.58: $\frac{x^2}{100}+\frac{y^2}{100}=\frac z4$ . Equal coefficients: round cross-sections. Its trace in $y=0$ is

x^2=25z=4pz\quad\Longrightarrow\quad p=\frac{25}{4}=6.25

Focus $(0,0,6.25)$ . Every incoming ray parallel to the axis reflects through the focus.

Trace in the

xz

-plane; rays at

x=5

and

x=10

hit at

z=1

and

z=4

Common mistakes

Mistake	Result	Fix
$x^2+y^2=9$ "is a circle" in space	misses every height	a missing variable is a ruling direction: a cylinder
Axis of a hyperboloid = a positive variable	wrong axis for one sheet	the odd sign: the minus for one sheet, the plus for two
Reading $16x^2+9y^2+16z^2=144$ as semi-axes 4, 3, 4	semi-axes 3, 4, 3 instead	divide to get 1 first
$9(x^2-2x+1)$ adds 1	it adds 9	subtract the coefficient times the square
$(y+2)^2$ read as centre $y=2$	the book's Example 2.59b	centre $y=-2$
Cone and hyperboloid confused	$=0$ versus $=1$	right side 0: cone
$z=x^2-y^2$ called elliptic	it is a saddle	opposite signs: hyperbolic paraboloid
$x^2=25z$ gives focal length 25	focus at 25	$4p=25$ , $p=6.25$

Practice and answers

Exercises from p203; the book's key: p835–836, p844–847.

Checkpoints 2.52–2.54

#	task	answer
2.52	graph the cylinder $z=y^2$	parabola $z=y^2$ in the $yz$ -plane, rulings along $x$
2.53	traces of $\frac{x^2}{3^2}+\frac{y^2}{2^2}-\frac{z^2}{5^2}=1$	ellipses ∥ $xy$ -plane; hyperbolas $\frac{x^2}{9}-\frac{z^2}{25}=1$ , $\frac{y^2}{4}-\frac{z^2}{25}=1$
2.54	$9x^2+y^2-z^2+2z-10=0$	$x^2+\frac{y^2}{9}-\frac{(z-1)^2}{9}=1$ : one sheet, centre $(0,0,1)$

Exercises

#	task	answer
303	$x^2+z^2=1$	cylinder, rulings along $y$
305	$z=\cos\left(\frac\pi2+x\right)$	$z=-\sin x$ : cylinder, rulings along $y$
307	$z=9-y^2$	cylinder, rulings along $x$
313, 315, 317	match: two sheets, elliptic paraboloid, one sheet	b, d, a
319	$-x^2+36y^2+36z^2=9$	$-\frac{x^2}{9}+\frac{y^2}{1/4}+\frac{z^2}{1/4}=1$ : one sheet, axis $x$
321	$-3x^2+5y^2-z^2=10$	$-\frac{x^2}{10/3}+\frac{y^2}{2}-\frac{z^2}{10}=1$ : two sheets, axis $y$
323	$5y=x^2-z^2$	$y=\frac{x^2}{5}-\frac{z^2}{5}$ : hyperbolic paraboloid, axis $y$
325	$x^2+5y^2+3z^2-15=0$	$\frac{x^2}{15}+\frac{y^2}{3}+\frac{z^2}{5}=1$ : ellipsoid
327	$x^2+5y^2-8z^2=0$	$\frac{x^2}{40}+\frac{y^2}{8}-\frac{z^2}{5}=0$ : cone, axis $z$
329	$6x=3y^2+2z^2$	$x=\frac{y^2}{2}+\frac{z^2}{3}$ : elliptic paraboloid, axis $x$
331	$x^2+z^2+4y=0$ , $z=0$	parabola $y=-\frac{x^2}{4}$
333	$-4x^2+25y^2+z^2=100$ , $x=0$	ellipse $\frac{y^2}{4}+\frac{z^2}{100}=1$
339	$x^2+2z^2+6x-8z+1=0$	$\frac{(x+3)^2}{16}+\frac{(z-2)^2}{8}=1$ : cylinder, rulings along $y$
341	$x^2+4y^2-4z^2-6x-16y-16z+5=0$	$\frac{(x-3)^2}{4}+(y-2)^2-(z+2)^2=1$ : one sheet, centre $(3,2,-2)$
343	$x^2+\frac{y^2}{4}-\frac{z^2}{3}+6x+9=0$	$(x+3)^2+\frac{y^2}{4}-\frac{z^2}{3}=0$ : cone, vertex $(-3,0,0)$
345	ellipsoid through $(2,0,0)$ , $(0,0,1)$ , $\left(\frac12,\sqrt{11},\frac12\right)$	$\frac{x^2}{4}+\frac{y^2}{16}+z^2=1$
347	cone $x^2-y^2-z^2=0$ and line $\frac{x-1}{2}=\frac{y+1}{3}=z$	$(1,-1,0)$ and $\left(\frac{13}{3},4,\frac53\right)$
349	equidistant from $(0,-1,0)$ and $y=1$	$x^2+z^2+4y=0$ : elliptic paraboloid
351	focus of $400z=x^2+y^2$	$(0,0,100)$
357	cylinder $9x^2+4y^2=18$ and ellipsoid $36x^2+16y^2+9z^2=144$	ellipses $\frac{x^2}{2}+\frac{y^2}{9/2}=1$ in $z=\pm2\sqrt2$

Level 3 · why it worksWhy the pictures behaveshort arguments

Why every trace is a conic

Put $z=k$ into the general equation. Every term keeps degree two or less in $x$ and $y$ :

Ax^2+By^2+Dxy+(G+Ek)\,x+(H+Fk)\,y+(Ck^2+Jk+K)=0

A second-degree equation in two variables is a conic (p196), or a point, a line, a pair of lines, or nothing. Slicing never raises the degree, so the slices of a quadric are conics.

Straight lines on curved surfaces

On $x^2+y^2-z^2=1$ , for every angle $t$ the whole line

\big(\cos t-s\sin t,\ \sin t+s\cos t,\ s\big)

lies on the surface: $(\cos t-s\sin t)^2+(\sin t+s\cos t)^2=1+s^2$ . Flip the sign of $s$ for a second family.

That is why cooling towers can be built from straight beams (p199). The saddle $z=x^2-y^2=(x-y)(x+y)$ is ruled too.

A cylinder need not stand straight

Example 2.55b, $z=2x^2-y$ , uses all three variables. Still, from any point on it, move along $\langle 0,1,-1\rangle$ :

2x^2-(y+t)=(2x^2-y)-t=z-t

The point stays on the surface. It is a cylinder whose rulings are slanted: the parabola $z=2x^2$ slid along $\langle 0,1,-1\rangle$ (Figure 2.78).

When a quadric falls apart

Exercise 353

$x^2+y^2+z^2+2xy+2xz+2yz+x+y+z=0$ is $(x+y+z)(x+y+z+1)=0$ : the parallel planes $x+y+z=0$ and $x+y+z=-1$ .

Other collapses

$x^2+y^2+z^2=-1$ : no points. $x^2+y^2=0$ : the $z$ -axis. $x^2-y^2=0$ : two planes.

The seventeen standard forms include these flat and empty cases; six are the true curved surfaces.

Objective	You can now
2.6.1 cylinders	spot the missing variable; rulings run along it
2.6.2 ellipsoids, paraboloids, hyperboloids	name the surface and its axis from the standard form
2.6.3 traces	set $x$ , $y$ or $z=k$ ; sketch from the ellipses, parabolas, hyperbolas

One equation, two meanings

Cylinders

Traces: slice it and look

Lab 1 · slice a surface

Six quadric surfaces

One picture

Reading an equation

Examples 2.55–2.59

Lab 2 · flip a sign

Lab 3 · complete the square

The focus of a dish

Common mistakes

Practice and answers

Why every trace is a conic

Straight lines on curved surfaces

A cylinder need not stand straight

When a quadric falls apart

Quadric Surfaces

In space, x2+y2=9x^2+y^2=9x2+y2=9 is a cylinder, not a circle

A curve plus parallel lines through it makes a cylinder

z=2x2−yz=2x^2-yz=2x2−y has no missing variable, so no ruling runs along an axis

Any curve works: find the missing variable and slide the curve along it

A trace is the curve a plane parallel to a coordinate plane cuts out

Second-degree equations give quadric surfaces, and their traces are conics

Slicing never raises the degree

Three elliptical traces frame the ellipsoid x222+y232+z252=1\frac{x^2}{2^2}+\frac{y^2}{3^2}+\frac{z^2}{5^2}=122x2​+32y2​+52z2​=1

Slices away from the centre shrink: at z=3z=3z=3 the trace is x24+y29=1625\frac{x^2}{4}+\frac{y^2}{9}=\frac{16}{25}4x2​+9y2​=2516​

Ellipses across, parabolas along: x2+y24=z5x^2+\frac{y^2}{4}=\frac z5x2+4y2​=5z​ is an elliptic paraboloid

One minus sign: ellipses across, hyperbolas along

A hyperboloid of one sheet is built from straight lines

The reflector x2100+y2100=z4\frac{x^2}{100}+\frac{y^2}{100}=\frac z4100x2​+100y2​=4z​ focuses energy at (0,0,6.25)(0,0,6.25)(0,0,6.25)

Right side 1: count the minus signs — none, one, two

Right side 0 gives a cone; a first-power variable gives a paraboloid

x2+y2−z2=cx^2+y^2-z^2=cx2+y2−z2=c: one sheet, then a cone, then two sheets as ccc goes 1, 0, −1

Name a quadric with three questions

Divide by 144 first: an ellipsoid with semi-axes 3, 4, 3

Complete the square: an elliptic paraboloid with vertex (1,−2,0)(1,-2,0)(1,−2,0)

Completing the square in zzz gives a hyperboloid of one sheet centred at (0,0,1)(0,0,1)(0,0,1)

Some second-degree equations fall apart into planes

A quadric is the conics it is made of

In space, $x^2+y^2=9$ is a cylinder, not a circle

$z=2x^2-y$ has no missing variable, so no ruling runs along an axis

Three elliptical traces frame the ellipsoid $\frac{x^2}{2^2}+\frac{y^2}{3^2}+\frac{z^2}{5^2}=1$

Slices away from the centre shrink: at $z=3$ the trace is $\frac{x^2}{4}+\frac{y^2}{9}=\frac{16}{25}$

Ellipses across, parabolas along: $x^2+\frac{y^2}{4}=\frac z5$ is an elliptic paraboloid

The reflector $\frac{x^2}{100}+\frac{y^2}{100}=\frac z4$ focuses energy at $(0,0,6.25)$

$x^2+y^2-z^2=c$ : one sheet, then a cone, then two sheets as $c$ goes 1, 0, −1

Complete the square: an elliptic paraboloid with vertex $(1,-2,0)$

Completing the square in $z$ gives a hyperboloid of one sheet centred at $(0,0,1)$