Calculus Volume 3 · Chapter 2 · Section 2.7

Cylindrical and Spherical Coordinates

Three numbers still name a point in space, but two of them can be angles: polar coordinates with a height, or a distance with two angles.

Idea 1 · cylindrical(r, θ, z): polar in the floor, z as before.x = r cos θ, y = r sin θ. Idea 2 · spherical(ρ, θ, φ): a distance and two angles.φ is measured down from the positive z-axis. Idea 3 · fit the shapeA cylinder is r = c; a sphere is ρ = c.Use the system where the shape holds one coordinate fixed.

Colour key r, ρ θ z φ key idea · 3D figures turn when you drag them · page links open the textbook

L1One point, three addresses
L1Cylindrical: polar plus a height
L1Spherical: a distance and two angles
L1One picture
L2The conversion formulas
L2Hold one coordinate fixed
L2Examples 2.60–2.67
L2Lab 1 · a point in cylindrical coordinates
L2Lab 2 · a point in spherical coordinates
L2Lab 3 · coordinate surfaces
L2Lab 4 · latitude and longitude
L2Common mistakes
L2Practice and answers
L3Where the formulas come from
L3Which θ? Let the signs decide
L3Half a cone
L3Where it is used next

Level 1 · see itTwo new ways to name a pointbefore the formulas

One point, three addresses

Rectangular coordinates give three distances along the axes. For round things — a water tank, a pipe, a dome — a distance and an angle say it more simply (p209). Example 2.60's point, three ways:

system	address	how to get there
rectangular	$(-2,\ 2\sqrt3,\ -2)$	along $x$ , along $y$ , down
cylindrical	$(4,\ \tfrac{2\pi}{3},\ -2)$	turn $\tfrac{2\pi}{3}$ , go out 4, down 2
spherical	$(2\sqrt5,\ \tfrac{2\pi}{3},\ 2.03)$	turn, tilt 2.03 rad from straight up, go $2\sqrt5$

rectangular cylindrical spherical

The point stays put; only the directions for reaching it change.

Grey: along the axes. Blue, amber, teal: out, turn, down. Dashed blue with the red angle: one distance, two angles.

Cylindrical: polar plus a height

Definition · p210

$(r,\theta)$ are the polar coordinates of the point's shadow in the $xy$ -plane; $z$ is the usual height.

x=r\cos\theta,\qquad y=r\sin\theta,\qquad z=z

Cylindrical coordinates are polar coordinates with $z$ carried along unchanged.

Example 2.60: $4\cos\tfrac{2\pi}{3}=-2$ , $4\sin\tfrac{2\pi}{3}=2\sqrt3$ , so $(4,\tfrac{2\pi}{3},-2)$ is $(-2,2\sqrt3,-2)$ .

The floor triangle gives

x

and

y

; the height is untouched.

Spherical: a distance and two angles

Definition · p215

$\rho$ : the distance from the origin
$\theta$ : the same angle as in cylindrical coordinates
$\varphi$ : the angle from the positive $z$ -axis, $0\le\varphi\le\pi$

x=\rho\sin\varphi\cos\theta,\quad y=\rho\sin\varphi\sin\theta,\quad z=\rho\cos\varphi

$\varphi$ is measured down from the north pole, not up from the floor.

Example 2.63:

(8,\tfrac\pi3,\tfrac\pi6)

. The standing triangle has legs

r=4

and

z=4\sqrt3

One picture

Cylindrical:

r=c

\theta=c

z=c

Spherical:

\rho=c

\theta=c

\varphi=c

Hold one coordinate fixed and it sweeps a surface; a point is where one surface of each kind meets.

Level 2 · compute itConverting, and naming surfacesand four labs

The conversion formulas

Theorem 2.15 · p210

\begin{aligned} x&=r\cos\theta & r^2&=x^2+y^2\\ y&=r\sin\theta & \tan\theta&=\tfrac yx\\ z&=z & z&=z \end{aligned}

Theorem 2.16 · p215

\begin{aligned} x&=\rho\sin\varphi\cos\theta & \rho^2&=x^2+y^2+z^2\\ y&=\rho\sin\varphi\sin\theta & \tan\theta&=\tfrac yx\\ z&=\rho\cos\varphi & \varphi&=\arccos\tfrac{z}{\rho} \end{aligned}

To and from cylindrical: $r=\rho\sin\varphi$ , $z=\rho\cos\varphi$ ; $\rho=\sqrt{r^2+z^2}$ .

tan θ = y/x is not enough

It holds for $\theta$ and $\theta+\pi$ alike (p210). Choose $\theta$ in the quadrant of $(x,y)$ : for $(-1,1)$ , $\arctan(-1)=-\tfrac\pi4$ , but $\theta=\tfrac{3\pi}{4}$ .

Every formula is a right triangle: one lying in the floor ( $x,y,r$ ), one standing up ( $r,z,\rho$ ).

Hold one coordinate fixed

Figure 2.92, Figure 2.99; Lab 3 draws each one.

equation	surface	in rectangular coordinates
$r=c$	cylinder of radius $c$ about the $z$ -axis	$x^2+y^2=c^2$
$\theta=c$	half-plane leaving the $z$ -axis at angle $c$	half of $y=x\tan c$
$z=c$	horizontal plane	$z=c$
$\rho=c$	sphere of radius $c$ about the origin	$x^2+y^2+z^2=c^2$
$\varphi=c$	half-cone about the $z$ -axis; $c=\tfrac\pi2$ is the $xy$ -plane	half of $z^2=(x^2+y^2)\cot^2 c$

Choose the system in which your shape is one coordinate held fixed.

Examples 2.60–2.67

Example	given	result
2.60 · p211	cylindrical $(4,\tfrac{2\pi}{3},-2)$	rectangular $(-2,2\sqrt3,-2)$
2.61 · p212	rectangular $(1,-3,5)$	cylindrical $(\sqrt{10},\ \arctan(-3)+2\pi,\ 5)\approx(\sqrt{10},5.03,5)$
2.62 · p213	$\theta=\tfrac\pi4$ ; $r^2+z^2=9$ ; $z=r$	half-plane; sphere of radius 3; cone
2.63 · p217	spherical $(8,\tfrac\pi3,\tfrac\pi6)$	rectangular $(2,2\sqrt3,4\sqrt3)$ ; cylindrical $(4,\tfrac\pi3,4\sqrt3)$
2.64 · p218	rectangular $(-1,1,\sqrt6)$	spherical $(2\sqrt2,\tfrac{3\pi}{4},\tfrac\pi6)$ ; cylindrical $(\sqrt2,\tfrac{3\pi}{4},\sqrt6)$
2.65 · p218	$\theta=\tfrac\pi3$ ; $\varphi=\tfrac{5\pi}{6}$ ; $\rho=6$ ; $\rho=\sin\theta\sin\varphi$	half-plane; half-cone $3x^2+3y^2-z^2=0$ , $z\le0$ ; sphere of radius 6; sphere $x^2+(y-\tfrac12)^2+z^2=\tfrac14$
2.66 · p221	Columbus, 40° N, 83° W	$(4000,\ -83^\circ,\ 50^\circ)$
2.67 · p222	bowling ball; submarine; conical tank; pipeline; football	spherical; rectangular; cylindrical; cylindrical; cylindrical

Lab 1 · a point in cylindrical coordinates

r4 θ120° z−2

presets

the cylinder through the point

Try: on the z-axis — with $r=0$ , every $\theta$ gives the same point. $\theta=180^\circ$ and $\theta=-180^\circ$ — the same half-plane. Going back, the signs of $x$ and $y$ pick the quadrant.

Lab 2 · a point in spherical coordinates

ρ8 θ60° φ30°

presets

the sphere and cone through the point

Try: in the floor — $\varphi=90^\circ$ , so $z=0$ and $r=\rho$ . below the floor — $\varphi>90^\circ$ , $z<0$ . $\rho\sin\varphi$ is the distance from the $z$ -axis: the cylindrical $r$ .

Lab 3 · coordinate surfaces

hold fixed

the constant cr = 2

Try: φ = c at 90° — the cone flattens into the $xy$ -plane (Exercise 397); past 90° it opens downward. θ = c — only half a plane: the other half is $\theta=c\pm180^\circ$ .

Lab 4 · latitude and longitude

Earth as a sphere of radius 4000 mi, the $z$ -axis through the North Pole, the $x$ -axis through the prime meridian (p221).

latitude40° N longitude83° W

cities

Try: latitude 90° — $\varphi=0^\circ$ , the North Pole. Sydney — south of the equator, so $\varphi>90^\circ$ . $\theta$ is the longitude with west negative; $\varphi=90^\circ-$ latitude.

Common mistakes

Mistake	Result	Fix
$\theta=\arctan\tfrac yx$ without the quadrant	$(-1,1,\sqrt6)$ gets $\theta=-\tfrac\pi4$ , a direction in quadrant IV	$(-1,1)$ is in quadrant II: $\theta=\tfrac{3\pi}{4}$
$\varphi$ measured up from the floor	Columbus at $\varphi=40^\circ$	$\varphi=90^\circ-\text{latitude}=50^\circ$
$x=\rho\cos\varphi\cos\theta$	$(8,\tfrac\pi3,\tfrac\pi6)$ gives $x\approx3.46$ , not 2	$r=\rho\sin\varphi$ , so $x=\rho\sin\varphi\cos\theta$
West longitude as positive $\theta$	Columbus at $\theta=83^\circ$ , in Asia	west is negative: $\theta=-83^\circ$
$\varphi=\tfrac{5\pi}{6}$ read as all of $3x^2+3y^2=z^2$	the upper half comes along	keep $z\le0$ : one nappe
$r=-\sqrt{10}$	a negative distance from the $z$ -axis	$r=\sqrt{x^2+y^2}\ge0$

Practice and answers

Exercises from p224; the book's key: p836–p837, p848–p850.

Checkpoints 2.55–2.61

#	task	answer
2.55	cylindrical $(5,\tfrac\pi6,4)$ to rectangular	$(\tfrac{5\sqrt3}{2},\ \tfrac52,\ 4)$
2.56	rectangular $(-8,8,-7)$ to cylindrical	$(8\sqrt2,\ \tfrac{3\pi}{4},\ -7)$
2.57	the surface $r=6$	cylinder of radius 6 about the $z$ -axis
2.58	spherical $(2,-\tfrac{5\pi}{6},\tfrac\pi6)$	rectangular $(-\tfrac{\sqrt3}{2},-\tfrac12,\sqrt3)$ ; cylindrical $(1,-\tfrac{5\pi}{6},\sqrt3)$
2.59	$\rho=13$ ; $\theta=\tfrac{2\pi}{3}$ ; $\varphi=\tfrac\pi4$	sphere of radius 13; half-plane; half-cone
2.60	Sydney, 34° S, 151° E	$(4000,\ 151^\circ,\ 124^\circ)$
2.61	a star map seen from Earth	spherical: origin at Earth's centre, $z$ to the North Pole, $x$ to the prime meridian

Exercises

#	task	answer	key step
363	cylindrical $(4,\tfrac\pi6,3)$ to rectangular	$(2\sqrt3,\ 2,\ 3)$	$4\cos\tfrac\pi6=2\sqrt3$
367	rectangular $(1,\sqrt3,2)$ to cylindrical	$(2,\ \tfrac\pi3,\ 2)$	$\tan\theta=\sqrt3$ , quadrant I
369	$(3,-3,7)$ to cylindrical	$(3\sqrt2,\ -\tfrac\pi4,\ 7)$	quadrant IV; $\tfrac{7\pi}{4}$ names the same angle
371	$r=4$	cylinder $x^2+y^2=16$	$r^2=x^2+y^2$
373	$r^2\cos2\theta+z^2+1=0$	hyperboloid of two sheets $-x^2+y^2-z^2=1$ , about the $y$ -axis	$r^2\cos2\theta=x^2-y^2$
375	$r=2\cos\theta$	cylinder $x^2-2x+y^2=0$ : centre $(1,0,0)$ , radius 1	multiply by $r$
377	$r=2\sec\theta$	plane $x=2$	$r\cos\theta=2$
383	$x^2+y^2-16x=0$ in cylindrical	$r=16\cos\theta$ , and $r=0$	$r^2=16r\cos\theta$
385	spherical $(3,0,\pi)$ to rectangular	$(0,0,-3)$	$\varphi=\pi$ : straight down
387	$(12,-\tfrac\pi4,\tfrac\pi4)$ to rectangular	$(6,\ -6,\ 6\sqrt2)$	$12\sin\tfrac\pi4=6\sqrt2=r$
391	rectangular $(0,3,0)$ to spherical	$(3,\ 90^\circ,\ 90^\circ)$	$x=0,\ y>0$ : $\theta=90^\circ$
395	$\rho=2\cos\varphi$	sphere $x^2+y^2+(z-1)^2=1$	multiply by $\rho$
399	$x^2+y^2-3z^2=0$ , $z\ne0$ , in spherical	$\varphi=\tfrac\pi3$ or $\varphi=\tfrac{2\pi}{3}$ : a cone	$\tan\varphi=\tfrac rz=\pm\sqrt3$
403	cylindrical $(1,\tfrac\pi4,3)$ to spherical	$(\sqrt{10},\ \tfrac\pi4,\ 0.3218)$	$\varphi=\arctan\tfrac13$
417	Washington, DC, 39° N, 77° W	$(4000,\ -77^\circ,\ 51^\circ)$	$\varphi=90^\circ-39^\circ$
419	Rio de Janeiro at $(4000,\ -43.17^\circ,\ 102.91^\circ)$	43.17° W, 12.91° S	latitude $=90^\circ-\varphi$

Exercise 419: $90^\circ-102.91^\circ=-12.91^\circ$ , so 12.91° S. The book's key prints 22.91° S, Rio's actual latitude, which would need $\varphi=112.91^\circ$ .

Level 3 · why it worksTwo triangles and a signshort arguments

Where the formulas come from

The standing triangle has its right angle under the point and $\varphi$ at the origin, between $\rho$ and the $z$ -axis (p216):

\rho^2=r^2+z^2,\qquad r=\rho\sin\varphi,\qquad z=\rho\cos\varphi

Put this $r$ into the floor triangle:

x=r\cos\theta=\rho\sin\varphi\cos\theta,\qquad y=r\sin\theta=\rho\sin\varphi\sin\theta

Why sin φ

$\varphi$ opens from the vertical, so the horizontal leg is the sine. Example 2.63: $8\sin\tfrac\pi6=4=r$ .

No quadrant rule for φ

$\arccos$ returns an angle in $[0,\pi]$ — exactly the range of $\varphi$ . Only $\theta$ needs the signs of $x$ and $y$ .

Which θ? Let the signs decide

where $(x,y)$ is	$\theta$ in $[0,2\pi)$	example
$x>0,\ y\ge0$	$\arctan\tfrac yx$	$(1,\sqrt3)$ : $\tfrac\pi3$
$x<0$	$\arctan\tfrac yx+\pi$	$(-1,1)$ : $-\tfrac\pi4+\pi=\tfrac{3\pi}{4}$
$x>0,\ y<0$	$\arctan\tfrac yx+2\pi$	$(1,-3)$ : $\approx5.03$
$x=0$	$\tfrac\pi2$ if $y>0$ , $\tfrac{3\pi}{2}$ if $y<0$	$(0,3)$ : $\tfrac\pi2$

$\theta$ and $\theta+2\pi k$ name the same half-plane, so the key's $-\tfrac\pi4$ for Exercise 369 is $\tfrac{7\pi}{4}$ . The pair $(x,y)$ knows its quadrant; the ratio $y/x$ has forgotten it.

Half a cone

\varphi=c:\qquad r=\rho\sin c,\quad z=\rho\cos c\quad\Longrightarrow\quad z=r\cot c,\ \ r\ge0

One value of $\varphi$ gives one nappe (p219). Squaring, as in Example 2.65, gives $3x^2+3y^2-z^2=0$ — both nappes — so $z\le0$ has to be kept by hand. Likewise $z=r$ in Example 2.62 is $\varphi=\tfrac\pi4$ , the upper nappe when $r\ge0$ ; $z^2=x^2+y^2$ is the whole double cone.

Squaring can add points: the sign it removes was information.

Where it is used next

Where	The new coordinates give
§5.5 Triple integrals	$dV=r\,dr\,d\theta\,dz$ and $dV=\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$
§6.6 Surface integrals	a cylinder or a sphere described by the two coordinates that vary on it
Physics	fields around a wire (cylindrical) or a point charge (spherical)
Maps and astronomy	latitude, longitude and star maps

A boundary that is a coordinate held fixed makes the limits of an integral constant.

situation	system
centre of gravity of a bowling ball	spherical, origin at the centre
a submarine in an ocean current	rectangular: no symmetry
pressure in a conical water tank	cylindrical, $z$ along the axis
oil flowing through a pipeline	cylindrical, $z$ along the pipe
leather for a football	cylindrical, $z$ along the ball's axis
Checkpoint 2.61: a star map seen from Earth	spherical, origin at Earth's centre, $z$ to the North Pole

Objective	You can now
2.7.1 cylindrical → rectangular	$x=r\cos\theta,\ y=r\sin\theta,\ z=z$
2.7.2 rectangular → cylindrical	$r=\sqrt{x^2+y^2}$ ; $\theta$ from $\tan\theta=\tfrac yx$ and the quadrant
2.7.3 spherical → rectangular	$(\rho\sin\varphi\cos\theta,\ \rho\sin\varphi\sin\theta,\ \rho\cos\varphi)$
2.7.4 rectangular → spherical	$\rho=\sqrt{x^2+y^2+z^2}$ ; $\theta$ by quadrant; $\varphi=\arccos\tfrac z\rho$

One point, three addresses

Cylindrical: polar plus a height

Spherical: a distance and two angles

One picture

The conversion formulas

Hold one coordinate fixed

Examples 2.60–2.67

Lab 1 · a point in cylindrical coordinates

Lab 2 · a point in spherical coordinates

Lab 3 · coordinate surfaces

Lab 4 · latitude and longitude

Common mistakes

Practice and answers

Where the formulas come from

Which θ? Let the signs decide

Half a cone

Where it is used next

Cylindrical and Spherical Coordinates

Round shapes get short equations in round coordinates

Cylindrical coordinates: polar coordinates in the floor, plus the height zzz

Theorem 2.15: the floor triangle converts both ways, but θ\thetaθ needs the quadrant

The signs of xxx and yyy pick θ\thetaθ: add π\piπ when x<0x<0x<0, 2π2\pi2π in quadrant IV

Fix one cylindrical coordinate: r=cr=cr=c is a cylinder, θ=c\theta=cθ=c a half-plane, z=cz=cz=c a plane

(4,2π3,−2)(4,\tfrac{2\pi}{3},-2)(4,32π​,−2) is the point (−2, 23, −2)(-2,\ 2\sqrt3,\ -2)(−2, 23​, −2)

Checkpoint 2.55: the same three steps put (5,π6,4)(5,\tfrac\pi6,4)(5,6π​,4) at (532,52,4)\big(\tfrac{5\sqrt3}{2},\tfrac52,4\big)(253​​,25​,4)

(1,−3,5)(1,-3,5)(1,−3,5) is (10, 5.03, 5)(\sqrt{10},\ 5.03,\ 5)(10​, 5.03, 5): quadrant IV sets θ\thetaθ

Checkpoint 2.56: in quadrant II, add π\piπ to the arctangent

Substitute r2=x2+y2r^2=x^2+y^2r2=x2+y2 to name the surface: half-plane, sphere, cone

z=rz=rz=r is only the upper half of the cone; z2=x2+y2z^2=x^2+y^2z2=x2+y2 is both halves

Spherical coordinates: a distance ρ\rhoρ and two angles, with φ\varphiφ measured down from the zzz-axis

Theorem 2.16: stand the triangle up, then lay it down

Two triangles give every formula, and φ\varphiφ needs no quadrant rule

Fix one spherical coordinate: ρ=c\rho=cρ=c is a sphere, θ=c\theta=cθ=c a half-plane, φ=c\varphi=cφ=c a half-cone

(8,π3,π6)(8,\tfrac\pi3,\tfrac\pi6)(8,3π​,6π​): stand up to get r,zr,zr,z, then lie down to get x,yx,yx,y

Checkpoint 2.58: a negative θ\thetaθ just turns the other way

(−1,1,6)(-1,1,\sqrt6)(−1,1,6​) is (22,3π4,π6)(2\sqrt2,\tfrac{3\pi}{4},\tfrac\pi6)(22​,43π​,6π​): ρ\rhoρ first, θ\thetaθ by quadrant, then φ\varphiφ

One spherical coordinate fixed: θ=π3\theta=\tfrac\pi3θ=3π​ half-plane, φ=5π6\varphi=\tfrac{5\pi}{6}φ=65π​ half-cone, ρ=6\rho=6ρ=6 sphere

In rectangular form, φ=5π6\varphi=\tfrac{5\pi}{6}φ=65π​ is 3x2+3y2−z2=03x^2+3y^2-z^2=03x2+3y2−z2=0 with z≤0z\le0z≤0

ρ=sin⁡θsin⁡φ\rho=\sin\theta\sin\varphiρ=sinθsinφ is a small sphere through the origin

On Earth, θ\thetaθ is longitude and φ=90∘−\varphi=90^\circ-φ=90∘− latitude

Pick the coordinate system that matches the symmetry

Two right triangles convert every system

Cylindrical coordinates: polar coordinates in the floor, plus the height $z$

Theorem 2.15: the floor triangle converts both ways, but $\theta$ needs the quadrant

The signs of $x$ and $y$ pick $\theta$ : add $\pi$ when $x<0$ , $2\pi$ in quadrant IV

Fix one cylindrical coordinate: $r=c$ is a cylinder, $\theta=c$ a half-plane, $z=c$ a plane

$(4,\tfrac{2\pi}{3},-2)$ is the point $(-2,\ 2\sqrt3,\ -2)$

Checkpoint 2.55: the same three steps put $(5,\tfrac\pi6,4)$ at $\big(\tfrac{5\sqrt3}{2},\tfrac52,4\big)$

$(1,-3,5)$ is $(\sqrt{10},\ 5.03,\ 5)$ : quadrant IV sets $\theta$

Checkpoint 2.56: in quadrant II, add $\pi$ to the arctangent

Substitute $r^2=x^2+y^2$ to name the surface: half-plane, sphere, cone

$z=r$ is only the upper half of the cone; $z^2=x^2+y^2$ is both halves

Spherical coordinates: a distance $\rho$ and two angles, with $\varphi$ measured down from the $z$ -axis

Two triangles give every formula, and $\varphi$ needs no quadrant rule

Fix one spherical coordinate: $\rho=c$ is a sphere, $\theta=c$ a half-plane, $\varphi=c$ a half-cone

$(8,\tfrac\pi3,\tfrac\pi6)$ : stand up to get $r,z$ , then lie down to get $x,y$

Checkpoint 2.58: a negative $\theta$ just turns the other way

$(-1,1,\sqrt6)$ is $(2\sqrt2,\tfrac{3\pi}{4},\tfrac\pi6)$ : $\rho$ first, $\theta$ by quadrant, then $\varphi$

One spherical coordinate fixed: $\theta=\tfrac\pi3$ half-plane, $\varphi=\tfrac{5\pi}{6}$ half-cone, $\rho=6$ sphere

In rectangular form, $\varphi=\tfrac{5\pi}{6}$ is $3x^2+3y^2-z^2=0$ with $z\le0$

$\rho=\sin\theta\sin\varphi$ is a small sphere through the origin

On Earth, $\theta$ is longitude and $\varphi=90^\circ-$ latitude