2.7 Cylindrical and Spherical Coordinates

← Chapter 2 §2.7 Cylindrical and Spherical Coordinates 柱坐标与球坐标 · Chapter 2 · Vectors in Space

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Access for free at openstax.org · CC BY-NC-SA 4.0

where $(x,y)$ is	$\theta$ in $[0,2\pi)$
$x>0,\ y\ge0$	$\arctan\tfrac yx$
$x<0$	$\arctan\tfrac yx+\pi$
$x>0,\ y<0$	$\arctan\tfrac yx+2\pi$
$x=0$	$\tfrac\pi2$ if $y>0$ , $\tfrac{3\pi}{2}$ if $y<0$

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	Takeaways; say them first, then click
2.7.1 Convert from cylindrical to rectangular coordinates.	$x=r\cos\theta,\ y=r\sin\theta,\ z=z$ ; round shapes get short equations
2.7.2 Convert from rectangular to cylindrical coordinates.	$r=\sqrt{x^2+y^2}$ ; $\theta$ from $\tan\theta=\tfrac yx$ and the quadrant; $r=c$ : cylinder
2.7.3 Convert from spherical to rectangular coordinates.	$(\rho\sin\varphi\cos\theta,\ \rho\sin\varphi\sin\theta,\ \rho\cos\varphi)$ : stand up, then lie down
2.7.4 Convert from rectangular to spherical coordinates.	$\rho=\sqrt{x^2+y^2+z^2}$ ; $\theta$ by quadrant; $\varphi=\arccos\tfrac z\rho$ ; $\rho=c$ : sphere

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	$r=\sqrt{x^2+y^2}\ge0$

#	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

Cylindrical and Spherical Coordinates

Round shapes get short equations in round coordinates

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

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

Drag $r$ , $\theta$ , $z$ : $x$ and $y$ follow the floor triangle

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

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

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

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

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

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

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

Any place on Earth: $\theta$ is its longitude, $\varphi=90^\circ-$ its latitude

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

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

Drag $\rho$ , $\theta$ , $\varphi$ : $\rho\sin\varphi$ is the distance from the $z$ -axis

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

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

Rectangular → spherical: $\rho$ first, $\theta$ by quadrant, then $\varphi$

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

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

Each coordinate held fixed sweeps out one surface

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

Pick the coordinate system that matches the symmetry

Two right triangles convert every system

Common mistakes

Practice and answers

#	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 $=90^\circ-\varphi$

Cylindrical and Spherical Coordinates

Round shapes get short equations in round coordinates

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

(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)

Drag rrr, θ\thetaθ, zzz: xxx and yyy follow the floor triangle

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)

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

(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

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

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

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

Any place on Earth: θ\thetaθ is its longitude, φ=90∘−\varphi=90^\circ-φ=90∘− its latitude