Section 13.9: Cylindrical and Spherical Coordinates
In the cylindrical coordinate system, a point P in space is represented by the ordered
triple (r, θ, z), where r and θ are polar coordinates of the projection of P onto the xy-plane
and z is the directed distance from the xy-plane to P .
Figure 1: A point expressed in cylindrical coordinates.
To convert from cylindrical to rectangular coordinates we use the relations
x = r cos θ
y = r sin θ
z = z.
To convert from rectangular to cylindrical coordinates we use the relations
p
y
tan θ =
r = x2 + y 2
z = z.
x
4π
Example: Convert the point 2, , 8 from cylindrical to rectangular coordinates.
3
Since r = 2, θ = 4π/3, and z = 8,
4π
1
x = r cos θ = 2 cos
=2 −
= −1,
3
2
√ !
√
4π
3
=2 −
= − 3,
y = r sin θ = 2 sin
3
2
z = z = 8.
√
Thus, the point is (−1, − 3, 8) in rectangular coordinates.
√
Example: Convert the point ( 3, 1, 4) from rectangular to cylindrical coordinates.
Since x =
√
3, y = 1, and z = 4,
p
√
x2 + y 2 = 3 + 1 = 2,
π
1
−1 y
−1
√
= ,
θ = tan
= tan
x
6
3
z = z = 4.
r=
π Thus, the point is 2, , 4 in cylindrical coordinates.
6
Example: Find an equation in cylindrical coordinates for the ellipsoid 4x2 + 4y 2 + z 2 = 1.
Since r2 = x2 + y 2 , it follows that
4x2 + 4y 2 + z 2 = 1
4r2 + z 2 = 1
z 2 = 1 − 4r2 .
In the spherical coordinate system, a point P in space is represented by the ordered
triple (ρ, θ, φ), where ρ ≥ 0 is the distance from the origin to P , θ is the same angle as in
cylindrical coordinates, and 0 ≤ φ ≤ π is the angle between the positive z-axis and the line
segment OP .
Figure 2: A point expressed in spherical coordinates.
The connection between rectangular and spherical coordinates can be seen in Figure 3.
If the point P has rectangular coordinates (x, y, z) and spherical coordinates (ρ, θ, φ), then
x = ρ sin φ cos θ
y = ρ sin φ sin θ
z = ρ cos φ
and
ρ=
p
x2 + y 2 + z 2 .
Figure 3: Relationship between rectangular and spherical coordinates.
π π
Example: Convert the point 4, ,
from spherical to rectangular coordinates.
4 6
Since ρ = 4, θ = π/4, and φ = π/6,
√
1
1
√
cos
=4
= 2,
x = ρ sin φ cos θ = 4 sin
6
4
2
2
π π √
1
1
√
y = r sin φ sin θ = 4 sin
sin
=4
= 2,
6
4
2
2
√ !
π √
3
z = ρ cos φ = 4 cos
=4
= 2 3.
6
2
π π √ √ √
Thus, the point is ( 2, 2, 2 3) in rectangular coordinates.
√
Example: Convert the point (1, −1, − 2) from rectangular to spherical coordinates.
√
Since x = 1, y = −1, and z = − 2,
p
√
ρ = x2 + y 2 + z 2 = 1 + 1 + 2 = 2,
y
7π
θ = tan−1
= tan−1 (−1) =
,
4
x
3π
z
1
=
φ = cos−1
= cos−1 − √
.
ρ
4
2
7π 3π
in spherical coordinates.
Thus, the point is 2, ,
4 4
Example: Find an equation in spherical coordinates for the surface 3x2 − x + 3y 2 + 3z 2 = 0.
Since ρ2 = x2 + y 2 + z 2 and x = ρ sin φ cos θ, it follows that
3x2 + 3y 2 + 3z 2 = x
3ρ2 = ρ sin φ cos θ
1
ρ =
sin φ cos θ.
3
p
Example: A solid lies above the cone 5z = x2 + y 2 and outside the sphere x2 + y 2 + z 2 = z.
Write a description of the solid in terms of spherical coordinates.
Since ρ2 = x2 + y 2 + z 2 and z = ρ cos φ, it follows that
x2 + y 2 + z 2 ≥ z
ρ2 ≥ ρ cos φ
ρ ≥ cos φ.
The first equation gives
p
x2 + y 2
x2 + y 2
ρ2 sin2 φ
1 − cos2 φ
1
1
cos φ ≥ √ .
26
5z
25z 2
25ρ2 cos2 φ
25 cos2 φ
26 cos2 φ
≥
≥
≥
≥
≥