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Spherical CoordinatesMATH 311, Calculus III
J. Robert Buchanan
Department of Mathematics
Spring 2019
Spherical CoordinatesAnother means of locating points in three-dimensional space isknown as the spherical coordinate system.
Coordinate DefinitionsIf the point P has Cartesian coordinates (x , y , z), the pointsspherical coordinates (ρ, φ, θ) are as follows:
ρ: the distance from the origin O to P.
ρ =√
x2 + y2 + z2 ≥ 0
φ: the angle between vector 〈x , y , z〉 and the positivez-axis.
0 ≤ φ = cos−1
(z√
x2 + y2 + z2
)≤ π
θ: the angle between vector 〈x , y ,0〉 and the positivex-axis.
θ = cos−1
(x√
x2 + y2
)
Converting from Spherical to Cartesian Coordinates
Given the spherical coordinates (ρ, φ, θ),
x = ρ sinφ cos θ
y = ρ sinφ sin θ
z = ρ cosφ.
Converting from Spherical to Cylindrical Coordinates
Given the spherical coordinates (ρ, φ, θ),
r = ρ sinφ
θ = θ
z = ρ cosφ.
Example (1 of 6)
If point P has spherical coordinates (ρ, φ, θ) = (2, π/4, π/3),find the coordinates of P in
1. Cartesian coordinates.2. Cylindrical coordinates.
Example (2 of 6)
1. Cartesian coordinates:
x = ρ sinφ cos θ = 2 sinπ
4cos
π
3=
1√2
y = ρ sinφ sin θ = 2 sinπ
4sin
π
3=
√32
z = ρ cosφ = 2 cosπ
4=√
2
2. Cylindrical coordinates:
r = ρ sinφ = 2 sinπ
4=√
2
θ = θ =π
3z = ρ cosφ = 2 cos
π
4=√
2
Example (3 of 6)
If point P has Cartesian coordinates (x , y , z) = (0,2√
3,−2),find the coordinates of P in
1. Spherical coordinates.2. Cylindrical coordinates.
Example (4 of 6)1. Spherical coordinates:
ρ =√
x2 + y2 + z2 =
√02 + (2
√3)2 + (−2)2 = 4
φ = cos−1
(z√
x2 + y2 + z2
)= cos−1
(−2
4
)=
2π3
θ = cos−1
(x√
x2 + y2
)= cos−1 (0) =
π
2
2. Cylindrical coordinates:
r =√
x2 + y2 =
√02 + (2
√3)2 = 2
√3
θ = cos−1
(x√
x2 + y2
)= cos−1 (0) =
π
2
z = z = −2
Example (5 of 6)
Find the equation in spherical coordinates of the hyperboloid of2 sheets: x2 − y2 − z2 = 1.
1 = x2 − y2 − z2
= ρ2 sin2 φ cos2 θ − ρ2 sin2 φ sin2 θ − ρ2 cos2 φ
= ρ2(
sin2 φ cos2 θ − sin2 φ sin2 θ − cos2 φ)
= ρ2(
sin2 φ[cos2 θ − sin2 θ
]− cos2 φ
)= ρ2
(sin2 φ cos 2θ − cos2 φ
)
Example (5 of 6)
Find the equation in spherical coordinates of the hyperboloid of2 sheets: x2 − y2 − z2 = 1.
1 = x2 − y2 − z2
= ρ2 sin2 φ cos2 θ − ρ2 sin2 φ sin2 θ − ρ2 cos2 φ
= ρ2(
sin2 φ cos2 θ − sin2 φ sin2 θ − cos2 φ)
= ρ2(
sin2 φ[cos2 θ − sin2 θ
]− cos2 φ
)= ρ2
(sin2 φ cos 2θ − cos2 φ
)
Example (6 of 6)
Find the equation in Cartesian coordinates of the surfacewhose equation in spherical coordinates is ρ = sinφ sin θ.
ρ = sinφ sin θ
ρ2 = ρ sinφ sin θ
x2 + y2 + z2 = yx2 + y2 − y + z2 = 0
x2 +
(y − 1
2
)2
+ z2 =14
Example (6 of 6)
Find the equation in Cartesian coordinates of the surfacewhose equation in spherical coordinates is ρ = sinφ sin θ.
ρ = sinφ sin θ
ρ2 = ρ sinφ sin θ
x2 + y2 + z2 = yx2 + y2 − y + z2 = 0
x2 +
(y − 1
2
)2
+ z2 =14
Spherical Coordinate Equations (1 of 3)Sphere: ρ = c > 0, a constant
x2 + y2 + z2 = c2
Spherical Coordinate Equations (2 of 3)
Plane: θ = θ0, with 0 ≤ θ0 ≤ 2π
Spherical Coordinate Equations (3 of 3)
Cone: φ = φ0, with 0 < φ0 < π
Volume Element in Spherical Coordinates
∆V ≈ (ρ sinφ∆θ)(ρ∆φ)(∆ρ)
= ρ2 sinφ∆ρ∆φ∆θ
dV = ρ2 sinφdρdφdθ
Iterated Integrals in Spherical Coordinates
The triple integral of f (ρ, φ, θ) over the solid region
Q = {(ρ, φ, θ) |g1(φ, θ) ≤ ρ ≤ g2(φ, θ), h1(θ) ≤ φ ≤ h2(θ), α ≤ θ ≤ β}
is∫∫∫Q
f (ρ, φ, θ) dV =
∫ β
α
∫ h2(θ)
h1(θ)
∫ g2(φ,θ)
g1(φ,θ)f (ρ, φ, θ)ρ2 sinφdρdφdθ.
Example
Evaluate the triple integral below by converting to sphericalcoordinates. ∫∫∫
Qe(x2+y2+z2)3/2
dV
where Q = {(x , y , z) | x2 + y2 + z2 ≤ 1}.
Solution
∫∫∫Q
e(x2+y2+z2)3/2dV =
∫ 2π
0
∫ π
0
∫ 1
0e(ρ2)3/2
ρ2 sinφdρdφdθ
= 2π∫ π
0
∫ 1
0ρ2eρ
3sinφdρdφ
= 2π∫ π
0
13
eρ3∣∣∣∣10
sinφdφ
= 2π∫ π
0
13
(e − 1) sinφdφ
=2(e − 1)π
3(− cosφ)|π0
=4(e − 1)π
3
Example
Find the volume of the solid that lies above the conez2 = x2 + y2 and below the sphere x2 + y2 + z2 = z.
Solution (1 of 2)
In spherical coordinates the equations of the cone and thesphere are
Cone: φ =π
4Sphere: ρ = cosφ.
Solution (2 of 2)
V =
∫∫∫Q
1 dV =
∫ 2π
0
∫ π/4
0
∫ cosφ
0(1)ρ2 sinφdρdφdθ
= 2π∫ π/4
0
∫ cosφ
0ρ2 sinφdρdφ
= 2π∫ π/4
0
13ρ3∣∣∣∣cosφ0
sinφdφ
=2π3
∫ π/4
0cos3 φ sinφdφ
= −2π3
∫ 1/√
2
1u3 du =
2π3
∫ 1
1/√
2u3 du
=π
6u4∣∣∣11/√
2=π
8
Example
Find the mass and center of mass of the solid hemisphere ofradius a if the density at any point in the solid is proportional toits distance from the base.
0
0.2
0.4
0.6
0.8
Solution (1 of 3)
I The distance of a point in the hemisphere from the base isthe z-coordinate of the point.
I In spherical coordinates z = ρ cosφ.I Without loss of generality, we may choose the
proportionality constant to be 1.
m =
∫∫∫Q
z dV
Myz =
∫∫∫Q
x z dV
Mxz =
∫∫∫Q
y z dV
Mxy =
∫∫∫Q
z2 dV
Solution (2 of 3)
m =
∫ 2π
0
∫ π/2
0
∫ a
0(ρ cosφ)ρ2 sinφdρdφdθ
= 2π∫ π/2
0
∫ a
0ρ3 cosφ sinφdρdφ
=πa4
2
∫ π/2
0cosφ sinφdφ =
πa4
2
∫ 1
0u du =
πa4
4
Mxy =
∫ 2π
0
∫ π/2
0
∫ a
0(ρ cosφ)2ρ2 sinφdρdφdθ
= 2π∫ π/2
0
∫ a
0ρ4 cos2 φ sinφdρdφ
=2πa5
5
∫ π/2
0cos2 φ sinφdφ
= −2πa5
5
∫ 0
1u2 du =
2πa5
5
∫ 1
0u2 du =
2πa5
15
Solution (3 of 3)
Myz =
∫ 2π
0
∫ π/2
0
∫ a
0(ρ sinφ cos θ)ρ2 sinφdρdφdθ
=
∫ a
0
∫ π/2
0
∫ 2π
0ρ3 sin2 φ cos θ dθ dφdρ
= 0
Mxz =
∫ 2π
0
∫ π/2
0
∫ a
0(ρ sinφ sin θ)ρ2 sinφdρdφdθ
=
∫ a
0
∫ π/2
0
∫ 2π
0ρ3 sin2 φ sin θ dθ dφdρ
= 0
Thus
(x , y , z) =
(Myz
m,Mxz
m,Mxy
m
)=
(0,0,
8a15
).
Homework
I Read Section 13.7.I Exercises: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49,
53, 57
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