3.36pt

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