Triple integrals in cylindrical and spherical coordinates

MATH 209— 

Calculus, 

III 

Volker Runde 

Triple 

integrals in 

cylindrical 

coordinates 



spherical 


MATH 209—Calculus, III 

Volker Runde 

University of Alberta 

Edmonton, Fall 2011

MATH 209— 

Calculus, 

III 

Volker Runde 









Cylindrical coordinates 

Rectangular versus cylindrical coordinates 

x = r cos θ, 

y = r sin θ, 

z = z, 

r 2 = x 2 + y 2 .

MATH 209— 

Calculus, 

III 

Volker Runde 









Integration in cyclindrical coordinations, I 

The setup 

Let E ⊂ R 3 be of type I, i.e., 

with 

E = {(x, y, z) : (x, y) ∈ D, u1(x, y) ≤ z ≤ u2(x, y)}, 

D = {(r, θ) : α ≤ θ ≤ β, h1(θ) ≤ r ≤ h2(θ)} 

in polar coordinates.

MATH 209— 

Calculus, 

III 

Volker Runde 









Integration in cyclindrical coordinations, II 

Theorem (change to cylindrical coordinates in a tiple integral) 

Given E and D as before: 

�� 

f (x, y, z) dV 

E 

�� 

u2(x,y) 

= 

f (x, y, z) dz dA 

D 

� β � h2(θ) 

= 

α 

h1(θ) 

u1(x,y) 

� u2(r cos θ,r sin θ) 

u1(r cos θ,r sin θ) 

f (r cos θ, r sin θ, z)r dz dr θ.

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, I 

Example 

A solid E lies inside the cylinder x 2 + y 2 = 1, below the plane 

z = 4, and above the paraboloid z = 1 − x 2 − y 2 . 

Its density ρ(x, y, z) at any point is proportional to the point’s 

distance from the z-axis. 

What is the mass of E? 

We have 

and 

E = {(x, y, z) : x 2 + y 2 ≤ 1, 1 − x 2 − y 2 ≤ z ≤ 4} 

= {(r, θ, z) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 1 − r 2 ≤ z ≤ 4} 

ρ(x, y, z) = C � x 2 + y 2 = Cr.

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, II 

Example (continued) 

Thus: 

�� 

m = C 

= C 

E 

� 2π � 1 

0 

0 

� x 2 + y 2 dV = C 

� 2π � 1 � 4 

r 2 (4 − (1 − r 2 )) dr dθ = 2Cπ 

� 

= 2Cπ r 3 � 

r 5 � 

+ � 

5 � 

0 

0 

1−r 2 

� 1 

0 

r=1 

r=0 

r 2 dz dr dθ 

3r 2 + r 4 dr 

� 

= 12Cπ 

. 

5

MATH 209— 

Calculus, 

III 

Volker Runde 









Spherical coordinates coordinates 

Rectangular versus spherical coordinates 

x = ρ sin φ cos θ, 

y = ρ sin φ sin θ, 

z = ρ cos φ, 

ρ 2 = x 2 + y 2 + z 2 .

MATH 209— 

Calculus, 

III 

Volker Runde 









Integration in spherical coordinates, I 

Definition 

A set of the form 

{(r, θ, φ) : r ∈ [a, b], θ ∈ [α, β], φ ∈ [c, d]} 

with a ≥ 0, β − α ≤ 2π, and d − c ≤ π is called a spherical 

wedge. 

Question 

How does one evaluate �� 

wedge? 

E 

f (x, y, z) dV if E is a spherical

MATH 209— 

Calculus, 

III 

Volker Runde 









Integration in spherical coordinates, II 

As in rectangular coordinates. . . 

Divide E into small spherical wedges Ej,k,ℓ, and pick a support 

point (x ∗ j,k,ℓ , y ∗ 

j,k,ℓ , z∗ j,k,ℓ ) ∈ Ej,k,ℓ. 

Then: 

�� 


with 

E 

= lim 

n,m,ν→∞ 

n� 

m� 

ν� 

j=1 k=1 ℓ=1 

∆Vj,k,ℓ = volume of Ej,k,ℓ. 

f (x ∗ ∗ 

j,k,ℓ , yj,k,ℓ , z∗ j,k,ℓ )∆Vj,k,ℓ

MATH 209— 

Calculus, 

III 

Volker Runde 









Integration in spherical coordinates, III 

Question 

What is the volume of Ej,k,ℓ, i.e., a very small spherical wedge? 

Answer 

∆Vj,k,ℓ ≈ ρ 2 j sin φℓ∆ρ∆θ∆φ.

MATH 209— 

Calculus, 

III 

Volker Runde 









Integration in spherical coordinates, IV 

Consequence 

�� 

E 

f (x, y, z) dV = lim 

n,m,ν→∞ 

n� 

m� 

ν� 

j=1 k=1 ℓ=1 

f (ρ ∗ j sin φ ∗ ℓ cos θ∗ k , ρ∗ j sin φ ∗ ℓ sin θ∗ k , ρ∗ j cos φ ∗ ℓ )ρ2 sin φ∆ρ∆θ∆φ 

� d � β � b 

= 

f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) 

c 

α 

a 

ρ 2 sin φ dρ dθ dφ.

MATH 209— 

Calculus, 

III 

Volker Runde 









Integration in spherical coordinates, V 

Theorem (change to spherical coordinates in a tiple integral) 

Let 

E := {(ρ, θ, φ) : α ≤ θ ≤ β, 

Then: 

�� 

E 


c ≤ φ ≤ d, ≤ g1(θ, φ) ≤ ρ ≤ g2(θ, φ)}. 

� d � β � g2(θ,φ) 

= 

f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) 

c 

α 

g1(θ,φ) 

ρ 2 sin φ dρ dθ dφ.

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, III 

Example 

Let 

Evaluate �� 

In spherical coordinates: 

E = {(x, y, z) : x 2 + y 2 + z 2 ≤ 1}. 

E 

e (x2 +y 2 +z 2 ) 3 2 dV . 

E = {(ρ, θ, φ) : ρ ∈ [0, 1], θ ∈ [0, 2π], φ ∈ [0, π]}.

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, IV 


Then: 

�� 

E 

e (x2 +y 2 +z 2 ) 3 2 dV 

� π 

= 

0 

� π 

= 

0 

� 2π � 1 

e 

0 0 

(ρ2 sin2 φ cos2 θ+ρ2 sin2 φ sin2 θ+ρ2 cos2 φ) 3 2 

ρ 2 sin φ dρ dθ dφ 

� 2π � 1 

e 

0 0 

(ρ2 sin2 φ+ρ2 cos2 φ) 3 2 2 

ρ sin φ dρ dθ dφ 

� π � 2π � 1 

= 

e 

0 0 0 

ρ3 

ρ 2 sin φ dρ dθ dφ = · · ·

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, V 


� π 

· · · = 

0 

� 2π � 1 

0 

0 

e ρ3 

ρ 2 sin φ dρ dθ dφ 

� π 

= 2π 

0 

� π 

= 2π sin φ 

0 

�� 1 

= 2π e 

0 

ρ3 

ρ 2 dρ 

� 1 

e 

0 

ρ3 

ρ 2 sin φ dρ dφ 

�� 1 

e 

0 

ρ3 

ρ 2 � 

dρ dφ 

� 1 

� �� π � 

sin φ dφ = 4π 

0 

= 4π 

3 

� 1 

0 

0 

e ρ3 

ρ 2 dρ 

e u du = 4π 

(e − 1). 

3

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, VI 

Example 

Let H be the upper hemisphere above the xy-plane and below 

the sphere x 2 + y 2 + z 2 = 1, i.e., 

H = {(x, y, z) : x 2 + y 2 + z 2 ≤ 1, z ≥ 0}. 

What is �� 

H x 2 + y 2 dV ? 

In spherical coordinates: 

� 

� 

H = (ρ, θ, φ) : ρ ∈ [0, 1], θ ∈ [0, 2π], φ ∈ 0, π 

�� 

. 

2

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, VII 


Then: 

�� 

H 

= 

= 

x 2 + y 2 dV 

� 2π 

0 

� 2π 

0 

� π 

2 

0 

� π 

2 

0 

� 1 

0 

� 1 

0 

= 

(ρ 2 cos 2 θ sin 2 φ + ρ 2 sin 2 θ sin 2 φ) 

ρ 2 sin φ dρ dφ dθ 

ρ 4 (cos 2 θ + sin 2 θ) sin 3 φ dρ dφ dθ 

� 2π 

0 

� π 

2 

0 

� 1 

0 

ρ 4 sin 3 φ dρ dφ dθ = · · ·

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, VIII 


· · · = 

� 2π 

0 

= 2π 

5 

� π 

2 

0 

= 2π 

� π 

2 

0 

= − 2π 

5 

� 1 

ρ 4 sin 3 φ dρ dφ dθ 

0 

�� 1 

0 

ρ 4 � 

dρ 

�� π 

2 

sin 3 φ dφ = 2π 

5 

� 0 

1 

0 

� π 

2 

0 

sin 3 φ dφ 

� 

sin φ(1 − cos 2 φ) dφ 

� 1 

1 − u 2 du = 2π 

1 − u 

5 0 

2 du 

= 2π 

� 

u − 

5 

u3 

� � 

u=1 � 

� 

3 � 

u=0 

= 4π 

15 .

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, IX 

Example 

Determine the volume V of the solid E that lies above the 

cone z = � x 2 + y 2 and below the sphere x 2 + y 2 + z2 = z. 

Note: 

x 2 + y 2 + z 2 = z ⇐⇒ x 2 + y 2 � 

+ z − 1 

�2 = 

2 

1 

4 . 

Project E onto the xy-plane and obtain D: 

� 

D = (x, y) : x 2 + y 2 ≤ 1 

� 

4 

� � 

= (r, θ) : r ∈ 0, 1 

� 

� 

, θ ∈ [0, 2π] . 

2

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, X 


Note: 

(x, y, z) ∈ E ⇐⇒ 

(x, y) ∈ D and � x 2 + y 2 ≤ z ≤ 

Hence, in cylindrical coordinates: 

E = 

� 

� 

(r, θ, z) : r ∈ 0, 1 

� 

, θ ∈ [0, 2π], 

2 

r ≤ z ≤ 

� 1 

4 − x 2 − y 2 + 1 

2 

� 

1 

4 − r 2 + 1 

� 

2

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, XI 


Then: 

�� 

V = 

E 

1 dV 

= 

= 2π 

� 2π 

� 1 

2 

0 0 

� 1 

2 

0 

= 2π 

� � 1 

4 −r 2 + 1 

2 

r dz dr dθ 

r 

�� 

1 

4 − r 2 + 1 

� 

− r r dr 

2 

�� 1 

� 1 

2 

2 

0 

r 

� 1 

4 − r 2 dr + 

0 

r 

2 − r 2 � 

dr .

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, XII 


On the side: 

and 

� 1 

2 

0 

r 

� 1 

2 

0 

r 

2 − r 2 dr = 

r 2 

4 

� 1 

4 − r 2 dr = − 1 

2 

− r 3 

3 

� 0 

1 

4 

= 1 

2 

� 

� 

� 

� 

r= 1 

2 

r=0 

√ u du 

� 1 

4 

0 

= 1 1 1 

− = 

16 24 48 

√ u du = u 3 

2 

3 

� 

� 

� 

� 

� 

u= 1 

4 

u=0 

= 1 

24 .

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, XIII 


Therefore: 

V = 2π 

� � 1 

2 

� 

1 

r 

0 4 − r 2 dr + 

� � 

1 1 

= 2π + 

24 48 

= π 

8 . 

� 1 

2 

0 

r 

2 − r 2 � 

dr

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, XIV 


Pass to spherical coordinates. 

For (x, y, z) on the sphere: 

ρ cos φ = z = x 2 + y 2 + z 2 

= ρ 2 cos 2 θ sin 2 φ + ρ 2 sin 2 θ sin 2 φ + ρ 2 cos 2 φ = ρ 2 . 

Hence, the sphere is: 

� 

� 

(ρ, θ, φ) : θ ∈ [0, 2π], φ ∈ 

0, π 

2 

� � 

, ρ = cos φ .

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, XV 


For (x, y, z) on the cone: 

ρ cos φ = z = � x 2 + y 2 

� 

= ρ2 cos2 θ sin2 φ + ρ2 sin2 θ sin2 φ = ρ sin φ. 

Hence, the cone is: 

{(ρ, θ, φ) : ρ ≥ 0, θ ∈ [0, 2π], φ ∈ [0, π], cos φ = sin φ} 

� 

= (ρ, θ, φ) : ρ ≥ 0, θ ∈ [0, 2π], φ = π 

� 

. 

4

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, XV 


It follows that 

� 

� 

E = (ρ, θ, φ) : θ ∈ [0, 2π], φ ∈ 0, π 

� 

� 

, 0 ≤ ρ ≤ cos φ . 

4 

Hence: 

�� 

V = 

E 

1 dV = 

= 2π 

� 

= 2π 

� π 

4 

0 

sin φ 

� 2π 

0 

� π 

4 

0 

ρ 3 

3 

� π 

4 

� cos φ 

0 0 

�� cos φ 

sin φ 

0 

� � 

ρ=cos φ � 

� 

� dφ = 2π 

3 

ρ=0 

ρ 2 sin φ dρ dφ dθ 

ρ 2 � 

dρ dφ 

� 1 

4 

0 

sin φ cos 3 φ dφ 

= · · ·

MATH 209— 

Calculus, 

III 

Volker Runde 









Examples, XVI 


· · · = 2π 

3 

= 2π 

3 

� 1 

4 

u= 1 

√ 

2 

� 1 

√2 

sin φ cos 

0 

3 φ dφ = − 2π 

u 

3 1 

3 du 

⎛ 

⎝ u4 

⎞ 

�u=1 

� 

� ⎠ 

4 � = 2π 

� � 

1 1 

− = 

3 4 16 

2π 

3 

3 π 

= 

16 8 .

Triple integrals in cylindrical and spherical coordinates

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