Triple Integrals: Physical Meaning, Calculation, and Examples, Lecture notes of Statistics

Triple integrals, their physical meaning, calculation over boxes, and provides examples. Triple integrals are used to calculate the volume or mass of a three-dimensional region with a given density function. The definition, calculation method, and an example of triple integrals over a box and a general region.

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3. Triple Integrals
3.1. Physical Meaning of Double Integrals
Consider a lamina occupying a region DR2with density function ρ(x, y). Then its total mass is
ZZD
ρ(x, y)dA.
Indeed, assume for simplicity that Dis a rectangle. We can split it into small subrectangles
R1, . . . , Rn, and choose a point (xi, yi)Riin each subrectangle. On each subrectangle Ri, the
density ρ(x, y) is almost constant: ρ(x, y)ρ(xi, yi). The mass of Riis approximately its area times
this almost constant density: ρ(xi, yi) Area(Ri). Thus, the total mass is approximately
n
X
i=1
ρ(xi, yi) Area(Ri)ZZD
ρ(x, y)dA.
3.2. Triple Integrals over Boxes
They are defined and calculated in much the same way as double integrals.
First, assume Bis a box, i.e. a region of the type
B= [a, b]×[c, d]×[p, q] = {axb, c yd, p zq}.
Assume a function f:BRis defined on this box. Let us split it into small sub-boxes
B1, B2, . . . , Bn. Pick a point (x1, y1, z1)B1,(x2, y2, z2)B2, . . . in each of these small boxes.
Then
ZZZB
fdV = lim
n
X
i=1
f(xi, yi, zi) Vol(Bi),
where the limit is taken as the sizes of boxes tend to zero. We can calculate it as an interated integral:
ZZZB
fdV =Zq
pZd
cZb
a
f(x, y, z)dxdy dz
We can take any order of integration, say first y, then z, then x. The result will be the same.
You cannot visualize it as the region under the graph of f, this would require 4D space. However,
you can visualize it as a mass of lamina occupying the region Bwith density ρ(x, y, z) at the point
(x, y, z). This is similar to a double integral as the mass of a lamina.
3.3. Example 1
Evaluate RRRByzdV , where B= [0,1] ×[0,1] ×[0,1]. We have:
ZZZB
yzdV =Z1
0Z1
0Z1
0
yzdz dydx =Z1
0"Z1
0
yz2
2
z=1
z=0
dy#dx =
Z1
0Z1
0
y
2dydx =Z1
0
y2
4
y=1
y=0
dx =Z1
0
1
4dx =x
4
x=1
x=0 =1
4.
1
pf2

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3. Triple Integrals

3.1. Physical Meaning of Double Integrals

Consider a lamina occupying a region D ⊆ R

2 with density function ρ(x, y). Then its total mass is

∫ ∫

D

ρ(x, y)dA.

Indeed, assume for simplicity that D is a rectangle. We can split it into small subrectangles

R 1 ,... , Rn, and choose a point (xi, yi) ∈ Ri in each subrectangle. On each subrectangle Ri, the

density ρ(x, y) is almost constant: ρ(x, y) ≈ ρ(xi, yi). The mass of Ri is approximately its area times

this almost constant density: ρ(xi, yi) Area(Ri). Thus, the total mass is approximately

n ∑

i=

ρ(xi, yi) Area(Ri) ≈

D

ρ(x, y)dA.

3.2. Triple Integrals over Boxes

They are defined and calculated in much the same way as double integrals.

First, assume B is a box, i.e. a region of the type

B = [a, b] × [c, d] × [p, q] = {a ≤ x ≤ b, c ≤ y ≤ d, p ≤ z ≤ q}.

Assume a function f : B → R is defined on this box. Let us split it into small sub-boxes

B 1 , B 2 ,... , Bn. Pick a point (x 1 , y 1 , z 1 ) ∈ B 1 , (x 2 , y 2 , z 2 ) ∈ B 2 ,... in each of these small boxes.

Then ∫ ∫ ∫

B

f dV = lim

n ∑

i=

f (xi, yi, zi) Vol(Bi),

where the limit is taken as the sizes of boxes tend to zero. We can calculate it as an interated integral:

B

f dV =

q

p

d

c

b

a

f (x, y, z)dxdydz

We can take any order of integration, say first y, then z, then x. The result will be the same.

You cannot visualize it as the region under the graph of f , this would require 4D space. However,

you can visualize it as a mass of lamina occupying the region B with density ρ(x, y, z) at the point

(x, y, z). This is similar to a double integral as the mass of a lamina.

3.3. Example 1

Evaluate

B

yzdV , where B = [0, 1] × [0, 1] × [0, 1]. We have:

B

yzdV =

0

[∫

1

0

[∫

1

0

yzdz

]

dy

]

dx =

0

[

0

yz

2

z=

z=

dy

]

dx =

0

[∫

1

0

y

dy

]

dx =

0

y

2

y=

y=

dx =

0

dx =

x

x=

x=

1

3.4. Triple Integrals over General Regions

Consider a bounded region E ⊆ R

3 and a function f : E → R. Let us define the triple integral of f

over E. Since E is bounded, enclose it into a box B. Extend f onto B by assigning it zero values

(inside B and outside E). By definition,

∫ ∫ ∫

E

f dV =

B

f dV.

The method of calculation is the same as for double integrals. Assume

E = {(x, y) ∈ D, u 1 (x, y) ≤ z ≤ u 2 (x, y)}.

Then

∫ ∫ ∫

E

f dV =

D

[

∫ (^) u 2 (x,y)

u 1 (x,y)

f (x, y, z)dz

]

dA

The same can be done with the roles of x, y, z interchanged.

3.5. Example 2

Find

E

zdV , where E = {x

2

  • y

2 ≤ 1 , 0 ≤ z ≤

x

2

  • y

2 }. Here, D = {x

2

  • y

2 ≤ 1 }, and

u 1 (x, y) = 0, u 2 (x, y) =

x^2 + y^2. This is equal to

D

[

x^2 +y^2

0

zdz

]

dA =

D

z

2

z=

x^2 +y^2

z=

dA =

D

x

2

  • y

2

dA.

Here, use polar coordinates: D = { 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π}, and dA = rdrdθ. Therefore, we have:

∫ (^2) π

0

0

r

2

rdrdθ =

∫ (^2) π

0

0

r

3 dr =

· 2 π ·

r

4

r=

r=

π

2