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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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3.1. 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
2 ≤ 1 , 0 ≤ z ≤
x
2
2 }. Here, D = {x
2
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
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
dθ
0
r
3 dr =
· 2 π ·
r
4
r=
r=
π
2