Higher Mathematics. Part 3
.pdfIn each of Ωi we select an arbitrary point Pi (ξi , ηi , ζi ) .Throughout Ωi , i = 1, 2, 3… n the density is almost constant and equal to μ(Pi ), i = 1, 2, ..., n .
The mass |
mi of this part of the body will then be given in the approximate |
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relation |
mi ≈ μ(Pi ) Vi . |
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The mass of the entire body will be |
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n |
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m ≈ ∑μ(Pi ) Vi . |
(6.1) |
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i=1 |
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Let λ be the largest diameter of Ωi , |
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i.e. |
λ = max d (Ωi ), |
i = 1, 2, ..., n. |
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1≤i≤ n |
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If (6.1) has a finite limit as λ → 0, |
which is independent on the way of |
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dividing |
Ω into partial regions |
and of |
selection of Pi Ωi , |
this limit is |
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considered to be the mass of the body |
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n |
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m = lim |
∑μ(Pi ) |
Vi . |
(6.2) |
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λ →0 |
i=1 |
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On the other hand, this limit is known as the triple integral of the function μ(P) over the domain Ω and is denoted by
∫∫∫μ(P)dV = lim |
n |
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∑μ(Pi ) |
Vi . |
(6.3) |
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Ω |
λ →0 |
i=1 |
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Hence |
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m = ∫∫∫μ(P)dV = ∫∫∫μ(x, y, z) dxdydz. |
(6.4) |
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Ω |
Ω |
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Here dx dy dz is an element of volume dV |
in Cartesian coordinates. |
Now we will give the formal definition of the triple integral. Consider a
closed cubable domain Ω in which a bounded function |
f (P(x, y, z)) , P Ω, is |
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defined. |
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I = 1, 2, 3 … n. Let their |
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We divide Ω into nonoverlaping cubable parts Ωi , |
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volumes be Vi , respectively. In each of |
Ωi we arbitrarily choose a point |
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Pi (ξi , ηi , ζi ) . Constructed sum |
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n |
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σ = ∑ f (ξi , ηi , ζi ) |
Vi |
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(6.5) |
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i=1 |
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is called integral sum for function f(x, y, z). |
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131 |
z
m
1 |
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c |
d |
y
a
b
x
Fig. 6.1
Substituting the iterated integral for the double integral we obtain
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d |
m |
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∫∫∫ f (x, y, z)dV =∫ dx∫ dy∫ f (x, y, z)dz |
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(6.14) |
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Ω |
a |
ñ |
l |
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When Ω is a rectangular parallelepiped, we could thus reduce the taking of the triple integral to taking three conventional integrals in a series. Formula (6.13) can be rewritten
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∫∫∫ f (x, y, z)dv =∫∫[∫ f (x, y, z)dz]dxd y, |
(6.15) |
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Ω |
D l |
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where D = {a ≤ x ≤ b, c ≤ y ≤ d} is a rectangle in the xy-plane. This rectangle is
an orthogonal projection of the parallelepiped on the xy-plane. |
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Ω so that a |
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z |
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Let’s consider a domain |
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S |
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z = z2(x, y) |
surface |
S, |
that bounds |
Ω, |
is intersected |
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2 |
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not more than at two points by any straight |
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line parallel to the z-axis (Fig. 6.2). |
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Let the surface S1 that bounds Ω from |
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S1 |
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below be described by the equation z = z1(x, y), |
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z = z1(x, y) |
and the surface S2 that bounds |
Ω from |
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y |
above |
be |
described |
by |
the |
equation |
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D |
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z = z2 (x, y). We project S1 |
and S2 on the |
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L |
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xy-plane into a domain D bounded by a |
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x |
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curve L. |
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The body Ω is bounded by a cylindrical |
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Fig. 6.2 |
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surface with a generatrix parallel to the z- |
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134 |
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where I is the Jacobian of the set of functions (6.18):
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∂x |
∂x |
∂x |
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∂u |
∂v |
∂w |
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J = |
∂y |
∂y |
∂y |
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≠ 0. |
(6.20) |
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∂u |
∂v |
∂w |
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∂z |
∂z |
∂z |
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∂u |
∂v |
∂w |
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6.3.2. Triple Integral in Cylindrical Coordinates
In cylindrical coordinates the position of a point M in space is determined by
three numbers ρ, ϕ, z, |
where ρ, |
ϕ are the polar coordinates of the projection |
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M onto the xy-plane and |
z is termed the cylindrical coordinates of M (Fig. 6.3). |
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z |
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z |
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M(ρ, ϕ, z) |
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y |
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x |
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Fig. 6.3
It is clear
0 ≤ ρ < ∞, 0 ≤ ϕ < 2π, − ∞ < z < +∞.
The relation of rectangular coordinates and cylindrical coordinates is apparent from the Fig. 6.3.
x = ρ cos ϕ, y = ρ sin ϕ, z = z.
Let’s find the Jacobian
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∂x |
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∂x |
∂x |
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∂ρ |
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∂ϕ |
∂z |
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cos ϕ |
−ρ sin ϕ |
0 |
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J = |
∂y |
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∂y |
∂y |
= |
= ρ. |
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sin ϕ |
ρ cos ϕ |
0 |
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∂ρ |
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∂ϕ |
∂z |
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0 |
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1 |
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∂z |
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∂z |
∂z |
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∂ρ |
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∂ϕ |
∂z |
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136
Since ρ ≥ 0, then
I |
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= ρ |
(6.21) |
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and for triple integrals the formula (6.19) for the transition from rectangular coordinates to cylindrical ones becomes
∫∫∫ f (x, y, z)dxdydz = ∫∫∫ f (ρ cos ϕ, ρ sin ϕ, z)ρdρdϕdz. |
(6.22) |
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G |
G′ |
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In cylindrical coordinates a volume element is dV = ρdρdϕdz.
Remark. Coordinates frequently pass to cylindrical, when the domain of integration derivates by a cylindrical surface, which projection on an appropriate plane is domain as a circle, annulus, sector and etc. Integrand function set as
f (x, y, z) = g(x2 + y2 , z).
6.3.3. Triple Integral in Spherical Coordinates
In spherical coordinates the position of a point M in space is determined by three numbers ρ, ϕ, θ, where ρ is the distance from the origin to the point M
(Fig. 6.4), ϕ is the angle between x-axis and the projection of the radius vector
OM of the point M on the xy-plane, θ is the angle between z-axis and the radius vector OM of M reckoned from z-axis.
z
z
M(ρ, ϕ, θ)
θρ
y
ϕ |
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x
x
Fig. 6.4
It is clear
0 ≤ ρ < +∞, 0 ≤ ϕ < 2π, 0 ≤ θ ≤ π.
The relation of rectangular coordinates and spherical coordinates is apparent from the Fig. 6.4.
137
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x = ρ sin θ cos ϕ, y = ρ sin θ sin ϕ, z = ρ cos θ. |
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(6.23) |
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Let’s find the Jacobian |
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∂x |
∂x |
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∂x |
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∂ρ |
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∂ϕ |
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∂θ |
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sin θ cos ϕ −ρ sin θ sin ϕ |
ρ cos θ cos ϕ |
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J = |
∂y |
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∂y |
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∂y |
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= −ρ2 sin θ. |
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= |
sin θ sin ϕ |
ρ sin θ cos ϕ |
ρ cos θ sin ϕ |
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∂ρ |
∂ϕ |
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∂θ |
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cos θ |
0 |
−ρ sin θ |
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∂z |
∂z |
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∂z |
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∂ρ |
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∂ϕ |
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∂θ |
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Therefore, |
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and becomes |
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J |
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= ρ2 sin θ |
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(6.24) |
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∫∫∫ f (x, y, z)dxdydz = |
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= ∫∫∫ |
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Ω |
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(6.25) |
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f (ρ sin θ cos ϕ, ρ sin θ sin ϕ, r cos θ)ρ2 sin θdρd θd ϕ |
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Ω*
In spherical coordinates a volume element is dV = ρ2 sin θdρdθdϕ .
Remark. The equation of a sphere x2 + y2 + z2 = R2 in spherical coordinates can be simplified to a form ρ = R .
The spherical coordinates are frequently used, when the domain of an integration is a full-sphere, or its part or a sector. The intergrand function has a
form f (x, y, z) = g(x2 + y2 + z2 ) . |
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For example, if the domain is a full-sphere x2 + y2 + z2 = R2 , |
then we have |
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2π |
π |
R |
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∫∫∫ f (x2 + y2 + z2 )dxdydz = ∫ dϕ∫sin θdθ∫ f (ρ2 )ρ2dρ. |
(6.26) |
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Ω |
0 |
0 |
0 |
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6.4. Application of Triple Integrals
We shall briefly discuss typical problems of application of triple integrals and give the necessary formulas without deriving them since their derivation is similar to that of the corresponding formulas for double integrals.
1. The volume of a body.
V = ∫∫∫ dxdydz. |
(6.27) |
Ω |
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138