If f x, y is bounded and the set of points of discontinuity of f x, y has a zero area e.g., the points of discontinuity lie on finitely many continuous curves in the x, y plane, then the
Trang 17.3.1-2 Classes of integrable functions.
Further on, it is assumed that D is a closed bounded domain.
1 If f (x, y) is continuous in D, then the double integral
D f (x, y) dx dy exists.
2 If f (x, y) is bounded and the set of points of discontinuity of f (x, y) has a zero area (e.g., the points of discontinuity lie on finitely many continuous curves in the x, y plane), then the double integral of f (x, y) over the domain D exists.
7.3.1-3 Properties of the double integral
1 Linearity If functions f (x, y) and g(x, y) are integrable in D, then
D
af (x, y) bg (x, y)
dx dy = a
D f (x, y) dx dy b
D g (x, y) dx dy,
where a and b are any numbers.
2 Additivity If the domain D is split into two subdomains D1and D2that do not have
common internal points and if the function f (x, y) is integrable in either subdomain, then
D f (x, y) dx dy =
D1
f (x, y) dx dy +
D2
f (x, y) dx dy.
3 Estimation theorem If m≤f (x, y)≤M in D, then
mS ≤
D f (x, y) dx dy ≤M S,
where S is the area of the domain D.
4 Mean value theorem If f (x, y) is continuous in D, then there exists at least one
internal point (¯x, ¯y)Dsuch that
D f (x, y) dx dy = f ( ¯x, ¯y) S.
The number f ( ¯x, ¯y) is called the mean value of the function f(x, y) in D.
5 Integration of inequalities If ϕ(x, y)≤f (x, y)≤g (x, y) in D, then
D ϕ (x, y) dx dy ≤
D f (x, y) dx dy≤
D g (x, y) dx dy.
In particular, if f (x, y)≥ 0in D, then
D f (x, y) dx dy ≥ 0
6 Absolute value theorem
D f (x, y) dx dy
≤
D
f (x, y)dx dy.
Trang 2z=f x y( , )
y
x
D z
O
Figure 7.4 A double integral of a nonnegative function f (x, y) over a domain D is equal to the volume of a
cylindrical body with base D in the plane z =0and bounded from above by the surface z = f (x, y).
7.3.1-4 Geometric meaning of the double integral
Let a function f (x, y) be nonnegative in D Then the double integral
D f (x, y) dx dy is
equal to the volume of a cylindrical body with base D in the plane z =0and bounded from
above by the surface z = f (x, y); see Fig 7.4.
7.3.2 Computation of the Double Integral
7.3.2-1 Use of iterated integrals
1◦ If a domain D is defined in the x, y plane by the inequalities a≤x≤b and y1(x)≤y≤y2(x)
(see Fig 7.5 a), then*
D f (x, y) dx dy =
b
a dx
y2x)
y1x) f (x, y) dy. (7.3.2.1)
The integral on the right-hand side is called an iterated integral.
y=y x1( )
x=x y2( )
x=x y1( )
y=y x2( )
a
c d
b
Figure 7.5 Computation of a double integral using iterated integrals: (a) illustration to formula (7.3.2.1),
(b) illustration to formula (7.3.2.2).
* It is assumed that in (7.3.2.1) and (7.3.2.2) the double integral on the right-hand side and the inner integral
on the right-hand side exist.
Trang 32◦ If D ={c≤y≤d , x1(y)≤x≤x2(y)}(see Fig 7.5 b), then
D f (x, y) dx dy =
d
c dy
x2y)
x1y) f (x, y) dx. (7.3.2.2)
Example 1 Compute the integral
I=
D
dx dy
(ax + by)2, where D ={0 ≤x≤ 1 , 1 ≤y≤ 3}is a rectangle, a >0, and b >0
Using formula (7.3.2.2), we get
D
dx dy
(ax + by)2 =
3
1 dy
1 0
dx
(ax + by)2.
Compute the inner integral:
1 0
dx
(ax + by)2 = –
1
a (ax + by)
x=1
x=0
= 1
a
1
by – 1
by + a
.
It follows that
I= 1
a
3 1
1
by – 1
by + a
dy= 1
ab ln 3(a + b)
a+ 3b .
3◦ Consider a domain D inscribed in a rectangle{a≤x≤b , c≤y≤d} Let the boundary
of D, within the rectangle, be intersected by straight lines parallel to the coordinate axes
at two points only, as shown in Fig 7.6 a Then, by comparing formulas (7.3.2.1) and
(7.3.2.2), we arrive at the relation
b
a dx
y2x)
y1 x) f (x, y) dy =
d
c dy
x2y)
x1y) f (x, y) dx,
which shows how the order of integration can be changed
c
D
D D D
1 2 3
d
b
Figure 7.6 Illustrations to the computation of a double integral in a simple (a) and a complex (b) domain.
4◦ In the general case, the domain D is first split into subdomains considered in Items1◦
and2◦, and then the property of additivity of the double integral is used For example, the
domain D shown in Fig 7.6 b is divided by the straight line x = a into three subdomains
D1, D2, and D3 Then the integral over D is represented as the sum of three integrals over
the resulting subdomains
Trang 47.3.2-2 Change of variables in the double integral.
1◦ Let x = x(u, v) and y = y(u, v) be continuously differentiable functions that map
one-to-one a domain D1in the u, v plane onto a domain D in the x, y plane, and let f (x, y)
be a continuous function in D Then
D f (x, y) dx dy =
D1
f x (u, v), y(u, v)
|J(u, v)| du dv,
where J(u, v) is the Jacobian (or Jacobian determinant) of the mapping of D1onto D:
J (u, v) = ∂ (x, y)
∂ (u, v) =
∂x
∂u ∂x ∂v
∂y
∂u ∂y ∂v
=
∂x
∂u
∂y
∂v – ∂x
∂v
∂y
∂u
The fraction before the determinant is a common notation for a Jacobian
The absolute value of the Jacobian characterizes the extension (contraction) of an
infinitesimal area element when passing from x, y to u, v.
2◦ The Jacobian of the mapping defining the change from the Cartesian coordinates x, y
to the polar coordinates ρ, ϕ,
is equal to
Example 2 Given a sphere of radius R and a right circular cylinder of radius a < R whose axis passes
through the sphere center, find the volume of the figure the cylinder cuts out of the sphere.
The volume of this figure is calculated as
V =
x2+y2 ≤a2
R2– x2– y2 dx dy.
Passing in the integral from x, y to the polar coordinates (7.3.2.3) and taking into account (7.3.2.4), we obtain
V =
2π 0
a 0
R2– ρ2ρ dρ dϕ= 4π
3
R3– (R2– a2)3/2
.
3◦ The Jacobians of some common transformations in the plane are listed in Table 7.1.
TABLE 7.1 Some common curvilinear coordinates in the plane and the respective Jacobians
Generalized polar coordinates ρ, ϕ x = aρ cos ϕ, y = bρ sin ϕ abρ
Elliptic coordinates u, v
(special system; u≥ 0 , 0 ≤v≤π) x = a cosh u cos v, y = a sinh u sin v a2(sinh2u+ sin2v)
Parabolic coordinates σ, τ x = aστ , y = 12a (τ2– σ2) a2(σ2+ τ2)
Bipolar coordinates σ, τ x= a sinh τ
cosh τ – cos σ , y =
a sin σ cosh τ – cos σ
a2
(cosh τ – cos σ)2
Trang 57.3.2-3 Differentiation of the double integral with respect to a parameter.
1◦ In various applications, situations may arise where the integrand function and the
integration domain depend on a parameter, t The derivative of such a double integral with respect to t is expressed as
d
dt
D(t) f (x, y, t) dx dy =
D(t)
∂
∂t f (x, y, t) dx dy +
L(t) (n⋅v )f (x, y, t) dl, (7.3.2.5)
where L(t) is the boundary of the domain D(t), n is outer unit normal to L(t), and v is the velocity of motion of the points of L(t).
2◦ If the boundary L(t) is specified by equations in parametric form
x = X(λ, t), y = Y (λ, t), α≤λ≤β, (7.3.2.6) then
n=
Y λ
X2
λ + Y λ2, –
X λ
X2
λ + Y λ2
, = 5
X t , Y t6
, (n⋅) = Y λ X t – X λ Y t
X2
λ + Y λ2 , dl=
X2
λ + Y λ2;
the subscripts λ and t denote the respective partial derivatives The last integral in (7.3.2.5)
becomes
L(t) (n⋅v )f (x, y, t) dl =
β
α f X (λ, t), Y (λ, t), t
(Y λ X t – X λ Y t ) dλ. (7.3.2.7)
Example 3 Let D(t) be a deformable plane domain bounded by an ellipse
x2
a2(t) +
y2
Let us rewrite the equation of the ellipse (7.3.2.8) in parametric form as
x = a(t) cos λ, y = b(t) sin λ, 0 ≤λ≤ 2π ,
which corresponds to the special case of (7.3.2.6) with X(λ, t) = a(t) cos λ and Y (λ, t) = b(t) sin λ Taking
into account the aforesaid and using formula (7.3.2.7), we obtain
L(t) (n⋅v )f (x, y, t) dl =
2π 0
f a (t) cos λ, b(t) sin λ, t
(a t bcos2λ + ab tsin2λ ) dλ. (7 3 2 9 )
Let us dwell on the simple special case of f (x, y, t) =1 , when the first integral on the right-hand side of (7.3.2.5) vanishes Evaluate the second integral by formula (7.3.2.9) to obtain
L(t) (n⋅v ) dl =
2π
0 (a t bcos2λ + ab tsin2λ ) dλ = a t b
2π
0 cos2λ dλ + ab t
2π
0 sin2λ dλ
Remark. Formula (7.3.2.10) is easy to derive directly from (7.3.2.5) noting that, for f (x, y, t) =1 , the
integral on the left-hand side of (7.3.2.5) gives the area of the ellipse, S = πa(t)b(t) Differentiating this formula
yields (7.3.2.10).
3◦ If the boundary L(t) is specified by an equation in implicit form,
F (x, y, t) =0, then one should take into account in (7.3.2.5) that
n=
F x
F2
x + F y2
, F y
F2
x + F y2
, v=
–F t
F x, –
F t
F y
, (n⋅v) = – 2Ft
F2
x + F y2
For an elliptic domain, specified by equation (7.3.2.8), we have
F (x, y, t) = x
2
a2(t) +
y2
b2(t) –1
Trang 67.3.3 Geometric and Physical Applications of the Double Integral
7.3.3-1 Geometric applications of the double integral
1 Area of a domain D in the x, y plane:
S=
D dx dy.
2 Area of a surface defined by an equation z = f (x, y) with (x, y)D(the surface is
projected onto a domain D in the x, y plane):
S =
D
∂f
∂x
2 +∂f
∂y
2 +1 dx dy
3 Area of a surface defined parametrically by equations x = x(u, v), y = y(u, v),
z = z(u, v), with (u, v)D1:
S=
D1
√
EG – F2du dv
Notation used:
E=
∂x
∂u
2 +
∂y
∂u
2 +
∂z
∂u
2 ,
G=
∂x
∂v
2 +
∂y
∂v
2 +
∂z
∂v
2 ,
F = ∂x
∂u
∂x
∂v + ∂y
∂u
∂y
∂v + ∂z
∂u
∂z
∂v
4 Area of a surface defined by a vector equation r = r(u, v) = x(u, v)i + y(u, v)j +
z (u, v)k, with (u, v)D1:
S =
D1
n (u, v)du dv.
Here, the unit normal is calculated as n(u, v) = r u ×r
v.
Remark The formulas from Items 3 and 4 are equivalent—they define one and the same surface in two forms, scalar and vector, respectively.
5 Calculation of volumes If a domain U of the three-dimensional space is defined by
5
(x, y)D , f (x, y)≤z≤g (x, y)6
, where D is a domain in the x, y plane, the volume of U
is calculated as
V =
D
g (x, y) – f (x, y)
dx dy
The three-dimensional domain U is a cylinder with base D bounded by the surface z = f (x, y) from below and the surface z = g(x, y) from above The lateral surface of this body consists
of segments of straight lines parallel to the z-axis.
Trang 77.3.3-2 Physical applications of the double integral.
Consider a flat plate that occupies a domain D in the x, y plane Let γ(x, y) be the surface density of the plate material (the case γ = const corresponds to a homogeneous plate).
1 Mass of a flat plate:
m=
D γ (x, y) dx dy.
2 Coordinates of the center of mass of a flat plate:
xc = 1
m
D xγ (x, y) dx dy, yc =
1
m
D yγ (x, y) dx dy,
where m is the mass of the plate.
3 Moments of inertia of a flat plate about the coordinate axes:
I x=
D y
2γ (x, y) dx dy, I
y =
D x
2γ (x, y) dx dy.
The moment of inertia of the plate about the origin of coordinates is calculated as I0= I x +I y
7.3.4 Definition and Properties of the Triple Integral
7.3.4-1 Definition of the triple integral
Let a function f (x, y, z) be defined in a domain U of the three-dimensional space Let us break up U into n subdomains (cells) that do not have common internal points Denote by
λ = λ( U n ) the diameter of the resulting partition U n, i.e., the maximum of the cell diameters (the diameter of a domain in space is the diameter of the minimal sphere enclosing the
domain) Select an arbitrary point, (x i , y i , z i ), i =1, 2, , n, in each cell and make up an integral sum
s n=
n
i=1
f (x i , y i , z i)ΔV i
where ΔV i is the volume of the ith cell If there exists a finite limit of the sums s n
as λ( U n) → 0 that depends on neither the partition U n nor the selection of the points
(x i , y i , z i ), then it is called the triple integral of the function f (x, y, z) over the domain U
U f (x, y, z) dx dy dz = lim λ→0s n.
7.3.4-2 Properties of the triple integral
The properties of triple integrals are similar to those of double integrals
1 Linearity If functions f (x, y, z) and g(x, y, z) are integrable in a domain U , then
U
af (x, y, z) bg(x, y, z)
dx dy dz
= a
U f (x, y, z) dx dy dz b
U g (x, y, z) dx dy dz,
where a and b are any numbers.
...Remark The formulas from Items and are equivalent—they define one and the same surface in two forms, scalar and vector, respectively.
5 Calculation of volumes If a domain U of the... corresponds to the special case of (7.3.2.6) with X(λ, t) = a(t) cos λ and Y (λ, t) = b(t) sin λ Taking
into account the aforesaid and using formula (7.3.2.7), we obtain... (7.3.2.5)
where L(t) is the boundary of the domain D(t), n is outer unit normal to L(t), and v is the velocity of motion of the points of L(t).
2◦