Multiple integrals (problems and exercises). Multiple integrals Coordinates of the center of mass of a plane figure

Def . Let ,
,

.

A set is called a closed interval or a closed bar in .

The set is called an open interval

or an open beam in .

Def . Measure of intervals And the quantity is called:

(More precisely
).

Def . If
such that
then the interval is called degenerate and
.

Properties of the gap measure:

A). Positivity:
, and
then and only when – degenerate.

b). Positive homogeneity: .

V). Additivity:

* For
such that
;

* For
And

.

G). Monotonicity of the measure: .

Def . The diameter of the beam (gap) is the value:

Note that
And
– this is not the same thing. For example, if – degenerate, then
,a
(generally speaking).

Wherein: * ;

* ;*
.

Def . Totality
subspans of the interval called interval partition , If: *;

*
; *
; *
; *
.

Magnitude
called the partition parameter P(wherein
).

Def . Splitting called partition refinement , if all elements of the partition obtained by partitioning the partition elements .

Indicated by:
. Reads: smaller or larger .

For the ratio “larger – smaller” the following is true:

*. transitivity – ; *.
;

*.


; *.

|
.

§. Definition of multiple integral

Let
– timber (gap) in ,
– partitioning the gap I. At each interval of the partition mark the point
.

We get
partition with marked points for
.

Magnitude
is called the Riemann integral sum for the function f (x) on the interval I by partition with marked points
.

Def :
=
=
.

Designating – many functions integrated on the beam I let's write:

Def : ε > 0 δ>0<.

If for the function f(x) on I and partitions
- denote by
– the largest and smallest value of the function f(x) on I k then the values
=
And
=
are called lower and upper Darboux sums.

§. Darboux criterion for the existence of a multiple integral.

T 0 . To function
was integrated on the beam (those.
) is necessary and sufficient so that

. Δ▲.

The integration of a function over a beam in Euclidean space is defined. How can one integrate a function over an arbitrary bounded set from Euclidean space?

Let us define the integral of the function f by many
.

Def : Let
And
– limited, i.e.
. Function
we call the characteristic function of the set M.

Then:
≡
.

The definition of a set integral does not depend on which beam containing M selected, i.e.

.

This means that the definition of the integral over a set is correct.

A necessary condition for integrability. To function f(x) on M be integrable it is necessary that f(x) was limited to M. Δ▲.

§. Properties of multiple integrals.

1  . Linearity: Many R M functions integrable on a set M – linear

space, and
– linear functional.

2  . Normalization condition:
. Another form of entry
essentially determines the measure of an arbitrary set from Euclidean space.

3  . If an integral over a set of Lebesgue measure zero exists, then it

equal to zero.

Note: A bunch of M is called the set of Lebesgue measure zero,

If

such that
And
.

4  . A.;b.;

V. If
And – separated from zero by M, That

5  .
And f=g p.v. (almost everywhere) on M, That
.

6  . Additivity: If
And
That

,

In general:
.

Δ. It follows from the equality: ▲

7  . Monotone:
And
That
.

8  . Integrating inequalities: if
ito

.

9  . Let


. In order to
, it is necessary and sufficient for there to be an interior point of the set M, wherein f (x) > 0 and continuous.

10  . Integrability of the integrable function module:
.

11  . Mean value theorem:
,
on M preserves the sign and
, That

.

If the set M– coherent and f(x) – continuous on
That
such that
.

12  . In order for the integral of a non-negative function to be equal to 0

necessary and sufficient to f(x) = 0 almost everywhere on M.

13  . Fubini's theorem. For double integral:

Let the area
- rectangle:. Then, provided that internal single integrals exist, to find the double integral, you can proceed to repeated integration (see Fig. a):

, or

E

Note: The external limits of integration must be constants; the internal limits of integration may depend on the variable over which integration is yet to be performed.

Formula (*) can be obtained using the set characteristic function D.

For multiple integral:

Let and some subsets of Euclidean spaces And . Let us define the Cartesian product of these sets, which is a subset of the Euclidean space
:.

Then Fubini's theorem for
has the form:
.

The theorem is also valid for beams X And Y, and for more complex configurations.

Examples:

1 0 . Calculate
, if the boundary of the area
given by the equations:

A).
;

2

–

.

Recipe: When setting integration limits in a double integral, it is recommended to start with the outer integration limits.

3

Passing to iterated integrals gives:
.

At the same time, in a triple integral, the placement of limits must begin with the internal limits of integration. Then project the area V to the plane xOy

setting limits in the area D– lying in a plane xOy.

4 0 . Change the order of integration in the iterated integral:
.

Multiple integral

integral of a function specified in some area on the plane, in three dimensions or n-dimensional space. Among K. and. distinguish between double integrals, triple integrals, etc. n-multiple integrals.

Let the function f(x, y) is given in some area D plane xOy. Let's split the area D on n partial areas d i, whose areas are equal s i , choose in each area d i point ( ξi, ηi) (cm. rice. ) and compose the integral sum

If, with an unlimited reduction in the maximum diameter of partial areas d i amounts S have a limit regardless of the choice of points ( ξi, ηi), then this limit is called the double integral of the function f(x, y) by region D and denote

The triple integral is defined similarly and, in general, n-multiple integral.

For the existence of a double integral it is sufficient, for example, that the region D was a closed squarable region (See Squarable region), and the function f(x, y) was continuous in D. K. and. have a number of properties similar to the properties of simple Integrals . To calculate K. and. usually lead it to an iterated integral (See Iterated integral). In special cases for K.’s information and. Green's formula and Ostrogradsky's formula can serve as integrals of lower dimension. K. and. have extensive applications: they are used to express the volumes of bodies, their masses, static moments, moments of inertia, etc.

Great Soviet Encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .

See what “Multiple integral” is in other dictionaries:

Integral of a function of several variables. It is determined using integral sums, similar to the definite integral of a function of one variable (see Integral calculus). Depending on the number of variables, there are double, triple, n... ... Big Encyclopedic Dictionary

Definite integral of a function of several variables. There are various concepts of K. and. (Riemann integral, Lebesgue integral, Lebesgue Stieltjes integral, etc.). The multiple Riemann integral is introduced on the basis of the Jordan measure. Let E be Jordan measurable... ... Mathematical Encyclopedia

In mathematical analysis, a multiple or multiple integral is a set of integrals taken from variables. For example: Note: a multiple integral is a definite integral; its calculation always results in a number. Contents 1... ...Wikipedia

Integral of a function of several variables. It is determined using integral sums, similar to the definite integral of a function of one variable (see Integral calculus). Depending on the number of variables, there are double, triple, n... ... encyclopedic Dictionary

Integral of a function of several variables. Determined using integral sums, similarly defined. integral of a function of one variable (see Integral calculus). Depending on the number of variables, there are double, triple, i... ... Natural science. encyclopedic Dictionary

Note: everywhere in this article where the sign is used, the (multiple) Riemann integral is meant, unless otherwise stated; Everywhere in this article where we talk about the measurability of a set, we mean Jordanian measurability, if not... ... Wikipedia

A multiple integral of the form where, which is the average value of the degree 2k of the modulus of a trigonometric sum. Vinogradov's theorem on the value of this integral, the mean value theorem, underlies estimates of Weyl sums. Literature Vinogradova inte... Wikipedia

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An integral in which integration over different variables is sequentially performed, i.e., an integral of the form (1) The function f(x, y) is defined on the set A lying in the direct product XX Y of the spaces X and Y, in which s are given finite measures mx and my,… … Mathematical Encyclopedia

An integral taken along any curve on a plane or in space. There are K. and. 1st and 2nd types. K. and. Type 1 arises, for example, when considering the problem of calculating the mass of a variable density curve; it is designated... ... Great Soviet Encyclopedia

Caution: When calculating improper integrals with singular points inside the integration interval, you cannot mechanically apply the Newton–Leibniz formula, since this can lead to errors.

General rule: The Newton–Leibniz formula is correct if the antiderivative of f(x) at the singular point of the latter is continuous.

Example 2.11.

Let us consider an improper integral with a singular point x = 0. The Newton–Leibniz formula, applied formally, gives

However, the general rule does not apply here; for f(x) = 1/x the antiderivative ln |x| is not defined at x = 0 and is infinitely large at this point, i.e. is not continuous at this point. It is easy to verify by direct verification that the integral diverges. Really,

The resulting uncertainty can be revealed in different ways as e and d tend to zero independently. In particular, setting e = d, we obtain the principal value of the improper integral equal to 0. If e = 1/n, and d =1/n 2, i.e. d tends to 0 faster than e, then we get

when and vice versa,

those. the integral diverges.n

Example 2.12.

Let us consider an improper integral with a singular point x = 0. The antiderivative of the function has the form and is continuous at the point x = 0. Therefore, we can apply the Newton–Leibniz formula:

A natural generalization of the concept of a definite Riemann integral to the case of a function of several variables is the concept of a multiple integral. For the case of two variables, such integrals are called double.

Consider in two-dimensional Euclidean space R´R, i.e. on a plane with a Cartesian coordinate system, a set E final area S.

Let us denote by ( i = 1, …, k) set partition E, i.e. such a system of its subsets E i, i = 1,. . ., k, that Ø for i ¹ j and (Fig. 2.5). Here we denote the subset E i without its border, i.e. internal points of the subset E i , which, together with its boundary Gr E i form a closed subset E i, . It is clear that the area S(E i) subsets E i coincides with the area of ​​its interior, since the area of ​​the boundary GrE i is equal to zero.

Let d(E i) denote set diameter E i, i.e. the maximum distance between two of its points. The quantity l(t) = d(E i) will be called fineness of partition t. If the function f(x),x = (x, y), is defined on E as a function of two arguments, then any sum of the form

X i О E i , i = 1, . . . , k, x i = (x i , y i),

depending both on the function f and the partition t, and on the choice of points x i О E i М t, is called integral sum of the function f .

If for a function f there exists a value that does not depend on either the partitions t or the choice of points (i = 1, ..., k), then this limit is called double Riemann integral from f(x,y) and is denoted

The function f itself is called in this case Riemann integrable.

Recall that in the case of a function with one argument as a set E over which integration is performed, the segment is usually taken , and its partition t is considered to be a partition consisting of segments. In other respects, as is easy to see, the definition of the double Riemann integral repeats the definition of the definite Riemann integral for a function of one argument.

The double Riemann integral of bounded functions of two variables has the usual properties of a definite integral for functions of one argument – linearity, additivity with respect to the sets over which integration is performed, preservation when integrating non-strict inequalities, product integrability integrated functions, etc.

The calculation of multiple Riemann integrals reduces to the calculation iterated integrals. Let us consider the case of the double Riemann integral. Let the function f(x,y) is defined on the set E lying in the Cartesian product of the sets X ´ Y, E М X ´ Y.

By repeated integral of the function f(x, y) is called an integral in which integration is sequentially performed over different variables, i.e. integral of the form

Set E(y) = (x: О E) М X is called cross section sets E corresponding to a given y, y О E y ; the set E y is called – projection set E on the Y axis.

For the iterated integral, the following notation is also used:

which, like the previous one, means that first, for a fixed y, y О E y , the function is integrated f(x, y) By x along the segment E(y), which is a section of the set E corresponding to this y. As a result, the inner integral defines some function of one variable - y. This function is then integrated as a function of one variable, as indicated by the outer integral symbol.

When changing the order of integration, we obtain a repeated integral of the form

where internal integration is carried out over y, and external - by x. How does this iterated integral relate to the iterated integral defined above?

If there is a double integral of the function f, i.e.

then both repeated integrals exist, and they are identical in magnitude and equal to double, i.e.

We emphasize that the condition formulated in this statement for the possibility of changing the order of integration in iterated integrals is only sufficient, but not necessary.

Other sufficient conditions the possibilities of changing the order of integration in iterated integrals are formulated as follows:

if at least one of the integrals exists

then the function f(x, y) Riemann integrable on the set E, both repeated integrals of it exist and are equal to the double integral. n

Let us specify the notation of projections and sections in the notation of iterated integrals.

If the set E is a rectangle

That E x = (x: a £ x £ b), E y = (y: c £ y £ d); wherein E(y) = E x for any y, y О E y . , A E(x) = Ey for any x , x О E x ..

Formal entry: " y y О E yÞ E(y) = ExÙ" x x О E xÞ E(x) = Ey

If the set E has curved border and allows representations

In this case, the repeated integrals are written as follows:

Example 2.13.

Calculate the double integral over a rectangular area, reducing it to iterative.

Since the condition sin 2 (x+ y) =| sin 2 (x + y)|, then checking the satisfiability of sufficient conditions for the existence of the double integral I in the form of the existence of any of the repeated integrals

there is no need to carry out this specifically and you can immediately proceed to calculating the repeated integral

If it exists, then the double integral also exists, and I = I 1 . Because the

So I = .n

Example 2.14.

Calculate the double integral over the triangular region (see Fig. 2.6), reducing it to repeated

Gr(E) = ( : x = 0, y = 0, x + y = 2).

First, let us verify the existence of the double integral I. To do this, it is enough to verify the existence of the repeated integral

those. the integrands are continuous on the intervals of integration, since they are all power functions. Therefore, the integral I 1 exists. In this case, the double integral also exists and is equal to any repeated one, i.e.

Example 2.15.

To better understand the connection between the concepts of double and iterated integrals, consider the following example, which may be omitted on first reading. A function of two variables f(x, y) is given

Note that for fixed x this function is odd in y, and for fixed y it is odd in x. As the set E over which this function is integrated, we take the square E = ( : -1 £ x £ 1, -1 £ y £ 1 ).

First we consider the iterated integral

Inner integral

is taken for fixed y, -1 £ y £ 1. Since the integrand for fixed y is odd in x, and integration over this variable is carried out over the segment [-1, 1], symmetrical with respect to point 0, then the internal integral is equal to 0. Obviously, that the outer integral over the variable y of the zero function is also equal to 0, i.e.

Similar reasoning for the second iterated integral leads to the same result:

So, for the function f(x, y) under consideration, repeated integrals exist and are equal to each other. However, there is no double integral of the function f(x, y). To see this, let us turn to the geometric meaning of calculating repeated integrals.

To calculate the iterated integral

a special type of partition of the square E is used, as well as a special calculation of integral sums. Namely, square E is divided into horizontal stripes (see Fig. 2.7), and each strip is divided into small rectangles. Each strip corresponds to a certain value of the variable y; for example, this could be the ordinate of the horizontal axis of the strip.

The calculation of integral sums is carried out as follows: first, the sums are calculated for each band separately, i.e. at fixed y for different x, and then these intermediate sums are summed for different bands, i.e. for different y. If the fineness of the partition tends to zero, then in the limit we obtain the above-mentioned repeated integral.

It is clear that for the second iterated integral

the set E is divided into vertical stripes corresponding to different x. Intermediate sums are calculated within each band in small rectangles, i.e. along y, and then they are summed for different bands, i.e. by x. In the limit, when the fineness of the partition tends to zero, we obtain the corresponding iterated integral.

To prove that a double integral does not exist, it is enough to give one example of a partition, the calculation of the integral sums for which, in the limit when the fineness of the partition tends to zero, gives a result different from the value of the repeated integrals. Let us give an example of such a partition corresponding to the polar coordinate system (r, j) (see Fig. 2.8).

In the polar coordinate system, the position of any point on the plane M 0 (x 0 , y 0), where x 0 , y 0 are the Cartesian coordinates of the point M 0, is determined by the length r 0 of the radius connecting it to the origin and the angle j 0 formed by this radius with a positive x-axis direction (the angle is counted counterclockwise). The connection between Cartesian and polar coordinates is obvious:

y 0 = r 0 × sinj 0 .

The partition is constructed as follows. First, square E is divided into sectors with radii emanating from the center of coordinates, and then each sector is divided into small trapezoids by lines perpendicular to the sector axis. The calculation of integral sums is carried out as follows: first along small trapezoids inside each sector along its axis (along r), and then across all sectors (along j). The position of each sector is characterized by the angle of its axis j, and the length of its axis r(j) depends on this angle:

if or , then ;

if , then ;

if , then

if , then .

Passing to the limit of the integral sums of a polar partition when the fineness of the partition tends to zero, we obtain a representation of the double integral in polar coordinates. Such a notation can be obtained in a purely formal way, replacing the Cartesian coordinates (x, y) with polar ones (r, j).

According to the rules of transition in integrals from Cartesian to polar coordinates, one should write, by definition:

In polar coordinates, the function f(x, y) will be written as follows:

Finally we have

Inner integral (improper) in the last formula

where the function r(j) is indicated above, 0 £ j £ 2p , is equal to +¥ for any j, because

Therefore, the integrand in the outer integral evaluated over j is not defined for any j. But then the outer integral itself is not defined, i.e. the original double integral is not defined.

Note that the function f(x, y) does not satisfy the sufficient condition for the existence of a double integral over the set E. Let us show that the integral

does not exist. Really,

Similarly, the same result is established for the integral

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Lectures 5-6

Topic2. Multiple integrals.

Double integral.

Control questions.

1. Double integral, its geometric and physical meaning

2. Properties of double integral.

3. Calculation of the double integral in Cartesian coordinates.

4. Change of variables in the double integral. Calculation of double integral in polar coordinates.

Let the function z = f (x , y) defined in a limited closed region D plane. Let's split the area D randomly on n elementary closed areas  1 , … , n, having areas   1 , …,   n and diameters d 1 , …, d n respectively. Let's denote d largest of the area diameters  1 , … ,  n. In every area  k choose an arbitrary point P k (x k ,y k) and compose integral sum functions f(x,y)

S =
(1)

Definition. Double integral functions f(x,y) by region D called the limit of the integral sum

, (2)

if it exists.

Comment. Cumulative sum S depends on how the area is divided D and selecting points P k (k=1, …, n). However, the limit
, if it exists, does not depend on how the area is partitioned D and selecting points P k .

A sufficient condition for the existence of a double integral. Double integral (1) exists if the function f(x,y) continuous in D except for a finite number of piecewise smooth curves and is limited in D. In what follows we will assume that all double integrals under consideration exist.

Geometric meaning of the double integral.

If f(x,y) ≥0 in area D, then double integral (1) is equal to the volume of the “cylindrical” body shown in the figure:

V =
(3)

The cylindrical body is limited below by the region D, from above - part of the surface z = f (x , y), from the sides - by vertical straight segments connecting the boundaries of this surface and region D.

Physical meaning of the double integral. Mass of a flat plate.

Let a flat plate be given D with a known density function γ( X,at), then breaking plate D into parts D i and choosing arbitrary points
, we obtain for the mass of the plate
, or, comparing with formula (2):

(4)

4. Some properties of the double integral.

Linearity. If WITH is a numerical constant, then

Additivity. If the area D “broken” into areas D 1 And D 2, then

3) Area of ​​the limited area D equal to

(5)

Calculation of double integral in Cartesian coordinates.

Let the area be given

Picture 1

D= { (x , y ): a ≤ x ≤ b , φ 1 (x ) ≤ y≤ φ 2 (x ) } (6)

Region D enclosed in a strip between straight lines x = a , y = b, bounded from below and above, respectively, by curves y = φ 1 (x ) And y = φ 2 (x ) .

Double integral (1) over a region D(4) is calculated by passing to the iterated integral:

(7)

This iterated integral is calculated as follows. First, the inner integral is calculated

by variable y, wherein x considered constant. The result will be a function of the variable x, and then the “outer” integral of this function over the variable is calculated x .

Comment. The process of transition to the repeated integral according to formula (7) is often called the placement of integration limits in the double integral. When setting integration limits, you need to remember two points. Firstly, the lower limit of integration should not exceed the upper one, and secondly, the limits of the outer integral should be constant, and the inner one should, in the general case, depend on the integration variable of the outer integral.

Let now the area D looks like

D= { (x , y ) : c ≤ y ≤ d , ψ 1 (y ) ≤ x ≤ ψ 2 (y ) } . (8)

Then

. (9)

Let's assume that the area D can be represented as (6) and (8) simultaneously. Then the equality holds

(10)

The transition from one iterated integral to another in equality (10) is called changing the order of integration in double integral.

Examples.

1) Change the order of integration in the integral

Solution. Using the form of the iterated integral, we find the region

D= { (x , y ): 0 ≤ x ≤ 1, 2 x ≤ y≤ 2 } .

Let's depict the area D. From the figure we see that this area is located in a horizontal strip between the straight lines y =0, y=2 and between lines x =0 And x=D

Sometimes, to simplify calculations, a change of variables is made:

,
(11)

If functions (11) are continuously differentiable and the determinant (Jacobian) is nonzero in the domain under consideration:

(12)

Ministry of Education and Science of the Russian Federation

Course work

Discipline: Higher mathematics

(Fundamentals of Linear Programming)

On the topic: MULTIPLE INTEGRALS

Completed by: ______________

Teacher:___________

Date ___________________

Grade _________________

Signature ________________

VORONEZH 2008

1 Multiple integrals

1.1 Double integral

1.2 Triple integral

1.3 Multiple integrals in curvilinear coordinates

1.4 Geometric and physical applications of multiple integrals

2 Curvilinear and surface integrals

2.1 Curvilinear integrals

2.2 Surface integrals

2.3 Geometric and physical applications

Bibliography

1 Multiple integrals

1.1 Double integral

Let us consider a closed region D in the Oxy plane, bounded by line L. Let us divide this region into n parts by some lines

Let a function z = f(x, y) be given in domain D. Let us denote by f(P 1), f(P 2),…, f(P n) the values ​​of this function at selected points and compose a sum of products of the form f(P i)ΔS i:

called the integral sum for the function f(x, y) in the domain D.

If there is the same limit of integral sums (1) for

Calculation of the double integral over the region D bounded by lines

1.2 Triple integral

The concept of a triple integral is introduced by analogy with a double integral.

Let a certain region V be given in space, bounded by a closed surface S. Let us define a continuous function f(x, y, z) in this closed region. Then we divide the region V into arbitrary parts Δv i, considering the volume of each part equal to Δv i, and compose an integral sum of the form

Limit at

The triple integral of the function f(x,y,z) over the region V is equal to the triple integral over the same region:

1.3 Multiple integrals in curvilinear coordinates

Let us introduce curvilinear coordinates on the plane, called polar. Let us select point O (pole) and the ray emanating from it (polar axis).

Rice. 2 Fig. 3

The coordinates of point M (Fig. 2) will be the length of the segment MO - the polar radius ρ and the angle φ between the MO and the polar axis: M(ρ,φ). Note that for all points of the plane, except the pole, ρ > 0, and the polar angle φ will be considered positive when measured in a counterclockwise direction and negative when measured in the opposite direction.

The relationship between the polar and Cartesian coordinates of point M can be set by aligning the origin of the Cartesian coordinate system with the pole, and the positive semi-axis Ox with the polar axis (Fig. 3). Then x=ρcosφ, y=ρsinφ. From here

Let us define in the region D bounded by the curves ρ=Φ 1 (φ) and ρ=Φ 2 (φ), where φ 1< φ < φ 2 , непрерывную функцию z = f(φ, ρ) (рис. 4).

In three-dimensional space, cylindrical and spherical coordinates are introduced.

The cylindrical coordinates of the point P(ρ,φ,z) are the polar coordinates ρ, φ of the projection of this point onto the Oxy plane and the applicate of this point z (Fig. 5).

Fig.5 Fig.6

Formulas for the transition from cylindrical to Cartesian coordinates can be specified as follows:

x = ρcosφ, y = ρsinφ, z = z. (8)

In spherical coordinates, the position of a point in space is determined by the linear coordinate r - the distance from the point to the origin of the Cartesian coordinate system (or the pole of the spherical system), φ - the polar angle between the positive semi-axis Ox and the projection of the point onto the Ox plane, and θ - the angle between the positive semi-axis of the axis Oz and segment OP (Fig. 6). Wherein

Let us set the formulas for the transition from spherical to Cartesian coordinates:

x = rsinθcosφ, y = rsinθsinφ, z = rcosθ. (9)

Then the formulas for transition to cylindrical or spherical coordinates in the triple integral will look like this:

where F 1 and F 2 are functions obtained by substituting their expressions through cylindrical (8) or spherical (9) coordinates into the function f instead of x, y, z.

1.4 Geometric and physical applications of multiple integrals

1) Area of ​​the flat region S:

Example 1.

Find the area of ​​figure D bounded by lines

It is convenient to calculate this area by counting y as an external variable. Then the boundaries of the region are given by the equations