Кратные интегралы. Учебное пособие
.pdfМИНИСТЕРСТВО НАУКИ И ВЫСШЕГО ОБРАЗОВАНИЯ РОССИЙСКОЙ ФЕДЕРАЦИИ
Белгородский государственный технологический университет им. В. Г. Шухова
Кратные интегралы
Утверждено ученым советом университета в качестве учебного пособия для студентов младших курсов технических направлений и специальностей
Белгород
2023
УДК 517.37 ББК 22.14
K78
Рецензенты:
Кандидат филологических наук, доцент Белгородского государственного национального исследовательского университета
(НИУ БелГУ) С.А. Алешкевич
Кандидат технических наук, доцент Белгородского государственного технологического университета им. В.Г. Шухова Г.Л. Окунева
Кратные интегралы: учебное пособие/ В. И. Дюкарева, K78 О. С. Дуганова, Ю. С. Некрасова, С. В. Рябцева – Белгород:
Изд-во БГТУ, 2023. – 59 с.
ISBN 978-5-361-01221-3
Учебное пособие подготовлено в соответствии с требованиями Федерального государственного стандарта высшего образования и рабочей программы дисциплины «Математика», содержит основные теоретические положения. Изложение сопровождается многочисленными обстоятельно разобранными примерами.
Издание предназначено для студентов младших курсов технических направлений и специальностей, изучающих математику на английском языке.
Данное издание публикуется в авторской редакции.
The manual was prepared with respect to the requirements of the Federal state higher education standard and the working program of a discipline «Mathematics». It contains the basic theoretical concepts. The presentation is accompanied by numerous thoroughly analyzed examples. Edition is for для undergraduate students of technical fields and specialties studying mathematics in English.
The manual is published in author reduction.
УДК 517.37 ББК 22.14
ISBN 978-5-361-01221-3 © Белгородский государственный технологический университет (БГТУ) им. В.Г. Шухова, 20223
INTRODUCTION
The manual is prepared with respect to the Federal State Standard of Higher Professional Education requirements.
The suggested manual is written on the base of the lectures performed by the authors during the last years Belgorod State Technological
University named after V.G. Shukhov. It corresponds to the requirements imposed by the standard for the acquisition of competencies approved by the competence plan. The theory is presented in modern language in an easy-to-understand form.
The structure of the manual is designed in strict accordance with the logic of the presentation of the material: the double integral, the triple integral, practices.
In addition to the theoretical material, the manual contains examples of the problems solving on topics and can be used both when studying a lecture course, in practical classes, and for students' independent work. Study manual is designed for a wide audience: students, teachers, graduate students, researchers.
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1.THE DOUBLE INTEGRAL
1.1. The Basic Concepts and Definitions
The double integral is the generalization of the integral to the case of a function of two variables.
Let a continuous function = ( ; ) be given in the closed domain D of a coordinate plane Oxy. Let’s split the domain D by n elementary domains( = 1, , ). We denote the areas of such domains by ∆ and the diameters (the largest distance between the points of the domain) by (fig. 1.1).
Fig. 1.1
Let’s choose an arbitrary multiply the value of a function the sum of such products
point ( , ) in each domain . We ( ; ) at such point by ∆ and constitute
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f (x ; y ) S |
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(1.1)
This sum is called the integral sum of a function ( ; ) in the domain the domain D.
Let’s consider the limit of an integral sum (1.1) when n tends to the infinity in such way that maximal diameter tends to zero (max → 0). If such limit exists and doesn’t depend on the way of the domain D dividing by parts and of the chois of the arbitrary points in them then it is called the double integral of a function ( ; ) in the domain D. We denote it as
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(or f (x; y)dS ). So, the double integral is defined by |
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the equality |
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f (x; y)dxdy = lim |
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(1.2)
Here f(x;y) is the integrable in the domain D function; D is the area of integrating; x and y are the variables of integrating ; dxdy (or dS)
is the area element.
Theorem 1.1. Sufficient condition for the integrability of the function. If the function = ( ; ) is continuous in the closed domain D then it is integrable in this domain.
Remarks.
1.The double integral can exist not only for continuous functions but in our course, we will deal only with the functions continuous in the domain of integrating.
2.It follows from the definition of the double integral that for integrable in the domain D function the integral sum limit exists and doesn’t depend on the way of the dividing of the domain by parts. So, we can divide the domain using parallel straight lines (fig.1.2).
Fig. 1.2
Therefore, we get elementary rectangles with an are ∆ = ∆ formula (1.2) can be written in view
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∙ ∆ and the
(1.3)
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1.2. The Geometrical Meaning of the Double Integral
Let’s consider a body limited by a surface = ( ; ) ≥ 0 on the top, by the closed domain D of a coordinate plane Oxy on the bottom, by the cylindric surface (the forming cylinders are parallel to Oz axis) from the sides and the directing is the bound of the domain D. So, we deal with the cylindric body. To find its volume we divide the domain D (projection of a surface = ( ; ) onto the plane Oxy) in arbitrary way by n elementary domains , the areas of which are equal to ∆ ( = 1, , ). Let’s consider the cylindrical columns with the bases limited by the pieces of the surface = ( ; ) (one of them is highlighted on a fig. 1.3)
Fig. 1.3
The totality of these columns is the body V. We denote the volume of a column with the base by ∆ and get = ∑ =1 ∆ . Let’s take an arbitrary point ( , ) in each domain and substitute each column by the straight cylinder with the similar base and the height = ( ; ). The volume of such cylinder is approximately equal to the volume ∆ of a cylindric
column that is ∆ ≈ ( |
; ) ∙ ∆ . At result, we get |
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The larger the number n and the smaller the diameter of the elementary area, the more accurate the equality (1.4). Evidently, we suppose that the number of the domains increases indefinitely ( → ∞) and each elementary domain
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shrinks to a point (max → 0). So, the volume can be expressed by the formula
V = lim |
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or according to (1.3) |
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(1.5)
So, the value of the double integral from nonnegative function is equal to the volume of the cylindric body. This is the geometrical meaning of the double integral.
1.3. The Physical Meaning of the Double Integral
Let’s find the mass of the flat plate taking into account that the surface density = ( ; ) is a continuous function of the coordinates of a point ( ; ). We split the flat plate D by n elementary parts ( = 1, , ) the
areas of which we denote by ∆ . Let’s take an arbitrary point ( , ) in each domain and calculate the density in it ( ; ). If the domains are
enough small then the density in each point ( ; ) differs slightly from
the value ( |
; ). When suppose approximately that the density in each |
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is = ∑ |
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≈ ( ; ) ∙ ∆ . The |
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Therefore |
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The accurate mass value we get as the limit of sum (1.6) when → ∞ and |
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m = lim (x ; y ) S |
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or according to (1.3)
m = (x; y)dxdy. D
(1.7)
So, the double integral of a function = ( ; ) numerically equals to the mass of the flat plate when suppose that ( ; ) is the density of this flat plate at a point ( ; ). This is the physical meaning of the double integral.
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1.4. The Main Properties of the Double Integral
The properties of the double integral are similar to the properties of a single variable function definite integral. We list the main properties of the double integral considering that the integral functions are integrable.
1.The constant c can be taken out of the double integral sign
c f (x; y)dxdy = c f (x; y)dxdy
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The double integral of an algebraic sum of two functions is equal to |
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( f (x; y) g(x; y))dxdy = f (x; y)dxdy g(x; y)dxdy. |
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If we divide the initial domain D by two domains D1 and D2 such |
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them line (fig. 1.4) then |
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f (x; y)dxdy = f (x; y)dxdy + f (x; y)dxdy
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Fig. 1.4
4.If in any domain D an inequality ( ; ) ≥ 0 takes place then
f (x; y)dxdy 0. D
If the functions ( ; ) and ( ; ) satisfy the inequality ( ; ) ≥ ( ; ) in the domain D then
f (x; y)dxdy g(x; y)dxdy.
D D
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5.We know that
S
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i=1n
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dS D
= S
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6. If the function ( ; ) is continuous in the closed domain D the area of which is S then
mS f (x; y)dxdy MS D
where m is the smallest value of the integral function and M is the largest value of the integral function in the domain D.
7. If the function ( ; ) is continuous in the closed domain D the area of which is S then in this domain a point (0; 0) exists such that
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The quantity f (x0 ; y0 ) = 1 f (x; y)dxdy is the average value
S D
of the function ( ; ) in the domain D.
1.5.The Calculation of the Double Integral in the Cartesian
Coordinates
Let’s show that the calculation of a double integral is reduced to the sequential calculation of two definite integrals. We find the double integral
D
f
(x; y)dxdy,
where ( ; ) ≥ 0 is continuous in the domain D. So,
according to the geometrical meaning the double integral expresses the volume of the cylindric body limited by the surface = ( ; ) on the top.
Let’s find this volume applying the parallel sections method. The volume of the cylindric body can be expressed by the definite integral
b V = S(x)dx,
a
where ( ) is an area of the section of a body by the plane perpendicular to Ox axis; = and = are the equations for the planes bound a given body.
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Let us first assume that the domain D is a curved trapezoid bound by the
straight lines = and = and the curves = 1( ) and |
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1.5). Such domain is called the correct relatively to the direction of Oy axis. Any straight line parallel to Oy axis intersects the domain bound no more than in two points. Let’s construct the section of the cylindric body by the plane perpendicular to Ox axis ( = , [ ; ]).
Fig. 1.5
In the section we the curved trapezoid ABCD bound by the lines = ( ; ) where = , = 0, = 1( ) and = 2( ) (fig. 1.6).
Fig. 1.6
An area ( ) of this trapezoid we find applying the definite integral
S(x) = 2 ( x) f (x; y)dy.
1 ( x)
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