Let us dwell in some detail on Ostrogradsky's work on multiple integrals.
Ostrogradsky's formula for transforming a triple integral into a double one, which we usually write in the form
where div A is the divergence of the field of vector A,
Аn is the scalar product of vector A and the unit vector of the external normal n of the boundary surface; in the mathematical literature it was often previously associated with the names of Gauss and Green.
In fact, in Gauss’s work on the attraction of spheroids, one can see only very special cases of formula (1), for example, with P=x, Q=R=0, etc. As for J. Green, in his work on the theory of electricity and there is no magnetism at all in formula (1); it derives another relationship between triple and double integrals, namely, Green's formula for the Laplace operator, which can be written in the form
Of course, we can derive formula (1) from (2), assuming
and in the same way it is possible to obtain formula (2) from formula (1), but Green did not think of doing this.
where on the left is the integral over the volume, and on the right is the integral over the boundary surface, and these are the direction cosines of the external normal.
Ostrogradsky's Paris manuscripts testify, with complete certainty, that both the discovery and the first message of the integral theorem (1) belong to him. It was first stated and proved, exactly as they are now doing, in the “Proof of a Theorem of the Integral Calculus,” presented to the Paris Academy of Sciences on February 13, 1826, after which it was formulated again in that part of the “Memoir on the Diffusion of Heat in Solids.” ”, which Ostrogradsky presented on August 6, 1827. “Memoir” was given for review to Fourier and Poisson, and the latter certainly read it, as evidenced by the entry on the first pages of both parts of the manuscript. Of course, the idea of attributing to himself the theorem, which he became acquainted with in the work of Ostrogradsky two years before presenting his work on the theory of elasticity, did not even occur to Poisson.
As for the relationship between the works on multiple integrals of Ostrogradsky and Green, we recall that in the “Note on the Theory of Heat” a formula was derived that embraces Green’s own formula as a very special case. The now unusual Cauchy symbolism used by Ostrogradsky in the “Note” until recently hid this important discovery from researchers. Of course, Greene retains the honor of the discovery and first publication in 1828 of the formula for Laplace operators that bears his name.
The discovery of a formula for transforming a triple integral into a double integral helped Ostrogradsky solve the problem of varying an n-fold integral, namely, to derive the general formula for transforming the integral from an expression of the type of divergence over a n-dimensional domain and the integral over the supersurface S bounding it with the equation L(x,y, z,…)=0. If we adhere to the previous notation, then the formula has the form
However, Ostrogradsky did not use the geometric images and terms that we use: the geometry of multidimensional spaces did not yet exist at that time.
In “Memoir on the Calculus of Variations of Multiple Integrals” two more important issues in the theory of such integrals are considered. First, Ostrogradsky derives a formula for changing variables in a multidimensional integral; secondly, for the first time he gives a complete and accurate description of the method of calculating an n-fold integral using n successive integrations over each of the variables within the appropriate limits. Finally, from the formulas contained in this memoir, the general rule of differentiation with respect to the parameter of a multidimensional integral is easily derived, when not only the integrand function, but also the boundary of the integration domain depends on this parameter. The named rule follows from the formulas in the memoir in such a natural way that later mathematicians even identified it with one of the formulas of this memoir.
Ostrogradsky devoted a special work to changing variables in multiple integrals. For the double integral, Euler derived the corresponding rule using formal transformations; for the triple integral, Lagrange derived it. However, although Lagrange's result is correct, his reasoning was not accurate: he seemed to proceed from the fact that the elements of volumes in the old and new variables - coordinates - are equal to each other. Ostrogradsky made a similar mistake at the beginning in the just mentioned derivation of the rule for replacing variables. In the article “On the Transformation of Variables in Multiple Integrals,” Ostrogradsky revealed Lagrange’s error, and also for the first time outlined that visual geometric method for transforming variables in a double integral, which, in a slightly more rigorous form, is also presented in our manuals. Namely, when replacing variables in the integral using formulas, the integration domain is divided by the coordinate lines of two systems u=const, v=const into infinitesimal curvilinear quadrangles. Then the integral can be obtained by first adding up those of its elements that correspond to an infinitely narrow curved strip, and then continuing to sum the elements in stripes until they are all exhausted. A simple calculation gives for the area, which, up to small higher order ones, can be considered as a parallelogram, the expression where, is chosen so that the area is positive. The result is the well-known formula
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
, and the corresponding largest distances between points in each of these parts will be denoted by d 1, d 2, ..., d n. Let us select a point P i in each part.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:
, (1)
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
and , which does not depend either on the method of partitioning the region D into parts or on the choice of points Pi in them, then it is called the double integral of the function f(x, y) over the region D and is denoted
. (2)
Calculation of the double integral over the region D bounded by lines
x = a, x = b(a< b), где φ 1 (х) и φ 2 (х) непрерывны на (рис. 1) сводится к последовательному вычислению двух определенных интегралов, или так называемого двукратного интеграла:
= 
(3)
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
, (4)Limit at
integral sums (11), independent of the method of partitioning the domain V and the choice of points Pi in each subdomain of this domain, is called the triple integral of the function f(x, y, z) over the domain V:
. (5)
The triple integral of the function f(x,y,z) over the region V is equal to the triple integral over the same region:
. (6)
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
, tg. 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).
(7)
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:
, (10)
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:
(11)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
And 
calculated using integration by parts: 
Previously, we proved the properties of a definite integral using its definition as the limit of sums. The basic properties of multiple integrals can be proved in exactly the same way. For simplicity, we will consider all functions to be continuous, so the integrals of them certainly make sense.
I. The constant factor can be taken out of the integral sign, and the integral of the finite sum of functions is equal to the sum of the integrals of the terms:
II. If a region is decomposed into a finite number of parts [for example, into two parts, then the integral over the entire region is equal to the sum of the integrals over all parts:
III. If in the area, then
In particular :
IV. If the sign in region (a) is preserved, then the mean value theorem holds, expressed by the formula
where is some point lying inside the region (a).
In particular, when we get
where is the area of the region.
Similar properties hold for the triple integral. Note that when defining a double and triple integral as the limit of a sum, it is always assumed that the region of integration is finite and the integrand function is in any case limited, that is, there is a positive number A such that at all points N of the region of integration. If these conditions are not met, then the integral can exist as an improper integral in the same way as was the case for a simple definite integral. We will deal with improper multiple integrals in §8.
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:
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) = (
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 = (
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
The concept of double integral
A double integral (DI) is a generalization of a definite integral (DI) of a function of one variable to the case of a function of two variables.
Let a continuous non-negative function $z=f\left(x,y\right)$ be defined in a closed domain $D$ located in the coordinate plane $xOy$. The function $z=f\left(x,y\right)$ describes a certain surface that is projected into the domain $D$. The region $D$ is bounded by a closed line $L$, the boundary points of which also belong to the region $D$. We assume that the line $L$ is formed by a finite number of continuous curves defined by equations of the form $y=\vartheta \left(x\right)$ or $x=\psi \left(y\right)$.
Let us divide the region $D$ into $n$ arbitrary sections of area $\Delta S_(i) $. In each of the sections we choose one arbitrary point $P_(i) \left(\xi _(i) ,\eta _(i) \right)$. At each of these points we calculate the value of the given function $f\left(\xi _(i) ,\eta _(i) \right)$. Let's consider the volume under that part of the surface $z=f\left(x,y\right)$ that is projected into the area $\Delta S_(i) $. Geometrically, this volume can be approximately represented as the volume of a cylinder with a base $\Delta S_(i) $ and height $f\left(\xi _(i) , \eta _(ii) \right)$, that is, equal to the product $f \left(\xi _(i) ,\eta _(i) \right)\cdot \Delta S_(i) $. Then the volume under the entire surface $z=f\left(x,y\right)$ within the region $D$ can be approximately calculated as the sum of the volumes of all cylinders $\sigma =\sum \limits _(i=1)^(n )f\left(\xi _(i) ,\eta _(i) \right)\cdot \Delta S_(i) $. This sum is called the integral sum for the function $f\left(x,y\right)$ in the domain $D$.
Let us call the diameter $d_(i) \left(\Delta S_(i) \right)$ of a section $\Delta S_(i) $ the largest distance between the extreme points of this section. Let $\lambda $ denote the largest of the diameters of all sections from the region $D$. Let $\lambda \to 0$ due to unlimited $n\to \infty $ refinement of the partitioning of the domain $D$.
Definition
If there is a limit of the integral sum $I=\mathop(\lim )\limits_(\lambda \to 0) \sigma $, then this number is called the CI of the function $f\left(x,y\right)$ over the domain $D $ and denote $I=\iint \limits _(D)f\left(x,y\right)\cdot dS $ or $I=\iint \limits _(D)f\left(x,y\right) \cdot dx\cdot dy $.
In this case, the region $D$ is called the region of integration, $x$ and $y$ are the integration variables, and $dS=dx\cdot dy$ is the area element.
From the definition follows the geometric meaning of DI: it gives the exact value of the volume of a certain curvilinear cylinder.
Application of double integrals
Body volume
In accordance with the geometric meaning of DI, the volume $V$ of some body bounded above by the surface $z=f\left(x,y\right)\ge 0$, below by the region $D$ on the plane $xOy$, on the sides by a cylindrical surface , the generators of which are parallel to the $Oz$ axis, and the guide is the contour of the region $D$ (line $L$), is calculated by the formula $V=\iint \limits _(D)f\left(x,y\right)\cdot dx\cdot dy $.
Let the body limit the surface $z=f_(2) \left(x,y\right)$ from above, and the surface $z=f_(1) \left(x,y\right)$ from below, and $f_( 2) \left(x,y\right)\ge f_(1) \left(x,y\right)$. The projection of both surfaces onto the plane $xOy$ is the same region $D$. Then the volume of such a body is calculated using the formula $V=\iint \limits _(D)\left(f_(2) \left(x,y\right)-f_(1) \left(x,y\right)\right )\cdot dx\cdot dy $.
Suppose that in the domain $D$ the function $f\left(x,y\right)$ changes sign. Then, to calculate the volume of the corresponding body, the region $D$ must be divided into two parts: part $D_(1) $, where $f\left(x,y\right)\ge 0$, and part $D_(2) $, where $f\left(x,y\right)\le 0$. In this case, the integral over the region $D_(1) $ will be positive and equal to the volume of that part of the body that lies above the plane $xOy$. The integral over the area $D_(2) $ will be negative and in absolute value equal to the volume of that part of the body that lies below the plane $xOy$.
Area of a flat figure
If we put $f\left(x,y\right)\equiv 1$ everywhere in the region $D$ on the coordinate plane $xOy$, then the CI is numerically equal to the area of the integration region $D$, that is, $S=\iint \limits _(D)dx\cdot dy $. In the polar coordinate system, the same formula takes the form $S=\iint \limits _(D^(*) )\rho \cdot d\rho \cdot d\phi $.
Area of an arbitrary surface
Let some surface $Q$, given by the equation $z=f_(1) \left(x,y\right)$, be projected onto the coordinate plane $xOy$ into the domain $D_(1)$. In this case, the surface area $Q$ can be calculated using the formula $S=\iint \limits _(D_(1) )\sqrt(1+\left(\frac(\partial z)(\partial x) \right)^ (2) +\left(\frac(\partial z)(\partial y) \right)^(2) ) \cdot dx\cdot dy $.
Quantity of substance
Let us assume that in the region $D$ some substance with surface density $\rho \left(x,y\right)$ is distributed on the plane $xOy$. This means that the surface density $\rho \left(x,y\right)$ is the mass of matter per elementary area $dx\cdot dy$ of the $D$ region. Under these conditions, the total mass of the substance can be calculated using the formula $M=\iint \limits _(D)\rho \left(x,y\right)\cdot dx\cdot dy $.
Note that the “substance” can be an electric charge, heat, etc.
Coordinates of the center of mass of a flat figure
The formulas for calculating the coordinate values of the center of mass of a flat figure are as follows:$ $$x_(c) =\frac(\iint \limits _(D)x\cdot \rho \left(x,y\right)\cdot dx\cdot dy )(M) $, $y_(c) =\frac(\iint \limits _(D)y\cdot \rho \left(x,y\right)\cdot dx\cdot dy )(M) $.
The quantities in the numerators are called the static moments $M_(y) $ and $M_(x) $ of the plane figure $D$ about the axes $Oy$ and $Ox$, respectively.
If the flat figure is homogeneous, that is, $\rho =const$, then these formulas are simplified and are expressed not through the mass, but through the area of the flat figure $S$: $x_(c) =\frac(\iint \limits _(D )x\cdot dx\cdot dy )(S) $, $y_(c) =\frac(\iint \limits _(D)y\cdot dx\cdot dy )(S) $.
Moments of inertia of the area of a plane figure
Let us consider a material flat figure on the plane $xOy$. Let us imagine it as a certain region $D$, over which a substance with a total mass $M$ is distributed with a variable surface density $\rho \left(x,y\right)$.
The value of the moment of inertia of the area of a flat figure relative to the $Oy$ axis: $I_(y) \; =\; \iint \limits _(D)x^(2) \cdot \; \rho (x,\; y)\; \cdot dx\; \cdot dy $. The value of the moment of inertia about the $Ox$ axis: $I_(x) \; =\; \iint \limits _(D)y^(2) \cdot \; \rho (x,\; y)\cdot\; dx\; \cdot dy $. The moment of inertia of a flat figure relative to the origin is equal to the sum of the moments of inertia relative to the coordinate axes, that is, $I_(O) =I_(x) +I_(y) $.
Triple integrals are introduced for functions of three variables.
Let us assume that a certain region $V$ of three-dimensional space is given, bounded by a closed surface $S$. We assume that the points that lie on the surface also belong to the region $V$. Suppose that some continuous function $f\left(x,y,z\right)$ is given in the domain $V$. For example, such a function, provided $f\left(x,y,z\right)\ge 0$, can be the volumetric distribution density of some substance, temperature distribution, etc.
Let us divide the region $V$ into $n$ arbitrary parts, the volumes of which are $\Delta V_(i) $. In each part we choose one arbitrary point $P_(i) \left(\xi _(i) ,\eta _(i) ,\varsigma _(i) \right)$. At each of these points we calculate the value of the given function $f\left(\xi _(i) ,\eta _(i) ,\varsigma _(i) \right)$.
Let us form the integral sum $\sum \limits _(i=1)^(n)f\left(\xi _(i) ,\eta _(i) ,\varsigma _(i) \right)\cdot \Delta V_ (i) $ and we will indefinitely refine $\left(n\to \infty \right)$ the division of the region $V$ so that the largest of the diameters $\lambda $ of all parts $\Delta V_(i) $ decreases indefinitely $ \left(\lambda \to 0\right)$.
Definition
Under the above conditions, the limit $I$ of this integral sum exists, is called the triple integral of the function $f\left(x,y,z\right)$ over the domain $V$ and is denoted $I\; =\; \iiiint \limits _(V)f\left(x,y,z\right)\; \cdot dV $ or $I\; =\; \iiiint \limits _(V)f\left(x,y,z\right)\cdot \; dx\cdot\; dy\; \cdot dz$.
