National Research Tomsk Polytechnic University ______
M.R. Cherkasov
MULTIPLE INTEGRALS
Course of Lectures
Tomsk 2009
Contents
MULTIPLE INTEGRALS
Lecture 1. Double Integral
1.1. Definition of a Double Integral 4
1.2. Calculation of a Double Integral 5
Lecture 2.
2.1. Exchange of Variables in a Double Integral 8
2.2. Double Integral in Polar Coordinates 9
2.3. Geometric Applications of a Double Integral 10
2.4. Physical Applications of a Double Integral 12
Lecture 3. Examples 12
Lecture 4. Triple Integral
4.1. Definition of a Triple Integral 19
4.2. Calculation of a Triple Integral 19
4.3. Exchange of Variables in a Triple Integral 21
4.4. Triple Integral in the Cylindrical Coordinates 21
4.5. Triple Integral in the Spherical Coordinates 23
Lecture 5.
5.1. Applications of a Triple Integral 23
5.2. Examples 25
LINE INTEGRALS
Lecture 6. Line Integral of the First Kind
6.1. Definition of a Line Integral of the First Kind 28
6.2. Calculation of a Line Integral of the First Kind 30
6.3. Applications of a Line Integral of the First Kind 31
6.4. Examples 32
Lecture 7. Line Integral of the Second Kind
7.1. Definition of a Line Integral of the Second Kind 34
7.2. Calculation of a Line Integral of the Second Kind 36
7.3. Relationalship Between the Line Integrals of the First and
the Second Kinds 37
Lecture 8.
8.1. Green Formula 37
8.2. Conditions for the Line Integral to be Independent on the Integration Path 38
8.3. Applications of a Line Integral of the Second Kind 39
8.4. Examples 40
SURFACE INTEGRALS
Lecture 9. Surface Integral of the First Kind
9.1. Definition of a Surface Integral of the First Kind 43
9.2. Calculation of a Surface Integral of the First Kind 45
9.3. Applications of a Surface Integral of the First Kind 46
9.4. Examples 47
Lecture 10. Surface Integral of the Second Kind
10.1. The Surface Orientation 50
10.2. Definition of a Surface Integral of the Second Kind 51
10.3. Properties of a Surface Integral of the Second Kind 52
10.4. Calculation of a Surface Integral of the Second Kind 53
10.5. Relationalship Between the Surface Integrals of the First
and the Second Kinds 53
Lecture 11.
11.1. Stokes Formula 54
11.2. Conditions for the Line Integral to be Independent on the Integration
Path in the Space 55
11.3. Ostrogradsky-Gauss Formula
11.4. Examples 57
SCALAR AND VECTOR FIELDS
Lecture 12. Scalar Fields
12.1. Definition of a Scalar Field 62
12.2. Directional Derivative 62
12.3. Gradient 64
12.4. Some Applications 65
12.5. Examples 65
Lecture 13. Vector Fields
13.1. Definition of a Vector Field 67
13.2. Flux of the Vector Field 68
13.3. Divergence of the Vector Field 68
13.4. Ostrogradsky-Gauss Formula in the Vector Form 69
13.5. Curl (Rotor) of the Vector Field 70
Lecture 14.
14.1. Stokes Formula 71
14.2. Potential Vector Fields 71
14.3. Solenoidal Vector Fields 73
14.4. Laplace (or Harmonic) Vector Fields 7 4
14.5. Hamilton and Laplace Operators 74
14.6. Examples 76
SERIES
Lecture 15. Numerical Series
15.1. The Convergence and the Sum of a Series 80
15.2. Geometric progression 80
15.3. Properties of a Numerical Series 81
15.4. The Necessity Condition for the Series Convergence 82
15.5. The Sufficient Tests for the Series Convergence 83
Lecture 16.
16.1. Alternating Series. Leibniz Theorem 86
16.2. Oscilating Series. Absolute and Conditional Convergence 86
16.3. Examples 87
Lecture 17. Functional Series
17.1. Definition of a Functional Series 91
17.2. Dominated Series 92
17.3. Power Series 92
17.4. Series in Powers of (x-a) 94
17.5. Teylor and Maclaurin Series 95
Lecture 18.
18.1. Expansion of Functions in the Maclaurin Series 96
18.2. Applications of Series 97
18.3. Examples 98
MULTIPLE INTEGRALS
LECTURE 1.
Double Integral
1.1. Definition of a Double Integral
Let be a closed domain in the -plane bounded by a line and let a continuous function be defined in .
Using arbitrary lines let us divide into subdomains and numerate them in an arbitrary order. Designating the area of subdomain by , and choosing in each subdomain an arbitrary point , , let us calculate the value of function at each of these points and compose the sum
, (1)
which is called the integral sum. Hereinafter, we shall call the maximum distances between points of a figure as its norm. For instance, the norm of a rectangular (or parallelepiped) is its diagonal, the norm of a ring is a diameter of its outer circumference. The norm of subdomain will be designated by .
Definition. If the limit of the integral sums sequence when the maximum of tends to zero as exists and does not depend on the way of partitioning the domain into subdomains, it is called the double integral of the function over the domain and is denoted by .
Thus, according to the definition,
, (2)
The quantity is called the differential area element in Cartesian coordinates.
Theorem 1. (Integrability Theorem) If is bounded on the closed domain and if it is continuous there except on a finite number of smooth curves, then is integrable on . In particular if is continuous on all , then it is integrable there.
The double integral is a definite integral and has properties similar those of a definite integral of a single variable function. Let us note ones the most important:
Property 1. The linearity of a double integral
(3)
,
where and are arbitrary numbers
Property 2. The additivity of a double integral
If domain is divided into two subdomains and such that , then
, (4)
Property 3. The mean-value theorem for a double integral
There exists in the domain such point that
, (5)
where is the area of the domain .
These properties may be proved on the base of definition of a double integral in Eq. 2.
Note. Let us consider a cylindrical solid bounded below by plane , above by the surface , and laterally by a cylindrical surface whose directrix is the boundary of the domain . Then the product is equal to the volume of a cylindrical column of the height and the area , and the sum gives an approximate value of the volume of the concerned cylindrical solid. The exact value of this volume one obtains through the limit as in Eq. (2). Thus, geometrically a double integral of a nonnegative function over a closed bounded domain determines the volume of the cylindrical solid restricted by the graph of this function and -plane.
1.2. Calculation of a Double Integral
Let us call the domain regular in direction of a coordinate axis, if any straight line parallel to this axis and passing through its interior point intersects its boundary at two points only, and regular, if it is regular in directions of both axes. An example of a domain regular in direction ofaxis, but not in direction of -axis, is shown in Fig. 1.
Let be a domain regular in the direction and bounded by lines: , as it is shown in Fig.2. Supposing functions being continuous on the segment , let us prove the following theorem.
Theorem 2. If the double integral exists and for all there exists a definite integral
, (6)
then the definite integral
(7a)
(which is called the twofold iterated integral) also exists and
. (7b)
To prove this theorem, let us consider a cylindrical solid bounded below by the -plane, above by the surface and laterally by a cylindrical surface with generatrix parallel to axis and the directrix being the boundary of the domain in which the solid is projected (Fig. 3.). The domain is projected into the segment on the -axis, i.e., . Considering from this segment as arbitrary but fixed, we find that the variable varies in limits . Hence the integral in Eq.(6) gives an area of a cross-section of the solid by the plane orthogonal to the axis and passing through the point , as one can see in Fig.(3).
Due to the arbitrariness of , the function is determined for all from segment . Integrating it over this segment as in Eq. (7a), we find that the integral in the left-hand side of this equation gives the volume of the concerned cylindrical solid. On the other hand, the same volume is determined by double integral of function over domain . Comparing results, we obtain `the equality Eq. (7b), which was to be proved.
Here in Eq. (7b) the integral over
is called the interior and the one over the
exterior. The exterior integral must have constant limits. The order of integration is important. You need to integrate with respect to first and then with respect to .
If the domain is regular in the direction and bounded by lines , then instead of Eq. (7b) one can write
. (8)
If the domain is regular, the equations Eq. (7b) and Eq. (8) are valid simultaneously so that
. (9)
This equality is known as a formula of changing the integration order.
Thus, the calculation of a double integral is reduced to sequential calculation of two definite integrals. Let us consider this procedure in detail, supposing the domain is regular and has the form shown in Fig.4. Projecting the domain on the axis, we find that . Taking an arbitrary but fixed value of from this segment, let us mentally draw from below the straight line parallel to the axis. This line penetrates the domain through the point , and goes out through the point . Thus, when is fixed, the variable varies in limits Taking into account the arbitrariness of fixed value of , one can see that the inequalities
(10)
entirely describe the domain and their extreme values are the limits of the twofold iterated integral in Eq. (7b).
Projecting in the same manner the domain on the axis, we obtain inequalities
(11)
which give the limits of the twofold iterated integral in Eq. (8). In practice, the choice of variable of the external integral depends on the structure of the domain boundaries. For example, if the domain has the form shown in Fig. 5, the choice of as the variable of the external integral will make us divide the domain into two parts by the line and consider two twofold iterated integrals instead of one when the variable is chosen. Let us note in conclusion that in the case of a rectangular domain of integration with sides parallel to the coordinate axes the limits of both the external and internal integrals are constant and the order of integration is insignificant.
LECTURE 2.
2.1. Change of Variables in a Double Integral
The change of variables in a double integral is produced with the aims: a) to simplify the domain of integration, b) to simplify the integrand.
Let the Cartesian coordinates and be functions of new variables and :
. (12)
The functions and are supposed being single-valued and continuous with their first order derivatives in a certain domain in which the domain is converted under transformation Eq. (12). Then the change of variables in a double integral is produced by means of the following formula
, (13)
where
(14)
is the functional determinant which is also called the Jacobian of transforming from Cartesian coordinates to curvilinear ones . The quantity is called the differential area element in the coordinate system. .
2.2 Double Integral in Polar Coordinates
The Cartesian coordinates are connected with the polar ones by the equations
(15)
where is the radius and is the polar angle. Calculating the partial derivatives and composing the Jacobian, we find
(16)
and in accordance with Eq. (13) obtain the double integral in polar coordinates:
. (17)
To transform this integral to a twofold iterated one it is necessary to convert the equations of boundaries of the domain to polar coordinates and to determine limits of variation of the variables and . Let the domain , in which the domain converts under transformation Eq. (15), in polar coordinates be bounded by rays and , and by curves and , as it is shown in Fig. 6. It is evident that . Considering from this segment as arbitrary but fixed, one can see that the ray penetrates into the domain through the line and goes out through the line . Therefore, along this ray varies in limits . Due to the arbitrariness of the fixed value of , these inequalities are true for all from segment . Thus, the domain is
(18)
and we obtain
. (19)
2.3. Geometric Applications of a Double Integral
a). The area of a closed bounded domain. Let be a domain in the -plane restricted by lines . In Cartesian coordinates the differential area element is . Integrating it over the domain one obtains its area , i.e.,
. (20)
If the domain is described by inequalities from Eqs. (10), then
. (21)
One can easily see that the right-hand side of this equation is a usual definite integral determining the area of a plane domain.
If the plane domain is given in polar coordinates by Eq. (18), the switching of variables in Eq. (21) gives:
. (22)
b). The volume of a cylindrical solid. If a cylindrical solid is bounded by surfaces: below by plane , above by surface and laterally by a cylindrical surface with the directrix being the boundary of the domain in which the solid is projected, its volume is determined by double integral
. (23)