Математика. Материалы курса

Ссылка на yuotube

Double integral

The lecture is devoted to the double integral. In some sense, this is a generalization of a certain integral. We considered this concept for a function of a single variable defined on a segment. So let’s start with the problem that leads to the concept of a double integral. Let the function be continuous and not negative on a closed bounded domain D, as a rule, the domain D is considered simply connected. A body appears in three-dimensional space. What is it? It is bounded by the XOY plane below, and above – by the graph of the function f. Remember that the function is not negative, so the graph is located above the XOY plane, and from the sides there is a cylindrical surface that passes through the boundary of the domain D in the XOY plane, forming parallel to the axis Oz, we will call such a body a curved cylinder.

The problem that leads to the definition of the double integral is the calculation of the volume of a curved cylinder. Let’s solve this problem. What will we start from? We will assume that the volume of a straight cylinder is well known to us, the volume of a circular cylinder is equal to the product of the area of the base and the height. How to start from this concept? What are we going to do? We divide the domain D, which lies in the XOY plane, by continuous curves into partial cells, partial domains D_i, and we will consider them as n. We introduce the notation delta S_i, it is the area of this cell, λ (lambda) _i is the diameter of the i-th cell. The diameter is the largest of all possible distances between pairs of points belonging to this cell, and λ is the largest of the diameters of all cells called the partition diameter. We take an arbitrary point (x_i, y_i) in this partial cell and build a straight cylinder.

What is a straight cylinder? This is a cylindrical body, whose upper and lower base are parallel. The base of this cylinder is the cell D_i, the height of the cylinder is equal to the value of the function at the point (x_i, y_i). The volume of such a cylinder is calculated using the formula product of the base area and height. Then the volume of a new body made up of straight cylinders of different heights is equal to the sum of the volumes of these partial cylinders. And so we got the final amount. It is clear that if the diameter of the partition tends to zero, the cell becomes smaller and smaller, the volume of this complex body will tend to the volume of the curved cylinder. Now let’s repeat the same reasoning for the function f. It is not necessary that the function is continuous or bounded, but it must be defined on a closed bounded domain D.

Again, divide the area into partial cells, take the points (x_i, y_i) and make the sum of Sigma as follows (see the video). This sum is called the integral sum of the function f, composed for a given partition of the domain into partial cells and the selection of points (x_i, y_i). If there is a finite limit to these integral sums when the diameter of the lambda partition tends to zero, and this limit does not depend on the way the domain D is partitioned and there is no choice of points (x_i, y_i), the value of this limit is called the double integral of the function f on the domain D, and the function is called integrable over the domain D. The value of this limit is denoted as follows (see in the video) notation of the double integral (see in the video), D is called the domain of integration, the integrand of two variables (see in the video), but the multiplier at the end of the entries dxdy, or sometimes ds in short is called the area element.

Let’s consider the following example. We calculate the double integral over the domain D of functions identically equal to dxdy. The integrand constant is equal to one. The integral sum in this case is the sum of the areas of partial cells, and the total area is the area of the area D each time equal to S and if the lambda tends to zero, the limit is also equal to S. Thus, the double integral over the domain dxdy is equal to the area of the domain D. The theorem on “The necessary condition for integrability”. If a function is integrable over domain D, it is bounded on it. Exactly the same theorem was formulated for a certain integral, if the function y = f (x) is integrable on a segment, it is bounded. It is an absolute analog and these are the same caveats that the inverse theorem is incorrect. The limitation of functions does not imply integrability. And a sufficient condition that the class of integrable functions gives us. Any continuous function on domain D, where D is closed and bounded, is integrable over this domain.

The properties of the double integral are virtually identical to the properties of the definite integral with some specifics. The linearity property, we talked about it, means the fulfillment of two conditions. The  property which is called uniformity when we can take a constant multiplier for the sign of the double integral and the second property of additivity: the double integral of the sum of functions is equal to the sum of the double integrals of these functions. If domain D is divided into two domains D₁, D₂, and these new domains do not share internal points, the integral over domain D can be represented as the sum of the double integrals of this function over domains D₁ и D₂. Let’s speak about the properties of the double integral related to inequalities. If the function is not negative on D, the double integral is also greater than or equal to zero, if the function is less than or equal to D, the double integral is less than or equal to zero.

What is the average value of a function when it is not a constant and the value is infinitely large? This is shown by the following theorem: if a function f is continuous on a closed and bounded domain D, then there is a point x₀ in this domain, the value of which is calculated as follows: the double integral over the domain on the area of the domain D. This value is called the average value of the function in domain D. Here is a simple proof. Again, we recall the theorem that if a function is continuous on a closed bounded domain, it has the largest and smallest values of m and M on it. And so m is the smallest, M is the largest value of the function. We apply the properties of the integral associated with inequalities and integrate them.