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Double integral. Tasks. Properties

Problems that lead to the concept of "double integral."

  1. Let a plane material plate be given in the plane, at each point of which the density is known. We need to find the mass of this plate. Since this plate has clear dimensions, it can be enclosed in a rectangle. The density of the plate can also be understood: at those points of the rectangle that do not belong to the plate, we will assume that the density is zero. We define a uniform division into an equal number of particles. Thus, the given shape will be divided into elementary rectangles. Consider one of these rectangles. We choose any point of this rectangle. Because of the small size of such a rectangle, we will assume that the density at each point of the given rectangle is a constant value. Then the mass of such a rectangular particle will be defined as the multiplication of the density at this point by the area of the rectangle. The area, as you know, is the multiplication of the length of the rectangle by the width. And on the coordinate plane - this change with some step. Then the mass of the entire plate will be the sum of the masses of such rectangles. If we go to the boundary in such a relation, then we can obtain an exact relation.
  2. We define a spatial body, which is bounded by the origin and some function. It is necessary to find the volume of this body. As in the previous case, we divide the area into rectangles. We will assume that at points that do not belong to the domain, the function will be 0. Consider one of the rectangular decompositions. Through the sides of this rectangle we draw planes that are perpendicular to the abscissa and ordinate axes. We obtain a parallelepiped, which is bounded from below by the plane relative to the axis of the applicator, and from above by the function that was specified in the condition of the problem. We select a point in the middle of the rectangle. Due to the small size of this rectangle, we can assume that the function within this rectangle has a constant value, and then you can calculate the volume of the rectangle. And the volume of the figure will be equal to the sums of all the volumes of such rectangles. To get the exact value, you need to go to the border.

As can be seen from the problems posed, in each example we come to the conclusion that different problems lead to consideration of double sums of the same type.

Properties of the double integral.

Let's put the problem. Suppose that in a certain closed region a function of two variables is given, for which the given function is continuous. Since the area is limited, you can put it in any rectangle that completely contains the properties of the point of the given area. We divide the rectangle into equal parts. We call the diameter of the breaking the largest diagonal from the resulting rectangles. Now we choose a point in the boundaries of one such rectangle. If we find a value at this point to add the sum, then such a sum will be called integral for the function in the given domain. We find the boundary of such an integral sum under the conditions that the diameter of the breakdown follows to 0, and the number of rectangles to infinity. If such a boundary exists and does not depend on how the region is divided into rectangles and from the choice of a point, then it is called a double integral.

The geometric content of the double integral: the double integral is numerically equal to the volume of the body, which was described in Problem 2.

Knowing the double integral (definition), you can set the following properties:

  1. The constant can be taken outside the integral sign.
  2. The integral of the sum (difference) is equal to the sum (difference) of the integrals.
  3. Of the functions less is the one whose double integral is smaller.
  4. The module can be introduced under the double integral sign.

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