Tasks that lead to the concept of “double integral”.
- Let a plane material plate be given in the plane, at each point of which the density is known. You 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 as follows: at those points of the rectangle that do not belong to the plate, we assume that the density is zero. We define a uniform partition into the same number of particles. Thus, the given figure will be divided into elementary rectangles. Consider one of these rectangles. Choose any point of this rectangle. Due to the small size of such a rectangle, we assume that the density at each point of this 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. Area, as you know, is the multiplication of the length of the rectangle by the width. And on the coordinate plane - this is a change in some steps. Then the mass of the entire plate will be the sum of the masses of such rectangles. If in this ratio we go to the boundary, then we can get the exact ratio.
- We define a spatial body, which is limited by the origin and some function. It is necessary to find the volume of the indicated body. As in the previous case, we divide the region into rectangles. We assume that at points that do not belong to the region, the function will be 0. Consider one of the rectangular partitions. Through the sides of this rectangle we draw planes that are perpendicular to the axes of abscissas and ordinates. We get a box that is bounded below by a plane relative to the axis of the applicate, and above by the function that was specified in the condition of the problem. Choose 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, then the volume of the rectangle can be calculated. And the volume of the figure will be equal to the sums of all volumes of such rectangles. To get the exact value, you need to go to the border.
As can be seen from the stated tasks, in each example we conclude that different tasks lead to the consideration of double sums of the same kind.
Properties of the double integral.
We pose the problem. Let a function of two variables be given in some closed domain, and the given function is continuous. Since the area is limited, you can place it in any rectangle that completely contains the properties of the point of a given area. Divide the rectangle into equal parts. We call the diameter of the break the largest diagonal of the resulting rectangles. Now we select a point within the boundaries of one such rectangle. If you find the value at this point, add the sum, then such a sum will be called the integral for the function in the given area. Find the boundary of such an integral sum, under the conditions that the diameter of the break follows to 0, and the number of rectangles to infinity. If such a boundary exists and does not depend on the method of dividing the region into rectangles and on 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 that was described in task 2.
Knowing the double integral (definition), you can set the following properties:
- The constant can be taken out of the sign of the integral.
- The integral of the sum (difference) is equal to the sum (difference) of the integrals.
- Of the functions less will be the one whose double integral is less.
- The module can be introduced under the double integral sign.