Practical lesson 2. Calculating a certain integral
The previous lesson was devoted to bivariate integration over the area that is a rectangle with sides parallel to the coordinate axes. Today the situation is much more interesting.
We are going to consider the case when the area is not a rectangle. The integration domain is the domain D, which is defined by the following conditions: 0 ≤ x ≤ 2 and 0 ≤ y ≤ 2-x. We get a triangle. How do we calculate this integral? In fact, there is several ways. It all depends on how we want to proceed to the repeated integral.
Let's consider the first method, when the external integral (we will write) in x. ∫(x+1) dx∫ydy. Note that the integrand is a product of functions of one variable, so each repeated integral contains only one variable. If we had the area D as a rectangle (there would be a number instead of 2), we would say that this is the product of definite integrals. Note that this happens only in an exceptional case, which we discussed in the last lesson. There are no definite integrals here.
How do we arrange the limits of integration? If we have an external integral for the variable x, then we have to answer between which vertical lines this figure of the integration area D is located? We see that it is located between the lines x=0 and x=2; these are constants. 0 is the lower limit of integration, and the upper limit is 2. We draw vertical lines that limit the integration area. Then we go to the borders for y. We need to write the limits of integration. We start moving in the constructed lane in the direction indicated by the Оу axis. First, we "stumble upon” the straight line y=0. The next line is the graph of the function y=2-x. This is the limits of integration: 0 and 2. Calculating the integral (see the video). We get 2 as the answer. Note that we integrate with both the external and repeated integrals for the variable x.
Let's consider the same example again, with the different order of integration. I write the outer integral in y, the inner integral in x. How do we set the limits of integration in this case? If the external integral is for y, it means that the external integral has the limits of the number; y is a constant, a horizontal line. We need to construct two horizontal lines, between which the integration area is located. They are set by the equations y=0 and y=2. Thus, the first integral has the integration limits from 0 to 2. How do we set the integration limits for the second integral? We draw an arrow in this lane in the direction indicated by the axis Ox. First, we approach the Oy axis: x=0. Then – y=2-x. This line is set by the equation y=2-x, y is an independent variable, and x is a function of the variable y, that is, we must express x via the equation y=2-x. We see that x=2-y. The limits of integration for the variable x, for the internal integral are from 0 to 2-y. Let's consider the calculation of this integral again (see the video). We get the same answer. In this case, the calculation is complete.
There is no big difference in these calculations, as you can see. It is difficult to say which method is preferable, but in some cases, it is important to choose the correct integration order.
Let's consider the situation where the area D is a triangle bounded by three straight lines (see the video). We take the external integral in x. It is our random choice. We act according to the scheme. Since the external integral is in x, we should consider which vertical lines this area is located between. In the first case: x=0, and in the second: x=2. This means that the integration limits are from 0 to 2. What about the internal integral? We draw an arrow. What do we have? At the beginning, we come across the line y=0. What happens next? We come across a certain function, y=h(x). Thus, I have to write from 0 to h(x), where the function h(x) on one segment is given by one formula, on the other segment it is set by another formula. What are we going to do? We have to divide the integration area with another vertical line and say that the double integral is equal to the sum of the double integrals over the integration areas D1 and D2. D1 is the triangle located to the left of this vertical line, and D2 is the triangle located to the right of it. We calculated a similar integral in the first example, but now we have to calculate two integrals. The problem is that the upper bound is set by different formulas on the segment [0; 1] and [1; 2]. We are not going to continue this calculation together; you can bring it to the end to gain your own confidence.
We are going to consider the solution of this problem in the second way. We take the external integral of the variable y. So far, we have just separated the variables into two integrals. The external integral in y means that the limits of integration are constants. Therefore, we must construct two horizontal lines between which there is the area of integration. The lower bound is y=0, and the upper bound is y=2. The limits of integration are 0 and 2.
To set the limits of integration of the internal integral, we draw an arrow inside the limits in the direction of the Ox axis. We have the equations of the straight lines where there is the dependence of x on y. Each straight line can be set in different ways. There is the dependence of x on y on the line x=y, and on the second line the dependence is x=2-y. We get y and 2-y. Let's start calculating the integral (see the video). The answer is 4/3.
To figure out how to integrate over an arbitrary domain, we need know how to solve problems devoted to changing the order of integration. You need to write this integral (see the video), changing the order of integration. There are no problems at this stage. The problem arises when we want to write the limits of integration, so first let's write down what the boundaries are. x changes from 0 to 1, and the borders at the top and bottom are the lines y=2x and y=2. Let's present this area (see the video). The integration area looks like this. Now, we can set the limits of integration by looking at this area. Where do we start? Since the external integral is in y, then we draw two horizontal lines. They are set by the equations y=0 and y=2; these are the integration limits for the external integral. Next, we draw an arrow inside, and see which lines limit the area here. These are x=0 and x=1/2y. We get the limits of the internal integral from 0 to 1/2y. The problem is solved.
The last and most interesting problem is when the integration area is a circle. In general, the most common situation is when the area of integration is a curved sector. Our task is to calculate this integral using the polar coordinates. What do we do? We use the substitution: x=pcosφ, y=psinφ. In addition, to avoid unnecessary transformations, we use the equation х2+у2=ρ2. Then the integral is the following (see the video). We get standard integrals that require some integration skills. This is for you to solve on your own.