Lecture 2. Calculation of the double integral
This lecture will be devoted to the calculation of the double integral.
We have already identified it, we know its properties.
How to find the value of the double integral?
This is a number, a limit of a integral sums.
We proceed to calculation.
Firstly, the range of integration can be a very different set in XOY plane.
The simplest case when D is a rectangle with sides parallel to the coordinate axes.
How to describe a set of points on the plane?
Let's look at different tasks.
Firstly, as a characteristic property of a point set on the plane, which coordinates satisfy the inequality (watch video).
Secondly, it is the Cartesian product of two segments (watch video).
Thirdly, we form the system of inequalities (watch video).
Thus, integration of the rectangular area.
Let's consider the first theorem: a function of two variables f (x, y) is defined on such area D.
What do we know about it: it is integrable in D, then there exists a double integral equal to the number, in addition, the most important condition is the following, for any point x of the fixed segment [a, b] there is a definite integral.
At this moment under the integral sign there is already a function of one variable y, which varies from c to d, and the integral in this case is a function of variable of x, so this is a condition.
In this case, the double integral is the integral of the resulting functions and it is of x in the segment [a, b].
Thus, the double integral reduces to calculation of definite integrals.
As a result: the integral of function I (x) is the integral of the integral, which we write as follows (watch the video), and call it a iterated integral of function f (x, y).
Thus, the theorem conclusion can be written in this way: the double integral of function f of the rectangular area D, under the theorem conditions is equal to iterated integral.
Note: here the outer integral of variable x changes from a to b, the inner integral is of the variable y, and integration limits of c and d.
It turns out that this theorem can be formulated.
Its analogue the outer integral is taken by y.
Then it is demanded that under the theorem conditions except a double integral you have the following: for every y in the segment [c, d] there is already a definite integral of the function of one variable x, respectively, of the segment [a, b], which is a function of variable y.
The double integral is again equal to definite integral of the function, and as a result we get iterated integral with another order of integration.
These theorems (1-2) can be summarized as follows: provided that the conditions that we are given a double integral is equal to each of the iterated integrals.
It is important to consider important cases because that leads to a very simple calculation: if the function of two variables is the product of two factors, the values are calculated as the product of two factors, each of which contains only one variable.
Integrating is again performed in a rectangular area.
Let's see: take the iterated integral, the outer integral of x.
We see: the internal functions of the inner integral are h(x), g(y). h (x) in the inner integral is a constant, so we can take outside the definite integral with limits of integration c, d.
We get: take out, but these integrals contain only one variable x, and th limits of integration are numbers, and the double integral is transformed into a product of two definite integrals on the segments (watch video).
The most difficult, the most interesting case is where integration is carried out in an area that is not a rectangle.
We’ll consider two cases: we have two functions of variable x, graph y1(x) is located below graph y2(x).
These functions are continuous on the segment [a, b], and we get a figure, the range of integration, the boundaries of which are the graphs of these functions, and vertical lines x = a, x = b.
We say that this area is proper in the direction of axis Oy.
What we mean: any vertical line located between boundaries x = a, x = b, crosses this area not more than in two points, at the inlet and outlet of the segment, sometimes it is degenerate.
In this case theorem sounds this way: again there is a double integral, expressed by a number and for any x, we demand: there is a definite integral with integration limits from y1(x), this is the lower limit of the area, to y2(x), this is the upper limit of the area.
Then, the double integral is calculated as the iterated.
Look (watch video).
The second case is when area D is bounded by two horizontal straight lines y = c and y = d, besides on the left and right by functions graphs of x1(y) and x2(y).
This areais in the correct direction Ox axis.
Why: any horizontal line, located between the already marked ones, crosses this area by the segment, sometimes degenerate.
Note (watch video), in this picture the area is not proper in the direction of axis Oy.
See, there is a vertical line that will cross this area not by the segment, but by the set which is a set of two segments with empty intersection
Theorem formulation sounds analogously: there is a double and, furthermore, function integrated over the segment [x1(y), x2(y)] for every fixed value y of segment [c, d], then the double integral is iterated.
Let's look at an example (watch the video), we will not calculate the double integral, because here function f is defined in a general way - just a function of two variables. Now it is important for us to form iterated integrals.
Area D is bounded by straight lines, let's build these lines (watch video).
We get a triangle.
Let's solve this problem in two ways: form iterated integrals with different order of integration.
So, the first way: when the external integration is by variable x.
We should remember: integration limits of the outer integral are always numbers, without exception.
Thus, variable x should change from number to number.
What do we do?
The first step: we define the limits of the outer integral, how variable x is changing from what number to what, we construct vertical lines that limit the integration area.
We had two lines x = 0, x = 2.
These are limits of the outer integral.
Further we define the limits of the inner integral.
We draw a vertical line between the already built ones and see: at the bottom this vertical line will be crossed on the green line, and on the top - on the blue line, in this picture (see video).
And the problem is: the graph of the lower function is defined by different formulas on different segments, so it is logical to separate the integration area into two parts.
Here the border of these parts is built (see video), and we can use a property of the double integral: the integral in area D is the sum of the integrals of areas D1 and D2 into which area D is split.
Thus, area D1is the first part, the whole triangle is divided into two parts, and D2 is the second part of this triangle.
See: we will repeat these two steps for area D1, and we repeat these two steps for area D2.
Therefore we look at this picture (watch video): the outer limits of integration by area D1.
Area D1 is located between vertical lines x = 0 and x = 1
The internal limits of integration: the bottom line y = 0, y = x, is for area D1.
We proceed to the second integral: the area, the second part of a larger triangle, is between vertical lines x = 1, x = 2.
We note the outer limits of integration and then move on to the inner limit, these are two straight lines.
Thus, the lower limit is y = x-1 and an upper limit is y = x. The problem is solved.
Let us consider the second method, remember if the integration of an external variable y, then y is changed in this case from a constant to a constant.
How to find these constants: build horizontal lines between which there is area D.
We see that there are two lines y = 0, y = 2, 0 and 2 are the integration limits.
The next step: define the limits of the inner integral, and we move from left to right on area D.
The first time we encounter the straight line x = y, we express x through y, x becomes a function of variable y, and then moving to the right, once again we encounter the following line x = y + 1, and as a result we get a double integral.
The double integral is equal to the iterated one.
Look, the solution to the same problem without calculating definite integrals leads to different results.
In the first case, the double integral is the sum of two iterated ones, and in the second case there is only one integral, so how efficiently, how quickly you solve the problem, depends on your seeing the way to solve this problem.
So every time you should consider carefully what this area is and how to decide the transition to an iterated integral more rationally.
And another important question: how to change variables in a double integral.
So, let variables x and y in the integral (watch video) be associated with the new variables u and v of the star function system (watch video).
In order to change variables, generally speaking, we need some conditions that we will not consider, because when solving problem we have to deal, as a rule, with continuous good function, and all these conditions are fulfilled.
The formula of passing to new variables u and v is according to the following formula (watch video).
See what we are doing in the double integral: in function f (x, y) x and y are replaced in accordance with the system (*)dudv appears in place of dxdy, but that's not all.
Look, here is factor | I (u, v) |, which is called the Jacobian of the mapping (*), this is a functional determinant, which is composed of partial derivatives of functions x and y of variables u and v (watch video).
The double integral is taken in area D *, which is obtained when mapping the function system (*).
Let's look at an example, new variables are ρ and φ.
Thus, transition from Cartesian coordinates to polar new variables ρ and φ.
We know the connection formula.
In order to carry out the replacement, we need to calcuate the Jacobian.
Generally speaking, this is a geometric sense Jacobian: if dxdy is an area cell, an area element, then the Jacobian is a distortion factor in the transition to the new coordinates.
Look, calculating the partial derivatives and then the determinant, we get ρ (look at the screen).
In such a case, except substituting x and y by ρcos φ, ρsin φ, dxdy - dρfφ, there is still factor ρ.
Element dxdy area becomes element area ρdρdφ, it is necessary to remember that.
Again, let's look at an example of how to switch to polar coordinates without calculations f (x, y), the function of two variables.
Let's form area D, this is a part of the unit circle centered in the origin of coordinates, which lies in the first quadrant (watch video).
Again there are two phases: an external integration of the transition to polar coordinates, we tend to take on the standard variable φ. φ should change from a constant to a constant. φ is equal to the constant, this is an equation of the half-ray centered at the pole, in the polar coordinate system, and with angle φ.
So to place the integration limits, we have to build two rays starting from the point of origin, between them there is an area of integration.
Look, we can clearly see that this area is located between rays φ = 0, φ = Π / 2, these are outer integration limits.
The next step, write down internal limits of integration, construct an arbitrary ray which is located between rays built during the first step and see in which points this ray intersects area D.
The first ρ = 0 point is a pole and the second point is on the circular curve of unit radius, it is defined by equation ρ = 1.
So, numbers 0 and 1 are limits of the inner integral.
First, pass to the double integral by the transition formula, x, and y are replaced by ρcos φ, ρsin φ, area element dxdy is replaced by ρdρdφ.
Then move to the iterated integral: outer integral dφ, the integration limits of the first step 0, Π / 2 and the inner integral 0, 1 (watch video). The problem is solved.