The explanation of why the sky is dreamed of depends on how it looked. If not...
If the curve is given by parametric equations, then the surface area obtained by rotating this curve around the axis is calculated by the formula 
. In this case, the “direction of drawing” of the line, about which so many copies were broken in the article, is indifferent. But, as in the previous paragraph, it is important that the curve is located higher abscissa axis - otherwise the function “responsible for the games” will take negative values and you will have to put a “minus” sign in front of the integral.
Example 3
Calculate the area of a sphere obtained by rotating a circle around the axis.
Solution: from the article on area and volume for a parametrically defined line you know that the equations define a circle with a center at the origin of radius 3.
Well sphere , for those who have forgotten, this is the surface ball(or spherical surface).
We adhere to the established solution scheme. Let's find derivatives: 
Let’s compose and simplify the “formula” root:
Needless to say, it turned out to be candy. Check out for comparison how Fichtenholtz butted heads with the area ellipsoid of revolution.
According to the theoretical remark, we consider the upper semicircle. It is “drawn” when the parameter value changes within the limits (it is easy to see that 
on this interval), thus:
Answer:
If you solve the problem in general view, then you get exactly the school formula for the area of a sphere, where is its radius.
It was such a painfully simple task, I even felt ashamed... I suggest you fix this bug =)
Example 4
Calculate the surface area obtained by rotating the first arc of the cycloid around the axis.
The task is creative. Try to derive or intuitively guess the formula for calculating the surface area obtained by rotating a curve around the ordinate axis. And, of course, the advantage of parametric equations should again be noted - they do not need to be modified in any way; there is no need to bother with finding other integration limits.
The cycloid graph can be viewed on the page Area and volume, if the line is specified parametrically. The surface of rotation will resemble... I don’t even know what to compare it with... something unearthly - round in shape with a pointed depression in the middle. For the case of rotation of a cycloid around an axis, an association instantly came to mind - an oblong rugby ball.
The solution and answer are at the end of the lesson.
We conclude our fascinating review with the case polar coordinates. Yes, just a review, if you look at textbooks on mathematical analysis (Fichtenholtz, Bokhan, Piskunov, and other authors), you can get a good dozen (or even much more) standard examples, among which you may well find the problem you need.
How to calculate surface area of revolution,
if the line is given in a polar coordinate system?
If the curve is given in polar coordinates equation, and the function has a continuous derivative on a given interval, then the surface area obtained by rotating this curve around the polar axis is calculated by the formula 
, where are the angular values corresponding to the ends of the curve.
In accordance with the geometric meaning of the problem, the integrand function 
, and this is achieved only under the condition (and are obviously non-negative). Therefore, it is necessary to consider angle values from the range , in other words, the curve should be located higher polar axis and its continuation. As you can see, the same story as in the two previous paragraphs.
Example 5
Calculate the surface area formed by rotating the cardioid around the polar axis.
Solution: the graph of this curve can be seen in Example 6 of the lesson about polar coordinate system. The cardioid is symmetrical about the polar axis, so we consider its upper half in the interval (which, in fact, is due to the above remark).
The surface of rotation will resemble a bullseye.
The solution technique is standard. Let's find the derivative with respect to "phi":
Let's compose and simplify the root:
I hope with regular trigonometric formulas no one had any difficulties.
We use the formula:
In between 
, hence: 
(I talked in detail about how to properly get rid of the root in the article Curve arc length).
Answer: 
An interesting and short task for you to solve on your own:
Example 6
Calculate the area of the spherical belt,
What is a ball belt? Place a round, unpeeled orange on the table and pick up a knife. Make two parallel cut, thereby dividing the fruit into 3 parts of arbitrary sizes. Now take the center, which has juicy flesh exposed on both sides. This body is called spherical layer, and the surface bounding it (orange peel) – ball belt.
Readers familiar with polar coordinates, easily presented a drawing of the problem: the equation specifies a circle with a center at the pole of radius , from which rays 
cut off less arc. This arc rotates around the polar axis and thus produces a spherical belt.
Now you can eat an orange with a clear conscience and a light heart, and on this tasty note we’ll end the lesson, don’t spoil your appetite with other examples =)
Solutions and answers:
Example 2:Solution
: calculate the surface area formed by the rotation of the upper branch around the abscissa axis. We use the formula 
.
In this case: 
;
Thus:
Answer:
Example 4:Solution
: use the formula 
. The first arc of the cycloid is defined on the segment .
Let's find derivatives:
Let's compose and simplify the root:
Thus, the surface area of rotation is:
In between 
, That's why 
First integralintegrate by parts
:
In the second integral we usetrigonometric formula
.
Answer:
Example 6:Solution
: use the formula:
Answer:
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How to calculate a definite integral
using the trapezoidal formula and Simpson's method?
Numerical methods are a fairly large section of higher mathematics and serious textbooks on this topic contain hundreds of pages. In practice, in tests Traditionally, some problems are proposed to be solved using numerical methods, and one of the common problems is the approximate calculation definite integrals. In this article I will look at two methods for approximate calculation of the definite integral - trapezoid method And Simpson method.
What do you need to know to master these methods? It may sound funny, but you may not be able to take integrals at all. And you don’t even understand what integrals are. From technical means You will need a micro calculator. Yes, yes, routine school calculations await us. Better yet, download mine semi-automatic calculator for the trapezoidal method and the Simpson method. The calculator is written in Excel and will reduce the time required for solving and completing problems by tens of times. For Excel dummies, a video manual is included! By the way, the first video recording with my voice.
First, let's ask ourselves: why do we need approximate calculations at all? It seems that you can find the antiderivative of the function and use the Newton-Leibniz formula, calculating the exact value of the definite integral. To answer the question, let’s immediately look at a demo example with a picture.
Calculate definite integral
Everything would be fine, but in this example the integral is not taken - before you is an untaken, so-called integral logarithm. Does this integral even exist? Let us depict in the drawing the graph of the integrand function: 
Everything is fine. Integrand continuous on the segment and the definite integral is numerically equal to the shaded area. There’s just one catch: the integral cannot be taken. And in such cases, numerical methods come to the rescue. In this case, the problem occurs in two formulations:
1) Calculate the definite integral approximately , rounding the result to a certain decimal place. For example, up to two decimal places, up to three decimal places, etc. Let's assume the approximate answer is 5.347. In fact, it may not be entirely correct (in reality, say, the more accurate answer is 5.343). Our task is only that to round the result to three decimal places.
2) Calculate the definite integral approximately, with a certain accuracy. For example, calculate a definite integral approximately with an accuracy of 0.001. What does it mean? This means that if the approximate answer is 5.347, then All the numbers must be reinforced concrete correct. More precisely, the answer 5.347 should differ from the truth in absolute value (in one direction or another) by no more than 0.001.
There are several basic methods for approximate calculation of the definite integral that occurs in problems:
Rectangle method. The integration segment is divided into several parts and a step figure is constructed ( histogram), which is close in area to the desired area: 
Do not judge strictly by the drawings, the accuracy is not ideal - they only help to understand the essence of the methods.
In this example, the integration segment is divided into three segments: 
. Obviously, the more frequent the partitioning (more smaller intermediate segments), the higher the accuracy. The rectangle method gives a rough approximation of the area, which is apparently why it is very rarely found in practice (I remember only one practical example). In this regard, I will not consider the rectangle method, and I will not even give a simple formula. Not because I’m lazy, but because of the principle of my workbook: what is extremely rare in practical problems is not considered.
Trapezoid method. The idea is similar. The integration segment is divided into several intermediate segments, and the graph of the integrand function approaches broken line line: 
Thus, our area (blue shading) is approximated by the sum of the areas of the trapezoids (red). Hence the name of the method. It is easy to see that the trapezoid method gives a much better approximation than the rectangle method (with the same number of partition segments). And, naturally, the more smaller intermediate segments we consider, the higher the accuracy will be. The trapezoid method is encountered from time to time in practical tasks, and several examples will be discussed in this article.
Simpson's method (parabola method). This is a more advanced method - the graph of the integrand is approximated not by a broken line, but by small parabolas. There are as many small parabolas as there are intermediate segments. If we take the same three segments, then Simpson's method will give an even more accurate approximation than the rectangle method or the trapezoid method.
I don’t see the point in constructing a drawing, since the visual approximation will be superimposed on the graph of the function (the broken line of the previous paragraph - and even then it almost coincided).
The problem of calculating a definite integral using Simpson's formula is the most popular task in practice. And the parabola method will be given considerable attention.
Example: Find the volume of a sphere with radius R.
In the cross sections of the ball, circles of variable radius y are obtained. Depending on the current x coordinate, this radius is expressed by the formula.
Then the cross-sectional area function has the form: Q(x) = .
We get the volume of the ball:
Example: Find the volume of an arbitrary pyramid with height H and base area S.
 |
When the pyramid is intersected by planes perpendicular to the height, in cross-section we obtain figures similar to the base. The similarity coefficient of these figures is equal to the ratio x/H , where x is the distance from the section plane to the top of the pyramid.
It is known from geometry that the ratio of the areas of similar figures is equal to the similarity coefficient squared, i.e.
From here we obtain the function of cross-sectional areas:
Finding the volume of the pyramid: 
Volume of bodies of rotation.
Consider the curve given by the equation y = f(x ). Let's assume that the function f(x ) is continuous on the interval [ a, b ]. If the corresponding curvilinear trapezoid with bases a and b rotate around the Ox axis, we get the so-called body of rotation.
y = f(x)
Surface area of a body of rotation.
M i B
Definition: Surface area of rotation curve AB around a given axis is the limit to which the areas of the surfaces of rotation of broken lines inscribed in the curve AB tend to when the largest of the lengths of the links of these broken lines tend to zero.
Let us divide the arc AB into n parts by points M 0, M 1, M 2, …, M n . The coordinates of the vertices of the resulting polyline have the coordinates x i and y i . By rotating the broken line around its axis, we obtain a surface consisting of the lateral surfaces of truncated cones, the area of which is equal to D P i . This area can be found using the formula:
Before moving on to the formulas for the area of a surface of revolution, we will give a brief formulation of the surface of revolution itself. A surface of revolution, or, what is the same thing, a surface of a body of revolution is a spatial figure formed by the rotation of a segment AB curve around the axis Ox(picture below).
Let us imagine a curved trapezoid bounded from above by the mentioned segment of the curve. A body formed by rotating this trapezoid around the same axis Ox, and is a body of revolution. And the area of the surface of revolution or the surface of a body of revolution is its outer shell, not counting the circles formed by rotation around the axis of straight lines x = a And x = b .
Note that a body of revolution and, accordingly, its surface can also be formed by rotating the figure not around the axis Ox, and around the axis Oy.
Calculating the area of a surface of revolution specified in rectangular coordinates
Let in rectangular coordinates on the plane the equation y = f(x) a curve is specified, the rotation of which around the coordinate axis forms a body of revolution.
The formula for calculating the surface area of revolution is as follows:
(1).
Example 1. Find the surface area of the paraboloid formed by rotation around its axis Ox arc of a parabola corresponding to the change x from x= 0 to x = a .
Solution. Let us express explicitly the function that defines the arc of the parabola:
Let's find the derivative of this function:
Before using the formula to find the area of a surface of revolution, let’s write that part of its integrand that represents the root and substitute the derivative we just found there:
Answer: The length of the arc of the curve is
.
Example 2. Find the surface area formed by rotation around an axis Ox astroid.
Solution. It is enough to calculate the surface area resulting from the rotation of one branch of the astroid, located in the first quarter, and multiply it by 2. From the astroid equation, we will explicitly express the function that we will need to substitute into the formula to find the surface area of rotation:
.
We integrate from 0 to a:
Calculation of the area of a surface of revolution specified parametrically
Let us consider the case when the curve forming the surface of revolution is given by parametric equations
Then the surface area of rotation is calculated by the formula
(2).
Example 3. Find the area of the surface of revolution formed by rotation around an axis Oy figure bounded by a cycloid and a straight line y = a. The cycloid is given by parametric equations
Solution. Let's find the intersection points of the cycloid and the straight line. Equating the cycloid equation 
and the equation of the line y = a, let's find
It follows from this that the boundaries of integration correspond to
Now we can apply formula (2). Let's find derivatives:
Let's write the radical expression in the formula, substituting the found derivatives:
Let's find the root of this expression:
.
Let's substitute what we found into formula (2):
.
Let's make a substitution:
And finally we find
Trigonometric formulas were used to transform expressions
Answer: The surface area of revolution is .
Calculating the area of a surface of revolution specified in polar coordinates
Let the curve, the rotation of which forms the surface, be specified in polar coordinates.
Greetings, dear students of the University of Argemona!
Today we will continue to learn how to materialize objects. Last time we rotated flat figures and got volumetric bodies. Some of them are very tempting and useful. I think that much of what a magician invents can be used in the future.
Today we will rotate curves. It is clear that in this way we can get some object with very thin edges (a cone or bottle for potions, a flower vase, a glass for drinks, etc.), because a rotating curve can create exactly this kind of objects. In other words, by rotating the curve we can get some kind of surface - closed on all sides or not. Why right now I remembered the leaky cup from which Sir Shurf Lonley-Lokley always drank.
So we will create a bowl with holes and a bowl without holes, and calculate the area of the created surface. I think that it (the surface area in general) will be needed for something - well, at least for applying special magic paint. On the other hand, the areas of magical artifacts may be required to calculate the magical forces applied to them or something else. We will learn to find it, and we will find where to apply it.
So, a piece of a parabola can give us the shape of a bowl. Let's take the simplest y=x 2 on the interval. It can be seen that when you rotate it around the OY axis, you get just a bowl. No bottom.
The spell for calculating the surface area of rotation is as follows:
Here |y| is the distance from the axis of rotation to any point on the curve that rotates. As you know, distance is a perpendicular.
A little more difficult with the second element of the spell: ds is the arc differential. These words don’t give us anything, so let’s not bother, but let’s move on to the language of formulas, where this differential is clearly presented for all cases known to us:
- Cartesian coordinate system;
- recording the curve in parametric form;
- polar coordinate system.
For our case, the distance from the axis of rotation to any point on the curve is x. We calculate the surface area of the resulting holey bowl:
To make a bowl with a bottom, you need to take another piece, but with a different curve: on the interval this is the line y=1.
It is clear that when it rotates around the OY axis, the bottom of the bowl will be in the form of a circle of unit radius. And we know how the area of a circle is calculated (using the formula pi*r^2. For our case, the area of the circle will be equal to pi), but let’s calculate it using a new formula - to check.
The distance from the axis of rotation to any point of this piece of the curve is also equal to x.
Well, our calculations are correct, which is good news.
And now homework.
1. Find the surface area obtained by rotating the broken line ABC, where A=(1; 5), B=(1; 2), C=(6; 2), around the OX axis.
Advice. Write down all segments in parametric form.
AB: x=1, y=t, 2≤t≤5
BC: x=t, y=2, 1≤t≤6
By the way, what does the resulting item look like?
2. Well, now come up with something yourself. I think three items will be enough.
5. Finding the surface area of bodies of revolution
Let the curve AB be the graph of the function y = f(x) ≥ 0, where x [a; b], and the function y = f(x) and its derivative y" = f"(x) are continuous on this segment.
Let us find the area S of the surface formed by the rotation of the curve AB around the Ox axis (Fig. 8).
Let's apply scheme II (differential method).
Through an arbitrary point x [a; b] draw a plane P perpendicular to the Ox axis. Plane П intersects the surface of rotation in a circle with radius y – f(x). The size S of the surface of the part of the figure of revolution lying to the left of the plane is a function of x, i.e. s = s(x) (s(a) = 0 and s(b) = S).
Let's give the argument x an increment Δx = dx. Through the point x + dx [a; b] we also draw a plane perpendicular to the Ox axis. The function s = s(x) will receive an increment of Δs, shown in the figure as a “belt”.
Let us find the differential area ds by replacing the figure formed between the sections with a truncated cone, the generatrix of which is equal to dl, and the radii of the bases are equal to y and y + dу. The area of its lateral surface is equal to: = 2ydl + dydl.
Rejecting the product dу d1 as an infinitesimal of a higher order than ds, we obtain ds = 2уdl, or, since d1 = dx.
Integrating the resulting equality in the range from x = a to x = b, we obtain
If the curve AB is given by the parametric equations x = x(t), y = y(t), t≤ t ≤ t, then the formula for the surface area of revolution takes the form
S=2 
dt.
Example: Find the surface area of a ball of radius R.
S=2 
= 
6. Finding the work of a variable force
Variable force work
Let material point M moves along the Ox axis under the action of a variable force F = F(x) directed parallel to this axis. The work done by a force when moving point M from position x = a to position x = b (a How much work must be done to stretch the spring by 0.05 m if a force of 100 N stretches the spring by 0.01 m? According to Hooke's law, the elastic force stretching the spring is proportional to this stretch x, i.e. F = kх, where k is the proportionality coefficient. According to the conditions of the problem, a force F = 100 N stretches the spring by x = 0.01 m; therefore, 100 = k 0.01, whence k = 10000; therefore, F = 10000x. Required work based on formula A= Find the work that must be expended to pump liquid over the edge from a vertical cylindrical tank of height N m and base radius R m (Fig. 13). The work spent on lifting a body of weight p to a height h is equal to p N. But the different layers of liquid in the tank are at different depths and the height of the rise (to the edge of the tank) of the different layers is not the same. To solve the problem, we apply scheme II (differential method). Let's introduce a coordinate system. 1) The work spent on pumping out a layer of liquid of thickness x (0 ≤ x ≤ H) from a reservoir is a function of x, i.e. A = A(x), where (0 ≤ x ≤ H) (A(0) = 0, A(H) = A 0). 2) Find the main part of the increment ΔA when x changes by the amount Δx = dx, i.e. we find the differential dA of the function A(x). Due to the smallness of dx, we assume that the “elementary” layer of liquid is located at the same depth x (from the edge of the reservoir). Then dA = dрх, where dр is the weight of this layer; it is equal to g АV, where g is the acceleration of gravity, is the density of the liquid, dv is the volume of the “elementary” layer of liquid (it is highlighted in the figure), i.e. dр = g. The volume of the indicated liquid layer is obviously equal to , where dx is the height of the cylinder (layer), is the area of its base, i.e. dv = . Thus, dр = . And 3) Integrating the resulting equality in the range from x = 0 to x = H, we find A 8. Calculation of integrals using the MathCAD package When solving some applied problems, it is necessary to use the operation of symbolic integration. In this case, the MathCad program can be useful both at the initial stage (it’s good to know the answer in advance or to know that it exists) and at the final stage (it’s good to check the result using an answer from another source or another person’s solution). When solving a large number of problems, you can notice some features of solving problems using the MathCad program. Let's try to understand with several examples how this program works, analyze the solutions obtained with its help and compare these solutions with solutions obtained by other methods. The main problems when using the MathCad program are as follows: a) the program gives the answer not in the form of familiar elementary functions, but in the form of special functions that are not known to everyone; b) in some cases “refuses” to give an answer, although there is a solution to the problem; c) sometimes it is impossible to use the result obtained because of its cumbersomeness; d) does not solve the problem completely and does not analyze the solution. In order to solve these problems, it is necessary to exploit the strengths and weaknesses of the program. With its help it is easy and simple to calculate integrals of fractional rational functions. Therefore, it is recommended to use the variable replacement method, i.e. Pre-prepare the integral for the solution. For these purposes, the substitutions discussed above can be used. It should also be borne in mind that the results obtained must be examined for the coincidence of the domains of definition of the original function and the obtained result. In addition, some of the solutions obtained require additional research. The MathCad program frees the student or researcher from routine work, but cannot free him from additional analysis both when setting a problem and when obtaining any results. This paper examined the main provisions related to the study of applications of a definite integral in a mathematics course. – an analysis of the theoretical basis for solving integrals was carried out; – the material was systematized and generalized. In the process of completing the course work, examples of practical problems in the field of physics, geometry, and mechanics were considered. Conclusion The examples of practical problems discussed above give us a clear idea of the importance of the definite integral for their solvability. It is difficult to name a scientific field in which the methods of integral calculus, in general, and the properties of the definite integral, in particular, would not be used. So, in the process of completing course work, we examined examples of practical problems in the field of physics, geometry, mechanics, biology and economics. Of course, this is far from an exhaustive list of sciences that use the integral method to search for an established value when solving a specific problem and establishing theoretical facts. The definite integral is also used to study mathematics itself. For example, when solving differential equations, which in turn make an irreplaceable contribution to solving practical problems. We can say that a definite integral is a certain foundation for the study of mathematics. Hence the importance of knowing how to solve them. From all of the above, it is clear why acquaintance with the definite integral occurs within the framework of secondary school, where students study not only the concept of the integral and its properties, but also some of its applications. Literature 1. Volkov E.A. Numerical methods. M., Nauka, 1988. 2. Piskunov N.S. Differential and integral calculus. M., Integral-Press, 2004. T. 1. 3. Shipachev V.S. Higher mathematics. M., Higher School, 1990.
