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 
. At the same time, the “drawing direction” 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 players" will take negative values and you will have to put a minus sign in front of the integral.
Example 3
Calculate the area of the sphere obtained by rotating the circle about the axis.
Decision: from the materials of the article about area and volume with a parametrically given line you know that the equations define a circle centered at the origin with radius 3.
well and sphere , for those who forgot, is the surface ball(or spherical surface).
We adhere to the developed solution scheme. Let's find derivatives: 
Let's compose and simplify the "formula" root:
Needless to say, it turned out candy. Check out for comparison how Fikhtengoltz butted heads with the square ellipsoid of revolution.
According to the theoretical remark, we consider the upper semicircle. It is "drawn" when changing the value of the parameter within (it is easy to see that 
on this interval), thus:
Answer:
If the problem is solved in general view, then exactly school formula the area of a sphere, where is its radius.
Something painfully simple problem, 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 deduce or intuit the formula for calculating the surface area obtained by rotating a curve around the y-axis. And, of course, the advantage of parametric equations should again be noted - they do not need to be modified somehow; no need to bother with finding other limits of integration.
The cycloid graph can be viewed on the page Area and volume if the line is set parametrically. The surface of rotation will resemble ... I don’t even know what to compare it with ... something unearthly - rounded with a pointed depression in the middle. Here, for the case of rotation of the cycloid around the axis, the association instantly came to mind - an oblong rugby ball.
Solution and answer at the end of the lesson.
We conclude our fascinating review with a case polar coordinates. Yes, it is an overview, if you look at the tutorials on mathematical analysis(Fikhtengolts, Bokhan, Piskunov, and other authors), you can get a dozen (or even much more) standard examples, among which it is quite possible that you will find the task you need.
How to calculate area surfaces of revolution,
if the line is given in polar coordinate system?
If the curve is set to 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 geometric sense problem integrand 
, and this is achieved only if ( and are known to be 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 extensions. As you can see, the same story as in the previous two paragraphs.
Example 5
Calculate the area of the surface formed by the rotation of the cardioid around the polar axis.
Decision: 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 on the gap (which, in fact, is also 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":
Compose and simplify the root:
I hope with supernumeraries trigonometric formulas no one had any problems.
We use the formula:
In between 
, hence: 
(I described in detail how to properly get rid of the root in the article Curve arc length).
Answer: 
An interesting and short task for independent solution:
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 middle, in which the juicy pulp is exposed on both sides. This body is called spherical layer, and its bounding surface (orange peel) - ball belt.
Readers familiar with polar coordinates, easily presented the drawing of the problem: the equation defines a circle centered at the pole of radius , from which rays 
cut off lesser arc. This arc rotates around the polar axis and thus a spherical belt is obtained.
Now you can with clear conscience and eat an orange with a light heart, on this delicious note we will complete the lesson, do not spoil your appetite with other examples =)
Solutions and answers:
Example 2:Decision
: calculate the area of the surface formed by the rotation of the upper branch around the x-axis. We use the formula 
.
In this case: 
;
Thus:
Answer:
Example 4:Decision
: use the formula 
. The first arc of the cycloid is defined on the segment .
Let's find derivatives:
Compose and simplify the root:
So the surface area of revolution is:
In between 
, That's why 
First integralintegrate by parts
:
In the second integral we usetrigonometric formula
.
Answer:
Example 6:Decision
: use the formula:
Answer:
Higher mathematics for correspondence students and not only >>>
(Go to main page)
How to calculate a definite integral
using the trapezoid formula and the Simpson method?
Numerical methods is a fairly large section of higher mathematics and serious textbooks on this topic have hundreds of pages. In practice, in tests, some tasks are traditionally proposed for solving by numerical methods, and one of the common tasks is approximate calculation definite integrals. In this article, I will consider two methods for the approximate calculation of a definite integral − trapezoidal method and simpson's method.
What do you need to know to master these methods? It sounds funny, but you may not be able to take integrals at all. And even do not understand what integrals are. Of the technical means, you will need a microcalculator. Yes, yes, we are waiting for routine school calculations. Better yet, download my semi-automatic calculator for the trapezoidal method and the Simpson method. The calculator is written in Excel and will allow you to reduce the time for solving and processing tasks tenfold. A video manual is included for Excel teapots! By the way, the first video with my voice.
First, let's ask ourselves the question, why do we need approximate calculations at all? It seems to be possible to find the antiderivative of the function and use the Newton-Leibniz formula, calculating the exact value of a certain integral. As an answer to the question, let us immediately consider demo with drawing.
Calculate a definite integral
Everything would be fine, but in this example the integral is not taken - before you is not taken, the so-called integral logarithm. Does this integral even exist? Let's depict the graph of the integrand in the drawing: 
Everything is fine. Integrand continuous on the segment and the definite integral is numerically equal to the shaded area. Yes, that's just one snag - the integral is not 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 say you get an approximate answer of 5.347. In fact, it may not be entirely correct (actually, let's say the more accurate answer is 5.343). Our task is only in that to round the result to three decimal places.
2) Calculate the definite integral approximately, with a certain precision. For example, calculate the definite integral approximately with an accuracy of 0.001. What does it mean? This means that if an approximate answer of 5.347 is obtained, then all figures must be reinforced concrete correct. To be more precise, the answer 5.347 should differ from the truth modulo (in one direction or another) by no more than 0.001.
There are several basic methods for the approximate calculation of a definite integral that occurs in problems:
Rectangle method. The segment of integration is divided into several parts and a step figure is constructed ( bar graph), which is close in area to the desired area: 
Do not judge strictly by the drawings, the accuracy is not perfect - they only help to understand the essence of the methods.
In this example, the segment of integration is divided into three segments: 
. Obviously, the more frequent the partition (the more smaller intermediate segments), the higher the accuracy. The method of rectangles gives a rough approximation of the area, apparently, therefore, it is very rare in practice (I remembered only one practical example). In this regard, I will not consider the method of rectangles, and will not even give a simple formula. Not because of laziness, but because of the principle of my solution book: what is extremely rare in practical tasks is not considered.
Trapezoidal method. The idea is similar. The integration segment is divided into several intermediate segments, and the graph of the integrand approaches broken line line: 
So 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, of course, 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 in this article several examples will be analyzed.
Simpson's method (parabola method). This is a more perfect way - the graph of the integrand is approached not by a broken line, but by small parabolas. How many intermediate segments - so many small parabolas. If we take the same three segments, then the Simpson method will give an even more accurate approximation than the rectangle method or the trapezoid method.
I don’t see the point in building a drawing, since visually the approximation will be superimposed on the graph of the function (the broken line of the previous paragraph - and even then it almost coincided).
The task of calculating a definite integral using the Simpson formula is the most popular task in practice. And the method of parabolas will be given considerable attention.
Greetings, dear students of Argemony University!
Today we will continue to study the materialization of objects. Last time we rotated flat figures and got three-dimensional bodies. Some of them are very tempting and useful. I think that much that the magician invents can be used in the future.
Today we will rotate the curves. It is clear that in this way we can get some kind of object with very thin edges (a cone or a bottle for potions, a vase for flowers, a glass for drinks, etc.), because a rotating curve can create just such 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 holey cup from which Sir Shurf Lonley-Lockley drank all the time.
So we will create a leaky bowl and a non-perforated one, and calculate the area of the created surface. I think that for some reason it (in general, the surface area) will be needed - well, at least for applying a special magic paint. And 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 how 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 it rotates around the OY axis, just a bowl is obtained. No bottom.
The spell to calculate 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 is rotating. 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 do not give us anything, so let's not bother, but switch to the language of formulas, where this differential is explicitly presented for all cases known to us:
- Cartesian coordinate system;
- records of 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 consider 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 obtained in the form of a circle of unit radius. And we know how the area of a circle is calculated (according to the formula pi * r ^ 2. For our case, the area of \u200b\u200bthe circle will be equal to pi), but we will calculate it using a new formula - for verification.
The distance from the axis of rotation to any point of this piece of the curve is also x.
Well, our calculations are correct, which pleases.
And now homework.
1. Find the surface area obtained by rotating the polyline ABC, where A=(1; 5), B=(1; 2), C=(6; 2), around the OX axis.
Advice. Record 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. Three items, I think, is enough.
Therefore, I will immediately move on to the basic concepts and practical examples.
Let's look at a simple picture
And remember: what can be calculated using definite integral?
First of all, of course, area of a curved trapezoid. Known since school days.
If this figure rotates around the coordinate axis, then we are already talking about finding body of revolution. It's also simple.
What else? Recently reviewed arc length problem .
And today we will learn how to calculate one more characteristic - one more area. Imagine that line revolves around the axis. As a result of this action, a geometric figure is obtained, called surface of revolution. In this case, it resembles such a pot without a bottom. And no cover. As the donkey Eeyore would say, a heartbreaking sight =)
To eliminate ambiguous interpretation, I will make a boring but important clarification:
from a geometric point of view, our "pot" has infinitely thin wall and two surfaces with the same area - external and internal. So, all further calculations imply the area only the outer surface.
In a rectangular coordinate system, the surface area of rotation is calculated by the formula:
or, more compactly: 
.
The same requirements are imposed on the function and its derivative as when finding curve arc length, but, in addition, the curve must be located higher axes . This is significant! It is easy to understand that if the line is located under axis, then the integrand will be negative: 
, and therefore a minus sign will have to be added to the formula in order to preserve the geometric meaning of the problem.
Consider an undeservedly overlooked figure:
Surface area of a torus
In a nutshell, tor is a donut. A textbook example, considered in almost all matan textbooks, is dedicated to finding volume torus, and therefore, for the sake of variety, I will analyze the rarer problem of its surface area. First with specific numerical values:
Example 1
Calculate the surface area of a torus obtained by rotating a circle 
around the axis.
Decision: how do you know the equation 
sets circle unit radius centered at . This makes it easy to get two functions: 
– sets the upper semicircle; 
– sets the lower semicircle: 
The essence is crystal clear: circle rotates around the x-axis and forms surface bagel. The only thing here, in order to avoid gross reservations, is to be careful in terminology: if you rotate a circle, bounded by a circle 
, then you get a geometric body, that is, the bagel itself. And now talk about square it surfaces, which obviously needs to be calculated as the sum of the areas:
1) Find the surface area, which is obtained by rotating the "blue" arc 
around the x-axis. We use the formula 
. As I have repeatedly advised, it is more convenient to carry out actions in stages:
We take a function 
and find it derivative:
And finally, we load the result into the formula: 
Note that in this case it turned out to be more rational double the integral of an even function in the course of the solution, rather than preliminarily discussing the symmetry of the figure with respect to the y-axis.
2) Find the surface area, which is obtained by rotating the "red" arc 
around the x-axis. All actions will differ in fact by only one sign. I will design the solution in a different style, which, of course, also has the right to life: 
3) Thus, the surface area of the torus:
Answer:
The problem could be solved in a general way - to calculate the surface area of the torus obtained by rotating the circle around the abscissa axis, and get the answer 
. However, for clarity and greater simplicity, I carried out the solution on specific numbers.
If you need to calculate the volume of the donut itself, please refer to the tutorial as a quick reference: 
According to the theoretical remark, we consider the upper semicircle. It is "drawn" when changing the value of the parameter within (it is easy to see that 
on this interval), thus:
Answer:
If we solve the problem in general terms, we get exactly the school formula for the area of a sphere, where is its radius.
Something painfully simple problem, 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 deduce or intuit the formula for calculating the surface area obtained by rotating a curve around the y-axis. And, of course, the advantage of parametric equations should again be noted - they do not need to be modified somehow; no need to bother with finding other limits of integration.
The cycloid graph can be viewed on the page Area and volume if the line is set parametrically. The surface of rotation will resemble ... I don’t even know what to compare it with ... something unearthly - rounded with a pointed depression in the middle. Here, for the case of rotation of the cycloid around the axis, the association instantly came to mind - an oblong rugby ball.
Solution and answer at the end of the lesson.
We conclude our fascinating review with a case polar coordinates. Yes, it’s a review, if you look into textbooks on mathematical analysis (by Fikhtengolts, Bokhan, Piskunov, and other authors), you can get a good dozen (or even noticeably more) standard examples, among which it is quite possible that you will find the problem you need.
How to calculate the surface area of revolution,
if the line is given in polar coordinate system?
If the curve is set to 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 
, and this is achieved only if ( and are known to be 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 extensions. As you can see, the same story as in the previous two paragraphs.
Example 5
Calculate the area of the surface formed by the rotation of the cardioid around the polar axis.
Decision: 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 on the gap (which, in fact, is also 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":
Compose and simplify the root:
I hope with supernumeraries
Let a body be given in space. Let its sections be constructed by planes perpendicular to the axis passing through the points x 
on her. The area of the figure formed in the section depends on the point X, which defines the section plane. Let this dependence be known and be given continuous on 
function. Then the volume of the part of the body located between the planes x=a and x=v calculated by the formula 
Example. Let's find the volume of a bounded body enclosed between the surface of a cylinder of radius :, a horizontal plane and an inclined plane z=2y and lying above the horizontal plane .
Obviously, the body under consideration is projected onto the axis of the segment 
, and for x 
the cross section of the body is a right triangle with legs y and z=2y, where y can be expressed in terms of x from the cylinder equation:
Therefore, the cross-sectional area S(x) is:
Applying the formula, we find the volume of the body:
Calculation of volumes of bodies of revolution
Let on the segment[ a,
b] is a continuous sign-constant function y=
f(x).
Volumes of a body of revolution formed by rotation around an axis Oh(or axes OU) curvilinear trapezoid bounded by a curve y=
f(x)
(f(x)
0) and direct y=0, x=a, x=b, are calculated according to the formulas:
,
( 19)
(20)
If a body is formed by rotation around an axis OU curvilinear trapezoid bounded by a curve 
and direct x=0,
y=
c,
y=
d, then the volume of the body of revolution is equal to
.
(21)
Example. Calculate the volume of a body obtained by rotating a figure bounded by lines around an axis Oh.
According to formula (19), the desired volume
Example. Let the line y=cosx be considered in the xOy plane on the segment 
.
E 
that line rotates in space around the axis, and the resulting surface of revolution limits some body of revolution (see Fig.). Find the volume of this body of revolution.
According to the formula, we get:
Surface area of rotation
,
, rotates around the Ox axis, then the surface area of rotation is calculated by the formula 
, where a and b- abscissas of the beginning and end of the arc.
If the arc of the curve given by a non-negative function 
,
, rotates around the Oy axis, then the surface area of rotation is calculated by the formula
,
where c and d are the abscissas of the beginning and end of the arc.
If the arc of the curve is given parametric equations
,
, and 
, then
If the arc is set to polar coordinates
, then
.
Example. Calculate the area of the surface formed by rotation in space around the axis of the part of the line y= 
located above the cutoff line.
As 
, then the formula gives us the integral
Let's make the change t=x+(1/2) in the last integral and get:
In the first of the integrals on the right side, we make the change z=t 2 -:
To calculate the second of the integrals on the right side, we denote it and integrate by parts, obtaining an equation for:
Moving to the left side and dividing by 2, we get
where, finally,
Applications of the definite integral to the solution of some problems of mechanics and physics
Variable force work. Consider the motion of a material point along the axis OX under the action of a variable force f, depending on the position of the point x on the axis, i.e. a force that is a function x. Then work A, necessary to move a material point from a position x = a into position x = b calculated by the formula:
To calculate liquid pressure force use Pascal's law, according to which the pressure of a liquid on a platform is equal to its area S multiplied by the immersion depth h, on the density ρ and the acceleration of gravity g, i.e.
.
1.
Moments and centers of mass of plane curves. If the arc of the curve is given by the equation y=f(x), a≤x≤b, and has a density 
, then static moments of this arc, M x and M y with respect to the coordinate axes Ox and Oy are
;
moments of inertia I X and I y relative to the same axes Ox and Oy are calculated by the formulas
a center of mass coordinates
and 
- by formulas
where l is the mass of the arc, i.e.
Example 1. Find the static moments and moments of inertia about the axes Ox and Oy of the catenary arc y=chx for 0≤x≤1.
If density is not specified, the curve is assumed to be uniform and 
. We have: Therefore,
Example 2 Find the coordinates of the center of mass of the circle arc x=acost, y=asint located in the first quadrant. We have:
From here we get:
In applications, the following is often useful. Theorem gulden. The surface area formed by the rotation of an arc of a plane curve around an axis that lies in the plane of the arc and does not intersect it is equal to the product of the length of the arc and the length of the circle described by its center of mass.
Example 3 Find the coordinates of the center of mass of the semicircle 
Because of the symmetry 
. When a semicircle rotates around the Ox axis, a sphere is obtained, the surface area of \u200b\u200bwhich is equal, and the length of the semicircle is equal to pa. By Gulden's theorem, we have 4 
From here 
, i.e. center of mass C has coordinates C 
.
2. Physical tasks. Some applications of the definite integral in solving physical problems are illustrated below in the examples.
Example 4 The speed of the rectilinear movement of the body is expressed by the formula (m / s). Find the path traveled by the body in 5 seconds from the start of the movement.
As path taken by the body with the speed v(t) for the time interval , is expressed by the integral
then we have:
P 
example. Let's find the area of the limited area lying between the axis and the line y=x 3 -x. Insofar as
the line crosses the axis at three points: x 1 \u003d -1, x 2 \u003d 0, x 3 \u003d 1.
The limited area between the line and the axis is projected onto a segment 
,
and on the segment 
,
line y=x 3 -x goes above the axis (i.e. line y=0, and on 
- below. Therefore, the area of the region can be calculated as follows:
P 
example. Find the area of the region enclosed between the first and second turns of the Archimedes spiral r=a 
(a>0) and a segment of the horizontal axis 
.
The first turn of the spiral corresponds to a change in the angle in the range from 0 to, and the second - from to. To bring an argument change 
to one gap, we write the equation of the second turn of the spiral in the form 
,
. Then the area can be found by the formula, putting 
and 
:
P 
example. Let's find the volume of the body bounded by the surface of rotation of the line y=4x-x 2 around the axis (with 
).
To calculate the volume of a body of revolution, we apply the formula
P 
example. Calculate the arc length of the line y=lncosx located between the straight lines and 
.
(we took as the value of the root , and not -cosx, since cosx > 0 when 
, the length of the arc is
Answer: 
.
Example. Calculate the area Q of the surface of revolution obtained by rotating the arc of the cycloid x=t-sint ; y=1-cost, with 
, around the axis.
D 
To calculate, we apply the formula:
We have: 
, so
To pass under the integral sign to a variable, we note that when 
we get 
, as well as 
In addition, we precompute
(so 
) and
We get:
Making the substitution , we arrive at the integral
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 \u003d f (x) and its derivative y "\u003d f" (x) are continuous on this segment.
Let's find the area S of the surface formed by the rotation of the curve AB around the Ox axis (Fig. 8).
We apply scheme II (differential method).
Through an arbitrary point x [a; b] draw the plane П, perpendicular to the axis Oh. The plane P intersects the surface of revolution along a circle with radius y - f(x). The value 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 Δх = dх. Through the point x + dx [a; b] also draw a plane perpendicular to the x-axis. The function s = s(x) will receive an increment of Δs, shown in the figure as a "belt".
Let us find the differential of the area ds, replacing the figure formed between the sections by a truncated cone, the generatrix of which is equal to dl, and the radii of the bases are equal to y and y + dy. Its lateral surface area is: = 2ydl + dydl.
Discarding the product dу d1 as an infinitesimal 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 area of the surface of revolution becomes
S=2 
dt.
Example: Find the surface area of a sphere of radius R.
S=2 
= 
6. Finding the work of a variable force
Variable force work
Let the material point M move along the Ox axis under the action of a variable force F = F(x) directed parallel to this axis. The work done by the force when moving point M from position x = a to position x = b (a What 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 that stretches the spring is proportional to this stretch x, i.e. F = kx, where k is the coefficient of proportionality. According to the condition of the problem, the force F = 100 N stretches the spring by x = 0.01 m; therefore, 100 = k 0.01, whence k = 10000; therefore, F = 10000x. The desired work based on the formula A= Find the work that must be expended to pump liquid over the edge from a vertical cylindrical tank of height H m and base radius R m (Fig. 13). The work expended on raising a body of weight p to a height h is equal to p H. But the different layers of the 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). We introduce a coordinate system. 1) The work expended on pumping out a layer of liquid of thickness x (0 ≤ x ≤ H) from the tank is a function of x, i.e. A \u003d A (x), where (0 ≤ x ≤ H) (A (0) \u003d 0, A (H) \u003d A 0). 2) We find the main part of the increment ΔA when x changes by Δx = dx, i.e. we find the differential dA of the function A(x). In view of the smallness of dx, we assume that the "elementary" liquid layer is at the same depth x (from the edge of the reservoir). Then dА = dрх, where dр is the weight of this layer; it is equal to g AV, where g is the acceleration of gravity, is the density of the liquid, dv is the volume of the “elementary” liquid layer (it is highlighted in the figure), i.e. dr = g. The volume of this 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 \u003d 0 to x \u003d H, we find A 8. Calculation of integrals using the MathCAD package When solving some applied problems, it is required to use the operation of symbolic integration. In this case, the MathCad program can be useful both at the initial stage (it is good to know the answer in advance or know that it exists) and at the final stage (it is good to check the result obtained using the answer from another source or the solution of another person). When solving a large number of problems, you can notice some features of solving problems using the MathCad program. Let's try to understand how this program works with a few examples, analyze the solutions obtained with its help and compare these solutions with the 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 far from known to everyone; b) in some cases "refuses" to give an answer, although the problem has a solution; c) sometimes it is impossible to use the result obtained because of its bulkiness; d) solves the problem incompletely and does not analyze the solution. In order to solve these problems, it is necessary to use 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 substitution 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 result obtained. In addition, some of the obtained solutions 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. In this paper, the main provisions related to the study of applications of a definite integral in the course of mathematics were considered. – an analysis of the theoretical basis for solving integrals was carried out; - the material was subjected to systematization and generalization. During the course work, examples of practical problems in the field of physics, geometry, mechanics were considered. Conclusion The examples of practical problems considered above give us a clear idea of the significance of a certain integral for their solvability. It is difficult to name a scientific area in which the methods of integral calculus, in general, and the properties of a definite integral, in particular, would not be applied. So in the process of doing the course work, we considered examples of practical problems in the field of physics, geometry, mechanics, biology and economics. Of course, this is by no means an exhaustive list of sciences that use the integral method to find a set value when solving a specific problem, and to establish theoretical facts. Also, the definite integral is used to study mathematics itself. For example, when solving differential equations, which in turn make an indispensable contribution to solving practical problems. We can say that the definite integral is a kind of 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 a definite integral occurs even within the framework of a secondary general education 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.
