5. Finding the surface area of ​​bodies of rotation

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 ​​rotation 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.

The required job 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 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 significance 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 looked at 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.

I. Volumes of bodies of revolution. Preliminarily study Chapter XII, paragraphs 197, 198 from the textbook by G. M. Fikhtengolts * Analyze in detail the examples given in paragraph 198.

508. Calculate the volume of a body formed by rotating an ellipse around the Ox axis.

Thus,

530. Find the surface area formed by rotation around the Ox axis of the sinusoid arc y = sin x from point X = 0 to point X = It.

531. Calculate the surface area of ​​a cone with height h and radius r.

532. Calculate the surface area formed

rotation of the astroid x3 -)- y* - a3 around the Ox axis.

533. Calculate the surface area formed by rotating the loop of the curve 18 ug - x (6 - x) z around the Ox axis.

534. Find the surface of the torus produced by the rotation of the circle X2 - j - (y-3)2 = 4 around the Ox axis.

535. Calculate the surface area formed by the rotation of the circle X = a cost, y = asint around the Ox axis.

536. Calculate the surface area formed by the rotation of the loop of the curve x = 9t2, y = St - 9t3 around the Ox axis.

537. Find the surface area formed by rotating the arc of the curve x = e*sint, y = el cost around the Ox axis

from t = 0 to t = —.

538. Show that the surface produced by the rotation of the cycloid arc x = a (q> -sin φ), y = a (I - cos φ) around the Oy axis is equal to 16 u2 o2.

539. Find the surface obtained by rotating the cardioid around the polar axis.

540. Find the surface area formed by the rotation of the lemniscate Around the polar axis.

Additional tasks for Chapter IV

Areas of plane figures

541. Find the entire area of ​​the region bounded by the curve And the axis Ox.

542. Find the area of ​​the region bounded by the curve

And the axis Ox.

543. Find the part of the area of ​​the region located in the first quadrant and bounded by the curve

l coordinate axes.

544. Find the area of ​​the region contained inside

loops:

545. Find the area of ​​the region bounded by one loop of the curve:

546. Find the area of ​​the region contained inside the loop:

547. Find the area of ​​the region bounded by the curve

And the axis Ox.

548. Find the area of ​​the region bounded by the curve

And the axis Ox.

549. Find the area of ​​the region bounded by the Oxr axis

straight and curve

Surface of revolution- a surface formed by rotation around a straight line (surface axis) of an arbitrary line (straight, flat or spatial curve). For example, if a straight line intersects the axis of rotation, then when it rotates, a conical surface will be obtained; if it is parallel to the axis, it will be cylindrical; if it intersects the axis, a single-sheet hyperboloid of revolution will be obtained. The same surface can be obtained by rotating a wide variety of curves. The area of ​​the surface of revolution formed by the rotation of a plane curve of finite length around an axis lying in the plane of the curve but not intersecting the curve is equal to the product of the length of the curve and the length of a circle with a radius equal to the distance from the axis to the center of mass of the curve. This statement is called Gylden's second theorem, or Pappus's centroid theorem.

The area of ​​the surface of revolution formed by the rotation of a curve around an axis can be calculated using the formula

For the case when the curve is specified in the polar coordinate system, the formula is valid

Mechanical applications of the definite integral (work of forces, static moments, center of gravity).

Calculation of work of forces

A material point moves along a continuously differentiable curve, while it is acted upon by a force directed tangentially to the trajectory in the direction of motion. Total work done by force F(s):

If the position of a point on the motion trajectory is described by another parameter, then the formula takes the form:

Calculation of static moments and center of gravity
Let on the coordinate plane Oxy some mass M be distributed with density p = p(y) on a certain set of points S (this can be an arc of a curve or a bounded flat figure). Let us denote s(y) - the measure of the specified set (arc length or area).

Definition 2. Number called kth moment mass M relative to the Ox axis.
At k = 0 M 0 = M - mass,
k = 1 M 1 - static moment,
k = 2 M 2 - moment of inertia.

Moments about the Oy axis are introduced similarly. In space, the concepts of moments of mass relative to coordinate planes are introduced in a similar way.
If p = 1, then the corresponding moments are called geometric. The coordinates of the center of gravity of a homogeneous (p - const) flat figure are determined by the formulas:

where M 1 y, M 1 x are the geometric static moments of the figure relative to the Oy and Ox axes; S is the area of ​​the figure.

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 and 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, exactly a review, if you look at the textbooks on mathematical analysis(Fichtengolts, Bokhan, Piskunov, other authors), you can get a good dozen (or even much more) standard examples, among which it is quite possible that you will find the task 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 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 , therefore: (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 independent decision:

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 exact value 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 cannot be taken - in front of you is an untaken integral, the 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 ( bar chart), 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 will not even give simple formula. Not because I’m lazy, but because of the principle of my solution book: 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 found 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.

This formula is called the formula for the volume of a body by the area of ​​parallel sections.

Example. Find the volume of the ellipsoid x 2 + y 2 + z 2 = 1. a 2b 2c 2

By cutting the ellipsoid with a plane parallel to the Oyz plane and at distances from it (-а ≤х ≤а), we obtain an ellipse (see Fig. 15):

The area of ​​this ellipse is

Therefore, according to formula (16), we have

Calculating surface area of ​​revolution

Let the AB curve be a graph of the function y = f (x) ≥ 0, where x [a,b], a function y = f (x) and its derivative y" = f" (x) are continuous on this segment.

Then the area S of the surface formed by the rotation of curve AB around the Ox axis is calculated by the formula

If the AB curve is given by the parametric equations х = x (t), у = у (t), t 1 ≤t ≤t 2, then the formula for the surface area of ​​rotation takes the form

S x = 2 π ∫ y (t )(x ′ (t ))2 + (y ′ (t ))2 dt .

Example Find the surface area of ​​a ball of radius R. Solution:

We can assume that the surface of the ball is formed by the rotation of the semicircle y = R 2 − x 2, - R ≤x ≤R, around the Ox axis. Using formula (19) we find

2 π ∫ R2 − x2 + x2 dx= 2 π Rx− R R = 4 π R2 .

−R

Example. Given a cycloid x = a (t − sin t), 0 ≤ t ≤ 2 π. y = a (1− cost) ,

Find the surface area formed by rotating it around the Ox axis. Solution:

When half of the cycloid arc rotates around the Ox axis, the surface area of ​​rotation is equal to

Calculating the arc length of a plane curve

Rectangular coordinates

Let in an arc, when the number of links of the broken line increases indefinitely, and the length of the largest rectangular coordinates is given a flat curve AB, the equation of which is y = f(x), where a ≤ x≤ b.

The length of the arc AB is understood as the limit to which the length of the broken line inscribed in this link tends to zero. Let us show that if the function y = f(x) and its derivative y′ = f′ (x) are continuous on the segment [a ,b ], then the curve AB has a length equal to

If the equation of the AB curve is given in parametric form

x = x(t) , α ≤ t ≤ β , y= y(t) ,

where x (t) and y (t) are continuous functions with continuous derivatives and x (α) = a, x (β) = b, then the length l of curve AB is found by the formula

This means l = 2π R. If the equation of a circle is written in the parametric form = R cost, y = R sint (0 ≤t ≤ 2π ), then

Polar coordinates

Let the curve AB be given by the equation in polar coordinates r =r (ϕ),α ≤ ϕ ≤ β. Let us assume that r (ϕ ) and r" (ϕ ) are continuous on the interval [α , β ].

If in the equalities x = r cosϕ, y = r sinϕ, connecting polar and Cartesian coordinates,

the angle ϕ is considered a parameter, then the curve AB can be set parametricallyx = r (ϕ) cos ϕ,

y = r(ϕ) sinϕ.

Applying formula (15), we obtain l = ∫ r 2 + r ′ 2 d ϕ .

Example Find the length of the cardioid r =a (1 + cosϕ ). Solution:

The cardioid r =a (1 + cosϕ) has the form shown in Figure 14. It is symmetrical about the polar axis. Let's find half the length of the cardioid:

Thus, 1 2 l = 4 a. This means l = 8a.