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 area surfaces 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 discussed the main provisions related to the study of applications definite integral in 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 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 learn 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.

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 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.

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

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.

Therefore, I will immediately move on to the basic concepts and practical examples.

Let's look at the simple picture

And remember: what can be calculated using definite integral?

First of all, of course, area of ​​a curved trapezoid. Familiar from school days.

If this figure rotates around the coordinate axis, then we are talking about finding volume of a body of revolution. Simple too.

What else? Was reviewed not long ago arc length problem .

And today we will learn how to calculate one more characteristic - another area. Imagine that line rotates around the axis. The result of this action is geometric figure, called surface of rotation. IN in this case it resembles a pot without a bottom. And without a lid. As Eeyore would say, a heartbreaking sight =)

To eliminate any ambiguous interpretation, I will make a boring but important clarification:

With geometric point our “pot” has a view infinitely thin wall and two surfaces with equal areas - external and internal. So, all further calculations imply the area only external surface.

In a rectangular coordinate system, the area of ​​the surface of revolution 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 you will have to add a minus sign to the formula in order to preserve the geometric meaning of the problem.

Let's look at an undeservedly overlooked figure:

Torus surface area

In a nutshell, torus is a donut. A textbook example, discussed in almost all textbooks on matan, is devoted to finding volume torus, and therefore, for the sake of variety, I will analyze the rarer problem of its surface area. First with specific numeric values:

Example 1

Calculate the surface area of ​​a torus obtained by rotating a circle around the axis.

Solution: as you know, the equation sets circle unit radius with center at point . In this case, it is easy to obtain two functions:

– sets the upper semicircle;
– sets the lower semicircle:

The point is crystal clear: circle rotates around the x-axis and forms surface bagel. The only thing here, in order to avoid crude reservations, is to be careful in terminology: if you rotate circle, bounded by a circle , then it will be geometric body, that is, the bagel itself. And now we’re talking about its area surfaces, which obviously needs to be calculated as the sum of areas:

1) Find the surface area that is obtained by rotating the “blue” arc around the abscissa axis. We use the formula . As I have repeatedly advised, it is more convenient to carry out actions in stages:

Let's take the function and find her 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 during the solution, rather than preliminarily reasoning about the symmetry of the figure relative to the ordinate axis.

2) Find the surface area that is obtained by rotating the “red” arc around the abscissa axis. All actions will differ in fact only by one sign. I’ll write the solution in a different style, which, of course, also has the right to life:


3) Thus, the surface area of ​​the torus is:

Answer:

The problem could be solved in general view– calculate the surface area of ​​a torus obtained by rotating a 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 textbook as a quick reference:

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 form, you will 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