Surface area of ​​a body of revolution. Finding the volume of a body by cross-sectional areas. Surface area of ​​revolution

Let a body be given in space. Let its sections be constructed by planes perpendicular to the axis passing through the points x
on it. The area of ​​the figure formed in the section depends on the point NS defining the section plane. Let this dependence be known and given continuous on function. Then the volume of the part of the body located between the planes x = a and x = in calculated by the formula

Example. Let us 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 into the segment
, and for x
the cross section of the body is 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 as follows:

Using the formula, we find the volume of the body:

Calculation of the volumes of bodies of revolution

Let on the segment [ a, b] a continuous constant-sign function is given y= f(x). Volumes of a body of revolution formed by rotation about an axis Oh(or axis OU) of a curved trapezoid bounded by a curve y= f(x) (f(x) 0) and straight y = 0, x = a, x =b, are calculated accordingly by the formulas:

, ( 19)

(20)

If the body is formed when rotating around an axis OU curved trapezoid bounded by a curve
and direct x=0, y= c, y= d, then the volume of the body of revolution is

. (21)

Example. Calculate the volume of a solid obtained by rotating a shape bounded by lines around an axis Oh.

According to formula (19), the required volume

Example. Let the line y = cosx be considered in the xOy plane on the segment .

NS This line rotates in space around an axis, and the resulting surface of revolution bounds some body of revolution (see figure). Let's find the volume of this body of revolution.

According to the formula, we get:

Surface area of ​​revolution

,
, 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 a curve given by a nonnegative 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 a curve arc is specified parametric equations
,
, and
, then

If the arc is specified in polar coordinates
, then

.

Example. Let us calculate the area of ​​the surface formed by the rotation in space around the axis of a part of the line y = located above the line segment.

As
, then the formula gives us the integral

We make the change t = x + (1/2) in the last integral and get:

In the first of the integrals on the right-hand side, we make the change z = t 2 -:

To calculate the second of the integrals on the right-hand side, we denote it and integrate it by parts, obtaining an equation for:

Moving to the left and dividing by 2, we get

whence finally

Applications of a definite integral to the solution of some problems in mechanics and physics

Variable force work. Consider the motion of a material point along the axis OX variable force f depending on the position of the point x on the axis, i.e. force as a function x... Then work A required to move a material point from a position x = a in position x = b calculated by the formula:

To calculate fluid pressure forces use Pascal's law, according to which the fluid pressure on the site is equal to its area S multiplied by the immersion depth h, on 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 with respect to the same axes Ox and Oy are calculated by the formulas

but center of mass coordinates and - according to the formulas

where l is the mass of the arc, i.e.

Example 1... Find static moments and moments of inertia about the axes Ox and Oy of the arc of the catenary y = chx at 0≤x≤1.

If no density is specified, it is assumed that the curve is uniform and
... We have: Consequently,

Example 2. Find the coordinates of the center of mass of the circular arc x = acost, y = asint, located in the first quarter. We have:

From here we get:

In applications, the following is often useful. Theorem Guilder... The area of ​​the surface formed by the rotation of the arc of a plane curve around an axis lying in the plane of the arc and not intersecting 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 a semicircle

Due to symmetry
... When the semicircle rotates around the Ox axis, a sphere is obtained, the surface area of ​​which is equal to, and the length of the semicircle is equal to n. By Gulden's theorem, we have 4

From here
, i.e. the center of mass C has coordinates C
.

2. Physical tasks. Some applications of the definite integral in solving physical problems are illustrated below in examples.

Example 4. Speed straight motion body is expressed by the formula (m / s). Find the path traversed by the body in 5 seconds from the start of the movement.

As body path with a speed v (t) over a period of time, is expressed by the integral

then we have:

NS
example. Find the area of ​​the bounded area lying between the axis and the line y = x 3 -x. Because the

the line crosses the axis at three points: x 1 = -1, x 2 = 0, x 3 = 1.

The bounded area between the line and the axis is projected onto a line segment
,and on the segment
,the line y = x 3 -x goes above the axis (that is, the line y = 0, and on - below. Therefore, the area of ​​the area can be calculated as follows:

NS
example. Let us find the area of ​​the area 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 cite a change in argument to one interval, 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
:

NS example. Let us find the volume of the body bounded by the surface of rotation of the line y = 4x-x 2 around the axis (for
).

To calculate the volume of a body of revolution, apply the formula

NS example. We calculate the length of the arc of the line y = lncosx located between the straight lines and
.

(we took the root as the value, not -cosx, since cosx> 0 for
, the arc length is

Answer:
.

Example. Let us calculate the area Q of the surface of revolution obtained by rotating the arc of the cycloid x = t-sint; y = 1-cost, for

, around the axis.

D To calculate, apply the formula:

We have:

, so

To pass under the integral sign to a variable, note that for

we get

, as well as

Moreover, let us preliminarily calculate

(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 = 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 AB curve around the Ox axis (Fig. 8).

Let us apply scheme II (differential method).

Through an arbitrary point x [a; b] draw plane P, perpendicular to the Ox axis. Plane P intersects the surface of revolution in 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 Δx = dx. Through the point x + dx [a; b] also draw a plane perpendicular to the Ox axis. The function s = s (x) will receive the increment Δs, shown in the figure in the form of a "belt".

Let us find the area differential ds, 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: = 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 takes the form

S = 2 dt.

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

S = 2 =

6. Finding Variable Force Work

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 stretching the spring is proportional to this extension x, i.e. F = kх, where k is the proportionality coefficient. 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 sought job based on the formula

A =

Find the work that needs to be spent in order to pump liquid over the edge from a vertical cylindrical tank with a height of H m and a base radius of R m (Figure 13).

The work expended on lifting a body with weight p to a height h is equal to p H. But different layers of liquid in the reservoir are at different depths and the height of rise (to the edge of the reservoir) of different layers is not the same.

To solve the problem, we will apply scheme II (differential method). Let's introduce a coordinate system.

1) The work spent on pumping out a layer of liquid with a thickness x (0 ≤ x ≤ H) from the 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 value 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 the liquid is located 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 АV, where g is the acceleration of gravity, is the density of the liquid, dv is the volume of the "elementary" layer of the liquid (it is highlighted in the figure), i.e. dр = 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 obtained 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 required to use the operation of symbolic integration. At the same time, the MathCad program can be useful both at the initial stage (it is good to know the answer in advance or to 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 with a few examples how this program works, analyze the solutions obtained with its help and compare these solutions with the solutions obtained in other ways.

The main problems when using MathCad are as follows:

a) the program gives the answer not in the form of the usual elementary functions, but in the form of special functions, which are not 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 due to 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 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 replacement method, i.e. prepare the integral for the solution in advance. 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 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.

In this paper, the main provisions related to the study of applications of a definite integral in the course of mathematics were considered.

- the analysis of the theoretical basis for the solution of integrals was carried out;

- the material was systematized and generalized.

In the course of the course work, examples of practical problems in the field of physics, geometry, mechanics were considered.

Conclusion

The examples of practical problems discussed above give us a clear idea of ​​the importance of definite 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 completing the course work, we considered examples of practical problems in the field of physics, geometry, mechanics, biology and economics. Of course, this is still 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.

Also, a certain integral is used to study mathematics itself. For example, when solving differential equations, which in turn make their irreplaceable contribution to solving problems of practical content. We can say that a definite integral is some foundation for the study of mathematics. Hence the importance of knowing the methods of solving 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 an integral and its properties, but also some of its applications.

Literature

1. Volkov E.A. Numerical methods. M., Science, 1988.

2. Piskunov NS Differential and integral calculus. M., Integral-Press, 2004.Vol. 1.

3. Shipachev V.S. Higher mathematics. M., Higher School, 1990.

Example: Find the volume of a sphere of 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 function of the cross-sectional areas 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 intersects with planes perpendicular to the height, in section we get 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 coefficient of similarity squared, i.e.

From here we obtain the function of the cross-sectional areas:

Find the volume of the pyramid:

The volume of bodies of revolution.

Consider the curve given by the equation y = f (x ). Suppose that the function f (x ) is continuous on the segment [ a, b ]. If the corresponding curvilinear trapezoid with bases a and b rotate around the Ox axis, then we get the so-called body of revolution.

y = f (x)

Surface area of ​​a body of revolution.

M i B

Definition: Surface area of ​​revolution curve AB around a given axis is called the limit to which the areas of the surfaces of rotation of the polygonal lines inscribed in the curve AB tend to, as the longest of the lengths of the links of these polygonal lines tends to zero.

We split 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 coordinates x i and y i ... When the polyline rotates around the axis, we obtain a surface consisting of the lateral surfaces of truncated cones, the area of ​​which is D P i ... This area can be found by the formula:

Therefore, I will go straight to the basic concepts and practical examples.

Let's look at a laconic picture

And remember: what can be calculated using definite integral ?

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

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

What else? Was considered not so long ago arc length problem .

And today we will learn how to calculate one more characteristic - one more area. Imagine that the 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 without a lid. As the donkey Eeyore would say, a heartbreaking sight =)

To exclude an 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 areas - external and internal. So, all further calculations imply the area only the outer surface.

In a rectangular coordinate system, the surface area of ​​revolution is calculated by the formula:

or, if more compact: .

The same requirements are imposed on the function and its derivative as in finding arc arc lengths , but in addition, the curve must be located higher axis. This is essential! 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:

Torus surface area

In a nutshell, torus is a donut... A textbook example, considered 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 asks circle unit radius centered at a point. That being said, it is easy to get two functions:

- sets the upper semicircle;
- sets the lower semicircle:

The essence is crystal clear: circle rotates around the abscissa axis and forms surface donut. The only thing here, in order to avoid gross reservations, should be careful in terminology: if you rotate a circle bounded by a circle , you get a geometric body, that is, the donut itself. And now talk about the area of ​​it surface, 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 abscissa axis. We use the formula ... As I have repeatedly advised, it is more convenient to carry out actions in stages:

Take the function and find her derivative :

Finally, we load the result into a 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 decision, rather than preliminarily reasoning about the symmetry of the figure relative to the ordinate axis.

2) Find the surface area, which is obtained by rotating the "red" arc around the abscissa axis. All actions will actually differ in only one sign. I will make 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 a general way - to 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 ran the solution on specific numbers.

If you need to calculate the volume of the donut itself, please refer to the textbook as an express reference:

According to the theoretical remark, we are considering the upper semicircle. It is "drawn" when the parameter value changes within (it is easy to see that on this interval), thus:

Answer:

If you solve the problem in general form, you get exactly the school formula for the area of ​​a sphere, where is its radius.

Something hurt a simple task, I even felt ashamed…. I suggest you fix this flaw =)

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 intuitively guess about the formula for calculating the surface area obtained by rotating a curve around the ordinate axis. And, of course, again the advantage of parametric equations should be noted - they do not need to be modified in any way; 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 defined parametrically ... The surface of rotation will resemble ... I do not even know what to compare with ... something unearthly - a rounded shape with a pointed recess in the middle. For the case of rotation of the cycloid around the axis, the association instantly came to mind - an oblong ball for playing rugby.

Solution and answer at the end of the lesson.

We conclude our fascinating review with a case polar coordinates ... Yes, just a review, if you look at textbooks on mathematical analysis (Fichtengolts, Bokhan, Piskunov, other authors), you can get a dozen (or even noticeably more) standard examples, among which there may well be the problem you need.

How to calculate the surface area of ​​revolution,
if the line is specified in a polar coordinate system?

If the curve is specified in polar coordinates equation, and the function has a continuous derivative at 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 certainly non-negative). Therefore, it is necessary to consider the values ​​of the angle from the range, in other words, the curve should be located higher polar axis and its continuation. As you can see, the story is the same as in the previous two paragraphs.

Example 5

Calculate the surface area formed by rotating the cardioid about the polar axis.

Solution: the graph of this curve can be viewed 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 bull's-eye.

The solution technique is standard. Let's find the derivative with respect to "phi":

Let's compose and simplify the root:

Hopefully with supernumerary