Applications of the definite integral are examples of solutions. Calculating the area of a flat figure

Lecture 21 Applications definite integral(2h)

a) figure area

As noted in Lecture 19, numerically equal to area curvilinear trapezoid, bounded curve at = f(x) , straight lines X = a, X = b and segment [ a, b] of the OX axis. At the same time, if f(x) £ 0 on [ a, b], then the integral should be taken with a minus sign.

If on given segment function at = f(x) changes sign, then to calculate the area of the figure enclosed between the graph of this function and the OX axis, one should divide the segment into parts, on each of which the function retains its sign, and find the area of each part of the figure. The desired area in this case is the algebraic sum of the integrals over these segments, and the integrals corresponding to the negative values of the function are taken in this sum with a minus sign.

If the figure is bounded by two curves at = f 1 (x) and at = f 2 (x), f 1 (x)£ f 2 (x), then, as follows from Fig. 9, its area is equal to the difference between the areas of curvilinear trapezoids a sun b and a AD b, each of which is numerically equal to the integral. Means,

Note that the area of the figure shown in Figure 10, a is found by the same formula: S = (Prove it!). Think about how to calculate the area of \u200b\u200bthe figure shown in Figure 10, b?

We talked only about curvilinear trapezoids adjacent to the OX axis. But similar formulas are also valid for figures adjacent to the y-axis. For example, the area of the figure shown in Figure 11 is found by the formula

Let the line y=f(x) limiting the curvilinear trapezoid can be given by the parametric equations , tО , and j(a)= a, j(b) = b, i.e. at= . Then the area of this curvilinear trapezoid is

b) Curve arc length

Let there be a curve at = f(x). Consider the arc of this curve corresponding to the change X on the segment [ a, b]. Let's find the length of this arc. To do this, we divide the arc AB into P parts with points A \u003d M 0, M 1, M 2, ..., M P= B (Fig. 14), corresponding to the points X 1 , X 2 , ..., x n Î [ a, b].

Denote D l i arc length, then l= . If the arc lengths D l i small enough, they can be considered approximately equal lengths corresponding segments connecting points M i-1,M i. These points have coordinates M i -1 (x i -1, f (x i-1)) , M i(x i, f(x i)). Then the lengths of the segments are equal respectively

Here the Lagrange formula is used. Let's put x i – x i-1=D x i, we get

Then l = , where

l = .

So the arc length of the curve at = f(x) corresponding to the change X on the segment [ a, b], is found by the formula

l = , (1)

If the curve is given parametrically, tО, i.e. y(t) = f(x(t)), then from formula (1) we obtain:

l=
.

So, if the curve is given parametrically, then the length of the arc of this curve corresponding to the change tн, is found by the formula

in) The volume of the body of revolution.

Fig.15

Consider a curvilinear trapezoid a AB b, bounded by a line at = f(x), straight X = a, X = b and segment [ a,b] of the OX axis (Fig. 15). Let this trapezoid rotate around the OX axis, the result will be a body of revolution. It can be proved that the volume of this body will be equal to

Similarly, you can derive the formula for the volume of a body obtained by rotating around the y-axis of a curvilinear trapezoid bounded by the graph of the function X= j( at), straight y = c , y = d and segment [ c,d] y-axis (Fig. 15):

Physical applications of the definite integral

In Lecture 19, we proved that, from a physical point of view, the integral is numerically equal to mass rectilinear thin inhomogeneous rod of length l= b – a, with variable linear density r = f(x), f(x) ³ 0, where X is the distance from the point of the rod to its left end.

Let us consider other physical applications of the definite integral.

Task 1. Find the work required to pump oil out of a vertical cylindrical tank with height H and base radius R. The density of the oil is r.

Decision. Let's build mathematical model this task. Let the axis OX pass along the axis of symmetry of the cylinder of height H and radius R, the beginning - in the center of the upper base of the cylinder (Fig. 17). Let's split the cylinder P small horizontal parts. Then where Ai- pumping work i th layer. This partition of the cylinder corresponds to the partition of the segment of the change in the layer height into P parts. Consider one of these layers located at a distance x i from the surface, width D X(or immediately dx). The pumping out of this layer can be considered as "raising" the layer to a height x i.

Then the work done to pump out this layer is equal to

Ai"R i x i, ,

where P i=rgV i= rgpR 2 dx, R i– weight, V i is the volume of the layer. Then Ai" R i x i= rgpR 2 dx.x i, where

, and hence .

Task 2. Find moment of inertia

a) a hollow thin-walled cylinder about an axis passing through its axis of symmetry;

b) a solid cylinder about an axis passing through its axis of symmetry;

c) thin rod length l about the axis passing through its middle;

d) thin rod length l about the axis passing through its left end.

Decision. As you know, the moment of inertia of a point about the axis is equal to J=mr 2 , and systems of points .

a) The cylinder is thin-walled, which means that the wall thickness can be neglected. Let the radius of the base of the cylinder R, its height H, and the mass density on the walls be equal to r.

Let's split the cylinder P parts and find where J i- moment of inertia i-th partition element.

Consider i-th partition element (an infinitesimal cylinder). All its points are at a distance R from the axis l. Let the mass of this cylinder t i, then t i= rV i» rs side= 2prR dx i, where x i O. Then J i» R 2 prR dx i, where

If r is a constant, then J= 2prR 3 N, and since the mass of the cylinder is M = 2prRН, then J= MR 2 .

b) If the cylinder is solid (filled), then we divide it into P vlo thin cylinders nestled one inside the other. If a P large, each of these cylinders can be considered thin-walled. This partition corresponds to the partition of the segment into P parts by points R i. Let's find the mass i-th thin-walled cylinder: t i= rV i, where

V i=pR i 2 H - pR i- 1 2 H \u003d pH (R i 2-R i -1 2) =

PH(R i-R i-1)(R i+R i -1).

Since the walls of the cylinder are thin, we can assume that R i+R i-1 » 2R i, and R i-R i-1=DR i, then V i» pH2R i DR i, where t i» rpН×2R i DR i,

Then finally

c) Consider a rod of length l, whose mass density is equal to r. Let the axis of rotation pass through its middle.

We model the rod as a segment of the OX axis, then the axis of rotation of the rod is the OY axis. Consider an elementary segment , its mass , the distance to the axis can be considered approximately equal to r i= x i. Then the moment of inertia of this section is , whence the moment of inertia of the entire rod is . Considering that the mass of the rod is , then

d) Now let the axis of rotation pass through the left end of the rod, i.e. the rod model is a segment of the OX axis. Then similarly, r i= x i, , where , and since , then .

Task 3. Find the pressure force of a fluid with density r on a right triangle with legs a and b, immersed vertically in a liquid so that the leg a is on the surface of the liquid.

Decision.

Let's build a task model. Let the top right angle triangle is at the origin, leg a coincides with the segment of the OY axis (the OY axis determines the surface of the liquid), the OX axis is directed downward, the leg b coincides with the segment of this axis. The hypotenuse of this triangle has the equation , or .

It is known that if on the horizontal area of the area S, immersed in a liquid of density r, is pressed by a column of liquid with a height h, then the pressure force is equal to (Pascal's law). Let's use this law.

Let us present some applications of the definite integral.

Calculating the area of a flat figure

The area of a curvilinear trapezoid bounded by a curve (where
), straight
,
and segment
axes
, is calculated by the formula

Area of a figure bounded by curves
and
(where
) straight
and
calculated by the formula

If the curve is given by parametric equations
, then the area of the curvilinear trapezoid bounded by this curve, straight lines
,
and segment
axes
, is calculated by the formula

where and are determined from the equations
,
, a
at
.

Area of a curved sector bounded by a curve defined in polar coordinates equation
and two polar radii
,
(
), is found by the formula

Example 1.27. Calculate the area of a figure bounded by a parabola
and direct
(Figure 1.1).

Decision. Let's find the points of intersection of the line and the parabola. To do this, we solve the equation

,
.

Where
,
. Then by formula (1.6) we have

Calculating the Arc Length of a Planar Curve

If the curve
on the segment
- smooth (that is, the derivative
is continuous), then the length of the corresponding arc of this curve is found by the formula

When specifying a curve parametrically
(
- continuously differentiable functions) the length of the arc of the curve corresponding to a monotonic change in the parameter from before , is calculated by the formula

Example 1.28. Calculate arc length of a curve
,
,
.

Decision. Let's find the derivatives with respect to the parameter :
,
. Then by formula (1.7) we obtain

2. Differential calculus of functions of several variables

Let each ordered pair of numbers
from some area
corresponds to a certain number
. Then called function of two variables and ,
-independent variables or arguments ,
-domain of definition functions, but the set all function values - its range and denote
.

Geometrically, the domain of a function is usually some part of the plane
bounded by lines that may or may not belong to this area.

Example 2.1. Find domain
functions
.

Decision. This function is defined at those points of the plane
, in which
, or
. Points of the plane for which
, form the boundary of the region
. The equation
defines a parabola (Fig. 2.1; since the parabola does not belong to the area
, it is shown as a dotted line). Further, it is easy to verify directly that the points for which
, located above the parabola. Region
is open and can be specified using the system of inequalities:

If variable give some boost
, a leave it constant, then the function
will receive an increment
called private increment function by variable :

Similarly, if the variable gets an increment
, a remains constant, then the function
will receive an increment
called private increment function by variable :

If limits exist:

they're called partial derivatives of a function
by variables and respectively.

Remark 2.1. The partial derivatives of functions of any number of independent variables are defined similarly.

Remark 2.2. Since the partial derivative with respect to any variable is a derivative with respect to this variable, provided that the other variables are constant, then all the rules for differentiating functions of one variable are applicable to finding partial derivatives of functions of any number of variables.

Example 2.2.
.

Decision. We find:

Example 2.3. Find Partial Derivatives of Functions
.

Decision. We find:

Full function increment
is called difference

Main part of total function increment
, linearly dependent on increments of independent variables
and
,is called the total differential of the function and denoted
. If a function has continuous partial derivatives, then the total differential exists and is equal to

where
,
- arbitrary increments of independent variables, called their differentials.

Similarly, for a function of three variables
the total differential is given by

Let the function
has at the point
first-order partial derivatives with respect to all variables. Then the vector is called gradient functions
at the point
and denoted
or
.

Remark 2.3. Symbol
is called the Hamilton operator and is pronounced "numbla".

Example 2.4. Find the gradient of a function at a point
.

Decision. Let's find partial derivatives:

,
,

and calculate their values at the point
:

,
,
.

Hence,
.

derivative functions
at the point
in the direction of the vector
called the limit of the ratio
at
:

, where
.

If the function
is differentiable, then the derivative in this direction is calculated by the formula:

where ,- angles, which vector forms with axes
and
respectively.

In the case of a function of three variables
the directional derivative is defined similarly. The corresponding formula has the form

where
- direction cosines of the vector .

Example 2.5. Find the derivative of a function
at the point
in the direction of the vector
, where
.

Decision. Let's find the vector
and its direction cosines:

,
,
,
.

Calculate the values of partial derivatives at the point
:

,
,
;
,
,
.

Substituting into (2.1), we obtain

Partial derivatives of the second order called partial derivatives taken from partial derivatives of the first order:

Partial derivatives
,
called mixed . The values of mixed derivatives are equal at those points where these derivatives are continuous.

Example 2.6. Find second order partial derivatives of a function
.

Decision. Calculate first partial derivatives of the first order:

,
.

Differentiating them again, we get:

,
,

,
.

Comparing the last expressions, we see that
.

Example 2.7. Prove that the function
satisfies the Laplace equation

Decision. We find:

,
.

Dot
called local maximum point (minimum ) functions
, if for all points
, other than
and belonging to a sufficiently small neighborhood of it, the inequality

(
).

The maximum or minimum of a function is called its extremum . The point at which the extremum of the function is reached is called extremum point of the function .

Theorem 2.1 (Necessary conditions for an extremum ). If point
is the extremum point of the function
, then at least one of these derivatives does not exist.

The points for which these conditions are met are called stationary or critical . Extreme points are always stationary, but a stationary point may not be an extreme point. For a stationary point to be an extremum point, sufficient extremum conditions must be satisfied.

Let us first introduce the following notation :

,
,
,
.

Theorem 2.2 (Sufficient conditions for an extremum ). Let the function
is twice differentiable in a neighborhood of a point
and dot
is stationary for the function
. Then:

1.If a
, then the point
is the extremum of the function, and
will be the maximum point at
(
)and the minimum point at
(
).

2.If a
, then at the point

there is no extremum.

3.If a
, then there may or may not be an extremum.

Example 2.8. Investigate a function for an extremum
.

Decision. Since in this case partial derivatives of the first order always exist, then to find the stationary (critical) points we solve the system:

,
,

where
,
,
,
. Thus, we got two stationary points:
,
.

,
,
.

For point
we get:, that is, there is no extremum at this point. For point
we get: and
, hence

at this point given function reaches a local minimum: .

Ministry of Education and Science of the Russian Federation

federal state autonomous educational institution

higher professional education

"Northern (Arctic) federal university named after M.V. Lomonosov"

Department of Math

COURSE WORK

By discipline Mathematics

Pyatysheva Anastasia Andreevna

Supervisor

Art. teacher

Borodkina T. A.

Arkhangelsk 2014

TASK FOR COURSE WORK

Applications of the definite integral

INITIAL DATA:

21. y=x 3 , y= ; 22.

INTRODUCTION

In this course work, I have been given the following tasks: calculate the areas of figures bounded by function graphs, bounded by lines given by equations, also bounded by lines given by equations in polar coordinates, calculate the lengths of arcs of curves given by equations in rectangular system coordinates given by parametric equations, given by equations in polar coordinates, as well as calculate the volumes of bodies bounded by surfaces, bounded by graphs of functions, and formed by the rotation of figures bounded by graphs of functions around the polar axis. I chose a term paper on the topic “Definite Integral. In this regard, I decided to find out how easy and fast you can use integral calculations, and how accurately you can calculate the tasks assigned to me.

INTEGRAL is one of the most important concepts of mathematics that arose in connection with the need, on the one hand, to find functions by their derivatives (for example, to find a function that expresses the path traveled by a moving point, according to the speed of this point), and on the other hand, to measure areas, volumes, lengths arcs, the work of forces for a certain period of time, etc.

Topic disclosure term paper I followed the following plan: the definition of a definite integral and its properties; curve arc length; area of a curvilinear trapezoid; surface area of rotation.

For any function f(x) continuous on the segment , there exists an antiderivative on this segment, which means that there exists indefinite integral.

If the function F(x) is some antiderivative of continuous function f(x), then this expression is known as the Newton-Leibniz formula:

The main properties of the definite integral:

If the lower and upper limits of integration are equal (a=b), then the integral is equal to zero:

If f(x)=1, then:

When rearranging the limits of integration, the definite integral changes sign to the opposite:

The constant factor can be taken out of the sign of a definite integral:

If the functions are integrable on, then their sum is integrable and the integral of the sum is equal to the sum integrals:

There are also basic integration methods, such as change of variable,:

Differential Fix:

The integration-by-parts formula makes it possible to reduce the calculation of the integral to the calculation of the integral, which may turn out to be simpler:

The geometric meaning of a definite integral is that for a continuous and non-negative function it is in the geometric sense the area of the corresponding curvilinear trapezoid.

In addition, using a definite integral, you can find the area of \u200b\u200bthe region bounded by curves, straight lines and, where

If a curvilinear trapezoid is bounded by a curve given by parametric lines x = a and x = b and the axis Ox, then its area is found by the formula, where they are determined from the equality:

. (12)

The main area, the area of \u200b\u200bwhich is found using a certain integral, is a curvilinear sector. This is the area bounded by two rays and a curve, where r and are polar coordinates:

If the curve is a graph of the function where, and the function of its derivative is continuous on this segment, then the surface area of the figure formed by the rotation of the curve around the Ox axis can be calculated by the formula:

. (14)

If a function and its derivative are continuous on a segment, then the curve has a length equal to:

If the curve equation is given in parametric form

where x(t) and y(t) are continuous functions with continuous derivatives and then the length of the curve is found by the formula:

If the curve is given by an equation in polar coordinates, where and are continuous on the segment, then the arc length can be calculated as follows:

If a curvilinear trapezoid rotates around the Ox axis, bounded by a continuous line segment and straight lines x \u003d a and x \u003d b, then the volume of the body formed by the rotation of this trapezoid around the Ox axis will be equal to:

If a curvilinear trapezoid is bounded by a graph of a continuous function and lines x = 0, y = c, y = d (c< d), то объем тела, образованного вращением этой трапеции вокруг оси Oy, будет равен:

If the figure is bounded by curves and (is “higher” than by straight lines x = a, x = b, then the volume of the body of revolution around the Ox axis will be equal to:

and around the y-axis (:

If the curvilinear sector is rotated around the polar axis, then the area of the resulting body can be found by the formula:

2. PROBLEM SOLVING

Task 14: Calculate the areas of figures bounded by function graphs:

1) Solution:

Figure 1 - Graph of functions

X changes from 0 to

x 1 = -1 and x 2 = 2 - integration limits (this can be seen in Figure 1).

3) Calculate the area of the figure using formula (10).

Answer: S = .

Task 15: Calculate the areas of the figures bounded by the lines given by the equations:

1) Solution:

Figure 2 - Graph of functions

Consider a function on the interval .

Figure 3 - Table of variables for the function

Since, then 1 arc will fit on this period. This arc consists of a central part (S 1) and side parts. The central part consists of the desired part and a rectangle (S pr):. Let's calculate the area of one central part of the arc.

2) Find the limits of integration.

and y = 6, hence

For an interval, the limits of integration.

3) Find the area of the figure using formula (12).

curvilinear integral trapezoid

Problem 16: Calculate the areas of figures bounded by lines given by equations in polar coordinates:

1) Solution:

Figure 4 - Graph of functions,

Figure 5 - Table of variable functions,

2) Find the limits of integration.

hence -

3) Find the area of the figure using formula (13).

Answer: S=.

Task 17: Calculate the lengths of arcs of curves given by equations in a rectangular coordinate system:

1) Solution:

Figure 6 - Graph of the function

Figure 7 - Table of function variables

2) Find the limits of integration.

varies from ln to ln, this is obvious from the condition.

3) Find the arc length using formula (15).

Answer: l =

Task 18: Calculate the lengths of arcs of curves given by parametric equations: 1)

1) Solution:

Figure 8- Function Graph

Figure 11 - Table of function variables

2) Find the limits of integration.

ts varies from, this is obvious from the condition.

Let's find the arc length using formula (17).

Task 20: Calculate the volumes of bodies bounded by surfaces:

1) Solution:

Figure 12 - Graph of functions:

2) Find the limits of integration.

Z changes from 0 to 3.

3) Find the volume of the figure using formula (18)

Task 21: Calculate the volumes of bodies bounded by function graphs, axis of rotation Ox: 1)

1) Solution:

Figure 13 - Graph of functions

Figure 15 - Function Graph Table

2) Find the limits of integration.

Points (0;0) and (1;1) are common for both graphs, therefore these are the limits of integration, which is obvious in the figure.

3) Find the volume of the figure using formula (20).

Task 22: Calculate the area of bodies formed by the rotation of figures bounded by function graphs around the polar axis:

1) Solution:

Figure 16 - Graph of the function

Figure 17 - Table of variables for the graph of the function

2) Find the limits of integration.

c changes from

3) Find the area of the figure using formula (22).

Answer: 3.68

CONCLUSION

In the process of completing my course work on the topic “Definite Integral”, I learned how to calculate areas different bodies, find the lengths of different arcs of curves, and calculate volumes. This representation about working with integrals, will help me in the future professional activity how to quickly and efficiently perform various activities. After all, the integral itself is one of the most important concepts of mathematics, which arose in connection with the need, on the one hand, to find functions by their derivatives (for example, to find a function that expresses the path traveled by a moving point, according to the speed of this point), and on the other hand, to measure areas, volumes, arc lengths, work of forces for a certain period of time, etc.

LIST OF USED SOURCES

1. Written, D.T. Lecture notes on higher mathematics: Part 1 - 9th ed. - M.: Iris-press, 2008. - 288 p.

2. Bugrov, Ya.S., Nikolsky, S.M. Higher Mathematics. Differential and integral calculus: V.2 - M.: Drofa, 2004. - 512 p.

3. V. A. Zorich, Mathematical Analysis. Part I. - Ed. 4th - M.: MTSNMO, 2002. - 664 p.

4. Kuznetsov D.A. "Collection of tasks for higher mathematics» Moscow, 1983

5. Nikolsky S. N. "Elements mathematical analysis". - M.: Nauka, 1981.

Area in polar coordinates

Consider a curvilinear sector OAB, limited by a line, given by the equation ρ=ρ(φ) in polar coordinates, two beams OA and OB, for which φ=α , φ=β .

We divide the sector into elementary sectors OM k-1 M k ( k=1, …, n, M 0 =A, Mn=B). Denote by Δφk angle between beams OM k-1 and OM k forming angles with the polar axis φk-1 and φk respectively. Each of the elementary sectors OM k-1 M k replace with a circular sector with radius ρ k \u003d ρ (φ "k), where φ" k- angle value φ from the interval [ φk-1 , φk], and central corner Δφk. The area of the last sector is expressed by the formula .

expresses the area of the "stepped" sector, which approximately replaces the given sector OAB.

Sector area OAB is called the limit of the area of the "stepped" sector at n→∞ and λ=max Δφ k → 0:

As , then

Curve arc length

Let on the interval [ a, b] a differentiable function is given y=f(x), whose graph is the arc . Line segment [ a,b] split into n parts dots x 1, x2, …, xn-1. These points will correspond to the points M1, M2, …, Mn-1 arcs, connect them with a broken line, which is called a broken line inscribed in an arc. The perimeter of this broken line is denoted by s n, i.e

Definition. The length of the arc of the line is the limit of the perimeter of the polyline inscribed in it, when the number of links M k-1 M k increases indefinitely, and the length of the largest of them tends to zero:

where λ is the length of the largest link.

We will count the length of the arc from some of its points, for example, A. Let at the point M(x,y) arc length is s, and at the point M"(x+Δx,y+Δy) arc length is s+Δs, where, i>Δs - arc length. From a triangle MNM" find the length of the chord: .

From geometric considerations follows that

that is, the infinitely small arc of the line and the chord that subtends it are equivalent.

Let's transform the formula expressing the length of the chord:

Passing to the limit in this equality, we obtain a formula for the derivative of the function s=s(x):

from which we find

This formula expresses the differential of the arc of a plane curve and has a simple geometric sense : expresses the Pythagorean theorem for an infinitesimal triangle MTN (ds=MT, ).

The differential of the arc of the space curve is given by

Consider an arc of a space line given by the parametric equations

where α ≤ t ≤ β, φ i (t) (i=1, 2, 3) are differentiable functions of the argument t, then

Integrating this equality over the interval [ α, β ], we obtain a formula for calculating the length of this line arc

If the line lies in a plane Oxy, then z=0 for all t∈[α, β], That's why

In case when flat line given by the equation y=f(x) (a≤x≤b), where f(x) is a differentiable function, the last formula takes the form

Let the flat line be given by the equation ρ=ρ(φ) (α≤φ≤β ) in polar coordinates. In this case we have parametric equations lines x=ρ(φ) cos φ, y=ρ(φ) sin φ, where the polar angle is taken as a parameter φ . Insofar as

then the formula expressing the length of the arc of the line ρ=ρ(φ) (α≤φ≤β ) in polar coordinates has the form

body volume

Let us find the volume of a body if the area of any cross section of this body perpendicular to a certain direction is known.

Let us divide this body into elementary layers by planes perpendicular to the axis Ox and defined by the equations x=const. For any fixed x∈ known area S=S(x) cross section of this body.

Elementary layer cut off by planes x=x k-1, x=x k (k=1, …, n, x 0 =a, xn=b), we replace it with a cylinder with a height ∆x k =x k -x k-1 and base area S(ξk), ξ k ∈.

The volume of the specified elementary cylinder is expressed by the formula Δvk =E(ξk)Δxk. Let's sum up all such products

which is the integral sum for the given function S=S(x) on the segment [ a, b]. It expresses the volume of a stepped body, consisting of elementary cylinders and approximately replacing the given body.

The volume of a given body is the limit of the volume of the specified stepped body at λ→0 , where λ - the length of the largest of the elementary segments ∆x k. Denote by V the volume of the given body, then by definition

On the other side,

Therefore, the volume of the body according to the given cross sections calculated by the formula

If the body is formed by rotation around an axis Ox curvilinear trapezoid bounded from above by an arc of a continuous line y=f(x), where a≤x≤b, then S(x)=πf 2 (x) and the last formula becomes:

Comment. The volume of a body obtained by rotating a curvilinear trapezoid bounded on the right by a function graph x=φ(y) (c ≤ x ≤ d), around the axis Oy calculated by the formula

Surface area of rotation

Consider the surface obtained by rotating the arc of the line y=f(x) (a≤x≤b) around the axis Ox(assume that the function y=f(x) has a continuous derivative). We fix the value x∈, the function argument will be incremented dx, which corresponds to the "elementary ring" obtained by rotating the elementary arc Δl. This "ring" is replaced by a cylindrical ring - the lateral surface of the body formed by the rotation of a rectangle with a base equal to the differential of the arc dl, and height h=f(x). Cutting the last ring and unfolding it, we get a strip with a width dl and length 2πy, where y=f(x).