Before proceeding to the formulas for the area of \u200b\u200ba surface of revolution, let us give a brief formulation of the surface of revolution itself. The surface of revolution, or, which is the same thing - the surface of the body of revolution - a spatial figure formed by the rotation of a segment AB curve around the axis Ox (picture below).
Imagine a curvilinear trapezoid bounded from above by the mentioned segment of the curve. The body formed by the rotation of this trapezoid around the same axis Ox, and there is a body of revolution. And the area of \u200b\u200bthe surface of revolution or the surface of the body of revolution is its outer shell, not counting the circles formed by rotation around the axis of straight lines x = a and x = b .
Note that the body of revolution and, accordingly, its surface can also be formed by rotating the figure not around the axis Ox, and around the axis Oy.
Calculating the area of \u200b\u200ba surface of revolution given in rectangular coordinates
Let in rectangular coordinates on the plane by the equation y = f(x) a curve is given, the rotation of which around the coordinate axis forms a body of revolution.
The formula for calculating the surface area of \u200b\u200brevolution is as follows:
(1).
Example 1.Find the surface area of \u200b\u200ba paraboloid formed by rotation about an axis Ox arc of the parabola corresponding to the change x from x \u003d 0 to x = a .
Decision. Let us express explicitly the function that defines the arc of the parabola:
Let's find the derivative of this function:
Before using the formula to find the area of \u200b\u200bthe surface of revolution, we write that part of its integrand, which is the root, and substitute the derivative we just found there:
Answer: the length of the arc of the curve is
.
Example 2.Find the surface area of \u200b\u200brotation about an axis Ox astroids.
Decision. It is enough to calculate the surface area resulting from the rotation of one branch of the astroid, located in the first quarter, and multiply it by 2. From the astroid equation, we explicitly express the function that we need to substitute in the formula to find the area of \u200b\u200brotation:
.
We perform integration from 0 to a:
Calculating the area of \u200b\u200ba surface of revolution given parametrically
Consider the case when the curve forming the surface of revolution is given by the parametric equations
Then the surface area of \u200b\u200brevolution is calculated by the formula
(2).
Example 3.Find the area of \u200b\u200ba surface of revolution formed by rotation about an axis Oy a figure bounded by a cycloid and a straight line y = a ... The cycloid is given by parametric equations
Decision. Let's find the intersection points of the cycloid and the line. Equating the cycloid equation 
and the equation of the straight line y = a , find
It follows from this that the boundaries of integration correspond to
Now we can apply formula (2). Find the derivatives:
Let's write the radical expression in the formula, substituting the found derivatives:
Let's find the root of this expression:
.
We substitute the found in the formula (2):
.
Let's make the substitution:
And finally we find
In the transformation of expressions, trigonometric formulas were used
Answer: the surface area of \u200b\u200brotation is.
Calculating the area of \u200b\u200ba surface of revolution given in polar coordinates
Let the curve, by rotation of which the surface is formed, be given in polar coordinates.
If the curve is given by parametric equations, then the surface area obtained by rotating this curve around the axis is calculated by the formula 
... At the same time, the "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 above the abscissa axis - otherwise the function "responsible for the players" will take negative values \u200b\u200band you will have to put a "minus" sign in front of the integral.
Example 3
Calculate the area of \u200b\u200ba sphere obtained by rotating a circle around an axis.
Decision: from article materials about area and volume with a parametrically specified lineyou know that the equations define a circle centered at the origin of radius 3.
well and sphere , for those who have forgotten, is the surface balls(or ball surface).
We adhere to the established solution scheme. Find the derivatives: 
Let's compose and simplify the "formula" root:
Needless to say, it turned out to be sweet. See for comparison how Fichtengolz butted the square ellipsoid of revolution.
According to the theoretical remark, we consider 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 \u200b\u200ba sphere, where is its radius.
Something hurt a simple task, I even felt ashamed…. I suggest you fix this flaw \u003d)
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, again the advantage of parametric equations should be noted - they do not need to be somehow modified; no need to bother 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 depression in the middle. For the case of cycloid rotation 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 an overview, if you look in 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 a problem you need.
How to calculate the surface area of \u200b\u200brevolution,
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 \u200b\u200bcorresponding to the ends of the curve.
In accordance with geometric meaning the integrand function 
, and this is achieved only under the condition (and are certainly non-negative). Therefore, it is necessary to consider the values \u200b\u200bof the angle from the range, in other words, the curve should be located above 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.
Decision: the graph of this curve can be viewed in Example 6 of the lesson about polar coordinate system... The cardioid is symmetric 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. Find the derivative with respect to "phi":
Let's compose and simplify the root:
Hopefully with supernumerary trigonometric formulas
no one had any difficulty.
We use the formula:
In between 
, hence: 
(I described in detail how to get rid of the root in detail in the article Curve arc length).
Answer: 
An interesting and short task for independent decision:
Example 6
Calculate the area of \u200b\u200bthe ball belt,
What is a ball belt? Place a round unpeeled orange on the table and pick up a knife. Do two parallel cut, thereby dividing the fruit into 3 parts of arbitrary sizes. Now take the middle, in which the juicy pulp is exposed on both sides. This body is called spherical layer, and its bounding surface (orange peel) is ball belt.
Readers familiar with polar coordinates, easily presented a drawing of the problem: the equation defines a circle centered at the pole of the radius, from which rays 
cut off less arc. This arc rotates around the polar axis and thus a spherical belt is obtained.
Now you can eat an orange with a clear conscience and a light heart, on this delicious note and finish the lesson, do not spoil your appetite with other examples \u003d)
Solutions and Answers:
Example 2:Decision
: calculate the surface area formed by the rotation of the upper branch around the abscissa axis. We use the formula 
.
In this case: 
;
In this way:
Answer:
Example 4:Decision
: use the formula 
... The first arc of the cycloid is defined on the segment .
Find the derivatives:
Let's compose and simplify the root:
Thus, the surface area of \u200b\u200brevolution is:
In between 
, so 
First integralintegrate by parts
:
In the second integral we usetrigonometric formula
.
Answer:
Example 6:Decision
: we use the formula:
Answer:
Higher mathematics for correspondence students and not only \u003e\u003e\u003e
(Go to home page)
How to calculate the definite integral
by the trapezoidal formula and the Simpson method?
Numerical methods are a fairly large section of higher mathematics and serious textbooks on this topic have hundreds of pages. In practice, in control works Traditionally, some problems are proposed for solving by numerical methods, and one of the common problems is approximate calculation definite integrals... In this article, I will look at two approximate calculation methods definite integral – trapezium method and simpson method.
What do you need to know to master these methods? It sounds funny, but you may not be able to take integrals at all. And even do not even understand what integrals are. From technical means, a microcalculator is required. Yes, we are waiting for routine school calculations. Better yet, download mine semi-automatic calculator for the trapezium method and the Simpson method... The calculator is written in Excel and will reduce the time for solving and processing problems by ten times. A video manual is included for excel teapots! By the way, the first video with my voice.
First, let's ask the question, why do we need approximate calculations at all? It seems to be possible to find antiderivative of function and use the Newton-Leibniz formula by calculating the exact value of the definite integral. As an answer to the question, let's go straight to the demo with a picture.
Calculate the definite integral
Everything would be fine, but in this example the integral is not taken - in front of you is an unbearable, so-called integral logarithm... Does this integral even exist? Let us depict the graph of the integrand in the drawing: 
It's okay. Integrand continuous on the segment and the definite integral is numerically equal to the shaded area. But there is only one catch - the integral is not taken. And in such cases, numerical methods come to the rescue. In this case, the task occurs in two formulations:
1) Calculate the definite integral approximately , rounding the result to a certain decimal place... For example, up to two decimal places, up to three decimal places, etc. Let's say the 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 thatto round the result to three decimal places.
2) Calculate a definite integral approximately, with a certain precision... For example, calculate a definite integral approximately with an accuracy of 0.001. What does it mean? This means that if an approximate answer of 5.347 is received, then all numbers should 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 the approximate calculation of a definite integral, which is found in problems:
Method of rectangles... The segment of integration is divided into several parts and a stepped figure is constructed ( bar chart), which is close in area to the required 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 often the partition (more smaller intermediate segments), the higher the accuracy. The method of rectangles gives a rough approximation of the area, apparently, therefore, it is very rare in practice (I remembered only one practical example). In this regard, I will not consider the method of rectangles, and I will not even give a simple formula. Not because laziness, but because of the principle of my reshebnik: what is extremely rare in practical problems is not considered.
Trapezium method... The idea is similar. The segment of integration is divided into several intermediate segments, and the graph of the integrand is approximated broken line line: 
Thus, our area (blue shading) is approximated by the sum of the areas of the trapeziums (red). Hence the name of the method. It is easy to see that the trapezoidal method gives a much better approximation than the rectangle method (with the same number of subdivisions). And, of course, the more smaller intermediate segments we consider, the higher the accuracy. The trapezoid method is encountered from time to time in practical exercises, and in this article several examples will be analyzed.
Simpson's method (parabola method)... This is a more perfect way - the graph of the integrand is approximated not by a broken line, but by small parabolas. How many intermediate segments - so many small parabolas. If we take the same three segments, then Simpson's method will give an even more accurate approximation than the rectangle method or the trapezoid method.
I see no reason to build a drawing, since visually the approximation will be superimposed on the function graph (the broken line of the previous paragraph - and that almost coincided).
The problem of calculating a definite integral using Simpson's formula is the most popular problem in practice. And the parabola method will receive considerable attention.
5. Finding the surface area of \u200b\u200bbodies of revolution
Let the curve AB be the graph of the function y \u003d f (x) ≥ 0, where x [a; b], and the function y \u003d f (x) and its derivative y "\u003d f" (x) are continuous on this segment.
Let us find the area S of the surface formed by the rotation of the AB curve around the Ox axis (Fig. 8).
Let's 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 \u003d s (x) (s (a) \u003d 0 and s (b) \u003d S).
Let's give the argument x the increment Δx \u003d dx. Through the point x + dx [a; b] also draw a plane perpendicular to the Ox axis. The function s \u003d s (x) will receive an increment of Δs, shown in the figure as 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 \u200b\u200bits lateral surface is: \u003d 2ydl + dydl.
Discarding the product dу d1 as an infinitesimal higher order than ds, we obtain ds \u003d 2уdl, or, since d1 \u003d dx.
Integrating the resulting equality in the range from x \u003d a to x \u003d b, we obtain
If the curve AB is given by the parametric equations x \u003d x (t), y \u003d y (t), t≤ t ≤ t, then the formula for the area of \u200b\u200bthe surface of revolution takes the form
S \u003d 2 
dt.
Example: Find the surface area of \u200b\u200ba ball of radius R.
S \u003d 2 
= 
6. Finding Variable Force Work
Variable force work
Let material point M moves along the Ox axis under the action of a variable force F \u003d F (x) directed parallel to this axis. The work done by force when moving point M from position x \u003d a to position x \u003d 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 a spring is proportional to this extension x, i.e. F \u003d kх, where k is the proportionality coefficient. According to the condition of the problem, the force F \u003d 100 N stretches the spring by x \u003d 0.01 m; therefore, 100 \u003d k 0.01, whence k \u003d 10000; therefore, F \u003d 10000x. The sought job based on the formula A \u003d Find the work that needs to be spent in order to pump out the 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 of 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 expended 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 \u003d A (x), where (0 ≤ x ≤ H) (A (0) \u003d 0, A (H) \u003d A 0). 2) Find the main part of the increment ΔA when x changes by the value Δx \u003d dx, i.e. we find the differential dA of the function A (x). In view of the smallness of dx, we assume that the "elementary" layer of the liquid is at the same depth x (from the edge of the reservoir). Then dА \u003d 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 \u003d g. The volume of this liquid layer is obviously equal to, where dx is the height of the cylinder (layer), is the area of \u200b\u200bits base, i.e. dv \u003d. Thus, dр \u003d. and 3) Integrating the obtained equality in the range from x \u003d 0 to x \u003d H, we find A 8. Calculation of integrals using the MathCAD package When solving some applied problems, it is required to use the operation of symbolic integration. 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 an answer from another source or a 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 the MathCad program 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 that 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 a student or researcher from routine work, but cannot free him from additional analysis, both when setting a problem and when obtaining any results. In this paper, the main provisions related to the study of applications of a definite integral in the course of mathematics were considered. - an analysis of the theoretical basis for solving integrals was carried out; - the material was 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 considered above give us a clear idea of \u200b\u200bthe significance of a 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 in 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 practical problems. 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 radiusR. 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 is:Q (x) \u003d. We get the volume of the ball: Example:Find the volume of an arbitrary pyramid with height H and base areaS. 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 ratiox / 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 equationy \u003d f (x ). Suppose the functionf (x ) is continuous on the segment [a, b ]. If the corresponding curvilinear trapezoid with bases a andb rotate around the Ox axis, then we get the so-called body of revolution.
Surface area of \u200b\u200ba body of revolution.
Definition:
Surface area of \u200b\u200brevolutioncurve AB around this 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 largest of the lengths of the links of these polygonal lines tends to zero. We split the arc AB inton parts by points M 0, M 1, M 2, ..., M n ... The coordinates of the vertices of the resulting polyline have coordinatesx 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 \u200b\u200bwhich isD P i ... This area can be found by the formula:
Let a body be given in space. Let its sections be constructed by planes perpendicular to the axis passing through the points x Example. Let's find the volume of a bounded body, enclosed between the surface of a cylinder of radius:, a horizontal plane and an inclined plane z \u003d 2y and lying above the horizontal plane. Obviously, the body under consideration is projected onto the axis into the segment Therefore, the cross-sectional area S (x) is: Using the formula, we find the volume of the body: Let on the segment [ a,
b] a continuous constant-sign function y=
f(x).
Volumes of a body of revolution formed by rotation about an axis Oh (or axis OU) of a curved trapezium bounded by a curve y=
f(x)
(f(x) If the body is formed when rotating around an axis OU curved trapezium bounded by a curve 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 \u003d cosx on the segment be considered in the xOy plane E According to the formula, we get: If the arc of a curve given by a non-negative function where c and d are the abscissas of the beginning and end of the arc. If a curve arc is specified parametric equations
If the arc is specified in polar coordinates
Example. Let us calculate the area of \u200b\u200bthe surface formed by the rotation in space around the axis of a part of the line y \u003d Because We make the change t \u003d 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 \u003d t 2 -: To calculate the second of the integrals on the right-hand side, we denote it and integrate it by parts, obtaining the equation for: Moving to the left and dividing by 2, we get whence finally Variable force work.
Consider the motion of a material point along the axis OXvariable force fdepending on the position of the point x on the axis, i.e. force as a function x... Then work Arequired 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 Smultiplied 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 \u003d f (x), a≤x≤b, and has a density moments of inertia I X and I y with respect to the same axes Ox and Oy are calculated by the formulas and center of mass coordinates
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 \u003d chx at 0≤x≤1. If no density is specified, it is assumed that the curve is uniform and Example 2. Find the coordinates of the center of mass of the circular arc x \u003d acost, y \u003d asint, located in the first quarter. We have: From here we get: The following is often useful in applications. Theorem
Guilder... The area of \u200b\u200bthe surface formed by the rotation of an 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 From here 2.
Physical tasks. Some applications of the definite integral in solving physical problems are illustrated below in examples. Example 4. The speed of the rectilinear movement of the body is expressed by the formula (m / s). Find the path traversed by the body in 5 seconds from the start of the movement. Because body path with a speed v (t) over a period of time, is expressed by the integral then we have: P the line crosses the axis at three points: x 1 \u003d -1, x 2 \u003d 0, x 3 \u003d 1. The bounded area between the line and the axis is projected onto a line segment P 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 P To calculate the volume of a body of revolution, apply the formula P (we took the root as the value, not -cosx, since cosx\u003e 0 for Answer: Example. Let us calculate the area Q of the surface of revolution obtained by rotating the arc of the cycloid x \u003d t-sint; y \u003d 1-cost, for D We have: To pass under the integral sign to a variable, note that for Moreover, let us preliminarily calculate (so that We get: Making the substitution, we arrive at the integral
y \u003d f (x)
M i B
on it. The area of \u200b\u200bthe figure formed in the section depends on the point xdefining 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 \u003d a and x \u003d in calculated by the formula 
, and for x 
the cross section of the body is a right-angled triangle with legs y and z \u003d 2y, where y can be expressed in terms of x from the cylinder equation:
Calculation of the volumes of bodies of revolution
0) and straight y \u003d 0, x \u003d a, x \u003db, are calculated accordingly by the formulas:
,
(19)
(20)
and direct x=0,
y=
c,
y=
d, then the volume of the body of revolution is
.
(21)
.
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.Surface area of \u200b\u200brevolution
,
, rotates around the Ox axis, then the surface area of \u200b\u200brotation is calculated by the formula 
where a and b - abscissas of the beginning and end of the arc.
,
, rotates around the Oy axis, then the surface area of \u200b\u200brotation is calculated by the formula
,
,
, and 
then
then
.
located above the line segment.
, then the formula gives us the integral
Applications of a definite integral to the solution of some problems in mechanics and physics
.
then static moments of this arc M x and M y relative to the coordinate axes Ox and Oy are
;
and 
- by formulas
... We have: Consequently,
... When the semicircle rotates around the Ox axis, a sphere is obtained, the surface area of \u200b\u200bwhich is equal, and the length of the semicircle is equal to n. By Gulden's theorem, we have 4 
, i.e. the center of mass C has coordinates C 
.
example. Find the area of \u200b\u200bthe bounded area lying between the axis and the line y \u003d x 3 -x. Insofar as
,
and on the segment 
,
line y \u003d x 3 -x goes above the axis (that is, line y \u003d 0, and on 
- below. Therefore, the area of \u200b\u200bthe area can be calculated as follows:
example. Let us find the area of \u200b\u200bthe region enclosed between the first and second turns of the Archimedes spiral r \u003d a 
(a\u003e 0) and a segment of the horizontal axis 
.
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 
:
example. Let us find the volume of the body bounded by the surface of rotation of the line y \u003d 4x-x 2 around the axis (for 
).
example. We calculate the length of the arc of the line y \u003d lncosx located between the straight lines and 
.
, the arc length is
.
, around the axis.
to calculate, apply the formula:
, so that
we get 
, as well as 
) and
