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 \u003d f (x) and its derivative y "\u003d f" (x) are continuous on this segment.
Let's find the area S of the surface formed by the rotation of the curve AB around the Ox axis (Fig. 8).
We apply scheme II (differential method).
Through an arbitrary point x [a; b] let's draw a plane P, perpendicular to the axis Ox. The plane P intersects the surface of revolution along 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 Δх = dх. Through the point x + dx [a; b] also draw a plane perpendicular to the x-axis. The function s = s(x) will receive an increment of Δs, shown in the figure as a "belt".
Let us find the differential of the area ds, replacing the figure formed between the sections by a truncated cone, the generatrix of which is equal to dl, and the radii of the bases are equal to y and y + dy. Its lateral surface area 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 becomes
S=2 
dt.
Example: Find the surface area of a sphere of radius R.
S=2 
= 
6. Finding the work of a variable force
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 that stretches the spring is proportional to this stretch x, i.e. F = kx, where k is the coefficient of proportionality. 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 desired work based on the formula A= Find the work that must be expended to pump liquid over the edge from a vertical cylindrical tank of height H m and base radius R m (Fig. 13). The work expended on raising a body of weight p to a height h is equal to p H. But the different layers of the liquid in the reservoir are at different depths and the height of the rise (to the edge of the reservoir) of the different layers is not the same. To solve the problem, we apply scheme II (differential method). We introduce a coordinate system. 1) The work expended on pumping out a layer of liquid of thickness x (0 ≤ x ≤ H) from the tank 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) We find the main part of the increment ΔA when x changes by Δx = 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" liquid layer is 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 AV, where g is the acceleration of gravity, is the density of the liquid, dv is the volume of the “elementary” liquid layer (it is highlighted in the figure), i.e. dr = 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 resulting 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. In this case, the MathCad program can be useful both at the initial stage (it is good to know the answer in advance or 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 how this program works with a few examples, analyze the solutions obtained with its help and compare these solutions with the 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 far from 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 because of its bulkiness; d) solves the problem incompletely 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 substitution 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 result obtained. In addition, some of the obtained solutions 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. – an analysis of the theoretical basis for solving integrals was carried out; - the material was subjected to systematization and generalization. During 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 the significance of a certain 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 doing the course work, we considered examples of practical problems in the field of physics, geometry, mechanics, biology and economics. Of course, this is by no means an exhaustive list of sciences that use the integral method to find a set value when solving a specific problem, and to establish theoretical facts. Also, the definite integral is used to study mathematics itself. For example, when solving differential equations, which in turn make an indispensable contribution to solving practical problems. We can say that the definite integral is a kind of 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 a definite integral occurs even within the average 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. I. Volumes of bodies of revolution. Preliminarily study chapter XII, p°p° 197, 198, according to the textbook by G. M. Fikhtengol'ts* Analyze in detail the examples given in p° 198. 508. Calculate the volume of the body formed by the rotation of the ellipse Around the x-axis. In this way, 530. Find the area of the surface formed by the rotation around the axis Ox of the arc of the sinusoid y \u003d sin x from the point X \u003d 0 to the point X \u003d It. 531. Calculate the surface area of a cone with height h and radius r. 532. Calculate the surface area formed by rotation of the astroid x3 -) - y* - a3 around the x-axis. 533. Calculate the area of the surface formed by the inversion of the loop of the curve 18 y-x(6-x)r around the x-axis. 534. Find the surface of the torus produced by the rotation of the circle X2 - j - (y-3)2 = 4 around the x-axis. 535. Calculate the area of the surface formed by the rotation of the circle X = a cost, y = asint around the Ox axis. 536. Calculate the area of the surface formed by the rotation of the loop of the curve x = 9t2, y = St - 9t3 around the axis Ox. 537. Find the area of the surface formed by the rotation of the arc of the curve x = e * sint, y = el cost around the axis Ox from t = 0 to t = -. 538. Show that the surface produced by the rotation of the arc of the cycloid x = a (q> - sin φ), y = a (I - cos φ) around the axis Oy, is equal to 16 u2 o2. 539. Find the surface obtained by rotating the cardioid around the polar axis. 540. Find the area of the surface formed by the rotation of the lemniscate Additional Tasks for Chapter IV Areas of plane figures 541. Find the entire area of a region bounded by a curve 542. Find the area of the region bounded by the curve And axis Oh. 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 area contained within loops: 545. Find the area of the region bounded by one loop of the curve: 546. Find the area of the area contained inside the loop: 547. Find the area of the region bounded by the curve 548. Find the area of the region bounded by the curve 549. Find the area of the region bounded by the Oxr axis straight and curve 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 Example 3 Calculate the area of the sphere obtained by rotating the circle about the axis. Decision: from the materials of the article about area and volume with a parametrically given line you know that the equations define a circle centered at the origin with radius 3. well and sphere
, for those who forgot, is the surface ball(or spherical surface). We adhere to the developed solution scheme. Let's find derivatives: Let's compose and simplify the "formula" root: Needless to say, it turned out candy. Check out for comparison how Fikhtengoltz butted heads with the square ellipsoid of revolution. According to the theoretical remark, we consider the upper semicircle. It is "drawn" when changing the value of the parameter within (it is easy to see that Answer: If we solve the problem in general terms, we get exactly the school formula for the area of a sphere, where is its radius. Something painfully simple problem, 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 deduce or intuit the formula for calculating the surface area obtained by rotating a curve around the y-axis. And, of course, the advantage of parametric equations should again be noted - they do not need to be modified somehow; 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 set parametrically. The surface of rotation will resemble ... I don’t even know what to compare it with ... something unearthly - rounded with a pointed depression in the middle. Here, for the case of rotation of the cycloid around the axis, the association instantly came to mind - an oblong rugby ball. Solution and answer at the end of the lesson. We conclude our fascinating review with a case polar coordinates. Yes, it’s a review, if you look into textbooks on mathematical analysis (by Fikhtengolts, Bokhan, Piskunov, and other authors), you can get a good dozen (or even noticeably more) standard examples, among which it is quite possible that you will find the problem you need. How to calculate the surface area of revolution, If the curve is set to 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 In accordance with the geometric meaning of the problem, the integrand Example 5 Calculate the area of the surface formed by the rotation of the cardioid around the polar axis. Decision: the graph of this curve can be seen in Example 6 of the lesson about polar coordinate system. The cardioid is symmetrical about the polar axis, so we consider its upper half on the gap (which, in fact, is also 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": Compose and simplify the root: We use the formula: In between Answer: An interesting and short task for an independent solution: Example 6 Calculate the area of the spherical belt, What is a ball belt? Place a round, unpeeled orange on the table and pick up a knife. Make two parallel cut, thereby dividing the fruit into 3 parts of arbitrary sizes. Now take the middle, in which the juicy pulp is exposed on both sides. This body is called spherical layer, and its bounding surface (orange peel) - ball belt. Readers familiar with polar coordinates, easily presented the drawing of the problem: the equation defines a circle centered at the pole of radius , from which rays Now you can eat an orange with a clear conscience and a light heart, on this tasty note we will finish the lesson, do not spoil your appetite with other examples =) Solutions and answers: Example 2:Decision
: calculate the area of the surface formed by the rotation of the upper branch around the x-axis. We use the formula Example 4:Decision
: use the formula Example 6:Decision
: use the formula: Higher mathematics for correspondence students and not only >>> (Go to main page) Numerical methods is a fairly large section of higher mathematics and serious textbooks on this topic have hundreds of pages. In practice, in tests, some tasks are traditionally proposed for solving by numerical methods, and one of the common tasks is the approximate calculation definite integrals. In this article, I will consider two methods for the approximate calculation of a definite integral − trapezoidal method and simpson's 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 understand what integrals are. Of the technical means, you will need a microcalculator. Yes, yes, we are waiting for routine school calculations. Better yet, download my semi-automatic calculator for the trapezoidal method and the Simpson method. The calculator is written in Excel and will allow you to reduce the time for solving and processing tasks tenfold. A video manual is included for Excel teapots! By the way, the first video with my voice. First, let's ask ourselves the question, why do we need approximate calculations at all? It seems to be possible to find the antiderivative of the function and use the Newton-Leibniz formula, calculating the exact value of a certain integral. As an answer to the question, let's immediately consider a demo example with a picture. Calculate a definite integral Everything would be fine, but in this example the integral is not taken - before you is not taken, the so-called integral logarithm. Does this integral even exist? Let's depict the graph of the integrand in the drawing: Everything is fine. Integrand continuous on the segment and the definite integral is numerically equal to the shaded area. Yes, that's just one snag - the integral is not taken. And in such cases, numerical methods come to the rescue. In this case, the problem occurs in two formulations: 1) Calculate the definite integral approximately , rounding the result to a certain decimal place. For example, up to two decimal places, up to three decimal places, etc. Let's say you get an approximate answer of 5.347. In fact, it may not be entirely correct (actually, let's say the more accurate answer is 5.343). Our task is only in that to round the result to three decimal places. 2) Calculate the definite integral approximately, with a certain precision. For example, calculate the definite integral approximately with an accuracy of 0.001. What does it mean? This means that if an approximate answer of 5.347 is obtained, then Everybody figures must be reinforced concrete correct. To be more precise, the answer 5.347 should differ from the truth modulo (in one direction or another) by no more than 0.001. There are several basic methods for the approximate calculation of a definite integral that occurs in problems: Rectangle Method. The segment of integration is divided into several parts and a step figure is constructed ( bar graph), which is close in area to the desired area: Do not judge strictly by the drawings, the accuracy is not perfect - they only help to understand the essence of the methods. In this example, the segment of integration is divided into three segments: Trapezoidal method. The idea is similar. The integration segment is divided into several intermediate segments, and the graph of the integrand approaches broken line line: So our area (blue shading) is approximated by the sum of the areas of the trapezoids (red). Hence the name of the method. It is easy to see that the trapezoid method gives a much better approximation than the rectangle method (with the same number of partition segments). And, of course, the more smaller intermediate segments we consider, the higher the accuracy will be. The trapezoid method is encountered from time to time in practical tasks, 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 approached 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 the Simpson method will give an even more accurate approximation than the rectangle method or the trapezoid method. I don’t see the point in building a drawing, since visually the approximation will be superimposed on the graph of the function (the broken line of the previous paragraph - and even then it almost coincided). The task of calculating a definite integral using the Simpson formula is the most popular task in practice. And the method of parabolas will be given considerable attention. Surface of revolution- a surface formed during rotation around a straight line (surface axis) of an arbitrary line (straight, flat or spatial curve). For example, if a straight line intersects the axis of rotation, then during its rotation a conical surface will be obtained, if it is parallel to the axis - cylindrical, if it intersects with the axis - a one-sheeted hyperboloid of revolution. The same surface can be obtained by rotating a wide variety of curves. The area of the surface of revolution formed by the rotation of a plane curve of finite length about an axis that lies in the plane of the curve but does not intersect the curve is equal to the product of the length of the curve and the length of a circle with a radius equal to the distance from the axis to the center of mass of the curve. This statement is called Hulden's second theorem, or Pappus' centroid theorem. The area of the surface of revolution formed by the rotation of a curve about an axis can be calculated by the formula For the case when the curve is given in the polar coordinate system, the formula is valid Mechanical applications of a definite integral (work of forces, static moments, center of gravity). Calculation of the work of forces A material point moves along a continuously differentiable curve, while a force acts on it, directed tangentially to the trajectory in the direction of movement. The total work done by the force F(s): If the position of a point on the motion trajectory is described by another parameter, then the formula takes the form: Calculation of static moments and center of gravity Definition 2. Number Moments about the Oy axis are introduced similarly. In space, the concepts of moments of mass with respect to coordinate planes are introduced in a similar way. where M 1 y , M 1 x - geometric static moments of the figure about the axes Oy and Ox; S is the area of the figure.
around the polar axis.
And axis Oh.
And axis Oh.
And axis Oh.
. At the same time, the “drawing direction” of the line, about which so many copies were broken in the article, is indifferent. But, as in the previous paragraph, it is important that the curve is located higher abscissa axis - otherwise, the function "responsible for the players" will take negative values and you will have to put a minus sign in front of the integral.
on this interval), thus:
if the line is given in polar coordinate system?
, where are the angular values corresponding to the ends of the curve.
, and this is achieved only if ( and are known to be 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 extensions. As you can see, the same story as in the previous two paragraphs.
I hope with supernumeraries trigonometric formulas no one had any problems.
, Consequently: 
(I described in detail how to properly get rid of the root in the article Curve arc length).
cut off lesser arc. This arc rotates around the polar axis and thus a spherical belt is obtained.
.
In this case: 
;
In this way:
Answer:
. The first arc of the cycloid is defined on the segment .
Let's find derivatives:
Compose and simplify the root:
So the surface area of revolution is:
In between 
, so 
First integralintegrate by parts
:
In the second integral we usetrigonometric formula
.
Answer:
Answer:
How to calculate a definite integral
using the trapezoid formula and the Simpson method?
. Obviously, the more frequent the partition (the 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 recalled only one practical example). In this regard, I will not consider the method of rectangles, and will not even give a simple formula. Not because of laziness, but because of the principle of my solution book: what is extremely rare in practical tasks is not considered.
Let some mass M be distributed on the Oxy coordinate plane with a density p = p(y) on some set of points S (this can be an arc of a curve or a bounded flat figure). Denote s(y) - the measure of the specified set (arc length or area).
is called the k-th moment of mass M about the Ox axis.
At k \u003d 0 M 0 \u003d M is the mass,
k \u003d 1 M 1 - static moment,
k \u003d 2 M 2 - moment of inertia.
If p = 1, then the corresponding moments are called geometric. The coordinates of the center of gravity of a homogeneous (p - const) flat figure are determined by the formulas:
