Greetings, dear students of Argemony University!

Today we will continue to study the materialization of objects. Last time we rotated flat figures and got three-dimensional bodies. Some of them are very tempting and useful. I think that much that the magician invents can be used in the future.

Today we will rotate the curves. It is clear that in this way we can get some kind of object with very thin edges (a cone or a bottle for potions, a vase for flowers, a glass for drinks, etc.), because a rotating curve can create just such objects. In other words, by rotating the curve, we can get some kind of surface - closed on all sides or not. Why right now I remembered the holey cup from which Sir Shurf Lonley-Lockley drank all the time.

So we will create a leaky bowl and a non-perforated one, and calculate the area of ​​the created surface. I think that for some reason it (in general, the surface area) will be needed - well, at least for applying a special magic paint. And on the other hand, the areas of magical artifacts may be required to calculate the magical forces applied to them or something else. We will learn how to find it, and we will find where to apply it.

So, a piece of a parabola can give us the shape of a bowl. Let's take the simplest y=x 2 on the interval . It can be seen that when it rotates around the OY axis, just a bowl is obtained. No bottom.

The spell to calculate the surface area of ​​rotation is as follows:

Here |y| is the distance from the axis of rotation to any point on the curve that is rotating. As you know, distance is a perpendicular.
A little more difficult with the second element of the spell: ds is the arc differential. These words do not give us anything, so let's not bother, but switch to the language of formulas, where this differential is explicitly presented for all cases known to us:
- Cartesian coordinate system;
- records of the curve in parametric form;
- polar coordinate system.

For our case, the distance from the axis of rotation to any point on the curve is x. We consider the surface area of ​​the resulting holey bowl:

To make a bowl with a bottom, you need to take another piece, but with a different curve: on the interval, this is the line y=1.

It is clear that when it rotates around the OY axis, the bottom of the bowl will be obtained in the form of a circle of unit radius. And we know how the area of ​​a circle is calculated (according to the formula pi * r ^ 2. For our case, the area of ​​\u200b\u200bthe circle will be equal to pi), but we will calculate it using a new formula - for verification.
The distance from the axis of rotation to any point of this piece of the curve is also x.

Well, our calculations are correct, which pleases.

And now homework.

1. Find the surface area obtained by rotating the polyline ABC, where A=(1; 5), B=(1; 2), C=(6; 2), around the OX axis.
Advice. Record all segments in parametric form.
AB: x=1, y=t, 2≤t≤5
BC: x=t, y=2, 1≤t≤6
By the way, what does the resulting item look like?

2. Well, now come up with something yourself. Three items, I think, is enough.

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 cross-sectional area function 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 crossing the pyramid 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 get the function of the cross-sectional areas:

Finding the volume of the pyramid:

The volume of bodies of revolution.

Consider the curve given by the equation y=f(x ). Let's assume 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 x-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 ​​rotation curve AB around a given axis is the limit to which the areas of the surfaces of revolution of broken lines inscribed in the curve AB tend to, when the largest of the lengths of the links of these broken lines tend to zero.

Let us 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 the coordinates x i and y i . When the broken line rotates around the axis, we obtain a surface consisting of lateral surfaces of truncated cones, the area of ​​which is equal to D P i . This area can be found using the formula:

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 “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.

Example 3

Calculate the area of ​​the sphere obtained by rotating the circle about the axis.

Solution: 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 on this interval), thus:

Answer:

If the problem is solved in general view, then you 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, we should again note the advantage parametric equations- 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 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 line is given in polar coordinate system?

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 , where are the angular values ​​corresponding to the ends of the curve.

In accordance with the geometric meaning of the problem, the integrand , 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.

Example 5

Calculate the area of ​​the surface formed by the rotation of the cardioid around the polar axis.

Solution: 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:

I hope with supernumeraries trigonometric formulas no one had any problems.

We use the formula:

In between , hence: (I described in detail how to properly get rid of the root in the article Curve arc length).

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 cut off lesser 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 tasty note we will finish the lesson, do not spoil your appetite with other examples =)

Solutions and answers:

Example 2:Solution : calculate the area of ​​the surface formed by the rotation of the upper branch around the x-axis. We use the formula .
In this case: ;

Thus:


Answer:

Example 4:Solution : use the formula . 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 , That's why

First integralintegrate by parts :

In the second integral we usetrigonometric formula .


Answer:

Example 6:Solution : use the formula:


Answer:

Higher mathematics for correspondence students and not only >>>

(Go to main page)

How to calculate a definite integral
using the trapezoid formula and the Simpson method?

Numerical methods is a fairly large section of higher mathematics and serious textbooks on this topic have hundreds of pages. In practice, in control work traditionally proposed for solving some problems by numerical methods, and one of the common problems is - 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 All 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 chart), 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:
. 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.

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 occurs occasionally in practical tasks, and several examples will be discussed in this article.

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.

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

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=2y and lying above the horizontal plane .

Obviously, the body under consideration is projected onto the axis of the segment
, and for x
the cross section of the body is a 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:

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

Calculation of volumes of bodies of revolution

Let on the segment[ a, b] is a continuous sign-constant function y= f(x). Volumes of a body of revolution formed by rotation around an axis Oh(or axes OU) curvilinear trapezoid bounded by a curve y= f(x) (f(x) 0) and direct y=0, x=a, x=b, are calculated according to the formulas:

, ( 19)

(20)

If a body is formed by rotation around an axis OU curvilinear trapezoid bounded by a curve
and direct x=0, y= c, y= d, then the volume of the body of revolution is equal to

. (21)

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

According to formula (19), the desired volume

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

E that line rotates in space around the axis, and the resulting surface of revolution limits some body of revolution (see Fig.). Find the volume of this body of revolution.

According to the formula, we get:

Surface area of ​​rotation

,
, 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 the curve given by a non-negative 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 the arc of the curve is given parametric equations
,
, and
, That

If the arc is set to polar coordinates
, That

.

Example. Calculate the area of ​​the surface formed by rotation in space around the axis of the part of the line y= located above the cutoff line.

Because
, then the formula gives us the integral

Let's make the change t=x+(1/2) in the last integral and get:

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

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

Moving to the left side and dividing by 2, we get

where, finally,

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

Variable force work. Consider the motion of a material point along the axis OX under the action of a variable force f, depending on the position of the point x on the axis, i.e. a force that is a function x. Then work A, necessary to move a material point from a position x = a into position x = b calculated by the formula:

To calculate liquid pressure force use Pascal's law, according to which the pressure of a liquid on a platform is equal to its area S multiplied by the immersion depth h, on the 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
, That 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 relative to the same axes Ox and Oy are calculated by the formulas

A center of mass coordinates And - by formulas

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

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

If density is not specified, the curve is assumed to be uniform and
. We have: Therefore,

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

From here we get:

In applications, the following is often useful. Theorem Gulden. The surface area formed by the rotation of an arc of a plane curve around an axis that lies in the plane of the arc and does not intersect 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 the semicircle

Because of the symmetry
. When a 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 pa. By Gulden's theorem, we have 4

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

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

Example 4 The speed of the rectilinear movement of the body is expressed by the formula (m / s). Find the path traveled by the body in 5 seconds from the start of the movement.

Because path taken by the body with the speed v(t) for the time interval , is expressed by the integral

then we have:

P
example. Let's find the area of ​​the limited area lying between the axis and the line y=x 3 -x. Because the

the line crosses the axis at three points: x 1 \u003d -1, x 2 \u003d 0, x 3 \u003d 1.

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

P
example. Find the area of ​​the region 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 bring an argument change to one gap, 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
:

P example. Let's find the volume of the body bounded by the surface of rotation of the line y=4x-x 2 around the axis (with
).

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

P example. Calculate the arc length of the line y=lncosx located between the straight lines and
.

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

Answer:
.

Example. Calculate the area Q of the surface of revolution obtained by rotating the arc of the cycloid x=t-sint ; y=1-cost, with

, around the axis.

D To calculate, we apply the formula:

We have:

, So

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

we get

, and

In addition, we precompute

(So
) And

We get:

Making the substitution , we arrive at the integral

Before proceeding to the formulas for the area of ​​a surface of revolution, we give a brief formulation of the surface of revolution itself. The surface of revolution, or, what is the same, the surface of a body of revolution is a spatial figure formed by the rotation of a segment AB curve around the axis Ox(picture below).

Let us 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 surface area of ​​rotation or the surface of a body of rotation is its outer shell, not counting the circles formed by rotation around the axis of 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 ​​a 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 ​​revolution is as follows:

(1).

Example 1 Find the surface area of ​​a paraboloid formed by rotation about an axis Ox the arc of the parabola corresponding to the change x from x= 0 to x = a .

Solution. We explicitly express the function that defines the arc of the parabola:

Let's find the derivative of this function:

Before using the formula for finding the area of ​​the surface of revolution, let's write the part of its integrand that is the root and substitute the derivative we just found there:

Answer: The arc length of the curve is

.

Example 2 Find the area of ​​the surface formed by rotation about an axis Ox astroids.

Solution. 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 will need to substitute in the formula to find the surface area of ​​rotation:

.

We perform integration from 0 to a:

Calculation of the surface area of ​​revolution given parametrically

Consider the case when the curve forming the surface of revolution is given by the parametric equations

Then the area of ​​the surface of revolution is calculated by the formula

(2).

Example 3 Find the area of ​​the surface of revolution formed by the rotation about an axis Oy figure bounded by a cycloid and a straight line y = a. The cycloid is given by the parametric equations

Solution. Find the intersection points of the cycloid and the line. Equating the cycloid equation and the equation of a straight line y = a, find

It follows from this that the limits of integration correspond to

Now we can apply formula (2). Let's find derivatives:

We write the radical expression in the formula, substituting the found derivatives:

Let's find the root of this expression:

.

Substitute the found in the formula (2):

.

Let's make a substitution:

And finally we find

In the transformation of expressions, trigonometric formulas were used

Answer: The area of ​​the surface of revolution is .

Calculating the area of ​​a surface of revolution given in polar coordinates

Let the curve whose rotation forms the surface be given in polar coordinates.