Using integrals to find the volumes of bodies of revolution

The practical usefulness of mathematics is due to the fact that without

Specific mathematical knowledge makes it difficult to understand the principles of the device and the use of modern technology. Every person in his life has to perform quite complex calculations, use commonly used equipment, find the necessary formulas in reference books, and create simple algorithms for solving problems. In modern society, more and more specialties that require a high level of education are associated with the direct application of mathematics. Thus, mathematics becomes a professionally significant subject for a student. The leading role belongs to mathematics in the formation of algorithmic thinking; it develops the ability to act according to a given algorithm and to construct new algorithms.

While studying the topic of using the integral to calculate the volumes of bodies of revolution, I suggest that students in elective classes consider the topic: “Volumes of bodies of revolution using integrals.” Below are methodological recommendations for considering this topic:

1. Area of ​​a flat figure.

From the algebra course we know that problems of a practical nature led to the concept of a definite integral..gif" width="88" height="51">.jpg" width="526" height="262 src=">

https://pandia.ru/text/77/502/images/image006_95.gif" width="127" height="25 src=">.

To find the volume of a body of rotation formed by the rotation of a curvilinear trapezoid around the Ox axis, bounded by a broken line y=f(x), the Ox axis, straight lines x=a and x=b, we calculate using the formula

https://pandia.ru/text/77/502/images/image008_26.jpg" width="352" height="283 src=">Y

3.Cylinder volume.

https://pandia.ru/text/77/502/images/image011_58.gif" width="85" height="51">..gif" width="13" height="25">..jpg" width="401" height="355">The cone is obtained by rotating the right triangle ABC (C = 90) around the Ox axis on which leg AC lies.

Segment AB lies on the straight line y=kx+c, where https://pandia.ru/text/77/502/images/image019_33.gif" width="59" height="41 src=">.

Let a=0, b=H (H is the height of the cone), then Vhttps://pandia.ru/text/77/502/images/image021_27.gif" width="13" height="23 src=">.

5.Volume of a truncated cone.

A truncated cone can be obtained by rotating a rectangular trapezoid ABCD (CDOx) around the Ox axis.

The segment AB lies on the straight line y=kx+c, where , c=r.

Since the straight line passes through point A (0;r).

Thus, the straight line looks like https://pandia.ru/text/77/502/images/image027_17.gif" width="303" height="291 src=">

Let a=0, b=H (H is the height of the truncated cone), then https://pandia.ru/text/77/502/images/image030_16.gif" width="36" height="17 src="> = .

6. Volume of the ball.

The ball can be obtained by rotating a circle with center (0;0) around the Ox axis. The semicircle located above the Ox axis is given by the equation

https://pandia.ru/text/77/502/images/image034_13.gif" width="13" height="16 src=">x R.

How to calculate the volume of a body of revolution
using a definite integral?

In general, there are a lot of interesting applications in integral calculus; using a definite integral, you can calculate the area of ​​a figure, the volume of a body of rotation, the length of an arc, the surface area of ​​rotation and much more. So it will be fun, please stay optimistic!

Imagine some flat figure on the coordinate plane. Introduced? ... I wonder who presented what... =))) We have already found its area. But, in addition, this figure can also be rotated, and rotated in two ways:

- around the abscissa axis;
- around the ordinate axis.

This article will examine both cases. The second method of rotation is especially interesting; it causes the most difficulties, but in fact the solution is almost the same as in the more common rotation around the x-axis. As a bonus I will return to problem of finding the area of ​​a figure, and I’ll tell you how to find the area in the second way - along the axis. It’s not so much a bonus as the material fits well into the topic.

Let's start with the most popular type of rotation.

flat figure around an axis

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

Solution: As in the problem of finding the area, the solution begins with a drawing of a flat figure. That is, on the plane it is necessary to construct a figure bounded by the lines , and do not forget that the equation specifies the axis. How to complete a drawing more efficiently and quickly can be found on the pages Graphs and properties of Elementary functions And . This is a Chinese reminder, and at this point I will not dwell further.

The drawing here is quite simple:

The desired flat figure is shaded in blue; it is the one that rotates around the axis. As a result of the rotation, the result is a slightly ovoid flying saucer that is symmetrical about the axis. In fact, the body has a mathematical name, but I’m too lazy to clarify anything in the reference book, so we move on.

How to calculate the volume of a body of revolution?

The volume of a body of revolution can be calculated using the formula:

In the formula, the number must be present before the integral. So it happened - everything that revolves in life is connected with this constant.

I think it’s easy to guess how to set the limits of integration “a” and “be” from the completed drawing.

Function... what is this function? Let's look at the drawing. The plane figure is bounded by the graph of the parabola at the top. This is the function that is implied in the formula.

In practical tasks, a flat figure can sometimes be located below the axis. This does not change anything - the integrand in the formula is squared: , thus the integral is always non-negative, which is very logical.

Let's calculate the volume of a body of rotation using this formula:

As I already noted, the integral almost always turns out to be simple, the main thing is to be careful.

Answer:

In your answer you must indicate the dimension - cubic units. That is, in our body of rotation there are approximately 3.35 “cubes”. Why cubic units? Because the most universal formulation. There could be cubic centimeters, there could be cubic meters, there could be cubic kilometers, etc., that’s how many green men your imagination can put in a flying saucer.

Find the volume of a body formed by rotation around the axis of a figure bounded by lines , ,

This is an example for you to solve on your own. Full solution and answer at the end of the lesson.

Let's consider two more complex problems, which are also often encountered in practice.

Calculate the volume of the body obtained by rotating around the abscissa axis of the figure bounded by the lines , , and

Solution: Let us depict in the drawing a flat figure bounded by the lines , , , , without forgetting that the equation defines the axis:

The desired figure is shaded in blue. When it rotates around its axis, it turns out to be a surreal donut with four corners.

Let us calculate the volume of the body of revolution as difference in volumes of bodies.

First, let's look at the figure circled in red. When it rotates around an axis, a truncated cone is obtained. Let us denote the volume of this truncated cone by .

Consider the figure that is circled in green. If you rotate this figure around the axis, you will also get a truncated cone, only a little smaller. Let's denote its volume by .

And, obviously, the difference in volumes is exactly the volume of our “donut”.

We use the standard formula to find the volume of a body of rotation:

1) The figure circled in red is bounded above by a straight line, therefore:

2) The figure circled in green is bounded above by a straight line, therefore:

3) Volume of the desired body of rotation:

Answer:

It is curious that in this case the solution can be checked using the school formula for calculating the volume of a truncated cone.

The decision itself is often written shorter, something like this:

Now let’s take a little rest and tell you about geometric illusions.

People often have illusions associated with volumes, which was noticed by Perelman (another) in the book Entertaining geometry. Look at the flat figure in the solved problem - it seems to be small in area, and the volume of the body of revolution is just over 50 cubic units, which seems too large. By the way, the average person drinks the equivalent of a room of 18 square meters of liquid in his entire life, which, on the contrary, seems too small a volume.

After a lyrical digression, it is just appropriate to solve a creative task:

Calculate the volume of a body formed by rotation about the axis of a flat figure bounded by the lines , , where .

This is an example for you to solve on your own. Please note that all cases occur in the band, in other words, ready-made limits of integration are actually given. Draw the graphs of trigonometric functions correctly, let me remind you of the lesson material about geometric transformations of graphs: if the argument is divided by two: , then the graphs are stretched twice along the axis. It is advisable to find at least 3-4 points according to trigonometric tables to complete the drawing more accurately. Full solution and answer at the end of the lesson. By the way, the task can be solved rationally and not very rationally.

Calculation of the volume of a body formed by rotation
flat figure around an axis

The second paragraph will be even more interesting than the first. The task of calculating the volume of a body of revolution around the ordinate axis is also a fairly common guest in test work. Along the way it will be considered problem of finding the area of ​​a figure the second method is integration along the axis, this will allow you not only to improve your skills, but also teach you to find the most profitable solution path. There is also a practical life meaning in this! As my teacher on mathematics teaching methods recalled with a smile, many graduates thanked her with the words: “Your subject helped us a lot, now we are effective managers and optimally manage staff.” Taking this opportunity, I also express my great gratitude to her, especially since I use the acquired knowledge for its intended purpose =).

I recommend it to everyone, even complete dummies. Moreover, the material learned in the second paragraph will provide invaluable assistance in calculating double integrals.

Given a flat figure bounded by the lines , , .

1) Find the area of ​​a flat figure bounded by these lines.
2) Find the volume of the body obtained by rotating a flat figure bounded by these lines around the axis.

Attention! Even if you only want to read the second point, be sure to read the first one first!

Solution: The task consists of two parts. Let's start with the square.

1) Let's make a drawing:

It is easy to see that the function specifies the upper branch of the parabola, and the function specifies the lower branch of the parabola. Before us is a trivial parabola that “lies on its side.”

The desired figure, the area of ​​which is to be found, is shaded in blue.

How to find the area of ​​a figure? It can be found in the “usual” way, which was discussed in class Definite integral. How to calculate the area of ​​a figure. Moreover, the area of ​​the figure is found as the sum of the areas:
- on the segment ;
- on the segment.

That's why:

Why is the usual solution bad in this case? Firstly, we got two integrals. Secondly, there are roots under integrals, and roots in integrals are not a gift, and besides, you can get confused in substituting the limits of integration. In fact, the integrals, of course, are not killer, but in practice everything can be much sadder, I just selected “better” functions for the problem.

There is a more rational solution: it consists of switching to inverse functions and integrating along the axis.

How to get to inverse functions? Roughly speaking, you need to express “x” through “y”. First, let's look at the parabola:

This is enough, but let’s make sure that the same function can be derived from the lower branch:

It's easier with a straight line:

Now look at the axis: please periodically tilt your head to the right 90 degrees as you explain (this is not a joke!). The figure we need lies on the segment, which is indicated by the red dotted line. In this case, on the segment the straight line is located above the parabola, which means that the area of ​​the figure should be found using the formula already familiar to you: . What has changed in the formula? Just a letter and nothing more.

! Note: The limits of integration along the axis should be set strictly from bottom to top!

Finding the area:

On the segment, therefore:

Please note how I carried out the integration, this is the most rational way, and in the next paragraph of the task it will be clear why.

For readers who doubt the correctness of integration, I will find derivatives:

The original integrand function is obtained, which means the integration was performed correctly.

Answer:

2) Let us calculate the volume of the body formed by the rotation of this figure around the axis.

I’ll redraw the drawing in a slightly different design:

So, the figure shaded in blue rotates around the axis. The result is a “hovering butterfly” that rotates around its axis.

To find the volume of a body of rotation, we will integrate along the axis. First we need to go to inverse functions. This has already been done and described in detail in the previous paragraph.

Now we tilt our head to the right again and study our figure. Obviously, the volume of a body of rotation should be found as the difference in volumes.

We rotate the figure circled in red around the axis, resulting in a truncated cone. Let us denote this volume by .

We rotate the figure circled in green around the axis and denote it by the volume of the resulting body of rotation.

The volume of our butterfly is equal to the difference in volumes.

We use the formula to find the volume of a body of revolution:

What is the difference from the formula in the previous paragraph? Only in the letter.

But the advantage of integration, which I recently talked about, is much easier to find , rather than first raising the integrand to the 4th power.

Answer:

Note that if the same flat figure is rotated around the axis, you will get a completely different body of rotation, with a different volume, naturally.

Given a flat figure bounded by lines and an axis.

1) Go to inverse functions and find the area of ​​a plane figure bounded by these lines by integrating over the variable.
2) Calculate the volume of the body obtained by rotating a flat figure bounded by these lines around the axis.

This is an example for you to solve on your own. Those interested can also find the area of ​​a figure in the “usual” way, thereby checking point 1). But if, I repeat, you rotate a flat figure around the axis, you will get a completely different body of rotation with a different volume, by the way, the correct answer (also for those who like to solve problems).

A complete solution to the two proposed points of the task is at the end of the lesson.

Yes, and don’t forget to tilt your head to the right to understand the bodies of rotation and the limits of integration!

I was about to finish the article, but today they brought an interesting example just for finding the volume of a body of revolution around the ordinate axis. Fresh:

Calculate the volume of a body formed by rotation around the axis of a figure bounded by curves and .

Solution: Let's make a drawing:

Along the way, we get acquainted with the graphs of some other functions. Here is an interesting graph of an even function...

Let T be a body of revolution formed by rotation around the abscissa axis of a curvilinear trapezoid located in the upper half-plane and limited by the abscissa axis, straight lines x=a and x=b and the graph of a continuous function y=f(x) .

Let's prove that this is the body of revolution is cubed and its volume is expressed by the formula

V=\pi \int\limits_(a)^(b) f^2(x)\,dx= \pi \int\limits_(a)^(b)y^2\,dx\,.

First, we prove that this body of revolution is regular if we choose the Oyz plane perpendicular to the axis of rotation as \Pi. Note that the section located at a distance x from the plane Oyz is a circle of radius f(x) and its area S(x) is equal to \pi f^2(x) (Fig. 46). Therefore, the function S(x) is continuous due to the continuity of f(x). Next, if S(x_1)\leqslant S(x_2), then this means that . But the projections of the sections onto the Oyz plane are circles of radii f(x_1) and f(x_2) with center O, and from f(x_1)\leqslant f(x_2) it follows that a circle of radius f(x_1) is contained in a circle of radius f(x_2) .

So, the body of revolution is regular. Therefore, it is cubed and its volume is calculated by the formula

V=\pi \int\limits_(a)^(b) S(x)\,dx= \pi \int\limits_(a)^(b)f^2(x)\,dx\,.

If a curvilinear trapezoid was bounded both below and above by the curves y_1=f_1(x), y_2=f_2(x), then

V= \pi \int\limits_(a)^(b)y_2^2\,dx- \pi \int\limits_(a)^(b)y_1^2\,dx= \pi\int\limits_(a )^(b)\Bigl(f_2^2(x)-f_1^2(x)\Bigr)dx\,.

Formula (3) can also be used to calculate the volume of a body of revolution in the case when the boundary of a rotating figure is specified by parametric equations. In this case, you have to use a change of variable under the definite integral sign.

In some cases it turns out to be convenient to decompose bodies of rotation not into straight circular cylinders, but into figures of a different type.

For example, let's find volume of a body obtained by rotating a curved trapezoid around the ordinate axis. First, let's find the volume obtained by rotating a rectangle with height y#, at the base of which lies the segment . This volume is equal to the difference in volumes of two straight circular cylinders

\Delta V_k= \pi y_k x_(k+1)^2- \pi y_k x_k^2= \pi y_k \bigl(x_(k+1)+x_k\bigr) \bigl(x_(k+1)- x_k\bigr).

But now it is clear that the required volume is estimated from above and below as follows:

2\pi \sum_(k=0)^(n-1) m_kx_k\Delta x_k \leqslant V\leqslant 2\pi \sum_(k=0)^(n-1) M_kx_k\Delta x_k\,.

It follows easily from here formula for the volume of a body of revolution around the ordinate axis:

V=2\pi \int\limits_(a)^(b) xy\,dx\,.

Example 4. Let's find the volume of a ball of radius R.

Solution. Without loss of generality, we will consider a circle of radius R with its center at the origin. This circle, rotating around the Ox axis, forms a ball. The equation of a circle is x^2+y^2=R^2, so y^2=R^2-x^2. Taking into account the symmetry of the circle relative to the ordinate axis, we first find half of the required volume

\frac(1)(2)V= \pi\int\limits_(0)^(R)y^2\,dx= \pi\int\limits_(0)^(R) (R^2-x^ 2)\,dx= \left.(\pi\!\left(R^2x- \frac(x^3)(3)\right))\right|_(0)^(R)= \pi\ !\left(R^3- \frac(R^3)(3)\right)= \frac(2)(3)\pi R^3.

Therefore, the volume of the entire ball is equal to \frac(4)(3)\pi R^3.

Example 5. Calculate the volume of a cone whose height h and base radius r.

Solution. Let us choose a coordinate system so that the Ox axis coincides with the height h (Fig. 47), and take the vertex of the cone as the origin of coordinates. Then the equation of straight line OA will be written in the form y=\frac(r)(h)\,x.

Using formula (3), we obtain:

V=\pi \int\limits_(0)^(h) y^2\,dx= \pi \int\limits_(0)^(h) \frac(r^2)(h^2)\,x ^2\,dx= \left.(\frac(\pi r^2)(h^2)\cdot \frac(x^3)(3))\right|_(0)^(h)= \ frac(\pi)(3)\,r^2h\,.

Example 6. Let's find the volume of the body obtained by rotating around the x-axis of the astroid \begin(cases)x=a\cos^3t\,\\ y=a\sin^3t\,.\end(cases)(Fig. 48).

Solution. Let's build an astroid. Let us consider half of the upper part of the astroid, located symmetrically relative to the ordinate axis. Using formula (3) and changing the variable under the definite integral sign, we find the limits of integration for the new variable t.

If x=a\cos^3t=0 , then t=\frac(\pi)(2) , and if x=a\cos^3t=a , then t=0 . Considering that y^2=a^2\sin^6t and dx=-3a\cos^2t\sin(t)\,dt, we get:

V=\pi \int\limits_(a)^(b) y^2\,dx= \pi \int\limits_(\pi/2)^(0) a^2\sin^6t \bigl(-3a \cos^2t\sin(t)\bigr)\,dt= \ldots= \frac(16\pi)(105)\,a^3.

The volume of the entire body formed by the rotation of the astroid will be \frac(32\pi)(105)\,a^3.

Example 7. Let us find the volume of the body obtained by rotating around the ordinate axis of a curvilinear trapezoid bounded by the x-axis and the first arc of the cycloid \begin(cases)x=a(t-\sin(t)),\\ y=a(1-\cos(t)).\end(cases).

Solution. Let's use formula (4): V=2\pi \int\limits_(a)^(b)xy\,dx, and replace the variable under the integral sign, taking into account that the first arc of the cycloid is formed when the variable t changes from 0 to 2\pi. Thus,

\begin(aligned)V&= 2\pi \int\limits_(0)^(2\pi) a(t-\sin(t))a(1-\cos(t))a(1-\cos( t))\,dt= 2\pi a^3 \int\limits_(0)^(2\pi) (t-\sin(t))(1-\cos(t))^2\,dt= \\ &= 2\pi a^3 \int\limits_(0)^(2\pi)\bigl(t-\sin(t)- 2t\cos(t)+ 2\sin(t)\cos( t)+ t\cos^2t- \sin(t)\cos^2t\bigr)\,dt=\\ &= \left.(2\pi a^3\!\left(\frac(t^2 )(2)+ \cos(t)- 2t\sin(t)- 2\cos(t)+ \sin^2t+ \frac(t^2)(4)+ \frac(t)(4)\sin2t+ \frac(1)(8)\cos2t+ \frac(1)(3)\cos^3t\right))\right|_(0)^(2\pi)=\\ &= 2\pi a^3 \!\left(2\pi^2+1-2+\pi^2+\frac(1)(8)+ \frac(1)(3)-1+2- \frac(1)(8) - \frac(1)(3)\right)= 6\pi^3a^3. \end(aligned)

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Topic: “Calculating the volumes of bodies of revolution using a definite integral”

Lesson type: combined.

The purpose of the lesson: learn to calculate the volumes of bodies of revolution using integrals.

Tasks:

consolidate the ability to identify curvilinear trapezoids from a number of geometric figures and develop the skill of calculating the areas of curvilinear trapezoids;

get acquainted with the concept of a three-dimensional figure;

learn to calculate the volumes of bodies of revolution;

promote the development of logical thinking, competent mathematical speech, accuracy when constructing drawings;

to cultivate interest in the subject, in operating with mathematical concepts and images, to cultivate will, independence, and perseverance in achieving the final result.

During the classes

I. Organizational moment.

Greetings from the group. Communicate lesson objectives to students.

I would like to start today's lesson with a parable. “Once upon a time there lived a wise man who knew everything. One man wanted to prove that the sage does not know everything. Holding a butterfly in his hands, he asked: “Tell me, sage, which butterfly is in my hands: dead or alive?” And he thinks: “If the living one says, I’ll kill her; if the dead one says, I’ll release her.” The sage, after thinking, replied: “Everything is in your hands.”

Therefore, let’s work fruitfully today, acquire a new store of knowledge, and we will apply the acquired skills and abilities in future life and in practical activities. “Everything is in your hands.”

II. Repetition of previously studied material.

Let's remember the main points of the previously studied material. To do this, let’s complete the task “Eliminate the extra word.”

(Students say an extra word.)

Right "Differential". Try to name the remaining words with one common word. (Integral calculus.)

Let's remember the main stages and concepts associated with integral calculus.

Exercise. Recover the gaps. (The student comes out and writes in the required words with a marker.)

Work in notebooks.

The Newton-Leibniz formula was derived by the English physicist Isaac Newton (1643-1727) and the German philosopher Gottfried Leibniz (1646-1716). And this is not surprising, because mathematics is the language spoken by nature itself.

Let's consider how this formula is used to solve practical problems.

Example 1: Calculate the area of ​​a figure bounded by lines

Solution: Let's construct graphs of functions on the coordinate plane . Let's select the area of ​​the figure that needs to be found.

III. Learning new material.

Pay attention to the screen. What is shown in the first picture? (The figure shows a flat figure.)

What is shown in the second picture? Is this figure flat? (The figure shows a three-dimensional figure.)

In space, on earth and in everyday life, we encounter not only flat figures, but also three-dimensional ones, but how can we calculate the volume of such bodies? For example: the volume of a planet, comet, meteorite, etc.

People think about volume both when building houses and when pouring water from one vessel to another. Rules and techniques for calculating volumes had to emerge; how accurate and justified they were is another matter.

The year 1612 was very fruitful for the residents of the Austrian city of Linz, where the famous astronomer Johannes Kepler lived, especially for grapes. People were preparing wine barrels and wanted to know how to practically determine their volumes.

Thus, the considered works of Kepler marked the beginning of a whole stream of research that culminated in the last quarter of the 17th century. design in the works of I. Newton and G.V. Leibniz of differential and integral calculus. From that time on, the mathematics of variables took a leading place in the system of mathematical knowledge.

Today you and I will engage in such practical activities, therefore,

The topic of our lesson: “Calculating the volumes of bodies of rotation using a definite integral.”

You will learn the definition of a body of rotation by completing the following task.

“Labyrinth”.

Exercise. Find a way out of the confusing situation and write down the definition.

IVCalculation of volumes.

Using a definite integral, you can calculate the volume of a particular body, in particular, a body of revolution.

A body of revolution is a body obtained by rotating a curved trapezoid around its base (Fig. 1, 2)

The volume of a body of revolution is calculated using one of the formulas:

1. around the OX axis.

2. , if the rotation of a curved trapezoid around the axis of the op-amp.

Students write down basic formulas in a notebook.

The teacher explains the solutions to the examples on the board.

1. Find the volume of the body obtained by rotating around the ordinate axis of a curvilinear trapezoid bounded by lines: x2 + y2 = 64, y = -5, y = 5, x = 0.

Solution.

Answer: 1163 cm3.

2. Find the volume of the body obtained by rotating a parabolic trapezoid around the x-axis y = , x = 4, y = 0.

Solution.

V. Math simulator.

2. The set of all antiderivatives of a given function is called

A) an indefinite integral,

B) function,

B) differentiation.

7. Find the volume of the body obtained by rotating around the abscissa axis of a curvilinear trapezoid bounded by lines:

D/Z. Consolidating new material

Calculate the volume of the body formed by the rotation of the petal around the x-axis y = x2, y2 = x.

Let's build graphs of the function. y = x2, y2 = x. Let's transform the graph y2 = x to the form y = .

We have V = V1 - V2 Let’s calculate the volume of each function:

Conclusion:

The definite integral is a certain foundation for the study of mathematics, which makes an irreplaceable contribution to solving practical problems.

The topic “Integral” clearly demonstrates the connection between mathematics and physics, biology, economics and technology.

The development of modern science is unthinkable without the use of the integral. In this regard, it is necessary to begin studying it within the framework of secondary specialized education!

VI. Grading.(With commentary.)

The great Omar Khayyam - mathematician, poet, philosopher. He encourages us to be masters of our own destiny. Let's listen to an excerpt from his work:

You say, this life is one moment.
Appreciate it, draw inspiration from it.
As you spend it, so it will pass.
Don't forget: she is your creation.

Definition 3. A body of revolution is a body obtained by rotating a flat figure around an axis that does not intersect the figure and lies in the same plane with it.

The axis of rotation may intersect the figure if it is the axis of symmetry of the figure.

Theorem 2.
, axis
and straight segments
And

rotates around an axis
. Then the volume of the resulting body of rotation can be calculated using the formula

(2)

Proof. For such a body, the cross section with abscissa is a circle of radius
, Means
and formula (1) gives the required result.

If the figure is limited by the graphs of two continuous functions
And
, and line segments
And
, and
And
, then upon rotation around the x-axis we obtain a body whose volume

Example 3. Calculate the volume of a torus obtained by rotating a circle bounded by a circle

around the abscissa axis.

R decision. The indicated circle is limited below by the graph of the function
, and from above –
. The difference of the squares of these functions:

Required volume

(the graph of the integrand is the upper semicircle, so the integral written above is the area of ​​the semicircle).

Example 4. Parabolic segment with base
, and height , rotates around the base. Calculate the volume of the resulting body (“lemon” by Cavalieri).

R decision. We will place the parabola as shown in the figure. Then its equation
, and
. Let's find the value of the parameter :
. So, the required volume:

Theorem 3. Let a curvilinear trapezoid bounded by the graph of a continuous non-negative function
, axis
and straight segments
And
, and
, rotates around an axis
. Then the volume of the resulting body of rotation can be found by the formula

(3)

The idea of ​​proof. We split the segment
dots

, into parts and draw straight lines
. The entire trapezoid will be decomposed into strips, which can be considered approximately rectangles with a base
and height
.

We cut the resulting cylinder by rotating such a rectangle along its generatrix and unfold it. We get an “almost” parallelepiped with dimensions:
,
And
. Its volume
. So, for the volume of a body of revolution we will have the approximate equality

To obtain exact equality, one must go to the limit at
. The sum written above is the integral sum for the function
, therefore, in the limit we obtain the integral from formula (3). The theorem has been proven.

Note 1. In Theorems 2 and 3 the condition
can be omitted: formula (2) is generally insensitive to the sign
, and in formula (3) it is enough
replaced by
.

Example 5. Parabolic segment (base
, height ) rotates around the height. Find the volume of the resulting body.

Solution. Let's place the parabola as shown in the figure. And although the axis of rotation intersects the figure, it - the axis - is the axis of symmetry. Therefore, we need to consider only the right half of the segment. Parabola equation
, and
, Means
. For volume we have:

Note 2. If the curvilinear boundary of a curvilinear trapezoid is given by parametric equations
,
,
And
,
then you can use formulas (2) and (3) with the replacement on
And
on
when it changes t from
before .

Example 6. The figure is limited by the first arc of the cycloid
,
,
, and the x-axis. Find the volume of the body obtained by rotating this figure around: 1) axis
; 2) axes
.

Solution. 1) General formula
In our case:

2) General formula
For our figure:

We invite students to carry out all the calculations themselves.

Note 3. Let a curved sector bounded by a continuous line
and rays
,

, rotates around a polar axis. The volume of the resulting body can be calculated using the formula.

Example 7. Part of a figure bounded by a cardioid
, lying outside the circle
, rotates around a polar axis. Find the volume of the resulting body.

Solution. Both lines, and therefore the figure they limit, are symmetrical about the polar axis. Therefore, it is necessary to consider only that part for which
. The curves intersect at
And

at
. Further, the figure can be considered as the difference of two sectors, and therefore the volume can be calculated as the difference of two integrals. We have:

Tasks for an independent decision.

1. A circular segment whose base
, height , rotates around the base. Find the volume of the body of rotation.

2. Find the volume of a paraboloid of revolution whose base , and the height is .

3. Figure bounded by an astroid
,
rotates around the abscissa axis. Find the volume of the resulting body.

4. Figure bounded by lines
And
rotates around the x-axis. Find the volume of the body of rotation.