Development of a lesson plan in mathematics on the topic “Flat figures and three-dimensional bodies” (grade 3). Mathematics lesson on the topic "flat and volumetric geometric bodies"

As with the problem of finding the area, you need confident drawing skills - this is almost the most important thing (since the integrals themselves will often be easy). Master literate and fast technology plotting can be done using teaching materials and Geometric Transformations of Graphs. But, in fact, I have already talked about the importance of drawings several times in class.

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

Imagine some flat figure on coordinate plane. Introduced? ... I wonder who presented what... =))) We have already found its area. But besides that this figure You can also rotate, and rotate 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

Example 1

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 Definite integral. How to calculate the area of ​​a figure. This is a Chinese reminder, and on at this moment I don't stop anymore.

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 rotation?

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 revolution 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 most universal formulation. There may be cubic centimeters, there may be cubic meters, maybe cubic kilometers, etc., that’s how many little green men your imagination can put in a flying saucer.

Example 2

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

This is an example for independent decision. Complete solution and the answer at the end of the lesson.

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

Example 3

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 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 revolution:

Answer:

It is curious that in in this case the solution can be verified using school formula to calculate 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 a liquid equivalent to a room with an area of ​​18 in his entire life. square meters, which, on the contrary, seems to be too small a volume.

In general, the education system in the USSR was truly the best. The same book by Perelman, published back in 1950, very well develops, as the humorist said, understanding and teaches you to look for original non-standard solutions problems. I recently re-read some of the chapters with great interest, I recommend it, it’s accessible even to humanists. No, you don’t need to smile that I offered a free time, erudition and broad horizons in communication are a great thing.

After lyrical digression it's appropriate to decide creative task:

Example 4

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 frequent guest in tests. 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 point in this. life meaning! As my teacher on methods of teaching mathematics recalled with a smile, many graduates thanked her with the words: “Your subject helped us a lot, now we effective managers and optimally manage our staff.” Taking this opportunity, I also express my great gratitude to her, especially since I use the knowledge gained in direct 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.

Example 5

Given a flat figure bounded by 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, first Necessarily read the first one!

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:

What's bad in this case? the usual way solutions? Firstly, we got two integrals. Secondly, integrals are roots, 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 are more rational way solutions: it consists in moving to inverse functions and integration 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 revolution.

The volume of our butterfly equal to the difference 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:

However, not a sickly butterfly.

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.

Example 6

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).

The 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!

Volumetric bodies. Look around you and you will find three-dimensional bodies everywhere. These are geometric shapes that have three dimensions: length, width and height. For example, to imagine a multi-story building, it is enough to say: “This house is three entrances long, two windows wide and six floors high.” Known to you from primary school cuboid and the cube are completely described by three dimensions. All objects around us have three dimensions, but not all of them can be named length, width and height. For example, for a tree we can specify only the height, for a rope - the length, for a hole - the depth. And for the ball? Does it also have three dimensions? We say that a body has three dimensions (is volumetric) if a cube or ball can be placed in it.

Geometry 11th grade

“Geometric bodies of rotation” - Visualization. Practical part. Job creative group. Repetition of the theory. People creative professions. Exchange of experience. Inspiration. Organizational moment. The only way to learn is to have fun. Museum of Geometric Solids. People who devoted themselves to science. Bodies. People of science are working. A sage was walking. Summing up. Cylindrical surface. People of working professions. Students' knowledge. Bodies of rotation. Basic knowledge.

“The Theorem of Three Perpendiculars” - Point. Perpendicularity of lines. Thinking. Theorem of three perpendiculars. Perpendicular to the plane of a parallelogram. Straight. Legs. Perpendicular. Theorem. Intersections of diagonals. Segment. Perpendicular to the plane of the triangle. Side of the rhombus. Sides of a triangle. Distance. Perpendiculars to lines. Think about it. MA segment. Construction tasks. Proof. Converse theorem. Tasks for using TTP.

“Sphere area” - Diameter of the ball (d=2R). Radius great circle is the radius of the ball. Layer=vsh.Seg.1-vsh.Seg.2. Segment height (h). The surface area of ​​a sphere with radius. Segment base. Vsh. sectors = 2/3PR2h. Center of the sphere (C). Volume of the ball ball segment and spherical layer. The area of ​​the first is expressed through the radius. times more area surface of a great circle. , and the surface area of ​​the sphere is 4РR2. the ball is described. The volume of the sphere is 288.

“In the world of polyhedra” - Polyhedra. Top of the cube. The world of polyhedra. Kepler-Poinsot bodies. Mathematics. Royal tomb. Euler characteristic. Tetrahedron. Geometry. Faros lighthouse. Convex polyhedra. Archimedes' bodies. Polyhedra in art. Fire. Stellated dodecahedron. Magnus Wenninger. Euler's theorem. Alexandria Lighthouse. Regular polyhedra. Five convex regular polyhedra. Developments of some polyhedra.

“Philosopher Pythagoras” - Knowledge of the basics of music. The word "philosopher". Life and scientific discoveries Pythagoras. Pythagoras met with Persian magicians. Mathematics. Flight direction. Motto. Egyptian temples. Thought. Founder modern mathematics. True. Immortal idea. Mnesarchus. Pythagoras.

“Problems in coordinates” - Find the length of vector a if it has coordinates: (-5; -1; 7). The simplest problems in coordinates. Dot product of vectors. Vector AB. Solving problems: (using cards). How to calculate the length of a vector from its coordinates. Lesson objectives. What's called scalar product vectors. The distance between points A and B. Vector A has coordinates (-3; 3; 1). M – the middle of the segment AB. Lesson plan. How to find the coordinates of the midpoint of a segment.

Geometric volumetric figures are solids, which occupy a non-zero volume in Euclidean (three-dimensional) space. These figures are studied by a branch of mathematics called “spatial geometry”. Knowledge about the properties of three-dimensional figures is used in engineering and the natural sciences. In the article we will consider the question of geometric three-dimensional figures and their names.

Geometric solids

Since these bodies have a finite dimension in three spatial directions, a system of three is used to describe them in geometry coordinate axes. These axes have the following properties:

1. They are orthogonal to each other, that is, perpendicular.
2. These axes are normalized, meaning the basis vectors of each axis are the same length.
3. Any of the coordinate axes is the result vector product two others.

Speaking of geometric volumetric figures and their names, it should be noted that they all belong to one of 2 large classes:

1. Class of polyhedra. These figures, based on the name of the class, have straight edges and flat faces. A face is a plane that limits a shape. The point where two faces join is called an edge, and the point where three faces join is called a vertex. Polyhedra include the geometric figure of a cube, tetrahedrons, prisms, and pyramids. For these figures, Euler's theorem is valid, which establishes a connection between the number of sides (C), edges (P) and vertices (B) for each polyhedron. Mathematically, this theorem is written as follows: C + B = P + 2.
2. Class of round bodies or bodies of revolution. These figures have at least one surface forming them that is curved. For example, a ball, a cone, a cylinder, a torus.

As for the properties of volumetric figures, the two most important of them should be highlighted:

1. The presence of a certain volume that a figure occupies in space.
2. The presence of each three-dimensional figure

Both properties for each figure are described by specific mathematical formulas.

Let us consider below the simplest geometric volumetric figures and their names: cube, pyramid, prism, tetrahedron and ball.

Cube figure: description

The geometric figure cube is a three-dimensional body formed by 6 square planes or surfaces. This figure is also called a regular hexahedron, since it has 6 sides, or a rectangular parallelepiped, since it consists of 3 pairs parallel sides, which are mutually perpendicular to each other. They call it a cube whose base is a square and whose height is equal to the side of the base.

Since a cube is a polyhedron or polyhedron, Euler's theorem can be applied to it to determine the number of its edges. Knowing that the number of sides is 6, and the cube has 8 vertices, the number of edges is: P = C + B - 2 = 6 + 8 - 2 = 12.

If we denote the length of a side of a cube by the letter “a,” then the formulas for its volume and surface area will look like: V = a 3 and S = 6*a 2, respectively.

Pyramid figure

A pyramid is a polyhedron that consists of a simple polyhedron (the base of the pyramid) and triangles that connect to the base and have one common top(top of the pyramid). The triangles are called the lateral faces of the pyramid.

The geometric characteristics of a pyramid depend on which polygon lies at its base, as well as on whether the pyramid is straight or oblique. A straight pyramid is understood to be a pyramid for which a straight line perpendicular to the base, drawn through the top of the pyramid, intersects the base at its geometric center.

One of simple pyramids is a quadrangular straight pyramid, at the base of which lies a square with side “a”, the height of this pyramid is “h”. For this pyramid figure, the volume and surface area will be equal: V = a 2 *h/3 and S = 2*a*√(h 2 +a 2 /4) + a 2, respectively. Applying for it, taking into account the fact that the number of faces is 5, and the number of vertices is 5, we obtain the number of edges: P = 5 + 5 - 2 = 8.

Tetrahedron figure: description

The geometric figure tetrahedron is understood as a three-dimensional body formed by 4 faces. Based on the properties of space, such faces can only represent triangles. Thus, a tetrahedron is a special case of a pyramid, which has a triangle at its base.

If all 4 triangles forming the faces of a tetrahedron are equilateral and equal to each other, then such a tetrahedron is called regular. This tetrahedron has 4 faces and 4 vertices, the number of edges is 4 + 4 - 2 = 6. Using standard formulas from plane geometry for the figure in question, we obtain: V = a 3 * √2/12 and S = √3*a 2, where a is the length of the side of the equilateral triangle.

It is interesting to note that in nature some molecules have the form regular tetrahedron. For example, a methane molecule CH 4, in which the hydrogen atoms are located at the vertices of the tetrahedron and are connected to the carbon atom by covalent chemical bonds. The carbon atom is located at the geometric center of the tetrahedron.

The tetrahedron shape, which is easy to manufacture, is also used in engineering. For example, the tetrahedral shape is used in the manufacture of anchors for ships. Note that space probe NASA's Mars Pathfinder, which landed on the surface of Mars on July 4, 1997, was also shaped like a tetrahedron.

Prism figure

This geometric figure can be obtained by taking two polyhedra, placing them parallel to each other in different planes of space, and connecting their vertices accordingly to each other. The result will be a prism, two polyhedra are called its bases, and the surfaces connecting these polyhedra will have the shape of parallelograms. A prism is called straight if it sides(parallelograms) are rectangles.

A prism is a polyhedron, so Euler's theorem is true for it. For example, if a hexagon lies at the base of a prism, then the number of sides of the prism is 8, and the number of vertices is 12. The number of edges will be equal to: P = 8 + 12 - 2 = 18. For a straight prism of height h, at the base of which lies a regular hexagon with side a, volume equals: V = a 2 *h*√3/4, surface area equals: S = 3*a*(a*√3 + 2*h).

Speaking about simple geometric volumetric figures and their names, we should mention the ball. A volumetric body called a ball is understood as a body that is limited to a sphere. In turn, a sphere is a collection of points in space equidistant from one point, which is called the center of the sphere.

Since the ball belongs to the class of round bodies, there is no concept of sides, edges and vertices for it. The surface area of ​​the sphere enclosing the ball is found by the formula: S = 4*pi*r 2, and the volume of the ball can be calculated by the formula: V = 4*pi*r 3 /3, where pi is the number pi (3.14), r is the radius of the sphere (ball).

Volumetric bodies can be obtained in a computer in various ways. The most commonly used method is to connect base bodies.  

The volumetric body of the bundle described above is, naturally, an idealized scheme.  

This volumetric body consists of parts called sections. The first section is enclosed between two adjacent level planes passing through adjacent iso-plasters and has the shape of a truncated elliptical cone. A volumetric body consisting of such sections serves geometric model reservoir layer. We will call this volumetric body a cone-ellipse model of a gas filling (CG model), which must be constructed in such a way that it turns out to be volumetrically isomorphic to the object, i.e. so that the volumes of the model section and the corresponding part of the reservoir are the same.  

If a volumetric body is formed by the rotation of a flat area A around an axis lying in its plane but not intersecting it, then it will have the shape of a ring. Let such a ring be wrapped with a wire, the turns of which are located in a plane passing through the axis of the ring; then the current function of the wire layer will be equal to φ (1 / 2π) π &, where π is full number turns, hell is the azimuthal angle measured around the axis of the ring.  

Models of volumetric bodies, tonally resolved according to this scheme, are shown in Fig. 1.5.4. Although the algorithm does not take into account falling shadows, the overall expressiveness of the image remains quite high due to the certainty of showing that a face belongs to one or another system of orthogonally oriented planes. If the three areas noted above are depicted in the figure different colors, then the effect will be even greater. Physical model such graphic solution shown in Fig. 1.5.5. It is based on the principle of illuminating an object with three sources different colors, located in accordance with the adopted system of orthogonal planes.  

For an existing solid body, set attributes, specifying the finite element type and material.  

In the case of volumetric bodies, this procedure must be done three times. The center of gravity can lie both inside and outside the body; for example, a semi-ring made of thick homogeneous wire has a center of gravity outside the body.  

When depicting three-dimensional bodies, students most often use the method of showing depth by creating a light silhouette on a dark background. Sometimes this method leads to a misconception about the nature of the volumetric-spatial form. The image in this case corresponds to the nature of the perception of the real form.  

Determining the center of gravity of volumetric bodies is associated with the concepts of plane and axis of symmetry. A plane of symmetry is a plane that divides a given body into two halves that are completely identical in size and shape. For this reason, the center of gravity of a symmetrical body lies in the plane of symmetry.