Presentations of formulas for the volumes of polyhedra and bodies of revolution. Bodies of revolution Volumes of bodies of revolution

Municipal budget educational institution

"Average comprehensive school No. 4"

Prepared by:

mathematic teacher

Fedina Lyubov Ivanovna .

Isilkul 2014

Lesson topic "Volumes of polyhedra and bodies of revolution"

Goals:

Summarize and systematize students’ knowledge on the topic of the lesson;

Strengthen students' computational and descriptive skills;

Develop thinking logical abilities, ability to work with geometric material, read drawings, work on them;

To develop a sense of responsibility, cohesion, conscious discipline, and the ability to work in a group;

Instill interest in the subject being studied.

Lesson type: lesson summary

Technology: personality-oriented, problem-research, critical thinking.

Form:

Equipment: ruler, pen, pencil, worksheets,
figures of cones, cylinders, prisms and pyramids, drawings of geometric bodies on A4 sheets + tape, Handout

Lesson plan.

Organizing time. State the topic and purpose of the lesson.

a) True or false;

b) Cluster on the topic “Volumes of bodies”;

d) Calculation of volumes of polyhedron models.

Solving stereometric problems.

Lesson summary.

Homework.

During the classes.

Don't be afraid that you don't know

- Be afraid that you won't learn.

Organizing time. State the topic and purpose of the lesson.

- Hello, the topic of our lesson is “Volumes of polyhedra and bodies of rotation.”

Think and try to formulate the purpose of the lesson: (students express the proposed formulation of the purpose of the lesson, at the end one of them makes a general conclusion).

Updating students' knowledge.

a) - Here are the presentation questions: “True or false?” , answer them using the signs “+” and “-”.

Presentation (Slide c1-4)

1. The volume of any polyhedron can be calculated using the formula: V = S base H .

2. It is not true that S of the ball = 4πR 2.

3. Is it true that if the volume of a cube is 64 cm 3, then the side is 8 cm?

4. Is it true that if the side of a cube is 5 cm, then the volume is 125 cm 3?

5. Is it true that the volume of a cone and a pyramid can be calculated using the formula:

V= S basic H.

6. It is not true that the height of a straight prism is equal to its lateral edge.

7. Is it true that all edges regular pyramid equilateral triangles?

8. Is it true that if a ball is inscribed in a rectangular parallelepiped, then the parallelepiped is a cube.

9. Is it true that the generatrix of a cylinder is greater than its height?

10.Can the axial section of a cylinder be a trapezoid?

11. Is it true that the volume of a cylinder is less than the volume of any prism described around it?

12. Is it true that if the axial sections of two cylinders are equal rectangles, then the volumes of the cylinders are also equal?

13. It is not true that the axial section of the cylinder is a square.

14. Is it true that the polyhedron called regular if the base is a regular polygon.

15. Is it true that if a cone is inscribed in a cylinder,V cone= V cylinder

Check your answers and write down what questions you found difficult.

b) Fill out the cluster on the topic “Volumes of bodies.”

Geometric bodies

Polyhedra

Bodies of rotation

prism

pyramid

cone

cylinder

ball

V= S basic H.

V= π R 3

V = S base H .

c) Solving problems from the presentation on the topic “Volumes”;

-Now let’s move on to the next stage of the lesson:

- Oral decision tasks based on ready-made drawings.

Presentation (slides 5 - 9)

Slide 5:

1. The volume of the parallelepiped is 6. Find the volume of the triangular pyramid ABCDA 1 IN 1 .(answer. 3)

Slide 6:

2. The cylinder and cone have a common base and a common height. Calculate the volume of the cylinder if the volume of the cone is 10. (answer: 30)

Slide 7:

3. A rectangular parallelepiped is described about a cylinder, the radius of the base and the height

which are equal to 1. Find the volume of the parallelepiped. (answer.4)

Slide 8:

4. Find the volume V of the part of the cylinder shown in the figure. Please indicate V/π in your answer. (answer.25)

Slide 9:

5.Find the volume V of the part of the cone shown in the figure. Please indicate V/π in your answer. (answer: 300)

d) Calculation of volumes of polyhedron models.

There are models of figures on the tables in front of you.

Your task:

Take the necessary measurements and calculate the volumes of these figures.

Check your results (the answers may be approximately equal).

3. Solving stereometric problems.

On the tables in front of you are envelopes with tasks of varying degrees of difficulty. Assess your knowledge and select two problems from the envelope and solve them yourself.

Pupils studying at “4” and “5” work at the board.

(Drawings of the figures are given on half of whatman paper. Students take the drawing, complete the missing conditions on it and solve the problem))

5. The generatrix and the radii of the larger and smaller base of the truncated cone are 13 cm, 11 cm, 6 cm, respectively. Calculate the volume of this cone. (answer: V = 892 cm 3)

6. Find the volume of a regular pyramid if the side edge is 3 cm and the side of the base is 4 cm. (answer. Answer: see 3)

7. The base of the pyramid is a square. The side of the base is 20 dm, and its height is 21 dm. Find the volume of the pyramid. (Answer: V = 2800 dm 3)

8. The diagonal of the axial section of the cylinder is 13 cm, the height is 5 cm. Find the volume of the cylinder. (Answer: cm 3)

9. The diagonal of the axial section of the cylinder is 10 cm, height is 8 cm. Find the volume of the cylinder. (answer: 72π cm 3)

10. The generatrix and the radii of the larger and smaller base of the truncated cone are 13 cm, 11 cm, 6 cm, respectively. Calculate the volume of this cone. (answer: 892 cm 3)

"5"

5. A regular quadrangular prism is inscribed in a cylinder. Find the ratio of the volumes of the prism and the cylinder. (answer: 2/π).

6. How many times will the area of ​​the lateral surface of the cone increase if its generatrix is ​​increased by 3 times? (answer.3)

4. Lesson summary.

Now it’s time to summarize the lesson and write down your homework.

So, answer the questions on pieces of paper:

Today I realized _______________.

Today I found out(a)______________.

I would like to ask___________ .

Homework. Select from envelope.

Hand in your notebooks.

Slide 1

Volumes and surfaces of bodies of revolution Mathematics teacher, Municipal Educational Institution Secondary School No. 8 x. Shuntuk of Maikopsk district of the Republic of Adygea Natalya Andreevna Gruner

Slide 2

Slide 3

contents 1. Types of bodies of revolution 2. Definitions of bodies of revolution: a) cylinder b) cone c) ball 3. Sections of bodies of revolution: a) cylinder b) cone c) ball 4. Volumes of bodies of revolution 5. Surface areas of bodies of revolution Complete the work

Slide 4

TYPES OF BODIES OF ROTATION Cylinder-body that describes a rectangle when rotating it about the side as an axis Cone-body that is obtained by rotation right triangle around its leg as an axis. A ball is a body obtained by rotating a semicircle around its diameter as an axis.

Slide 5

DEFINITION OF A CYLINDER A cylinder is a body that consists of two circles that do not lie in the same plane and are combined by parallel translation, and all the segments connecting the corresponding points of these circles. The circles are called the bases of the cylinder, and the segments connecting the corresponding points of the circles' circumferences form the cylinder.

Slide 6

DEFINITION OF A CONE A cone is a body that consists of a circle that is the base of the cone, a point not lying in the plane of this circle, the vertex of the cone and all the segments connecting the vertex of the cone with the points of the base.

Slide 7

SECTIONS OF A CYLINDER The cross section of a cylinder with a plane parallel to its axis is a rectangle. Axial section is a section of a cylinder with a plane passing through its axis. A section of a cylinder with a plane parallel to the bases is a circle.

Slide 8

DEFINITION OF A BALL A ball is a body that consists of all points in space located at a distance not greater than a given one from a given point. This point is called the center of the ball, and this distance is the radius of the ball.

Slide 9

SECTION OF A CONE The section of a cone by a plane passing through its vertex is isosceles triangle. The axial section of a cone is the section passing through its axis. A section of a cone by a plane parallel to its bases is a circle with its center on the axis of the cone.

Slide 10

SECTIONS OF A BALL A section of a sphere by a plane is a circle. The center of this ball is the base of the perpendicular drawn from the center of the ball onto the cutting plane. The section of a ball by the diametrical plane is called a great circle.

Slide 11

VOLUME OF BODIES OF ROTATION figure formula rule cylinder V=S*H The volume of a cylinder is equal to the product of the area of ​​the base and the height. cone V=1/3*S*H The volume of a cone is equal to one third of the product of the area of ​​the base and the height. ball V=4/3*P*R3 Volume of the ball Theorem. The volume of a sphere of radius R is equal to: Ball segment Ball segment. Volume of the spherical segment. Ball sector V=2/3*P*R2*N Ball segment. Volume of the spherical segment.

Slide 12

SURFACE AREA OF BODIES OF ROTATION figure rule The area of ​​the lateral surface of a cylinder is equal to the product of the circumference of the base and the height. The area of ​​the lateral surface of the cone is equal to half the product of the circumference of the base and the length of the generatrix. The surface area of ​​a sphere is calculated by the formula S=4*P*R*R

Slide 13

Volume of a sphere Theorem. The volume of a sphere of radius R is equal to: Proof. Let's consider a ball of radius R with a center at point O and choose the Ox axis in an arbitrary way (Fig.). The section of the ball by a plane perpendicular to the Ox axis and passing through the point M of this axis is a circle with a center at the point M. Let us denote the radius of this circle by r, and its area by S(x), where x is the abscissa of point M. Let us express S (x) through x and R. From the right triangle OMC we find: (2.6.1) Since, then (2.6.2) Note that this formula is true for any position of the point M on the diameter AB, i.e. For all x , satisfying the condition. Using the basic formula to calculate volumes tel at, we get The theorem is proven.

Slide 14

Ball segment. Volume of the spherical segment. A spherical segment is a part of a ball cut off from it by a plane. Any plane intersecting a ball splits it into two segments. Segment volume

Slide 15

Ball sector. Volume of the spherical sector. A spherical sector, a body that is obtained from a spherical segment and a cone. Sector volume V=2/3Р2H

Slide 16

Problem No. 1. The tank has cylinder shape to the bases of which equal spherical segments are attached. The radius of the cylinder is 1.5 m, and the height of the segment is 0.5 m. How long must the generatrix of the cylinder be in order for the capacity of the tank to be 50 m3?

“Cylinder geometry grade 11” - 3. Cylinder axis. 2. 3. Obtaining a cylinder. 4. Radius of the base. Geometry 11th grade. 2. The concept of a cylindrical surface. 1.Lesson development 2.Lesson materials. 4. Section by a plane perpendicular to the axis. Theoretical material Problems. Geometry 11th grade Topic: Cylinder. 1. Examples of cylinders. 1.

“Lesson Volume of a cylinder” - Cylindrical surface. Oral exercises on the topic. B. Axial section - ……………. N. D1. Any axial sections of the cylinder ..... among themselves. Lesson plan. A1. D. A. Straight cylinder.

“Cylinder surface” - Film by: A. Shevchenko R. Trushenkov. "The concept of a cylinder." L1. Formative. Axial section. L. Algebra & Geometria Entertainment. Cylinder axis. Cylinder base.

“Cylinder cone ball” - Definition of a cylinder. Types of bodies of rotation. Volumes of bodies of rotation. Volumes and surfaces of bodies of rotation. Definition of a ball. The section of a ball by the diametrical plane is called a great circle. Volume of the spherical segment. Volume of the spherical sector. Table of contents. Definition of a cone. Sections of a cylinder. Sections of a ball. Given: Proof.

“Cylinder volume” - Cylinders from life. Volume of a cylinder Volume of a cone. Tower cylinders. Volume of a cone. Cylinder: history. The volume of a cylinder is equal to the product of the area of ​​the base and the height. Cylinder volume. Huge size cones. Volume of a truncated cone. Cone: history. A bucket is an example of a truncated cone. Vodovzvodnaya Tower (Moscow) Own house of the architect K. Melnikov (Moscow) Sforza Castle (Milan).

Volumes and surfaces of bodies of revolution

Mathematics teacher, Municipal Educational Institution Secondary School No. 8

X. Shuntuk Maikopsk district of the Republic of Adygea

Gruner Natalya Andreevna

900igr.net

1. Types of bodies of revolution 2. Definitions of bodies of revolution: a) cylinder

3.Sections of bodies of revolution:

a) cylinder

4. Volumes of bodies of revolution 5. Surface areas of bodies of revolution

To finish work

TYPES OF BODIES OF ROTATION

A cylinder is a body that describes a rectangle when rotating it about a side as an axis

A cone is a body that is obtained by rotating a right triangle around its leg as an axis

A ball is a body obtained by rotating a semicircle around its diameter as an axis

DEFINITION OF A CYLINDER

A cylinder is a body that consists of two circles that do not lie in the same plane and are combined by parallel translation, and all the segments connecting the corresponding points of these circles.

The circles are called the bases of the cylinder, and the segments connecting the corresponding points of the circles' circumferences form the cylinder.

DEFINITION OF CONE

A cone is a body that consists of a circle that is the base of the cone, a point not lying in the plane of this circle, the vertex of the cone and all the segments connecting the vertex of the cone with the points of the base.

CYLINDER SECTIONS

The cross section of a cylinder with a plane parallel to its axis is a rectangle.

Axial section is a section of a cylinder by a plane passing through its axis

The cross section of a cylinder with a plane parallel to the bases is a circle.

DEFINITION OF BALL

A ball is a body that consists of all points in space located at a distance not greater than a given one from a given point. This point is called the center of the ball, and this distance is the radius of the ball.

CONE SECTION

The section of a cone by a plane passing through its vertex is an isosceles triangle.

The axial section of a cone is the section passing through its axis.

A section of a cone by a plane parallel to its bases is a circle with its center on the axis of the cone.

SECTIONS OF THE BALL

The section of a sphere by a plane is a circle. The center of this ball is the base of the perpendicular drawn from the center of the ball onto the cutting plane.

The section of a ball by the diametrical plane is called a great circle.

VOLUME OF BODIES OF ROTATION

The volume of a cylinder is equal to the product of the area of ​​the base and the height.

Ball segment

The volume of a cone is equal to one third of the product of the area of ​​the base and the height.

Volume of a sphere Theorem. The volume of a sphere of radius R is equal to:

V=2/3 *P* R 2 *N

Ball segment. Volume of the spherical segment.

SURFACE AREA OF BODIES OF ROTATION

The lateral surface area of ​​a cylinder is equal to the product of the circumference of the base and its height.

The area of ​​the lateral surface of the cone is equal to half the product of the circumference of the base and the length of the generatrix.

The surface area of ​​a sphere is calculated by the formula S=4* P *R*R

Volume of a sphere Theorem. The volume of a sphere of radius R is equal to .

Proof. Consider a ball of radius R centered at a point ABOUT and select the axis Oh in any way (Fig.). Section of a ball by a plane perpendicular to the axis Oh and passing through the point M this axis is a circle with center at the point M. Let us denote the radius of this circle by r, and its area through S(x), Where X- abscissa of the point M. Let's express S(x) through X And R. From a right triangle Compulsory medical insurance we find:

Because , then (2.6.2)

Note that this formula is true for any position of the point M on diameter AB, i.e. For everyone X, satisfying the condition. Applying the basic formula for calculating the volumes of bodies at

, we get

The theorem has been proven.

Ball segment. Volume of the spherical segment.

• A spherical segment is a part of a ball cut off from it by a plane. Any plane intersecting a ball splits it into two segments.
• Segment volume

Ball sector. Volume of the spherical sector.

• A spherical sector, a body that is obtained from a spherical segment and a cone.

• Sector volume

• V=2/3 P R 2 H

Task No. 1.

• The tank has the shape cylinder, to the bases to which equal spherical segments are attached. The radius of the cylinder is 1.5 m, and the height of the segment is 0.5 m. How long must the generatrix of the cylinder be in order for the capacity of the tank to be 50 m3?

Ball segments.

answer: ~6.78.

Task No. 2.

• O is the center of the ball.
• O 1 is the center of the cross-sectional circle of the ball. Find the volume and surface area of ​​the sphere.

Given: a ball cross-section with center O 1. R sec. =6cm. Angle OAB=30 0 . V ball =? S spheres = ?

• Solution :

V=4/3 P R 2 S=4 P R 2

V ∆ OO 1 A : angle O 1 =90 0 ,ABOUT 1 A=6,

angle OAB=30 0 . tg 30 0 =OO 1 / ABOUT 1 A OO 1 =O 1 A* tg30 0 .OO 1 =6*√3 ÷ 3 =2 √3

OA= R=OO 1 ( According to the St., the leg lies opposite the angle of 30 0 ).

OA=2√3 ÷2 =√3

V=4 P(√3) 2 ÷ 3=(4*3,14*3) ÷ 3=12,56

S= 4P(√3) 2 =4*3,14*3=37,68

Answer :V=12 ,56; S=37 ,68.

Task № 3

The semi-cylindrical vault of the basement is 6m. length and 5.8 m. in diameter. Find the complete surface of the basement.

Given: Cylinder ABCD-axial section. BP=6m. D= 5.8m. S p.pod.= ?

• Solution:

• S p.pod. =(S p ÷ 2)+ S ABCD
• S p ÷ 2= (2P Rh+2 P R 2)÷2=2(P Rh+ P R 2)÷2= P Rh+ P R 2
• R=d÷2=5.8 ÷ 2=2.9 m.
• S p ÷ 2=3.14*2.9+3.14*(2.9) 2 =

54,636+26,4074=81,0434

ABCD-rectangular (by definition of axial section)

S ABCD = AB * AD = 5.8 * 6 = 34.8 m 2

S p.pod. =34.8+81.0434≈116m2.

Answer: S p.pod. ≈116m2.