V cylinder \u003d S main. h
Example 2 Given a right circular cone ABC equilateral, BO = 10. Find the volume of the cone.
Solution
Find the radius of the base of the cone. C \u003d 60 0, B \u003d 30 0,
Let OS = a, then BC = 2 a. According to the Pythagorean theorem:
Answer: .
Example 3. Calculate the volumes of the figures formed by the rotation of the areas bounded by the specified lines.
y2=4x; y=0; x=4.
Limits of integration a = 0, b = 4.
V= 
|
=32π
Tasks
Option 1
1. The axial section of the cylinder is a square, the diagonal of which is 4 dm. Find the volume of the cylinder.
2. The outer diameter of the hollow sphere is 18 cm, the wall thickness is 3 cm. Find the volume of the walls of the sphere.
X figure bounded by lines y 2 =x, y=0, x=1, x=2.
Option 2
1. The radii of three balls are 6 cm, 8 cm, 10 cm. Determine the radius of the ball, the volume of which is equal to the sum of the volumes of these balls.
2. The area of the base of the cone is 9 cm 2, its total surface area is 24 cm 2. Find the volume of the cone.
3. Calculate the volume of the body formed by rotation around the O axis X figure bounded by lines y 2 =2x, y=0, x=2, x=4.
1. Write the properties of volumes of bodies.
2. Write a formula for calculating the volume of a body of revolution around the Oy axis.
Let a right circular cylinder be given, the horizontal plane of projections is parallel to its base. When a cylinder is intersected by a plane in general position (we assume that the plane does not intersect the bases of the cylinder), the intersection line is an ellipse, the section itself has the shape of an ellipse, its horizontal projection coincides with the projection of the base of the cylinder, and the front also has the shape of an ellipse. But if the cutting plane makes an angle equal to 45 ° with the cylinder axis, then the section, which has the shape of an ellipse, is projected by a circle onto that projection plane to which the section is inclined at the same angle.
If the cutting plane intersects the side surface of the cylinder and one of its bases (Fig. 8.6), then the line of intersection has the shape of an incomplete ellipse (part of an ellipse). The horizontal projection of the section in this case is part of the circle (projection of the base), and the frontal is part of the ellipse. The plane can be located perpendicular to any projection plane, then the section will be projected onto this projection plane by a straight line (part of the trace of the secant plane).
If the cylinder is intersected by a plane parallel to the generatrix, then the lines of intersection with the lateral surface are straight, and the section itself has the shape of a rectangle if the cylinder is straight, or a parallelogram if the cylinder is inclined.
As you know, both the cylinder and the cone are formed by ruled surfaces.
The line of intersection (line of cut) of the ruled surface and the plane in the general case is a certain curve, which is constructed from the points of intersection of the generators with the secant plane.
Let it be given straight circular cone. When crossing it with a plane, the line of intersection can take the form of: a triangle, an ellipse, a circle, a parabola, a hyperbola (Fig. 8.7), depending on the location of the plane.
A triangle is obtained when the cutting plane, crossing the cone, passes through its vertex. In this case, the lines of intersection with the lateral surface are straight lines intersecting at the top of the cone, which, together with the line of intersection of the base, form a triangle projected onto the projection planes with distortion. If the plane intersects the axis of the cone, then a triangle is obtained in the section, in which the angle with the vertex coinciding with the vertex of the cone will be maximum for the triangle sections of the given cone. In this case, the section is projected onto the horizontal projection plane (it is parallel to its base) by a straight line segment.
The line of intersection of a plane and a cone will be an ellipse if the plane is not parallel to any of the generators of the cone. This is equivalent to the fact that the plane intersects all generators (the entire lateral surface of the cone). If the cutting plane is parallel to the base of the cone, then the intersection line is a circle, the section itself is projected onto the horizontal projection plane without distortion, and onto the frontal plane - as a straight line segment.
The line of intersection will be a parabola when the secant plane is parallel to only one generatrix of the cone. If the cutting plane is parallel to two generators at the same time, then the line of intersection is a hyperbola.
A truncated cone is obtained if a right circular cone is intersected by a plane parallel to the base and perpendicular to the axis of the cone, and the upper part is discarded. In the case when the horizontal projection plane is parallel to the bases of the truncated cone, these bases are projected onto the horizontal projection plane without distortion by concentric circles, and the frontal projection is a trapezoid. When a truncated cone is intersected by a plane, depending on its location, the cut line may take the form of a trapezoid, ellipse, circle, parabola, hyperbola, or part of one of these curves, the ends of which are connected by a straight line.
TEXT EXPLANATION OF THE LESSON:
We continue to study the section of solid geometry "Body of revolution".
The bodies of revolution include: cylinders, cones, balls.
Let's remember the definitions.
Height is the distance from the top of a figure or body to the base of the figure (body). Otherwise, a segment connecting the top and bottom of the figure and perpendicular to it.
Remember, to find the area of a circle, multiply pi by the square of the radius.
The area of the circle is equal.
Recall how to find the area of a circle, knowing the diameter? Because
let's put it into the formula:
A cone is also a body of revolution.
A cone (more precisely, a circular cone) is a body that consists of a circle - the base of the cone, a point that does not lie in the plane of this circle - the top of the cone and all segments connecting the top of the cone with the points of the base.
Let's get acquainted with the formula for finding the volume of a cone.
Theorem. The volume of a cone is equal to one third of the base area multiplied by the height.
Let's prove this theorem.
Given: a cone, S is the area of its base,
h is the height of the cone
Prove: V=
Proof: Consider a cone with volume V, base radius R, height h, and apex at point O.
Let us introduce the axis Ox through OM, the axis of the cone. An arbitrary section of a cone by a plane perpendicular to the x-axis is a circle centered at the point
M1 - the point of intersection of this plane with the axis Ox. Let us denote the radius of this circle as R1, and the cross-sectional area as S(x), where x is the abscissa of the point M1.
From the similarity of right-angled triangles OM1A1 and OMA (ے OM1A1 = ے OMA - straight lines, ےMOA-common, which means that the triangles are similar in two angles) it follows that
The figure shows that OM1=x, OM=h
or whence by the property of proportion we find R1 = .
Since the section is a circle, then S (x) \u003d πR12, we substitute the previous expression instead of R1, the area of \u200b\u200bthe section is equal to the ratio of the product of pi er square by square x to the square of height:
Let's apply the basic formula
calculating the volumes of bodies, with a=0, b=h, we get the expression (1)
Since the base of the cone is a circle, the area S of the base of the cone will be equal to pi er square
in the formula for calculating the volume of a body, we replace the value of pi er square by the area of \u200b\u200bthe base and we get that the volume of the cone is equal to one third of the product of the area of \u200b\u200bthe base and the height
The theorem has been proven.
Corollary of the theorem (formula for the volume of a truncated cone)
The volume V of a truncated cone, whose height is h, and the areas of the bases S and S1, is calculated by the formula
Ve is equal to one third of ash multiplied by the sum of the areas of the bases and the square root of the product of the areas of the base.
Problem solving
A right triangle with legs 3 cm and 4 cm rotates around the hypotenuse. Determine the volume of the resulting body.
When the triangle rotates around the hypotenuse, we get a cone. When solving this problem, it is important to understand that two cases are possible. In each of them, we apply the formula for finding the volume of a cone: the volume of a cone is equal to one third of the product of the base and the height
In the first case, the drawing will look like this: a cone is given. Let radius r = 4, height h = 3
The area of the base is equal to the product of π times the square of the radius
Then the volume of the cone is equal to one third of the product of π times the square of the radius times the height.
Substitute the value in the formula, it turns out that the volume of the cone is 16π.
In the second case, like this: given a cone. Let radius r = 3, height h = 4
The volume of a cone is equal to one third of the base area multiplied by the height:
The area of the base is equal to the product of π times the square of the radius:
Then the volume of the cone is equal to one third of the product of π times the square of the radius times the height:
Substitute the value in the formula, it turns out that the volume of the cone is 12π.
Answer: The volume of the cone V is 16 π or 12 π
Problem 2. Given a right circular cone with a radius of 6 cm, angle BCO = 45 .
Find the volume of the cone.
Solution: A ready-made drawing is given for this task.
Let's write the formula for finding the volume of a cone:
We express it in terms of the radius of the base R:
We find h \u003d BO by construction, - rectangular, because angle BOC=90 (the sum of the angles of a triangle), the angles at the base are equal, so the triangle ΔBOC is isosceles and BO=OC=6 cm.
Municipal educational institution
Alekseevskaya secondary school
"Education Centre"
Lesson Development
Subject: DIRECT CIRCULAR CONE.
SECTION OF A CONE BY PLANES
Mathematic teacher
academic year
Subject: DIRECT CIRCULAR CONE.
SECTION OF A CONE BY PLANES.
The purpose of the lesson: to analyze the definitions of a cone and subordinate concepts (vertex, base, generators, height, axis);
consider sections of the cone passing through the vertex, including axial ones;
to promote the development of spatial imagination of students.
Lesson objectives:
Educational: to study the basic concepts of a body of revolution (cone).
Developing: to continue the formation of the skills of analysis, comparison; ability to highlight the main thing, to formulate conclusions.
Educational: fostering students' interest in learning, instilling communication skills.
Lesson type: lecture.
Teaching methods: reproductive, problematic, partially search.
Equipment: table, models of bodies of revolution, multimedia equipment.
During the classes
I. Organizing time.
In the previous lessons, we have already got acquainted with the bodies of revolution and dwelled on the concept of a cylinder in more detail. On the table you see two drawings and, working in pairs, formulate the correct questions on the topic covered.
P. Checking homework.
Work in pairs using a thematic table (a prism inscribed in a cylinder and a prism described near the cylinder).
For example, in pairs and individually, students can ask the following questions:
What is a circular cylinder (cylinder generatrix, cylinder bases, cylinder side surface)?
What prism is called inscribed near a cylinder?
Which plane is called tangent to the cylinder?
What shapes are polygons? ABC, A1 B1 C1 , ABCDEandA1 B1 C1 D1 E1 ?
- What kind of prism is a prism ABCDEABCDE? (Straightmy.)
- Prove that it is a straight prism.
(optionally 2 pairs of students at the blackboard do the work)
III. Updating of basic knowledge.
According to the material of planimetry:
Thales' theorem;
Properties of the midline of a triangle;
Area of a circle.
According to the material of stereometry:
concept homothety;
The angle between a line and a plane.
IV.Learning new material.
(educational - methodical set"Live mathematics », Attachment 1.)
After the presentation of the material, a work plan is proposed:
1. Definition of a cone.
2. Definition of a right cone.
3. Elements of a cone.
4. Development of the cone.
5. Obtaining a cone as a body of revolution.
6. Types of sections of the cone.
Students will find answers to these questions on their own.children in paragraphs 184-185, accompanying them with drawings.
Valeological pause: Tired? Let's have a rest before the next practical stage of work!
Massage of the reflex zones on the auricle, responsible for the work of internal organs;
· Massage of reflex zones on the palms of the hands;
Gymnastics for the eyes (squint and sharply open your eyes);
Stretching the spine (raise your arms up, pull yourself up with your right, and then with your left hand)
Breathing exercises aimed at saturating the brain with oxygen (inhale sharply through the nose 5 times)
A thematic table is compiled (together with the teacher), accompanying the completion of the table with questions and material received from various sources (textbook and computer presentation)
"Cone. Frustum".
Thematictable 1. Cone (straight, circular) is called the body obtained by rotation right triangle around the straight line containing the leg. Dot M - vertex cone, circle with center O – basecone, line segment MA=l aboutdeveloping cones, segment MO= H - cone height, line segment OA= R - base radius, segment sun= 2 R - base diametervaniya, triangle MVS -axial section, < BMC - corner at the top of the axial section, < MBO - cornerthe slope of the generatrix to the planebase bones _________________________________________ 2.
Cone development- sector < BMBl = a - sweep angle. Sweep arc length BCV1 =2π R = la . Lateral surface area S. = π R l Total surface area (sweep area) S= π R ( l + R ) |
cone called the body, which consists of a circle - grounds cone, a point not lying in the plane of this circle, - peaks cone and all segments connecting the top of the cone with the points of the base - generators ______________________________
|
3. Sections of a cone by planes |
|
Section of a cone by a plane passing through through the top of the cone, - isosceles triangle AMB: AM=VM - generators of the cone, AB - chord; Axial section- isosceles triangle AMB: AM=BM - generators of the cone, AB - diameter of the base. | |
The section of the cone by a plane perpendicular to the axis of the cone, - a circle; at an angle to the axis of the cone - ellipse. |
|
truncated cone called the part of the cone enclosed between the base and the section of the cone parallel to the base. Circles with centers 01 and O2 - top and bottom base truncated cone, d andR - base radii, line segment AB= l - generatrix, ά - generatrix slope angleto the plane bottom base, line segment 01O2 -height(the distance between flatgrounds), trapezoid ABCD - axial section. |
|
v.Fixing the material.
Front work.
· Orally (using a ready-made drawing) No. 9 and No. 10 are solved.
(two students explain the solution of problems, the rest can make brief notes in notebooks)
No. 9. The radius of the base of the cone is 3m, the height of the cone is 4m. find the generatrix.
(Solution:l=√ R2 + H2 =√32+42=√25=5m.)
No. 10 Forming a cone l inclined to the base plane at an angle of 30°. Find the height.
(Solution:H = l sin 30◦ = l|2.)
|
· Solve the problem according to the finished drawing.
The height of the cone is h. Through generators MA and MB a plane is drawn that makes an angle a with the plane of the base of the cone. Chord AB constricts an arc with a degree measure R.
1. Prove that the section of a cone by a plane MAV- isosceles triangle.
2. Explain how to construct the linear angle of a dihedral angle formed by the secant plane and the plane of the base of the cone.
3. Find MS.
4. Make (and explain) a plan for calculating the length of the chord AB and sectional area MAV.
5. Show in the figure how you can draw a perpendicular from a point O to the section plane MAV(justify the construction).
· Repetition:
studied material from planimetry:
Definition of an isosceles triangle;
Properties of an isosceles triangle;
Area of a triangle
studied material from stereometry:
Determining the angle between planes;
A method for constructing a linear angle of a dihedral angle.
Self test
1. Draw bodies of revolution formed by the rotation of the flat figures shown in the figure.
2. Indicate the rotation of which flat figure produced the depicted body of revolution. (b)
Diagnostic work consists of two parts, including 19 tasks. Part 1 contains 8 tasks basic level Difficulty with short answers. Part 2 contains 4 tasks of an increased level of complexity with a short answer and 7 tasks of increased and high levels Difficulties with a detailed answer.
For execution diagnostic work in mathematics, 3 hours 55 minutes (235 minutes) are allotted.
Answers to tasks 1-12 are written as an integer or a final decimal fraction. Write the numbers in the answer fields in the text of the work, and then transfer them to the answer sheet No. 1. When completing tasks 13-19, you need to write down the complete solution and the answer to the answer sheet No. 2.
All forms are completed in bright black ink. The use of gel, capillary or fountain pens is allowed.
When completing assignments, you can use a draft. Draft entries do not count towards the assessment of the work.
The points you get for completed tasks are summed up.
We wish you success!
Task Conditions
- Find if
- To obtain an enlarged image of a light bulb on the screen in the laboratory, a converging lens with a main focal length = 30 cm is used. The distance from the lens to the light bulb can vary from 40 to 65 cm, and the distance from the lens to the screen - in the range from 75 to 100 cm. The image on the screen will be clear if the ratio is met. Specify the greatest distance from the lens that the light bulb can be placed so that its image on the screen is clear. Express your answer in centimeters.
- The ship passes along the river to the destination for 300 km and after parking returns to the point of departure. Find the speed of the current, if the speed of the ship in still water is 15 km / h, the parking lasts 5 hours, and the ship returns to the point of departure 50 hours after leaving it. Give your answer in km/h.
- Find the smallest value of a function on a segment
- a) Solve the equation
b) Find all the roots of this equation that belong to the segment - Given a right circular cone with a vertex M. Axial section of the cone - a triangle with an angle of 120 ° at the apex M. The cone generator is . Through the dot M a section of the cone is drawn perpendicular to one of the generators.
a) Prove that the resulting triangle is an obtuse triangle.
b) Find the distance from the center O the base of the cone to the plane of the section. - Solve the Equation
- Circle with center O touches the side AB isosceles triangle abc, side extensions AC and continuation of the foundation sun at the point N. Dot M- middle of base Sun.
a) Prove that MN=AC.
b) Find OS, if the sides of the triangle ABC are 5, 5 and 8. - Business project "A" assumes an increase in the amounts invested in it by 34.56% annually during the first two years and by 44% annually over the next two years. Project "B" assumes growth by a constant integer n percent annually. Find the smallest value n, under which for the first four years the project "B" will be more profitable than the project "A".
- Find all values of the parameter , , for each of which the system of equations
has the only solution - Anya plays a game: two different natural numbers are written on the board
and , both are less than 1000. If both are natural numbers, then Anya makes a move - she replaces the previous ones with these two numbers. If at least one of these numbers is not a natural number, then the game ends.
a) Can the game go on for exactly three moves?
b) Are there two initial numbers such that the game will last at least 9 moves?
c) Anya made the first move in the game. Find the largest possible ratio of the product of the obtained two numbers to the product