Given a right circular cone with a vertex. Lesson "Volume of a cone

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.

Control questions:

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.

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 sectional area 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.

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 axis of the cylinder, then the section, which has the shape of an ellipse, is projected by a circle onto that plane of projections 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.

Diagnostic work consists of two parts, including 19 tasks. Part 1 contains 8 tasks of a basic level of complexity with a short answer. Part 2 contains 4 tasks of an increased level of complexity with a short answer and 7 tasks of an increased and high level of complexity with a detailed answer.
3 hours 55 minutes (235 minutes) are allotted to perform diagnostic work in mathematics.
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.
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