The formula for the volume of a cone which contains the diameter. All formulas for volumes of geometric bodies

The volume of a cone is expressed by the same formula as the volume of a pyramid: V = 1 / 3 S h,

where V is the volume of the cone, S is the area of the base of the cone, h- his high.

Finally V = 1 / 3 πR 2 h, where R is the radius of the base of the cone.

Obtaining the formula for the volume of a cone can be explained by the following reasoning:

Let a cone be given (fig). Let's inscribe a regular pyramid into it, that is, we'll build a pyramid inside the cone whose vertex coincides with the vertex of the cone, and the base is a regular polygon inscribed in the base of the cone.

The volume of this pyramid will be expressed by the formula: V’ = 1 / 3 S’ h, where V is the volume of the pyramid,

S’ is the area of its base, h- height of the pyramid.

If we take a polygon with a very large number of sides as the base of the pyramid, then the area of the base of the pyramid will differ very little from the area of the circle, and the volume of the pyramid will differ very little from the volume of the cone. If we neglect these differences in size, then the volume of the cone is expressed by the following formula:

V=1/3S h, where V is the volume of the cone, S is the area of the base of the cone, h- height of the cone.

Replacing S through πR 2, where R is the radius of the circle, we get the formula: V = 1 / 3 πR 2 h, expressing the volume of the cone.

Note. In the formula V = 1 / 3 S h a sign of exact and not approximate equality is placed, although based on the reasoning carried out we could consider it approximate, but in senior high school it is proven that equality

V=1/3S h exact, not approximate.

Volume of an arbitrary cone

Theorem. The volume of an arbitrary cone is equal to one third of the product of the area of the base and the height, those.

V = 1/3 QH, (1)

where Q is the area of the base, and H is the height of the cone.

Consider a cone with vertex S and base Ф (Fig.).

Let the area of the base Φ be equal to Q, and the height of the cone be equal to H. Then there are sequences of polygons Φ n and F' n with areas Q n and Q' n such that

F n⊂ Ф n⊂ Ф' n and $\lim_(n \rightarrow \infty)$ Q’ n= $\lim_(n \rightarrow \infty)$ Q n= Q.

It is obvious that a pyramid with a top S and a base F' n will be inscribed in a given cone, and a pyramid with vertex S and base Ф n- described around the cone.

The volumes of these pyramids are respectively equal

V n= 1 / 3 Q n H, V' n= 1 / 3 Q' n H

$\lim_(n \rightarrow \infty)$ V n= $\lim_(n \rightarrow \infty)$ V’ n= 1 / 3 QH

then formula (1) is proven.

Consequence. The volume of a cone, the base of which is an ellipse with semi-axes a and b, is calculated by the formula

V = 1/3π ab H (2)

In particular, volume of a cone whose base is a circle of radius R, calculated by the formula

V = 1 / 3 π R 2 H (3)

where H is the height of the cone.

As is known, the area of an ellipse with semi-axes A And b equal to π ab, and therefore formula (2) is obtained from (1) with Q = π ab. If a = b= R, then formula (3) is obtained.

Volume of a right circular cone

Theorem 1. The volume of a right circular cone with height H and base radius R is calculated by the formula

V = 1 / 3 π R 2 H

This cone can be considered as a body obtained by rotating a triangle with vertices at points O(0; 0), B(H; 0), A(H; R) around the axis Oh(rice.).

Triangle OAB is a curvilinear trapezoid corresponding to the function

y = R / H X, X∈ . Therefore, using the well-known formula, we get

$$ V=\pi\int_(0)^(H)(\frac(R)(H)x)^2dx=\\=\frac(\pi R^2)(H^2)\cdot\frac (x^3)(3)\left|\begin(array)(c)H\\\\ 0\end(array)\right.=\\=\frac(1)(3)\pi R^2H $$

Consequence. The volume of a right circular cone is equal to one third of the product of the area of the base and the height, i.e.

where Q - base area, and H - cone height.

Theorem 2. The volume of a truncated cone with base radii r and R and height H is calculated by the formula

V = 1 / 3 πH( r 2 + R 2 + r R).

A truncated cone can be obtained by rotating around an axis Oh trapezoid O ABC (fig.).

The straight line AB passes through the points (0; r) and (H; R), so it has the equation

$$ y=\frac(R-r)(H)x + r $$

we get

$$ V=\pi\int_(0)^(H)(\frac(R-r)(H)x + r)^2dx $$

To calculate the integral, we make the replacement

$$ u=\frac(R-r)(H)x + r, du=\frac(R-r)(H)dx $$

Obviously when X varies from 0 to H, variable And varies from r to R, and therefore

$$ V=\pi\int_(r)^(R)u^2\frac(H)(R-r)du=\\=\frac(\pi H)(R-r)\cdot\frac(u^3) (3)\left|\begin(array)(c)R\\\\ r\end(array)\right.=\\=\frac(\pi H)(3(R-r))(R^3- r^3)=\\=\frac(1)(3)\pi H(R^2 + r^2 + Rr) $$

Geometry is a difficult science, but useful. At school we all learned how to calculate the volumes of three-dimensional bodies, but not everyone remembers the formulas for these calculations well. This article will help you brush up on how to find the volume of a cone. This three-dimensional figure is formed by the circular rotation of a right triangle. You can calculate its volume in different ways, depending on what initial data you have.

Instructions:

In most cases, the radius of the base circle and the height are used for the calculation. The formula for the volume of a cone in this case looks like: V= πRh, Where π=3.14, R– radius of the base, h– height of the figure. Simply put, with this formula we calculate the area of the base and multiply it by the height. However, calculating the volume of a cone may take a different form if you know other parameters of your figure.
If you know the length of the side of the cone and the radius of the base, to find the volume of the figure you will need to find out what its height is. This will help us Pythagorean theorem , because the radius of the base in this case is leg right triangle, and the side side, respectively, hypotenuse. In order to find the length of the second leg, which is the height of the cone, we will use the well-known formula a^2+b^2=c^2 .
But how to find the volume of a cone if neither the length of the side nor the radius of the base are known? In this case, you need to know degree of angle at the apex of the cone and its height. With this data in hand, you can calculate the radius of the base. Do not forget that a cone is a figure formed by the rotation of a right triangle around one of its legs. If the vertex angle is divided in two, you will get the degree of one of the two acute angles of this triangle. Using the definitions of trigonometric functions, we can find out the length of the side opposite this angle, that is, in our case, the radius of the base. In this case it will be equal l*sin(α), Where l– the length from the top of the cone to the base, the height, accordingly, will be equal to l*cos(α), using these values, we derive the following formula for the radius of the base R= h/cos(α)*sin(α) or, equivalently, R = h*tan(α).

A sphere whose volume is 8π is inscribed in a cube. Find the volume of the cube.

Solution

Let a be the side of the cube. Then the volume of the cube is V = a 3.

Since the ball is inscribed in a cube, the radius of the ball is equal to half the edge of the cube, i.e. R = a/2 (see figure).

The volume of the ball is equal to V w = (4/3)πR 3 and equal to 8π, therefore

(4/3)πR 3 = 8π,

And the volume of the cube is equal to V = a 3 = (2R) 3 = 8R 3 = 8*6 = 48.

Task B9 (Typical options 2015)

The volume of the cone is 32. Through the middle of the height, parallel to the base of the cone, a section is drawn, which is the base of a smaller cone with the same vertex. Find the volume of the smaller cone.

Solution

Let's consider the tasks:

72353. The volume of the cone is 10. A section is drawn through the middle of the height parallel to the base of the cone, which is the base of a smaller cone with the same vertex. Find the volume of the smaller cone.

Let us immediately note that the original and cut off cone are similar and if we consider the cut off cone relative to the original one, we can say this: the smaller cone is similar to the larger one with a coefficient equal to one half or 0.5. We can write:

One could write:

One could think so!

Let's consider the original cone relative to the cut-off one. We can say that the larger cone is similar to the cut off one with a coefficient equal to two, let’s write:

Now look at the solution without using similarity properties.

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

Consider the lateral projection (side view) with the indicated cross-section:

Let the radius of the larger cone be equal to R, the height equal to H. The section (the base of the smaller cone) passes through the middle of the height, which means its height will be equal to H/2. And the radius of the base is equal to R/2, this follows from the similarity of triangles.

Let's write down the volume of the original cone:

The volume of the cut off cone will be equal to:

Such detailed solutions are presented so that you can see how the reasoning can be built. Act in any way - the main thing is that you understand the essence of the decision. Even if the path you chose is not rational, the result (the correct result) is important.

Answer: 1.25

318145. In a cone-shaped vessel, the liquid level reaches half its height. The volume of liquid is 70 ml. How many milliliters of liquid must be added to completely fill the container?

This task is similar to the previous one. Even though we are talking about a liquid here, the principle of the solution is the same.

We have two cones - this is the vessel itself and the “small” cone (filled with liquid), they are similar. It is known that the volumes of such bodies are related as follows:

The initial cone (vessel) is similar to a cone filled with liquid with a coefficient equal to 2, since it is said that the liquid level reaches half the height. You can write in more detail:

We calculate:

Thus, you need to add:

History of the definition of a cone

Geometry as a science emerged from the practical requirements of construction and observations of nature. Gradually, experimental knowledge was generalized, and the properties of some bodies were proven through others. The ancient Greeks introduced the concept of axioms and proofs. An axiom is a statement obtained through practical means and does not require proof.

In his book, Euclid gave a definition of a cone as a figure that is obtained by rotating a right triangle around one of its legs. He also owns the main theorem that determines the volume of a cone. This theorem was proven by the ancient Greek mathematician Eudoxus of Cnidus.

Another mathematician of ancient Greece, Apollonius of Perga, who was a student of Euclid, developed and expounded the theory of conic surfaces in his books. He owns the definition of a conical surface and a secant to it. Schoolchildren today study Euclidean geometry, which has preserved the basic theorems and definitions from ancient times.

Basic definitions

A right circular cone is formed by rotating a right triangle around one leg. As you can see, the concept of a cone has not changed since the time of Euclid.

The hypotenuse AS of the right triangle AOS, when rotated around the leg OS, forms the lateral surface of the cone, therefore it is called the generator. The leg OS of the triangle turns simultaneously into the height of the cone and its axis. Point S becomes the vertex of the cone. The leg AO, having described a circle (base), turned into the radius of a cone.

If you draw a plane from above through the vertex and axis of the cone, you can see that the resulting axial section is an isosceles triangle, in which the axis is the height of the triangle.

Where C- circumference of the base, l— length of the cone generatrix, R— radius of the base.

Formula for calculating the volume of a cone

To calculate the volume of a cone, use the following formula:

where S is the area of the base of the cone. Since the base is a circle, its area is calculated as follows:

This implies:

where V is the volume of the cone;

n is a number equal to 3.14;

R is the radius of the base corresponding to the segment AO in Figure 1;

H is the height equal to the segment OS.

Truncated cone, volume

There is a straight circular cone. If you cut off the upper part with a plane perpendicular to the height, you get a truncated cone. Its two bases have the shape of a circle with radii R1 and R2.

If a right cone is formed by rotating a right triangle, then a truncated cone is formed by rotating a rectangular trapezoid around a straight side.

The volume of a truncated cone is calculated using the following formula:

V=n*(R 1 2 +R 2 2 +R 1 *R 2)*H/3.

Cone and its section by plane

The ancient Greek mathematician Apollonius of Perga wrote the theoretical work Conic Sections. Thanks to his work in geometry, definitions of curves appeared: parabola, ellipse, hyperbola. Let's look at what the cone has to do with it.

Let's take a straight circular cone. If the plane intersects it perpendicular to the axis, then a circle is formed in the section. When a secant intersects a cone at an angle to the axis, an ellipse is obtained in the section.

A cutting plane perpendicular to the base and parallel to the axis of the cone forms a hyperbola on the surface. A plane cutting the cone at an angle to the base and parallel to the tangent to the cone creates a curve on the surface, which is called a parabola.

The solution of the problem

Even the simple task of how to make a bucket of a certain size requires knowledge. For example, you need to calculate the dimensions of a bucket so that it has a volume of 10 liters.

V=10 l=10 dm 3 ;

The development of the cone has the form shown schematically in Figure 3.

L is the generatrix of the cone.

To find out the surface area of the bucket, which is calculated using the following formula:

S=n*(R 1 +R 2)*L,

it is necessary to calculate the generator. We find it from the volume value V=n*(R 1 2 +R 2 2 +R 1 *R 2)*H/3.

Hence H=3V/n*(R 1 2 +R 2 2 +R 1 *R 2).

A truncated cone is formed by rotating a rectangular trapezoid, in which the side is the generatrix of the cone.

L 2 =(R 2- R 1) 2 +H 2.

Now we have all the data to build a drawing of a bucket.

Why are fire buckets cone shaped?

Who ever wondered why fire buckets have a seemingly strange conical shape? And this is not just like that. It turns out that a conical bucket when extinguishing a fire has many advantages over a regular one, shaped like a truncated cone.

Firstly, as it turns out, the fire bucket fills with water faster and does not spill when carried. A cone with a larger volume than a regular bucket allows you to transfer more water at a time.

Secondly, water from it can be thrown over a greater distance than from a regular bucket.

Thirdly, if the conical bucket falls from your hands and falls into the fire, then all the water is poured onto the source of the fire.

All of these factors save time - the main factor when extinguishing a fire.

Practical use

Schoolchildren often have questions about why they need to learn how to calculate the volume of various geometric bodies, including a cone.

And design engineers are constantly faced with the need to calculate the volume of conical parts of machine parts. These are drill tips, parts of lathes and milling machines. The cone shape will allow drills to easily enter the material without requiring initial marking with a special tool.

The volume of a cone is a pile of sand or earth poured onto the ground. If necessary, by taking simple measurements, you can calculate its volume. Some may be confused by the question of how to find out the radius and height of a pile of sand. Armed with a tape measure, we measure the circumference of the mound C. Using the formula R=C/2n we find out the radius. Throwing a rope (tape measure) over the vertex, we find the length of the generatrix. And calculating the height using the Pythagorean theorem and volume is not difficult. Of course, this calculation is approximate, but it allows you to determine whether you were deceived by bringing a ton of sand instead of a cube.

Some buildings are shaped like a truncated cone. For example, the Ostankino TV tower is approaching the shape of a cone. It can be imagined as consisting of two cones placed on top of each other. The domes of ancient castles and cathedrals represent a cone, the volume of which ancient architects calculated with amazing accuracy.

If you look closely at the surrounding objects, many of them are cones:

funnels for pouring liquids;
horn-loudspeaker;
parking cones;
lampshade for floor lamp;
the usual Christmas tree;
wind musical instruments.

As can be seen from the examples given, the ability to calculate the volume of a cone and its surface area is necessary in professional and everyday life. We hope that the article will come to your aid.

The bodies of rotation studied in school are the cylinder, cone and ball.

If in a problem on the Unified State Exam in mathematics you need to calculate the volume of a cone or the area of a sphere, consider yourself lucky.

Apply formulas for volume and surface area of a cylinder, cone and sphere. All of them are in our table. Learn by heart. This is where knowledge of stereometry begins.

Sometimes it's good to draw the view from above. Or, as in this problem, from below.

2. How many times is the volume of a cone circumscribed about a regular quadrangular pyramid greater than the volume of a cone inscribed in this pyramid?

It's simple - draw the view from below. We see that the radius of the larger circle is times larger than the radius of the smaller one. The heights of both cones are the same. Therefore, the volume of the larger cone will be twice as large.

Another important point. We remember that in the problems of part B of the Unified State Examination in mathematics, the answer is written as an integer or a final decimal fraction. Therefore, there should not be any or in your answer in part B. There is no need to substitute the approximate value of the number either! It must definitely shrink! It is for this purpose that in some problems the task is formulated, for example, as follows: “Find the area of the lateral surface of the cylinder divided by.”

Where else are the formulas for volume and surface area of bodies of revolution used? Of course, in problem C2 (16). We will also tell you about it.