Volume of the smaller cone formula. Cone volume

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- its height.

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 write it in correct pyramid, i.e., we will build a pyramid inside the cone, the top of which coincides with the top of the cone, and the base is regular polygon, inscribed at the base of the cone.

The volume of this pyramid is 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 as the base of the pyramid a polygon with very a large number sides, 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, not approximate equality is placed, although based on the reasoning carried out we could consider it approximate, but in high school high school it is proved that the 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. Direct volume 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 curved trapezoid, corresponding function

y = R / H X, X∈ . Therefore, using 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

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

Volume of a cone. Now we have reached cones and cylinders. In addition to those that have already been published, there will be about nine articles, we will consider all types of tasks. If during the year open bank New tasks will be added, of course, they will also be posted on the blog. This article presents the theory and examples in which it is used. It’s not enough to know the formula for the volume of a cone; by the way, here it is:

We can write:

To solve some examples, you need to understand how volumes relate similar bodies. It is to understand, and not just learn the formula:

That is, if we increase (decrease) the linear dimensions of the body by k times, then the ratio of the volume of the resulting body to the volume of the original will be equal to k 3 .

PLEASE NOTE! It doesn’t matter how you define the volumes:

The fact is that in the process of solving problems when considering similar bodies, some may get confused with the coefficient k. The question may arise - What is it equal to?

(depending on the value specified in the condition)

It all depends on “which side” you look at. It is important to understand this! Let's look at an example: given a cube, the edge of the second cube is three times larger:

IN in this case, the similarity coefficient is three (the edge is increased three times), which means the ratio will look like this:

That is, the volume of the resulting (larger) cube will be 27 times larger.

You can look from the other side.

Given a cube, the edge of the second cube is three times smaller:

The similarity coefficient is equal to one third (reducing the edge by three times), which means the ratio will look like:

That is, the volume of the resulting cube will be 27 times less.

Conclusion! Indices are not important when denoting volumes; it is important to understand how bodies are viewed relative to each other.

It is clear that:
— if the original body increases, then the coefficient will be greater than one.
— if the original body decreases, then the coefficient will be less than one.
The following can be said about volume ratios:
- if in the problem we divide the volume larger body by a smaller one, then we get the cube of the similarity coefficient, and the coefficient itself will be greater than one.
— if we divide the volume of a smaller body by a larger one, we will get the cube of the similarity coefficient, and the coefficient itself will be less than one.

The most important thing to remember is that when it comes to the VOLUME of similar bodies, the similarity coefficient has a THIRD degree, and not a second degree, as is the case with areas.

One more point concerning.

The condition contains such a concept as the generatrix of a cone. This is a segment connecting the top of the cone with the points of the base circle (indicated by the letter L in the figure).

It is worth noting here that we will analyze problems only with a straight cone (hereinafter simply a cone). The generators of a right cone are equal

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

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 described around the correct quadrangular pyramid, is greater than the volume of the 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. Remember that in the problems of part B Unified State Exam options in mathematics the answer is written as an integer or finite number decimal. 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.

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 edges 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. A section is drawn through the middle of the height parallel to the base of the cone, which is the base smaller cone with the same top. 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:

So 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: