Each student in the study of stereometry in high school came across a cone. Two important characteristics of this spatial figure are surface area and volume. In this article, we show how to find the volume of a round cone.
Round cone as a figure of rotation of a right triangle
Before proceeding directly to the topic of the article, a cone should be described from a geometric point of view.
Let there be some right triangle. If it is rotated around any of the legs, then the result of this action will be the desired figure shown in the figure below.
Here the leg AB is part of the axis of the cone, and its length corresponds to the height of the figure. The second leg (segment CA) is the radius of the cone. During rotation, he will describe the circle bounding the base of the figure. Hypotenuse BC is called the generatrix of the figure, or its generatrix. Point B is the vertex of the cone, which is its only one.
Given the properties of the triangle ABC, we can write the relationship between the generatrix g, radius r and height h in the form of the following equality:
g 2 = h 2 + r 2
This formula is useful in solving many geometric problems with the figure in question.
Cone Volume Formula
The volume of any spatial figure is called the region of space, which is limited by the surface of this figure. There are two such surfaces for the cone:
- Lateral, or conical. She is educated by all generators.
- Base. In this case, it is a circle.
We get the formula for determining the volume of the cone. To do this, we cut it mentally into many layers parallel to the base. Each of the layers has a thickness dx, which tends to zero. The area S x of the layer, which is located at a distance x from the top of the figure, is equal to the following expression:
S x = pi * r 2 * x 2 / h 2
The validity of this expression can be checked intuitively by substituting the values x = 0 and x = h. In the first case, we get an area equal to zero, in the second case it will be equal to the area of the round base.
To determine the volume of the cone, it is necessary to add small "bulbs" of each layer, that is, you should use the integral calculus:
V = ∫ 0 h (pi * r 2 * x 2 / h 2 * dx) = pi * r 2 / h 2 * ∫ 0 h (x 2 * dx)
Calculating this integral, we arrive at the final formula for the round cone:
V = 1/3 * pi * r 2 * h
It is interesting to note that this formula is completely analogous to that used to calculate the volume of an arbitrary pyramid. This coincidence is not accidental, because any pyramid goes into a cone with an increase in the number of its edges to infinity.
Volume calculation task
It is useful to give an example of solving the problem, which will demonstrate the use of the derived formula for volume V.
A round cone is given, in which the base area is 37 cm 2 , and the generatrix of the figure is three times the radius. What is the volume of the cone?
We are entitled to use the formula for volume if we know two quantities: height h and radius r. We find the formulas that define them in accordance with the condition of the problem.
The radius r can be calculated by knowing the area of the circle S o , we have:
S o = pi * r 2 =>
r = √ (S o / pi)
Using the condition of the problem, we write the equality for the generator g:
g = 3 * r = 3 * √ (S o / pi)
Knowing the formulas for r and g, we calculate the height h:
h = √ (g 2 - r 2 ) = √ (9 * S o / pi - S o / pi) = √ (8 * S o / pi)
We found all the necessary parameters. Now it's time to substitute them in the formula for V:
V = 1/3 * pi * r 2 * h = 1/3 * pi * S o / pi * √ (8 * S o / pi) = S o / 3 * √ (8 * S o / pi)
It remains to substitute the base area S o and calculate the value of the volume: V = 119.75 cm 3 .