The area of ​​each base of the cylinder is π r 2, the area of ​​both bases will be 2π r 2 (fig.).

The area of ​​the lateral surface of a cylinder is equal to the area of ​​a rectangle whose base is 2π r, and the height is equal to the height of the cylinder h, i.e. 2π rh.

The total surface of the cylinder will be: 2π r 2 + 2π rh= 2π r(r+ h).

The area of ​​the lateral surface of the cylinder is taken to be sweep area its lateral surface.

Therefore, the area of ​​the lateral surface of a right circular cylinder is equal to the area of ​​the corresponding rectangle (Fig.) and is calculated by the formula

S b.c. = 2πRH, (1)

If we add the area of ​​its two bases to the area of ​​the lateral surface of the cylinder, we obtain the total surface area of ​​the cylinder

S full =2πRH + 2πR 2 = 2πR (H + R).

Volume of a straight cylinder

Theorem. Volume of a straight cylinder equal to the product area of ​​its base to height , i.e.

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

Since the area of ​​the base of the cylinder is Q, then there are sequences of circumscribed and inscribed polygons with areas Q n and Q' n such that

\(\lim_(n \rightarrow \infty)\) Q n= \(\lim_(n \rightarrow \infty)\) Q’ n= Q.

Let us construct a sequence of prisms whose bases are the described and inscribed polygons discussed above, and whose side edges are parallel to the generatrix of the given cylinder and have length H. These prisms are circumscribed and inscribed for the given cylinder. Their volumes are found by the formulas

V n=Q n H and V' n= Q' n H.

Hence,

V= \(\lim_(n \rightarrow \infty)\) Q n H = \(\lim_(n \rightarrow \infty)\) Q’ n H = QH.

Consequence.
The volume of a right circular cylinder is calculated by the formula

V = π R 2 H

where R is the radius of the base and H is the height of the cylinder.

Since the base of a circular cylinder is a circle of radius R, then Q = π R 2, and therefore

Exists a large number of problems related to the cylinder. In them you need to find the radius and height of the body or the type of its section. Plus, sometimes you need to calculate the area of ​​a cylinder and its volume.

Which body is a cylinder?

In the school curriculum, a circular cylinder, that is, one at the base, is studied. But the elliptical appearance of this figure is also distinguished. From the name it is clear that its base will be an ellipse or an oval.

The cylinder has two bases. They are equal to each other and are connected by segments that combine the corresponding points of the bases. They are called the generators of the cylinder. All generators are parallel to each other and equal. They make up the lateral surface of the body.

In general, a cylinder is an inclined body. If the generators make a right angle with the bases, then we speak of a straight figure.

Interestingly, a circular cylinder is a body of revolution. It is obtained by rotating a rectangle around one of its sides.

Main elements of the cylinder

The main elements of the cylinder look like this.

1. Height. It is the shortest distance between the bases of the cylinder. If it is straight, then the height coincides with the generatrix.
2. Radius. Coincides with the one that can be drawn at the base.
3. Axis. This is a straight line that contains the centers of both bases. The axis is always parallel to all generators. In a straight cylinder it is perpendicular to the bases.
4. Axial section. It is formed when a cylinder intersects a plane containing an axis.
5. Tangent plane. It passes through one of the generatrices and is perpendicular to the axial section, which is drawn through this generatrix.

How is a cylinder connected to a prism inscribed in it or described around it?

Sometimes there are problems in which you need to calculate the area of ​​a cylinder, but some elements of the associated prism are known. How do these figures relate?

If a prism is inscribed in a cylinder, then its bases are equal polygons. Moreover, they are inscribed in the corresponding bases of the cylinder. The lateral edges of the prism coincide with the generators.

The described prism has regular polygons at its base. They are described around the circles of the cylinder, which are its bases. The planes that contain the faces of the prism touch the cylinder along their generators.

On the area of ​​the lateral surface and base for a right circular cylinder

If you unwrap the side surface, you will get a rectangle. Its sides will coincide with the generatrix and the circumference of the base. Therefore, the lateral area of ​​the cylinder will be equal to the product of these two quantities. If you write down the formula, you get the following:

S side = l * n,

where n is the generator, l is the circumference.

Moreover, the last parameter is calculated using the formula:

l = 2 π * r,

here r is the radius of the circle, π is the number “pi” equal to 3.14.

Since the base is a circle, its area is calculated using the following expression:

S main = π * r 2 .

On the area of ​​the entire surface of a right circular cylinder

Since it is formed by two bases and a side surface, you need to add these three quantities. That is, the total area of ​​the cylinder will be calculated by the formula:

S floor = 2 π * r * n + 2 π * r 2 .

It is often written in a different form:

S floor = 2 π * r (n + r).

On the areas of an inclined circular cylinder

As for the bases, all the formulas are the same, because they are still circles. And here side surface no longer produces a rectangle.

To calculate the area of ​​the lateral surface of an inclined cylinder, you will need to multiply the values ​​of the generatrix and the perimeter of the section, which will be perpendicular to the selected generatrix.

The formula looks like this:

S side = x * P,

where x is the length of the cylinder generatrix, P is the perimeter of the section.

By the way, it is better to choose a section such that it forms an ellipse. Then the calculations of its perimeter will be simplified. The length of the ellipse is calculated using a formula that gives an approximate answer. But it is often sufficient for the tasks of a school course:

l = π * (a + b),

where “a” and “b” are the semi-axes of the ellipse, that is, the distance from the center to its nearest and farthest points.

The area of ​​the entire surface must be calculated using the following expression:

S floor = 2 π * r 2 + x * R.

What are some sections of a right circular cylinder?

When a section passes through an axis, its area is determined as the product of the generatrix and the diameter of the base. This is explained by the fact that it has the shape of a rectangle, the sides of which coincide with the designated elements.

To find the cross-sectional area of ​​a cylinder that is parallel to the axial one, you will also need a formula for a rectangle. In this situation, one of its sides will still coincide with the height, and the other will be equal to the chord of the base. The latter coincides with the section line along the base.

When the section is perpendicular to the axis, it looks like a circle. Moreover, its area is the same as that of the base of the figure.

It is also possible to intersect at some angle to the axis. Then the cross-section results in an oval or part of it.

Sample problems

Task No. 1. Given a straight cylinder whose base area is 12.56 cm 2 . It is necessary to calculate the total area of ​​the cylinder if its height is 3 cm.

Solution. It is necessary to use the formula for the total area of ​​a circular straight cylinder. But it lacks data, namely the radius of the base. But the area of ​​the circle is known. It is easy to calculate the radius from this.

He turns out to be equal square root from the quotient that is obtained by dividing the area of ​​the base by pi. After dividing 12.56 by 3.14, the result is 4. The square root of 4 is 2. Therefore, the radius will have this value.

Answer: S floor = 50.24 cm 2.

Task No. 2. A cylinder with a radius of 5 cm is cut by a plane parallel to the axis. The distance from the section to the axis is 3 cm. The height of the cylinder is 4 cm. You need to find the cross-sectional area.

Solution. The cross-sectional shape is rectangular. One of its sides coincides with the height of the cylinder, and the other is equal to the chord. If the first quantity is known, then the second one needs to be found.

To do this, additional construction must be made. At the base we draw two segments. They will both start at the center of the circle. The first will end at the center of the chord and equal to the known distance to the axis. The second is at the end of the chord.

You will get a right triangle. The hypotenuse and one of the legs are known in it. The hypotenuse coincides with the radius. The second leg is equal to half the chord. The unknown leg multiplied by 2 will give the desired chord length. Let's calculate its value.

In order to find the unknown leg, you will need to square the hypotenuse and the known leg, subtract the second from the first and take the square root. The squares are 25 and 9. Their difference is 16. After taking the square root, 4 remains. This is the desired leg.

The chord will be equal to 4 * 2 = 8 (cm). Now you can calculate the cross-sectional area: 8 * 4 = 32 (cm 2).

Answer: S cross is equal to 32 cm 2.

Task No. 3. It is necessary to calculate the area axial section cylinder. It is known that a cube with an edge of 10 cm is inscribed in it.

Solution. The axial section of the cylinder coincides with a rectangle that passes through the four vertices of the cube and contains the diagonals of its bases. The side of the cube is the generatrix of the cylinder, and the diagonal of the base coincides with the diameter. The product of these two quantities will give the area that you need to find out in the problem.

To find the diameter, you will need to use the knowledge that the base of the cube is a square, and its diagonal forms an equilateral right triangle. Its hypotenuse is the desired diagonal of the figure.

To calculate it, you will need the formula of the Pythagorean theorem. You need to square the side of the cube, multiply it by 2 and take the square root. Ten to the second power is one hundred. Multiplied by 2 is two hundred. The square root of 200 is 10√2.

The section is again a rectangle with sides 10 and 10√2. Its area can be easily calculated by multiplying these values.

Answer. S section = 100√2 cm 2.

Represents geometric body, bounded by two parallel planes and a cylindrical surface.

The cylinder consists of a side surface and two bases. The formula for the surface area of ​​a cylinder includes a separate calculation of the area of ​​the base and the side surface. Since the bases in the cylinder are equal, its total area will be calculated by the formula:

We will consider an example of calculating the area of ​​a cylinder after we know all the necessary formulas. First we need the formula for the area of ​​the base of a cylinder. Since the base of the cylinder is a circle, we will need to apply:
We remember that in these calculations the constant number Π = 3.1415926 is used, which is calculated as the ratio of the circumference of a circle to its diameter. This number is a mathematical constant. We will also look at an example of calculating the area of ​​the base of a cylinder a little later.

Cylinder side surface area

The formula for the area of ​​the lateral surface of a cylinder is the product of the length of the base and its height:

Now let's look at a problem in which we need to calculate the total area of ​​a cylinder. In the given figure, the height is h = 4 cm, r = 2 cm. Let us find the total area of ​​the cylinder.
First, let's calculate the area of ​​the bases:
Now let's look at an example of calculating the area of ​​the lateral surface of a cylinder. When expanded, it represents a rectangle. Its area is calculated using the above formula. Let's substitute all the data into it:
The total area of ​​a circle is the sum of double the area of ​​the base and the side:

Thus, using the formulas for the area of ​​the bases and the lateral surface of the figure, we were able to find the total surface area of ​​the cylinder.
The axial section of the cylinder is a rectangle in which the sides are equal to the height and diameter of the cylinder.

The formula for the axial cross-sectional area of ​​a cylinder is derived from the calculation formula:

Find the area of ​​the axial section perpendicular to the bases of the cylinder. One of the sides of this rectangle is equal to the height of the cylinder, the second - to the diameter of the base circle. Accordingly, the cross-sectional area in this case will be equal to the product of the sides of the rectangle. S=2R*h, where S is the cross-sectional area, R is the radius of the base circle, given by the conditions of the problem, and h is the height of the cylinder, also given by the conditions of the problem.

If the section is perpendicular to the bases, but does not pass through the axis of rotation, the rectangle will not be equal to the diameter of the circle. It needs to be calculated. To do this, the problem must say at what distance from the axis of rotation the section plane passes. For ease of calculations, construct a circle at the base of the cylinder, draw a radius and plot on it the distance at which the section is located from the center of the circle. From this point, draw perpendiculars to their intersection with the circle. Connect the intersection points to the center. You need to find the chords. Find the size of half a chord using the Pythagorean theorem. It will be equal to the square root of the difference between the squares of the radius of the circle from the center to the section line. a2=R2-b2. The entire chord will, accordingly, be equal to 2a. Calculate the cross-sectional area, which is equal to the product of the sides of the rectangle, that is, S=2a*h.

The cylinder can be cut without passing through the plane of the base. If the cross section is perpendicular to the axis of rotation, then it will be a circle. Its area in this case is equal to the area of ​​the bases, that is, calculated by the formula S = πR2.

Helpful advice

To more accurately imagine the section, make a drawing and additional constructions for it.

Sources:

• cylinder cross section area

The line of intersection of a surface with a plane belongs to both the surface and the cutting plane. The line of intersection of a cylindrical surface with a cutting plane parallel to the straight generatrix is ​​a straight line. If the cutting plane is perpendicular to the axis of the surface of revolution, the section will be a circle. In general, the line of intersection of a cylindrical surface with a cutting plane is a curved line.

You will need

• Pencil, ruler, triangle, patterns, compass, meter.

Instructions

On the frontal plane of projections П₂, the section line coincides with the projection of the cutting plane Σ₂ in the form of a straight line.
Designate the points of intersection of the generatrices of the cylinder with the projection Σ₂ 1₂, 2₂, etc. to points 10₂ and 11₂.

On the plane P₁ is a circle. Points 1₂, 2₂, etc. marked on the section plane Σ₂. using a projection connection line are projected onto the outline of this circle. Mark their horizontal projections symmetrically relative to the horizontal axis of the circle.

Thus, the projections of the desired section are determined: on the P₂ plane – a straight line (points 1₂, 2₂…10₂); on the P₁ plane – a circle (points 1₁, 2₁…10₁).

Using two, construct the natural size of the section of this cylinder by the frontal projecting plane Σ. To do this, use the projection method.

Draw the plane П₄ parallel to the projection of the plane Σ₂. On this new x₂₄ axis, mark point 1₀. Distances between points 1₂ – 2₂, 2₂ – 4₂, etc. from the frontal projection of the section, place it on the x₂₄ axis, draw thin lines of the projection connection perpendicular to the x₂₄ axis.

IN this method the P₄ plane is replaced by the P₁ plane, therefore, from the horizontal projection, transfer the dimensions from the axis to the points to the axis of the P₄ plane.

For example, on P₁ for points 2 and 3 this will be the distance from 2₁ and 3₁ to the axis (point A), etc.

Laying aside the indicated distances from the horizontal projection, you get points 2₀, 3₀, 6₀, 7₀, 10₀, 11₀. Then, for greater accuracy of construction, the remaining intermediate points are determined.

By connecting all the points with a pattern curve, you obtain the required natural size of the section of the cylinder by the frontal projecting plane.

Sources:

• how to replace a plane

Tip 3: How to find the axial cross-sectional area of ​​a truncated cone

To solve this problem, you need to remember what a truncated cone is and what properties it has. Be sure to make a drawing. This will allow you to determine what geometric figure the section represents. It is quite possible that after this, solving the problem will no longer be difficult for you.

Instructions

A round cone is a body obtained by rotating a triangle around one of its legs. Straight lines emanating from the apex cone and intersecting its base are called generators. If all generators are equal, then the cone is straight. At the base of the round cone lies a circle. The perpendicular dropped to the base from the vertex is the height cone. At the round straight cone the height coincides with its axis. The axis is a straight line connecting to the center of the base. If the horizontal cutting plane of a circular cone, then its upper base is a circle.

Since it is not specified in the problem statement that it is the cone that is given in this case, we can conclude that this is a straight truncated cone, the horizontal section of which is parallel to the base. Its axial section, i.e. vertical plane, which through the axis of the round cone, is an equilateral trapezoid. All axial sections round straight cone are equal to each other. Therefore, to find square axial sections, you need to find square trapezoid, the bases of which are the diameters of the bases of a truncated cone, and the lateral sides are its constituents. Frustum height cone is also the height of the trapezoid.

The area of ​​a trapezoid is determined by the formula: S = ½(a+b) h, where S – square trapezoid; a – the size of the lower base of the trapezoid; b – the size of its upper base; h – the height of the trapezoid.

Since the condition does not specify which ones are given, it is possible that the diameters of both bases of the truncated cone known: AD = d1 – diameter of the lower base of the truncated cone;BC = d2 – diameter of its upper base; EH = h1 – height cone.Thus, square axial sections truncated cone is defined: S1 = ½ (d1+d2) h1

Sources:

• area of ​​a truncated cone

The cylinder is a spatial figure and consists of two equal grounds, which represent circles and a lateral surface connecting the lines delimiting the bases. To calculate square cylinder, find the areas of all its surfaces and add them up.