Formulas

CYLINDER VOLUME

CONE VOLUME

VOLUME OF THE TRUNCATED CONE

BALL VOLUME

V=1/3∏H(R2+r2+Rr)

V=4/3 ∙ ∏R 3

Formulas for calculating volume: ball, spherical sector, spherical layer, spherical sector and sphere area

• The area of ​​a sphere is:

S=4 π R 2 ,

where R is the radius of the sphere

• The volume of the ball is:

V = 1 ⅓ π R 3 = 4/3 π R 3

where R is the radius of the ball

• The volume of the spherical segment is equal to:

V = π h 2 (R - ⅓ h) ,

where R is the radius of the ball and h is the height of the segment

• The volume of the spherical layer is equal to:

V = V 1 – V 2 ,

where V 1 is the volume of one spherical segment, and V 2 is the volume of the second spherical segment

• The volume of the spherical sector is equal to:

V = ⅔ π R 2 h ,

where R is the radius of the ball and h is the height of the ball segment

Theoretical dictation

Option 1

Fill in the missing words in the text .

• Any section of a sphere by a plane is a circle. The center of this circle is …………………… the perpendicular dropped from the center of the ball to the cutting plane.

2. The center of the ball is its ………………….……. symmetry.

3. The axial section of the ball is ………………………….

4. The lines of intersection of the two spheres are…………………

5. Planes equidistant from the center intersect the ball in ……………... circles.

6. Near any regular pyramid, a sphere can be described, and its center lies on ……………… .. of the pyramid.

base

center

a circle

circle

equal

altitude

Theoretical dictation

Option 2

plane

circle

altitude

perpendicular

touch

altitude

Card #1

A plane perpendicular to the diameter of the sphere divides its parts 3cm and 9cm. Find the volume of the sphere?

288 P cm³

Card #2

Two equal spheres are located so that the center of one lies on the surface of the other. How is the volume of the common part of the balls related to the volume of the whole ball?

5 / 16

Card #3

What part of the volume of the sphere is the volume of the spherical segment, whose height is equal to 0.1 of the diameter of the ball, equal to 20 cm?

Task #1

The volume of a ball of radius R is equal to V . Find: the volume of a ball of radius: a) 2 R b) 0.5 R

Task #2

What is the volume of the spherical sector if the radius of the base circle is 60 cm, and the radius of the ball is 75 cm.

QUICKLY AND BRIEFLY WRITE ANSWERS TO THE QUESTIONS:

• How many spheres can be held:

a) through the same circle;

b) through a circle and a point not belonging to its plane?

2. How many spheres can be drawn through four points that are vertices:

a) a square

b) an isosceles trapezoid;

3. Is it true that one big circle passes through any two points of the sphere?

4. Through what two points of the sphere can several great circles be drawn?

5. How should two equal circles be located so that a sphere of the same radius can pass through them?

endlessly

one

endlessly

endlessly

None

diametrically opposed

have a common center

Theoretical dictation

Option 2

Fill in the missing words in the text.

• Any diametral plane of the ball is its ………………… symmetry.

2. The axial section of the sphere is………………..

3. The center of the ball described near the regular pyramid lies on …………………. pyramids.

4. The radius of the sphere drawn to the point of contact between the sphere and the plane ………………...……………………..to the tangent plane.

5. The tangent plane has only one common point with the ball …………………….

6. A sphere can be inscribed in any regular pyramid, and its center lies on ……………… .…….pyramids.

plane

circle

altitude

perpendicular

touch

altitude

Lv.52

Level 1 Option 1

1. At a distance of 12 cm from the center of the ball, a section is drawn, the radius of which is 9 cm. Find the volume of the sphere and its surface area.

2. A sphere of radius 3 cm has a cent at the point O (4; -2; 1). Write an equation for the sphere into which this sphere will pass if it is symmetric about the OXY plane. Find the volume of the sphere enclosed by the given sphere.

Level 1 Option 2

1. Through a point lying on a sphere, a section of radius 3 cm is drawn at an angle of 60 ° to the radius of the sphere drawn to this point. Find the area of ​​the sphere and the volume of the sphere.

2. A sphere of radius 3 has a center at the point O (-2;5;3). Write an equation for the sphere into which this sphere will go if it is symmetric about the plane OX Z . Find the area of ​​this sphere.

Test independent work lvl.52

Level 2 Option 1

1. A section is drawn at a distance of 2√7 cm from the center of the ball. The chord of this section is 4 cm, subtracting the angle 90°. Find the volume of the sphere and its surface area.

2. A sphere centered at the point O (2; 1; -2) passes through the origin. Write an equation for the sphere into which this sphere will pass if it is symmetric about the abscissa axis. Find the volume of the sphere bounded by the resulting sphere.

Level 2 Option 2

1. At a distance of 4 cm from the center of the ball, a section was drawn. A chord removed from the center of this section by √5cm, subtracting an angle of 120°. Find the volume of the sphere and its surface area.

2. A sphere centered at point O (-1;-2;2) passes through the origin. Write an equation for the sphere into which the given sphere will pass with symmetry about the Z = 1 plane. Find the area of ​​the sphere.

Independent work

Option 2

• Ball diameter ½ dm. Calculate the volume of a sphere and the area of ​​a sphere.

2. A volleyball has a radius of 12 dm. How much air is in the ball?

Option 1

• ball radius ¾ dm. Calculate the volume of a sphere and the area of ​​a sphere.

2. A soccer ball has a diameter of 30 dm. How much air is in the ball?

Independent work

Option 1

Option 2

• solve problems :

• Write down the formulas for the area of ​​a sphere, the volume of a sphere and its parts.
• solve problems :

№ 1. The volume of the sphere is 36 Pcm³. Find the area of ​​the sphere bounding the given sphere.

№ 2. A section is drawn in a sphere of radius 15 cm, the area of ​​which is 81 cm². Find the volume of the smaller spherical segment cut off by the section plane.

№ 3. Find the volume of a spherical sector if the radius of the sphere is 6cm and the height of the corresponding segment is one sixth of the diameter of the sphere.

№ 1. The surface area of ​​the sphere is 144P cm². Find the volume of this sphere.

№ 2. A section is drawn at a distance of 9 m from the center of the ball, the circumference of which is 24P cm. Find the volume of the smaller spherical segment cut off by the section plane.

№ 3. Find the volume of a spherical sector if the radius of the sphere is 6cm and the height of the cone forming the sector is one third of the diameter of the sphere.

113.04=4πR³/3 = R³=27, R=3. S=4πR², S=4π3²=36π. Answer: 3.36π. Given: ball; S=64π cm² Find: R, V Solution: S=4πR², 64π=4πR², = R=4 V=4πR³/3, V=4π4³/3=256π/3. Answer: 4.256π/3. 3. Given: spherical segment, rbase=60 cm, Rball=75 cm. Find: Vspherical segment. Solution: V=πh²(R-⅓h) O ₁ С=√R²-r²=√75²-60²=45 h= OS-OS ₁ =75-45=30 V=π 30² (75-⅓ 30) =58500π. Answer: 58500π. "width="640"

Problem solving with self-testing.

Given: ball; V=113.04 cm³,

Find: R, S.

Solution: V=4πR³/3, = 113.04=4πR³/3 = R³=27, R=3.

S=4πR², S=4π3²=36π.

Answer: 3.36π.

Given: ball; S=64π cm²

Find: R, V

Solution: S=4πR², 64π=4πR², = R=4

V=4πR³/3, V=4π4³/3=256π/3.

Answer: 4.256π/3.

3. Given: spherical segment, r main = 60 cm, R ball = 75 cm.

Find: Vspherical segment.

Solution: V=πh²(R-⅓h) O ₁ C=√R²-r²=√75²-60²=45

h= OS-OS ₁ =75-45=30 V=π 30² (75-⅓ 30)=58500π.

Answer: 58500π.

Reflection

Show your mood with an emoji.

Take the emoticon that matches your mood at the end of the lesson and, as you leave, attach it to the board with a magnetic base.

Homework

• Homework

• Repeat the formulas for the volumes of a ball, a spherical segment, a spherical layer, a spherical sector. #723, #724, #755

Literature and Internet resources

Textbook on geometry 10-11 class Atanasyan L.S., 2008

Gavrilova N.F. Lesson developments in geometry Grade 11

The radius of a ball (denoted as r or R) is the line segment that connects the center of the ball to any point on its surface. As in the case of a circle, the radius of a ball is an important quantity that is needed to find the ball's diameter, circumference, surface area and/or volume. But the radius of the ball can also be found from a given value of the diameter, circumference, and other quantities. Use a formula in which you can substitute these values.

Steps

Formulas for calculating the radius

Calculate the radius from the diameter. The radius is half the diameter, so use the formula d = D/2. This is the same formula used to calculate the radius and diameter of a circle.

• For example, given a ball with a diameter of 16 cm. The radius of this ball: r = 16/2 = 8 cm. If the diameter is 42 cm, then the radius is 21 cm (42/2=21).

1. Calculate the radius from the circumference of the circle. Use the formula: r = C/2π. Since the circumference is C = πD = 2πr, then divide the formula for calculating the circumference by 2π and get the formula for finding the radius.

   • For example, given a ball with a circumference of 20 cm. The radius of this ball is: r = 20/2π = 3.183 cm.
   • The same formula is used to calculate the radius and circumference of a circle.

2. Calculate the radius from the volume of the sphere. Use the formula: r = ((V/π)(3/4)) 1/3. The volume of the ball is calculated by the formula V = (4/3)πr 3 . Separating r on one side of the equation, you get the formula ((V / π) (3/4)) 3 \u003d r, that is, to calculate the radius, divide the volume of the ball by π, multiply the result by 3/4, and raise the result to the power 1/3 (or take the cube root).

   • For example, given a ball with a volume of 100 cm 3. The radius of this sphere is calculated as follows:
     • ((V/π)(3/4)) 1/3 = r
     • ((100/π)(3/4)) 1/3 = r
     • ((31.83)(3/4)) 1/3 = r
     • (23.87) 1/3 = r
     • 2.88 cm= r

3. Calculate the radius from the surface area. Use the formula: r = √(A/(4 π)). The surface area of ​​the ball is calculated by the formula A \u003d 4πr 2. By isolating r on one side of the equation, you get the formula √(A/(4π)) = r, that is, to calculate the radius, you need to take the square root of the surface area divided by 4π. Instead of taking the root, the expression (A/(4π)) can be raised to the power of 1/2.

   • For example, given a sphere with a surface area of ​​1200 cm 3 . The radius of this sphere is calculated as follows:
     • √(A/(4π)) = r
     • √(1200/(4π)) = r
     • √(300/(π)) = r
     • √(95.49) = r
     • 9.77 cm= r

   Definition of basic quantities

   1. Remember the basic quantities that are relevant to calculating the radius of the ball. The radius of a ball is a segment that connects the center of the ball to any point on its surface. The radius of a sphere can be calculated from given values ​​of diameter, circumference, volume, or surface area.

      Use the values ​​of these quantities to find the radius. The radius can be calculated from given values ​​of diameter, circumference, volume, and surface area. Moreover, these values ​​can be found from a given value of the radius. To calculate the radius, simply convert the formulas to find the given values. Below are the formulas (in which there is a radius) to calculate the diameter, circumference, volume and surface area.

   Finding the radius from the distance between two points

   1. Find the coordinates (x, y, z) of the center of the ball. The radius of a sphere is equal to the distance between its center and any point lying on the surface of the sphere. If the coordinates of the center of the ball and any point lying on its surface are known, you can find the radius of the ball using a special formula by calculating the distance between two points. First, find the coordinates of the center of the ball. Keep in mind that since the ball is a three-dimensional figure, the point will have three coordinates (x, y, z), and not two (x, y).

      • Consider an example. Given a ball centered with coordinates (4,-1,12) . Use these coordinates to find the radius of the ball.

   2. Find the coordinates of a point on the surface of the sphere. Now you need to find the coordinates (x, y, z) any point on the surface of the sphere. Since all points lying on the surface of the ball are located at the same distance from the center of the ball, any point can be chosen to calculate the radius of the ball.

      • In our example, let's assume that some point lying on the surface of the ball has coordinates (3,3,0) . By calculating the distance between this point and the center of the ball, you will find the radius.

   3. Calculate the radius using the formula d \u003d √ ((x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2). Having learned the coordinates of the center of the ball and the point lying on its surface, you can find the distance between them, which is equal to the radius of the ball. The distance between two points is calculated by the formula d \u003d √ ((x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2), where d is the distance between the points, (x 1, y 1 ,z 1) are the coordinates of the center of the ball, (x 2 ,y 2 ,z 2) are the coordinates of a point lying on the surface of the ball.

      • In this example, instead of (x 1, y 1, z 1), substitute (4, -1,12), and instead of (x 2, y 2, z 2) substitute (3,3,0):
        • d \u003d √ ((x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2)
        • d = √((3 - 4) 2 + (3 - -1) 2 + (0 - 12) 2)
        • d = √((-1) 2 + (4) 2 + (-12) 2)
        • d = √(1 + 16 + 144)
        • d = √(161)
        • d=12.69. This is the desired radius of the ball.

   4. Keep in mind that in general cases r = √((x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2). All points lying on the surface of the ball are located at the same distance from the center of the ball. If in the formula for finding the distance between two points "d" is replaced by "r", you get a formula for calculating the radius of the ball from the known coordinates (x 1, y 1, z 1) of the center of the ball and the coordinates (x 2, y 2, z 2 ) any point lying on the surface of the sphere.

      • Square both sides of this equation and you get r 2 = (x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2 . Note that this equation corresponds to the equation of a sphere r 2 = x 2 + y 2 + z 2 centered at (0,0,0).
   • Don't forget about the order in which the math operations are performed. If you do not remember this order, and your calculator knows how to work with parentheses, use them.
   • This article talks about calculating the radius of a ball. But if you're having trouble learning geometry, it's best to start by calculating the quantities associated with a ball in terms of a known radius.
   • π (Pi) is the letter of the Greek alphabet, which means a constant equal to the ratio of the diameter of a circle to the length of its circumference. Pi is an irrational number that is not written as a ratio of real numbers. There are many approximations, for example, the ratio 333/106 will allow you to find the number Pi with an accuracy of up to four digits after the decimal point. As a rule, they use the approximate value of pi, which is 3.14.

Write a program to calculate the area of ​​a circle S and the volume of the ball V based on given radius R. Implement the program as a Windows application.

Mathematical statement of the problem

Before starting the development of the application, it is necessary to carry out a mathematical formulation of the problem, that is, to determine the formulas by which the calculation will be made, as well as the input data and outgoing results.

The area of ​​a circle is calculated by the formula:

S = π · R²

The input value here is the radius of the circle R, the result is the area of ​​the circle - S.
The volume of a sphere is calculated by the formula:

V = 4/3 π R³

The input value here is, again, the radius of the circle R, the result is the volume of the ball (although, as you know, the “ball” has no volume).
Both formulas contain the constant π , equal to 3.14159.
Thus, we will draw a sequence of stages for solving the problem (Figure 1).

Rice. 1. Stages of problem solving

Performance

1. Create an application of type VCL Form Application .

Launch Visual Application Development System Embracadero RAD Studio Delphi 2010 and create a Windows application. A detailed example of creating an application using the Windows Form Application template is described.

The initial view of the application form before the start of design is shown in Figure 2.

Rice. 2. View of the program window

2. Standard tab of the Tool Palette toolbar.

In this application, you need to use several components, which are listed below:

• type component T Label A that represents the line of text that is displayed on the form;
• type component TButton A that represents a button on the form;
• type component TEdi t , which is the text input string.

All these components are located on the Tool Palette toolbar on the Standard tab (see Fig. 3.).

Rice. 3. Standard tab on the component palette

3. TLabel component

3.1. Placement of the TLabel component on the form

To do this, click on the TLabel component (Fig. 4), and then click in the upper left corner of the form, as shown in Fig. 5.

Rice. 4. TLabel component on the tool palette

Rice. 5. Component of type TLabel on the main form of the program

3.2. Setting text in TLabel

To perform any actions with the TLabel component, it must first be selected using the mouse or by selecting it in the Object Inspector panel. After that, set the Caption property of the TLabel component to the value " R=» (Fig. 6).

Rice. 6. Caption Property

As a result, the text "Label1" on the form will change to the text "R =".
The Object Inspector allows you to view many other properties of this component. In our case, we will be interested in the Name property, which contains the value of the name of the variable (object). By default, this value is "Label1". This means that at the time of writing the program code, the properties of this component can be accessed with the prefix "Label". For example, in order to change the Caption property in the program, you need to type the following line:

Label1.Caption:= "R=" ;

In the same way, we put components on the form with the names Label2 and Label3 just below the previous component. Set the value of the Caption property to "S = " and "V = " respectively.

The form of the application should look like the following (Fig. 7).

Rice. 7. Application form after placing the components Label1, Label2, Label3

The transfer and processing of all other components from the Tool Palette is carried out in a similar way.

4. TEdit component

From the Tool Palette, from the Standard tab, add the TEdit component representing the input line. By using this component, we will get the values ​​of the radius of the circle entered by the user from the keyboard. After adding the component to the form, the Delphi system creates a variable component called Edit1 (Name property).

Clear the component's Text property.

5. TButton component

Add a component from the Tool Palette palette TButton , which is a regular button, after clicking on which the area of ​​the circle and the volume of the ball will be calculated. In the application, Delphi will automatically add a variable component named Button1 .

We set the Caption property of the component to the value " Calculate".

The application form in design mode will look like the one shown in Fig. eight.

Rice. 8. Application form after adding TEdit and TButton components

6. Programming the click event on the button "Calculate"

The next step in the application being developed is programming an event in Delphi that occurs when Button1 is clicked. The mouse click event on the button is called OnClick.

Delphi 2010 automatically generates a code snippet where you need to write your own event handling code. The code generated by the system looks like this:

procedurebegin end ;

The first task is to determine the input data, results or intermediate variables that will be used in the program.

According to the conditions of the problem in our program, we describe three variables with the appropriate designation:

• R is the radius of the circle;
• S - area of ​​a circle;
• V is the volume of the sphere.

All variables must be of real type.
The program also uses one constant - the number Pi. Let's call it Pi . Note that Delphi has a built-in function called Pi , but this will not be used in our application. Thus, the description of variables and constants before the word begin will be as follows:

const Pi = 3.1415 ; // Pi var R:real; // Circle Radius S:real; // Area of ​​a circle v:real; // volume of the ball

Between the begin and end statements, enter the following lines of the main program code:

// 1. Reading the value of the radius of the circle from Edit1.Text R:= StrToFloat(Edit1.Text); S:= Pi*R*R; // 3. Calculate the volume of the ball V:= 4 /3 * Pi * R * R * R; // 4. Outputting results with precision // 3 decimal places Label2.Caption:="S=" +FloatToStrF(S,ffFixed,8 ,3 ); Label3.Caption:="V=" +FloatToStrF(V,ffFixed,8 ,3 );

Let's explain some functions (methods) used in the program code. The StrToFloat function converts the string value of Edit1.Text to a real number. For example, after executing the following code

x:= StrToFloat( "-3.675" );

the value of x becomes -3.675 .

In paragraphs 2 and 3, the usual calculations of the area of ​​a circle and the volume of a ball occur using the arithmetic operations of the Pascal language.

In paragraph 4, the output of the results is carried out. Since the program is implemented as a Windows application, to display the result, it is enough to fill in the value of the Caption property in the Label2 (area) and Label3 (volume) components.

The FloatToStrF function performs the inverse conversion to the StrToFloat function, that is, it converts a real number into a string. For example, to convert the number 2.87 to a string with an accuracy of 4 decimal places, you would write:

v:= 2.87; str:= FloatToStrF(v, ffFixed, 8 , 4 );

where v is a variable of real type; str is a string type variable; ffFixed - conversion format. The constant 8 means that a total output width of 8 characters is used. The constant 4 means precision after the decimal point.

The general listing of the procedure for handling the OnClick event of the Button1 component looks like this.

procedure TForm1.Button1Click(Sender: TObject); const Pi = 3.1415 ; // Pi var R:real; // Circle Radius S:real; // Area of ​​a circle v:real; // volume of the ball begin // 1. Reading the radius value// circles from Edit1.Text R:= StrToFloat(Edit1.Text); // 2. Calculate the area of ​​the circle S:= Pi*R*R; // 3. Calculate the volume of the ball V:= 4/3 * Pi * R * R * R; // 4. Outputting results with precision // 3 decimal places Label2.Caption:="S=" +FloatToStrF(S,ffFixed,8 ,3 ); Label3.Caption:="V=" +FloatToStrF(V,ffFixed,8 ,3 ); end ;

7. Setting the name of the application

To change the name of the application instead of the incomprehensible " Form1", You need to set the value in the Caption property of the main form to" Calculation of the area of ​​a circle and the volume of a ball«.

8. Application execution result

After starting the application (program) for execution, a window is displayed asking you to enter the radius of the circle R . Enter the value 2.5. The window with the result of the program execution is shown in Figure 9.

Rice. 9. Application execution result

Results

When solving this problem, the components of the following types were used:

• TLabel - a component of the "label" type, representing a regular text string for display on the form;
• TButton - a component that represents a regular button on the form;
• TEdit is a component that implements an input line designed to receive information entered by the user from the keyboard.

To design the program interface, the Tool Palette toolbar and the Object Inspector were used.

Also considered are two additional functions that convert a string to a number and vice versa, namely:

• the StrToFloat function, which converts a string representing a number to a real number (for example, '3,678' => 3.678 ) taking into account the regional settings of Windows ;
• function FloatToStrF , which converts a real number into a string form according to the specified format (for example, 2.88 => ‘2,880’ ) taking into account the regional settings of Windows .

where V is the desired ball volume, π - 3.14 , R - radius.

Thus, with a radius of 10 centimeters ball volume equals:

V

3.14×103

= 4186,7

cubic centimeters.

In geometry ball is defined as a certain body, which is a collection of all points in space that are located from the center at a distance not exceeding a given one, called the radius of the ball. The surface of a sphere is called a sphere, and it is formed by rotating a semicircle about its diameter, which remains motionless.

This geometric body is very often encountered by design engineers and architects, who often have to calculate the volume of a sphere. For example, in the design of the front suspension of the vast majority of modern cars, so-called ball bearings are used, in which, as you might guess from the name itself, balls are one of the main elements. With their help, the hubs of the steered wheels and levers are connected. From how right it will be computed their volume largely depends not only on the durability of these units and the correctness of their work, but also on traffic safety.

In technology, such parts as ball bearings are widely used, with the help of which the axles are fastened in the fixed parts of various units and assemblies and their rotation is ensured. It should be noted that when calculating them, designers need find the volume of a sphere(or rather, balls placed in a cage) with a high degree of accuracy. As for the production of metal balls for bearings, they are made from metal wire using a complex technological process that includes the stages of forming, hardening, rough grinding, finishing lapping and cleaning. By the way, those balls that are included in the design of all ballpoint pens are made using exactly the same technology.

Quite often, balls are also used in architecture, and there they are most often decorative elements of buildings and other structures. In most cases, they are made of granite, which often requires a lot of manual labor. Of course, it is not required to observe such a high precision in the manufacture of these balls as those used in various units and mechanisms.

Such an interesting and popular game as billiards is unthinkable without balls. For their production, various materials are used (bone, stone, metal, plastics) and various technological processes are used. One of the main requirements for billiard balls is their high strength and ability to withstand high mechanical loads (primarily shock). In addition, their surface must be an exact sphere in order to ensure smooth and even rolling on the surface of billiard tables.

Finally, not a single New Year or Christmas tree can do without such geometric bodies as balls. These decorations are made in most cases from glass by blowing, and in their production the greatest attention is paid not to dimensional accuracy, but to the aesthetics of the products. At the same time, the technological process is almost completely automated and Christmas balls are only packed manually.

Volume of a ball Theorem The volume of a ball of radius R is equal to 4/3 πR 3 R x B O C M A Proof Consider a ball of radius R centered at point O and choose the axis Ox arbitrarily. The section of the ball by a plane perpendicular to the axis Ox and passing through the point M of this axis is a circle centered at the point M. Let us denote the radius of this circle as R, and its area as S(x), where x is the abscissa of the point M. Express S( x) through x and R. From the right triangle OMC we find R = OC²-OM² = R²-x² Since S (x) = p r ², then S (x) = p (R²-x²). Note that this formula is true for any position of the point M on the diameter AB, i.e., for all x satisfying the condition –R x R. Applying the basic formula for calculating the volumes of bodies with a = –R, b = R, we obtain: R R R R R V = p (R²-x²) dx = p R² dxp - x²dx = p R²x - px³/3 = 4/3 pR³. -R -R -R -R -R Theorem proven x

Volumes of a spherical segment, spherical layer and spherical sector A) A spherical segment is a part of a ball cut off from it by some plane. In Figure 1, the secant plane α, passing through t.B, divides the ball into 2 spherical segments. The circle obtained in the section is called the base of each of these segments, and the lengths of the segments AB and BC of diameter AC, perpendicular to the secant plane, are called the heights of the segments. x АВ=h α О А С Spherical segment Fig.1

If the radius of the ball is equal to R, and the height of the segment is equal to h (in Fig. 1 h =AB), then the volume V of the spherical segment is calculated by the formula: V = ph² (R-1/3h). B) A spherical layer is a part of a ball enclosed between 2 parallel cutting planes (Fig. 2). The circles obtained in the section of the ball by these planes are called the bases of the spherical layer, and the distance between the planes is called the height of the spherical layer. The volume of the spherical layer can be calculated as the difference between the volumes of 2 spherical segments. A B C x Fig.2 Spherical layer

C) A spherical sector is a body obtained by rotating a circular sector with an angle less than 90 degrees around a straight line containing one of the radii limiting the circular sector (Fig. 3). The spherical sector consists of a spherical segment and a cone. If the radius of the ball is equal to R, and the height of the spherical segment is equal to h, then the volume V of the spherical sector is calculated by the formula: V = 2/3 pR² h h O R r Fig.3 Spherical sector

Area of ​​a sphere Unlike the lateral surface of a cylinder or cone, a sphere cannot be unfolded onto a plane, and, therefore, the method of determining and calculating the surface area using a sweep is not suitable for it. To determine the area of ​​the sphere, we use the concept of the circumscribed polyhedron. Let a polyhedron circumscribed near a sphere have n faces. We will increase n indefinitely in such a way that the largest size of each face of the described polyhedra tends to zero. For the area of ​​the sphere, we take the limit of the sequence of areas of surfaces of polyhedra circumscribed around the sphere as the largest size of each face tends to zero => ">