What is the radius of the ball? Ball volume

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

Mathematical formulation of the problem

Before starting to develop an 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 output results.

The area of ​​a circle is calculated using 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 the ball 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 solving the problem

Performance

1. Creating an application of the VCL Form Application type.

Launch a 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 starting design is shown in Figure 2.

Rice. 2. View of the program window

2. Standard tab of the Tool Palette.

This application requires the use of several components, which are listed below:

• component type TLabel, representing a line of text that is displayed on the form;
• component type TButton, representing a button on the form;
• component type TEdi t , which is the text input string.

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

Rice. 3. Standard tab on the component palette

3. TLabel component

3.1. Placing a TLabel Component on a Form

To do this, you need to 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 a TLabel component, you must first select it using the mouse or by selecting it in the Object Inspector panel. After this, 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 when writing program code, the properties of this component can be accessed with the prefix “Label“. For example, in order to change the Caption property in a program, you need to type the following line:

In the same way, we place components on the form with the names Label2 and Label3 just below the previous component. Set the Caption property values ​​to “S = ” and “V = “, respectively.

The application form should look something like this (Fig. 7).

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

Transferring and processing all other components from the Tool Palette is carried out in the same way.

4. TEdit component

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

Clear the Text property of the component.

5. TButton Component

We add a TButton component from the Tool Palette, 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.

Set the Caption property of the component to the value “Calculate“.

The application form in design mode will look like shown in Fig. 8.

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

6. Programming a click event on the “Calculate” button

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 a button is called OnClick.

Delphi 2010 automatically creates a piece of program code in which you need to enter your own event processing code. The code generated by the system looks like:

The first task is to determine the inputs, outputs, or intermediate variables that will be used in the program.

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

• R – radius of the circle;
• S - area of ​​a circle;
• V – volume of the ball.

All variables must be of real type.
The program also uses one constant - the number Pi. Let's denote it by the name Pi. It should be noted 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:

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

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

the x value will become -3.675.

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

In paragraph 4, the results are displayed. 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 to a string. For example, to convert the number 2.87 to a string with a precision of 4 decimal places, you need to write:

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

The general listing of the procedure for processing the OnClick event of the Button1 component looks like this:

7. Setting the application name

To change the name of the application instead of the incomprehensible “Form1” you need to set 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 launching 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

To solve this problem, the following types of components were used:

• TLabel is a “label” type component that represents a regular text string for display on the form;
• TButton - a component representing a regular button on a 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, we used the Tool Palette and the Object Inspector.

We also consider two additional functions that convert a string to a number and back, namely:

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

The radius of a ball (denoted as r or R) is the segment that connects the center of the ball with any point on its surface. As with a circle, the radius of a ball is an important quantity 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 diameter, circumference and other quantity. Use a formula into which you can substitute these values.

Steps

Formulas for calculating radius

Calculate the radius from the diameter. The radius is equal to half the diameter, so use the formula g = D/2. This is the same formula that is 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. Use the formula: r = C/2π. Since the circumference of a circle 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. Isolating r on one side of the equation, you get the formula ((V/π)(3/4)) 3 = r, that is, to calculate the radius, divide the volume of the ball by π, multiply the result by 3/4, and raise the resulting result to a power 1/3 (or take the cube root).

   • For example, given a ball with a volume of 100 cm 3 . The radius of this ball 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: g = √(A/(4 π)). The surface area of ​​the ball is calculated by the formula A = 4πr 2. Isolating r on one side of the equation gives you the formula √(A/(4π)) = r, which is to calculate the radius by taking 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 ball is calculated as follows:
     • √(A/(4π)) = r
     • √(1200/(4π)) = r
     • √(300/(π)) = r
     • √(95.49) = r
     • 9.77 cm= r

   Determination of basic quantities

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

      Use the values ​​of these quantities to find the radius. Radius can be calculated from given values ​​of diameter, circumference, volume, and surface area. Moreover, the indicated values ​​can be found from a given radius value. To calculate the radius, simply convert the formulas to find the values ​​shown. Below are the formulas (which include radius) for calculating 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 ball is equal to the distance between its center and any point lying on the surface of the ball. 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 a ball is a three-dimensional figure, the point will have three coordinates (x, y, z), rather than two (x, y).

      • Let's look at an example. Given a ball with center coordinates (4,-1,12) . Use these coordinates to find the radius of the ball.

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

      • In our example, let us 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 = √((x 2 - x 1) 2 + (y 2 - y 1) 2 + (z 2 - z 1) 2). Having found out the coordinates of the center of the ball and a 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 = √((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) – coordinates of the center of the ball, (x 2 , y 2 , z 2) – coordinates of a point lying on the surface of the ball.

      • In the example under consideration, 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 = √((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 ball’s center and the coordinates (x 2,y 2,z 2 ) any point lying on the surface of the ball.

      • 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 with its center at coordinates (0,0,0).
   • Don't forget about the order of performing mathematical operations. If you don't remember this order, and your calculator can 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 using a known radius value.
   • π (Pi) is a letter of the Greek alphabet that denotes 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 Pi to within four decimal places. As a rule, they use the approximate value of Pi, which is 3.14.

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 with a center at point O and choose the Ox axis arbitrarily. A section of a ball by a plane perpendicular to the Ox axis and passing through point M of this axis is a circle with a center at point M. Let us denote the radius of this circle by R, and its area by S(x), where x is the abscissa of point M. Let us express S( x) through x and R. From the right triangle OMC we find R = OC²-OM² = R²-x² Since S (x) = n r², then S (x) = n (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 The theorem is proven x

Volumes of a spherical segment, spherical layer and spherical sector A) A spherical segment is a part of a sphere cut off from it by some plane. In Figure 1, the secant plane α, passing through point 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 segments AB and BC of diameter AC perpendicular to the cutting plane are called the heights of the segments. x AB=h α O A C Ball 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 = рh² (R-1/3h). · B) The spherical layer is the part of the 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 the height of the spherical layer. The volume of the spherical layer can be calculated as the difference in the volumes of two spherical segments. A B C x Fig. 2 Ball 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 Ball sector

Area of ​​a Sphere Unlike the lateral surface of a cylinder or cone, a sphere cannot be turned onto a plane, and, therefore, the method of determining and calculating the surface area using a development is not suitable for it. To determine the area of ​​a sphere, we use the concept of a circumscribed polyhedron. Let a polyhedron described around a sphere have n faces. We will increase n without limit in such a way that the largest size of each face of the described polyhedra tends to zero. For the area of ​​a sphere we take the limit of the sequence of surface areas of polyhedra described around the sphere as the largest size of each face tends to zero => ">

Formulas

CYLINDER VOLUME

CONE VOLUME

VOLUME OF A TRUNCUMATED CONE

BALL VOLUME

V=1/3∏H(R2+r2+Rr)

V=4/3 ∙ ∏R 3

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

• The area of ​​the 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 .

• Every section of a ball by a plane is a circle. The center of this circle is the …………………… perpendicular dropped from the center of the ball onto the secant plane.

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

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

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

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

6. A sphere can be described around any regular pyramid, with its center lying on ……………….. of the pyramid.

base

center

circle

circle

equal

height

Theoretical dictation

Option 2

plane

circle

height

perpendicular

touch

height

Card No. 1

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

288 P cm³

Card No. 2

Two equal balls are positioned so that the center of one lies on the surface of the other. How does the volume of the total part of the balls relate to the volume of the whole ball?

5 / 16

Card No. 3

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

Task No. 1

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

Task No. 2

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

WRITE QUICKLY AND BRIEFLY ANSWERS TO THE QUESTIONS:

• How many spheres can be drawn:

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) square;

b) isosceles trapezoid;

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

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

5. How should two equal circles be positioned 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 diametrical plane of a ball is its ………………… symmetry.

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

3. The center of a sphere circumscribed about a regular pyramid lies on …………………. pyramids.

4. The radius of the sphere drawn to the point of contact of 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 into any regular pyramid, with its center lying on the ……………… .…….pyramid.

plane

circle

height

perpendicular

touch

height

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 3cm has its center at point O (4;-2;1). Write an equation for the sphere into which this sphere will go when it is symmetrical about the OXY plane. Find the volume of a sphere bounded by a given sphere.

Level 1 Option 2

1. Through a point lying on the 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 point O (-2;5;3). Write an equation for the sphere into which this sphere will go when it is symmetrical about the OX Z plane. Find the area of ​​this sphere.

Test independent work level 52

Level2 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 equal to 4 cm, subtending an angle of 90°. Find the volume of the sphere and its surface area.

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

Level2 Option 2

1. A section is made at a distance of 4 cm from the center of the ball. A chord distant from the center of this section by √5 cm, subtending an angle of 120°. Find the volume of the sphere and its surface area.

2. A sphere with center at point O (-1;-2;2) passes through the origin. Write an equation for the sphere into which this sphere will go when it is symmetric about the plane Z = 1. Find the area of ​​the sphere.

Independent work

Option 2

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

2. A volleyball has a radius of 12 cm. What volume of air is contained in the ball?

Option 1

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

2. A soccer ball has a diameter of 30 dm. What volume of air is contained 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 ball and its parts.
• Solve problems :

№ 1. The volume of the sphere is 36 Psm³. Find the area of ​​the sphere enclosing this ball.

№ 2. A sphere of radius 15 cm has a section whose area is 81 cm². Find the volume of the smaller spherical segment cut off by the cutting plane.

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

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

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

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

Solving problems with self-test.

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 base = 60 cm, R ball = 75 cm.

Find: Vspherical segment.

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

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

Answer: 58500π.

Reflection

Express your mood with an emoticon.

Take a smiley face that matches your mood at the end of the lesson and, when leaving, attach it to the board with a magnetic base.

Homework

• Homework

• Repeat the formulas for the volumes of a sphere, a spherical segment, a spherical layer, and a spherical sector. No. 723, No. 724, No. 755

Literature and Internet resources

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

Gavrilova N.F. Lesson developments in geometry grade 11