Calculation of the area of ​​a figure bounded by a parametrically defined curve. Calculation of volumes of bodies of rotation using a definite integral Rotation around the axis oy volume

Sections: Mathematics

Lesson type: combined.

The purpose of the lesson: learn to calculate the volumes of bodies of revolution using integrals.

Tasks:

• consolidate the ability to identify curvilinear trapezoids from a number of geometric figures and develop the skill of calculating the areas of curvilinear trapezoids;
• get to know the concept volumetric figure;
• learn to calculate the volumes of bodies of revolution;
• promote the development of logical thinking, competent mathematical speech, accuracy when constructing drawings;
• to cultivate interest in the subject, in operating with mathematical concepts and images, to cultivate will, independence, and perseverance in achieving the final result.

During the classes

I. Organizational moment.

Greetings from the group. Communicate lesson objectives to students.

Reflection. Calm melody.

– I would like to start today’s lesson with a parable. “Once upon a time there lived a wise man who knew everything. One man wanted to prove that the sage does not know everything. Holding a butterfly in his hands, he asked: “Tell me, sage, which butterfly is in my hands: dead or alive?” And he himself thinks: “If the living one says, I will kill her; the dead one will say, I will release her.” The sage, after thinking, replied: "All in your hands". (Presentation.Slide)

– Therefore, let’s work fruitfully today, acquire a new store of knowledge, and we will apply the acquired skills and abilities in future life and in practical activities. "All in your hands".

II. Repetition of previously studied material.

– Let’s remember the main points of the previously studied material. To do this, let's complete the task “Eliminate the extra word.”(Slide.)

(The student goes to I.D. uses an eraser to remove the extra word.)

- Right "Differential". Try to name the remaining words with one common word. (Integral calculus.)

– Let's remember the main stages and concepts associated with integral calculus..

“Mathematical bunch”.

Exercise. Recover the gaps. (The student comes out and writes in the required words with a pen.)

– We will hear an abstract on the application of integrals later.

Work in notebooks.

– The Newton-Leibniz formula was derived by the English physicist Isaac Newton (1643–1727) and the German philosopher Gottfried Leibniz (1646–1716). And this is not surprising, because mathematics is the language spoken by nature itself.

– Let’s consider how this formula is used to solve practical problems.

Example 1: Calculate the area of ​​a figure bounded by lines

Solution: Let's build graphs of functions on the coordinate plane . Let's select the area of ​​the figure that needs to be found.

III. Learning new material.

– Pay attention to the screen. What is shown in the first picture? (Slide) (The figure shows a flat figure.)

– What is shown in the second picture? Is this figure flat? (Slide) (The figure shows a three-dimensional figure.)

– In space, on earth and in Everyday life We encounter not only flat figures, but also three-dimensional ones, but how can we calculate the volume of such bodies? For example, the volume of a planet, comet, meteorite, etc.

– People think about volume both when building houses and when pouring water from one vessel to another. Rules and techniques for calculating volumes had to emerge; how accurate and reasonable they were is another matter.

Message from a student. (Tyurina Vera.)

The year 1612 was very fruitful for the residents of the Austrian city of Linz, where the famous astronomer Johannes Kepler lived, especially for grapes. People were preparing wine barrels and wanted to know how to practically determine their volumes. (Slide 2)

– Thus, the considered works of Kepler laid the foundation for a whole stream of research that culminated in the last quarter of the 17th century. design in the works of I. Newton and G.V. Leibniz of differential and integral calculus. From that time on, the mathematics of variables took a leading place in the system of mathematical knowledge.

– Today you and I will engage in such practical activities, therefore,

The topic of our lesson: “Calculating the volumes of bodies of rotation using a definite integral.” (Slide)

– You will learn the definition of a body of rotation by completing the following task.

“Labyrinth”.

Labyrinth (Greek word) means going underground. A labyrinth is an intricate network of paths, passages, and interconnecting rooms.

But the definition was “broken,” leaving clues in the form of arrows.

Exercise. Find a way out of the confusing situation and write down the definition.

Slide. “Map instruction” Calculation of volumes.

With help definite integral you can calculate the volume of a particular body, in particular, a body of rotation.

A body of revolution is a body obtained by rotating a curved trapezoid around its base (Fig. 1, 2)

The volume of a body of rotation is calculated using one of the formulas:

1. around the OX axis.

2. , if the rotation of a curved trapezoid around the axis of the op-amp.

Each student receives an instruction card. The teacher emphasizes the main points.

– The teacher explains the solutions to the examples on the board.

Let's consider an excerpt from the famous fairy tale by A. S. Pushkin “The Tale of Tsar Saltan, of his glorious and mighty son Prince Guidon Saltanovich and of the beautiful Princess Swan” (Slide 4):

…..
And the drunken messenger brought
On the same day the order is as follows:
“The king orders his boyars,
Without wasting time,
And the queen and the offspring
Secretly throw into the abyss of water.”
There is nothing to do: boyars,
Worrying about the sovereign
And to the young queen,
A crowd came to her bedroom.
They declared the king's will -
She and her son have an evil share,
We read the decree aloud,
And the queen at the same hour
They put me in a barrel with my son,
They tarred and drove away
And they let me into the okiyan -
This is what Tsar Saltan ordered.

What should be the volume of the barrel so that the queen and her son can fit in it?

– Consider the following tasks

1. Find the volume of the body obtained by rotating around the ordinate axis of a curvilinear trapezoid bounded by lines: x 2 + y 2 = 64, y = -5, y = 5, x = 0.

Answer: 1163 cm 3 .

Find the volume of the body obtained by rotating a parabolic trapezoid around the abscissa axis y = , x = 4, y = 0.

IV. Consolidating new material

Example 2. Calculate the volume of the body formed by the rotation of the petal around the x-axis y = x 2 , y 2 = x.

Let's build graphs of the function. y = x 2 , y 2 = x. Schedule y2 = x convert to the form y= .

We have V = V 1 – V 2 Let's calculate the volume of each function

– Now, let’s look at the tower for the radio station in Moscow on Shabolovka, built according to the design of the remarkable Russian engineer, honorary academician V. G. Shukhov. It consists of parts - hyperboloids of rotation. Moreover, each of them is made of straight metal rods connecting adjacent circles (Fig. 8, 9).

- Let's consider the problem.

Find the volume of the body obtained by rotating the hyperbola arcs around its imaginary axis, as shown in Fig. 8, where

cube units

Group assignments. Students draw lots with tasks, draw drawings on whatman paper, and one of the group representatives defends the work.

1st group.

Hit! Hit! Another blow!
The ball flies into the goal - BALL!
And this is a watermelon ball
Green, round, tasty.
Take a better look - what a ball!
It is made of nothing but circles.
Cut the watermelon into circles
And taste them.

Find the volume of the body obtained by rotation around the OX axis of the function limited

Error! The bookmark is not defined.

– Please tell me where we meet this figure?

House. task for 1 group. CYLINDER (slide) .

"Cylinder - what is it?" – I asked my dad.
The father laughed: The top hat is a hat.
To have a correct idea,
A cylinder, let's say, is a tin can.
Steamboat pipe - cylinder,
The pipe on our roof too,

All pipes are similar to a cylinder.
And I gave an example like this -
Kaleidoscope My love,
You can't take your eyes off him,
And it also looks like a cylinder.

- Exercise. Homework: graph the function and calculate the volume.

2nd group. CONE (slide).

Mom said: And now
My story will be about the cone.
Stargazer in a high hat
Counts the stars all year round.
CONE - stargazer's hat.
That's what he is like. Understood? That's it.
Mom was standing at the table,
I poured oil into bottles.
-Where is the funnel? No funnel.
Look for it. Don't stand on the sidelines.
- Mom, I won’t budge.
Tell us more about the cone.
– The funnel is in the form of a watering can cone.
Come on, find her for me quickly.
I couldn't find the funnel
But mom made a bag,
I wrapped the cardboard around my finger
And she deftly secured it with a paper clip.
The oil is flowing, mom is happy,
The cone came out just right.

Exercise. Calculate the volume of a body obtained by rotating around the abscissa axis

House. task for the 2nd group. PYRAMID(slide).

I saw the picture. In this picture
There is a PYRAMID in the sandy desert.
Everything in the pyramid is extraordinary,
There is some kind of mystery and mystery in it.
And the Spasskaya Tower on Red Square
It is very familiar to both children and adults.
If you look at the tower, it looks ordinary,
What's on top of it? Pyramid!

Exercise. Homework: graph the function and calculate the volume of the pyramid

– Volumes different bodies we calculated based on the basic formula for the volumes of bodies using the integral.

This is another confirmation that the definite integral is some foundation for the study of mathematics.

- Well, now let's rest a little.

Find a pair.

Mathematical domino melody plays.

“The road that I myself was looking for will never be forgotten...”

Research work. Application of the integral in economics and technology.

Tests for strong students and mathematical football.

Math simulator.

2. The set of all antiderivatives of a given function is called

A) an indefinite integral,

B) function,

B) differentiation.

7. Find the volume of the body obtained by rotating around the abscissa axis of a curvilinear trapezoid bounded by lines:

D/Z. Calculate the volumes of bodies of revolution.

Reflection.

Reception of reflection in the form syncwine(five lines).

1st line – topic name (one noun).

2nd line – description of the topic in two words, two adjectives.

3rd line – description of the action within this topic in three words.

The 4th line is a phrase of four words that shows the attitude to the topic (a whole sentence).

The 5th line is a synonym that repeats the essence of the topic.

1. Volume.
2. Definite integral, integrable function.
3. We build, we rotate, we calculate.
4. A body obtained by rotating a curved trapezoid (around its base).
5. Body of rotation (volumetric geometric body).

Conclusion (slide).

• A definite integral is a certain foundation for the study of mathematics, which makes an irreplaceable contribution to solving practical problems.
• The topic “Integral” clearly demonstrates the connection between mathematics and physics, biology, economics and technology.
• Development modern science is unthinkable without using the integral. In this regard, it is necessary to begin studying it within the framework of secondary specialized education!

Grading. (With commentary.)

The great Omar Khayyam - mathematician, poet, philosopher. He encourages us to be masters of our own destiny. Let's listen to an excerpt from his work:

You will say, this life is one moment.
Appreciate it, draw inspiration from it.
As you spend it, so it will pass.
Don't forget: she is your creation.

When we figured out the geometric meaning of a definite integral, we came up with a formula that can be used to find the area of ​​a curvilinear trapezoid bounded by the x-axis and straight lines x = a, x = b, as well as a continuous (non-negative or non-positive) function y = f(x). Sometimes it is more convenient to specify the function that limits the figure in parametric form, i.e. express the functional dependence through the parameter t. In this material, we will show how you can find the area of ​​a figure if it is limited by a parametrically defined curve.

After explaining the theory and deriving the formula, we will look at several typical examples to find the area of ​​such figures.

Basic formula for calculation

Let us assume that we have a curvilinear trapezoid, the boundaries of which are the straight lines x = a, x = b, the O x axis and a parametrically defined curve x = φ (t) y = ψ (t), and the functions x = φ (t) and y = ψ (t) are continuous on the interval α; β, α< β , x = φ (t) будет непрерывно возрастать на нем и φ (α) = a , φ (β) = b .

Definition 1

To calculate the area of ​​a trapezoid under such conditions, you need to use the formula S (G) = ∫ α β ψ (t) · φ " (t) d t.

We derived it from the formula for the area of ​​a curvilinear trapezoid S (G) = ∫ a b f (x) d x by substitution method x = φ (t) y = ψ (t):

S (G) = ∫ a b f (x) d x = ∫ α β ψ (t) d (φ (t)) = ∫ α β ψ (t) φ " (t) d t

Definition 2

Taking into account the monotonic decrease of the function x = φ (t) on the interval β; α, β< α , нужная формула принимает вид S (G) = - ∫ β α ψ (t) · φ " (t) d t .

If the function x = φ (t) is not one of the basic elementary ones, then we will need to remember the basic rules for increasing and decreasing a function on an interval to determine whether it will be increasing or decreasing.

In this paragraph we will analyze several problems using the formula derived above.

Example 1

Condition: find the area of ​​the figure formed by the line given by equations of the form x = 2 cos t y = 3 sin t.

Solution

We have parametrically given line. Graphically it can be displayed as an ellipse with two semi-axes 2 and 3. See illustration:

Let's try to find the area 1 4 of the resulting figure, which occupies the first quadrant. The region is in the interval x ∈ a; b = 0 ; 2. Next, multiply the resulting value by 4 and find the area of ​​the whole figure.

Here is the progress of our calculations:

x = φ (t) = 2 cos t y = ψ (t) = 3 sin t φ α = a ⇔ 2 cos α = 0 ⇔ α = π 2 + πk , k ∈ Z , φ β = b ⇔ 2 cos β = 2 ⇔ β = 2 πk , k ∈ Z

With k equal to 0, we get the interval β; α = 0 ; π 2. The function x = φ (t) = 2 cos t will decrease monotonically on it (for more details, see the article on the main elementary functions and their properties). This means that you can apply the formula for calculating the area and find the definite integral using the Newton-Leibniz formula:

- ∫ 0 π 2 3 sin t · 2 cos t " d t = 6 ∫ 0 π 2 sin 2 t d t = 3 ∫ 0 π 2 (1 - cos (2 t) d t = = 3 · t - sin (2 t) 2 0 π 2 = 3 π 2 - sin 2 π 2 2 - 0 - sin 2 0 2 = 3 π 2

This means that the area of ​​the figure given by the original curve will be equal to S (G) = 4 · 3 π 2 = 6 π.

Answer: S(G) = 6π

Let us clarify that when solving the problem above, it was possible to take not only a quarter of the ellipse, but also its half - the upper or lower one. One half will be located on the interval x ∈ a; b = - 2 ; 2. In this case we would have:

φ (α) = a ⇔ 2 cos α = - 2 ⇔ α = π + π k, k ∈ Z, φ (β) = b ⇔ 2 cos β = 2 ⇔ β = 2 π k, k ∈ Z

Thus, with k equal to 0, we get β; α = 0 ; π. The function x = φ (t) = 2 cos t will decrease monotonically on this interval.

After this, we calculate the area of ​​half the ellipse:

- ∫ 0 π 3 sin t · 2 cos t " d t = 6 ∫ 0 π sin 2 t d t = 3 ∫ 0 π (1 - cos (2 t) d t = = 3 · t - sin (2 t) 2 0 π = 3 π - sin 2 π 2 - 0 - sin 2 0 2 = 3 π

It is important to note that you can only take the top or bottom, but not the right or left.

You can create a parametric equation for a given ellipse, the center of which will be located at the origin. It will look like x = a · cos t y = b · sin t . Proceeding in the same way as in the example above, we obtain a formula for calculating the area of ​​the ellipse S e l and p with a = πab.

You can define a circle whose center is located at the origin using the equation x = R · cos t y = R · sin t , where t is a parameter and R is the radius of this circle. If we immediately use the formula for the area of ​​an ellipse, then we will get a formula with which we can calculate the area of ​​a circle with radius R: S k r y r a = πR 2 .

Let's look at one more problem.

Example 2

Condition: find what the area of ​​the figure will be equal to, which is limited by a parametrically defined curve x = 3 cos 3 t y = 2 sin 3 t.

Solution

Let us immediately clarify that this curve has the shape of an elongated astroid. Typically the astroid is expressed using an equation of the form x = a · cos 3 t y = a · sin 3 t .

Now let's look in detail at how to construct such a curve. Let's build based on individual points. This is the most common method and is applicable for most tasks. More complex examples require differential calculus to identify a parametrically defined function.

We have x = φ (t) = 3 cos 3 t, y = ψ (t) = 2 sin 3 t.

These functions are defined for all real values ​​of t. For sin and cos it is known that they are periodic and their period is 2 pi. Having calculated the values ​​of the functions x = φ (t) = 3 cos 3 t, y = ψ (t) = 2 sin 3 t for some t = t 0 ∈ 0; 2 π π 8 , π 4 , 3 π 8 , π 2 , . . . , 15 π 8, we get points x 0; y 0 = (φ (t 0) ; ψ (t 0)) .

Let's make a table of the total values:

t 0	0	π 8	π 4	3 π 8	π 2	5 π 8	3 π 4	7 π 8	π
x 0 = φ (t 0)	3	2 . 36	1 . 06	0 . 16	0	- 0 . 16	- 1 . 06	- 2 . 36	- 3
y 0 = ψ (t 0)	0	0 . 11	0 . 70	1 . 57	2	1 . 57	0 . 70	0 . 11	0

t 0	9 π 8	5 π 4	11 π 8	3 π 2	13 π 8	7 π 4	15 π 8	2π
x 0 = φ (t 0)	- 2 . 36	- 1 . 06	- 0 . 16	0	0 . 16	1 . 06	2 . 36	3
y 0 = ψ (t 0)	- 0 . 11	- 0 . 70	- 1 . 57	- 2	- 1 . 57	- 0 . 70	- 0 . 11	0

After this, mark the required points on the plane and connect them with one line.

Now we need to find the area of ​​that part of the figure that is located in the first coordinate quarter. For it x ∈ a; b = 0 ; 3:

φ (α) = a ⇔ 3 cos 3 t = 0 ⇔ α = π 2 + πk , k ∈ Z , φ (β) = b ⇔ 3 cos 3 t = 3 ⇔ β = 2 πk , k ∈ Z

If k is equal to 0, then we get the interval β; α = 0 ; π 2 , and the function x = φ (t) = 3 cos 3 t will decrease monotonically on it. Now we take the area formula and calculate:

- ∫ 0 π 2 2 sin 3 t · 3 cos 3 t " d t = 18 ∫ 0 π 2 sin 4 t · cos 2 t d t = = 18 ∫ 0 π 2 sin 4 t · (1 - sin 2 t) d t = 18 ∫ 0 π 2 sin 4 t d t - ∫ 0 π 2 sin 6 t d t

We have obtained definite integrals that can be calculated using the Newton-Leibniz formula. Antiderivatives for this formula can be found using the recurrent formula J n (x) = - cos x · sin n - 1 (x) n + n - 1 n J n - 2 (x) , where J n (x) = ∫ sin n x d x .

∫ sin 4 t d t = - cos t · sin 3 t 4 + 3 4 ∫ sin 2 t d t = = - cos t · sin 3 t 4 + 3 4 - cos t · sin t 2 + 1 2 ∫ sin 0 t d t = = - cos t sin 3 t 4 - 3 cos t sin t 8 + 3 8 t + C ⇒ ∫ 0 π 2 sin 4 t d t = - cos t sin 3 t 4 - 3 cos t sin t 8 + 3 8 t 0 π 2 = 3 π 16 ∫ sin 6 t d t = - cos t · sin 5 t 6 + 5 6 ∫ sin 4 t d t ⇒ ∫ 0 π 2 sin 6 t d t = - cos t · sin 5 t 6 0 π 2 + 5 6 ∫ 0 π 2 sin 4 t d t = 5 6 3 π 16 = 15 π 96

We calculated the area of ​​a quarter of a figure. It is equal to 18 ∫ 0 π 2 sin 4 t d t - ∫ 0 π 2 sin 6 t d t = 18 3 π 16 - 15 π 96 = 9 π 16.

If we multiply this value by 4, we get the area of ​​the entire figure - 9 π 4.

In exactly the same way, we can prove that the area of ​​the astroid, given by the equations x = a · cos 3 t y = a · sin 3 t, can be found by the formula S a stroid = 3 πa 2 8, and the area of ​​the figure , which is limited by the line x = a · cos 3 t y = b · sin 3 t , is calculated using the formula S = 3 πab 8 .

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Before moving on to the formulas for the area of ​​a surface of revolution, we will give a brief formulation of the surface of revolution itself. A surface of revolution, or, what is the same thing, a surface of a body of revolution is a spatial figure formed by the rotation of a segment AB curve around the axis Ox(picture below).

Let us imagine a curved trapezoid bounded from above by the mentioned segment of the curve. A body formed by rotating this trapezoid around the same axis Ox, and is a body of revolution. And the area of ​​the surface of revolution or the surface of a body of revolution is its outer shell, not counting the circles formed by rotation around the axis of straight lines x = a And x = b .

Note that a body of revolution and, accordingly, its surface can also be formed by rotating the figure not around the axis Ox, and around the axis Oy.

Calculating the area of ​​a surface of revolution specified in rectangular coordinates

Let in rectangular coordinates on the plane the equation y = f(x) a curve is specified, the rotation of which around the coordinate axis forms a body of rotation.

The formula for calculating the surface area of ​​revolution is as follows:

(1).

Example 1. Find the surface area of ​​the paraboloid formed by rotation around its axis Ox arc of a parabola corresponding to the change x from x= 0 to x = a .

Solution. Let us express explicitly the function that defines the arc of the parabola:

Let's find the derivative of this function:

Before using the formula to find the area of ​​a surface of revolution, let’s write that part of its integrand that represents the root and substitute the derivative we just found there:

Answer: The length of the arc of the curve is

.

Example 2. Find the surface area formed by rotation around an axis Ox astroid.

Solution. It is enough to calculate the surface area resulting from the rotation of one branch of the astroid, located in the first quarter, and multiply it by 2. From the astroid equation, we will explicitly express the function that we will need to substitute into the formula to find the surface area of ​​rotation:

.

We integrate from 0 to a:

Calculation of the area of ​​a surface of revolution specified parametrically

Let us consider the case when the curve forming the surface of revolution is given by parametric equations

Then the surface area of ​​rotation is calculated by the formula

(2).

Example 3. Find the area of ​​the surface of revolution formed by rotation around an axis Oy figure bounded by a cycloid and a straight line y = a. The cycloid is given by parametric equations

Solution. Let's find the intersection points of the cycloid and the straight line. Equating the equation of a cycloid and the equation of a straight line y = a, let's find

It follows from this that the boundaries of integration correspond to

Now we can apply formula (2). Let's find derivatives:

Let's write the radical expression in the formula, substituting the found derivatives:

Let's find the root of this expression:

.

Let's substitute what we found into formula (2):

.

Let's make a substitution:

And finally we find

Trigonometric formulas were used to transform expressions

Answer: The surface area of ​​revolution is .

Calculating the area of ​​a surface of revolution specified in polar coordinates

Let the curve, the rotation of which forms the surface, be specified in polar coordinates.

Let us consider examples of the application of the resulting formula, which allows us to calculate the areas of figures limited by parametrically specified lines.

Example.

Calculate the area of ​​a figure bounded by a line whose parametric equations have the form .

Solution.

In our example, the parametrically defined line is an ellipse with semi-axes of 2 and 3 units. Let's build it.

Let us find the area of ​​the quarter of the ellipse located in the first quadrant. This area lies in the interval . We calculate the area of ​​the entire figure by multiplying the resulting value by four.

What we have:

For k = 0 we get the interval . On this interval the function monotonically decreasing (see section). We apply the formula to calculate the area and find the definite integral using the Newton-Leibniz formula:

Thus, the area of ​​the original figure is equal to .

Comment.

A logical question arises: why did we take a quarter of the ellipse and not half? It was possible to see the upper (or lower) half of the figure. She is in the interval . For this case we would get

That is, for k = 0 we get the interval . On this interval the function monotonically decreasing.

Then the area of ​​half the ellipse is found as

But you won’t be able to take the right or left half of the ellipse.

The parametric representation of an ellipse centered at the origin and semi-axes a and b has the form . If we act in the same way as in the analyzed example, we get formula for calculating the area of ​​an ellipse .

A circle with a center at the origin of radius R is specified through the parameter t by a system of equations. If you use the resulting formula for the area of ​​an ellipse, you can immediately write formula for finding the area of ​​a circle radius R: .

Let's solve one more example.

Example.

Calculate the area of ​​a figure bounded by a curve specified parametrically.

Solution.

Looking ahead a little, the curve is an “elongated” astroid. (Astroid has the following parametric representation).

Let us dwell in detail on the construction of the curve that bounds the figure. We will build it point by point. Typically, such a construction is sufficient to solve most problems. In more complex cases, a detailed study of a parametrically defined function using differential calculus will undoubtedly be required.

In our example.

These functions are defined for all real values ​​of the parameter t, and from the properties of sine and cosine we know that they are periodic with a period of two pi. Thus, calculating the function values ​​for some (For example ), we get a set of points .

For convenience, let's put the values ​​in the table:

We mark the points on the plane and CONSISTENTLY connect them with a line.

Let us calculate the area of ​​the region located in the first coordinate quadrant. For this area .

At k=0 we get the interval , on which the function decreases monotonically. We apply the formula to find the area:

We calculate the resulting definite integrals using the Newton-Leibniz formula, and find the antiderivatives for the Newton-Leibniz formula using a recurrent formula of the form , Where .

Therefore, the area of ​​the quarter figure is , then the area of ​​the entire figure is equal to .

Similarly, it can be shown that astroid area is located as , and the area of ​​the figure bounded by the line is calculated by the formula.