Physical applications of a specific integral operation of variable force. Mechanical applications of a specific integral. Surface area of \u200b\u200brotation

A certain integral (OI) is widely used in practical applications of mathematics and physics.

In particular, in geometry with the help of OI, they find areas of simple figures and complex surfaces, the volumes of rotation and bodies of the arbitrary shape, the lengths of the curves on the plane and in space.

In physics and theoretical mechanics, the OI are used to calculate the static moments, masses and centers of the masses of material curves and surfaces, to calculate the operation of variable force on the curvilinear pathway, etc.

Plaza Flat Figure

Let some flat figure in the Cartesian rectangular coordinate system $ xoy $ is limited to the curve $ y \u003d y_ (1) \\ left (x \\ right) $, below - curve $ y \u003d y_ (2) \\ left (x \\ right) $ , and on the left and right vertical direct $ x \u003d a $ and $ x \u003d b $, respectively. In general, the area of \u200b\u200bsuch a figure is expressed using OI $ s \u003d \\ int \\ limits _ (a) ^ (b) \\ left (y_ (1) \\ left (x \\ right) -y_ (2) \\ left (x \\ right ) \\ Right) \\ CDOT DX $.

If some flat figure in the Cartesian rectangular coordinate system $ xoy $ on the right is limited to the curve $ x \u003d x_ (1) \\ left (y \\ right) $, on the left - curve $ x \u003d x_ (2) \\ left (y \\ right) $, and below and on top of horizontal direct $ y \u003d C $ and $ y \u003d d $, respectively, then the area of \u200b\u200bsuch a figure is expressed using OI $ s \u003d \\ int \\ limits _ (C) ^ (d) \\ left (x_ (1 ) \\ Left (Y \\ Right) -X_ (2) \\ left (Y \\ Right) \\ Right) \\ CDOT DY $.

Let the flat figure (the curvilinear sector), considered in the polar coordinate system, is formed by a chart of a continuous function $ \\ rho \u003d \\ rho \\ left (\\ phi \\ right) $, as well as two rays under the corners of $ \\ phi \u003d \\ alpha $ and $ \\ phi \u003d \\ beta $, respectively. The formula for calculating the area of \u200b\u200bsuch a curved sector has the form: $ s \u003d \\ Frac (1) (2) \\ CDOT \\ INT \\ Limits _ (\\ Alpha) ^ (\\ Beta) \\ RHO ^ (2) \\ Left (\\ Phi \\ Right ) \\ Cdot D \\ Phi $.

Arc Length Krivoy

If on the segment of $ \\ left [\\ alpha, \\; \\ Beta \\ Right] $ The curve is set by the $ \\ rho \u003d \\ rho \\ left equation (\\ phi \\ right) $ in the polar coordinate system, then its arc length is calculated using the $ L \u003d \\ int \\ limits _ (\\ alpha) ^ (\\ beta) \\ sqrt (\\ rho ^ (2) \\ left (\\ phi \\ right) + \\ rho "^ (2) \\ left (\\ phi \\ right)) \\ Cdot D \\ Phi $.

If on the segment of the $ \\ left $ the curve is set by the $ y \u003d y \\ left equation (x \\ right) $, then its arc length is calculated using OI $ L \u003d \\ int \\ Limits _ (a) ^ (b) \\ SQRT (1 + y "^ (2) \\ Left (X \\ Right)) \\ CDOT DX $.

If on the segment of $ \\ left [\\ alpha, \\; \\ Beta \\ Right] $ The curve is set parametrically, that is, $ x \u003d x \\ left (t \\ right) $, $ y \u003d y \\ left (t \\ right) $, then its arc length is calculated using OI $ l \u003d \\ Calculation of body volume on parallel cross sections

Let it be necessary to find the volume of the spatial body, the coordinates of the points of which satisfy the conditions of $ A \\ LE X \\ LE b $, and for which the area of \u200b\u200bthe cross sections of $ s \\ left (x \\ right) $ planes perpendicular to the axis of $ ox $ are known.

The formula for calculating the volume of such a body has a form $ v \u003d \\ int \\ limits _ (a) ^ (b) S \\ Left (X \\ Right) \\ CDOT DX $.

Volume volume of rotation

Let a non-negative continuous function of $ Y \u003d Y \\ left (x \\ right) $ forming a curvilinear trapezium (KTR) are set on a segment of $ \\ left $. If you rotate this CTR around the axis of $ ox $, the body is formed, called the body of rotation.

The calculation of the volume of the rotation body is a special case of calculating the volume of the body according to the well-known areas of its parallel sections. The corresponding formula has a form $ v \u003d \\ int \\ limits _ (a) ^ (b) s \\ left (x \\ right) \\ cdot dx \u003d \\ pi \\ cdot \\ int \\ limits _ (a) ^ (b) y ^ ( 2) \\ Left (X \\ Right) \\ CDOT DX $.

Let some flat figure in the Cartesian rectangular coordinate system $ xoy $ is limited to the curve $ y \u003d y_ (1) \\ left (x \\ right) $, below - curve $ y \u003d y_ (2) \\ left (x \\ right) $ where $ y_ (1) \\ left (x \\ right) $ and $ y_ (2) \\ left (x \\ right) $ is non-negative continuous functions, and on the left and right vertical straight $ x \u003d a $ and $ x \u003d b $ respectively. Then the volume of the body formed by the rotation of this figure around the axis $ ox $ is expressed by the $ V \u003d \\ PI \\ CDOT \\ INT \\ Limits _ (a) ^ (b) \\ left (y_ (1) ^ (2) \\ left (x \\ Right) -y_ (2) ^ (2) \\ left (X \\ Right) \\ Right) \\ CDOT DX $.

Let some flat figure in the decartular rectangular coordinate system $ xoy $ on the right limited curve $ x \u003d x_ (1) \\ left (y \\ right) $, on the left - curve $ x \u003d x_ (2) \\ left (y \\ right) $ , where $ x_ (1) \\ left (y \\ right) $ and $ x_ (2) \\ left (y \\ right) $ is non-negative continuous functions, and below and from above horizontal direct $ y \u003d C $ and $ y \u003d D $, respectively. Then the volume of the body formed by the rotation of this figure around the axis $ oy $ is expressed as $ v \u003d \\ pi \\ cdot \\ int \\ limits _ (C) ^ (d) \\ left (x_ (1) ^ (2) \\ left (y \\ Right) -X_ (2) ^ (2) \\ Left (Y \\ Right) \\ Right) \\ CDot DY $.

Surface surface of the body of rotation

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Let the $ \\ left $ be given a non-negative function $ y \u003d y \\ left (x \\ right) $ with a continuous derivative of $ y "\\ left (x \\ right) $. This feature forms the CRT. If you rotate this CTT around the axis $ ox $, then she sacrows the body of rotation, and the arc of the CRT is its surface. The surface area of \u200b\u200bsuch a rotation body is expressed by the $ Q \u003d 2 \\ CDOT \\ PI \\ CDOT \\ int \\ Limits _ (a) ^ (b) y \\ left (a) ^ (b) y \\ left ( X \\ Right) \\ Cdot \\ sqrt (1 + y "^ (2) \\ left (x \\ right)) \\ CDOT DX $.

Suppose that the curve is $ x \u003d \\ phi \\ left (Y \\ Right) $, where $ \\ phi \\ left (y \\ right) $ - set on the segment $ C \\ le y \\ le d $ non-negative function, rotate around the axis $ Oy $. In this case, the surface area of \u200b\u200bthe formed rotation body is expressed as $ Q \u003d 2 \\ CDOT \\ PI \\ CDOT \\ INT \\ Limits _ (C) ^ (D) \\ Phi \\ Left (Y \\ Right) \\ CDOT \\ SQRT (1+ \\ PHI "^ (2) \\ Left (Y \\ Right)) \\ CDOT DY $.

Physical applications OI

To calculate the path traveled at the time of time $ T \u003d T $, at a variable speed of the $ V \u003d V \\ left (t \\ right) of the material point, which started movement at the time of $ T \u003d T_ (0) $, UI $ s \u003d \\ int \\ Limits _ (T_ (0)) ^ (T) V \\ Left (T \\ Right) \\ CDOT DT $.
To calculate the operation of the variable, the $ F \u003d F \\ Left (X \\ Right) $ applied to the material point moving along the straight path along the $ ox $ axis from the point $ x \u003d a $ to the point $ x \u003d b $ (direction of force Coincides with the direction of movement) use OI $ a \u003d \\ int \\ limits _ (a) ^ (b) F \\ Left (x \\ Right) \\ CDOT DX $.
Static moments relative to the coordinate axes of the material curve $ y \u003d y \\ left (x \\ right) $ on the range $ \\ left $ are expressed by $ m_ (x) \u003d \\ rho \\ cdot \\ int \\ limits _ (a) ^ (b) y \\ left (x \\ right) \\ Cdot \\ SQRT (1 + y "^ (2) \\ left (x \\ right)) \\ CDOT DX $ and $ m_ (y) \u003d \\ rho \\ cdot \\ int \\ limits _ (a ) ^ (b) x \\ cdot \\ sqrt (1 + y "^ (2) \\ left (x \\ right)) \\ CDOT DX $, where the linear density of the $ \\ rho $ of this curve is considered constant.
The center of the mass material curve is a point in which the entire mass is conventionally focused in such a way that the static moments of the point relative to the coordinate axes are equal to the corresponding static moments of the entire curve in general.

Formulas for calculating the coordinates of the center of mass of a flat curve have a form $ x_ (C) \u003d \\ FRAC (\\ int \\ limits _ (a) ^ (b) x \\ cdot \\ sqrt (1 + y "^ (2) \\ left (x \\ \u003d \\ FRAC (\\ int \\ limits _ (a) ^ (b) y \\ left (x \\ right) \\ cdot \\ sqrt (1 + y "^ (2) \\ left (x \\ right)) \\ CDOT DX) ( \\ int \\ limits _ (a) ^ (b) \\ sqrt (1 + y "^ (2) \\ left (x \\ right)) \\ CDOT DX) $.

The static moments of the material flat figure in the form of the KRT relative to the coordinate axes are expressed by the formulas $ M_ (x) \u003d \\ FRAC (1) (2) \\ CDOT \\ RHO \\ CDOT \\ INT \\ Limits _ (a) ^ (b) Y ^ (2) \\ Left (X \\ Right) \\ CDOT DX $ and $ M_ (Y) \u003d \\ Rho \\ Cdot \\ int \\ limits _ (a) ^ (b) X \\ CDOT Y \\ Left (X \\ Right) \\ CDOT DX $.
The coordinates of the center of the masses of the material flat figure in the form of a CRT formed by the curve $ y \u003d y \\ left (x \\ right) $ in the range of $ \\ left $ are calculated by $ x_ (C) \u003d \\ FRAC formulas (\\ int \\ limits _ (a ) ^ (b) x \\ cdot y \\ left (x \\ right) \\ cdot dx) (\\ int \\ limits _ (a) ^ (b) y \\ left (x \\ right) \\ CDOT DX) $ and $ y_ ( C) \u003d \\ FRAC (\\ FRAC (1) (2) \\ Cdot \\ int \\ limits _ (a) ^ (b) y ^ (2) \\ left (x \\ right) \\ CDOT DX) (\\ int \\ limits _ (a) ^ (b) Y \\ Left (X \\ Right) \\ CDOT DX) $.

Topic 6.10. Geometric and physical applications of a specific integral

1. The area of \u200b\u200bthe curvilinear trapezion bounded by the curve y \u003d f (x) (F (x)\u003e 0), straight x \u003d a, x \u003d b and the segment [a, b] axis oh, is calculated by the formula

2. The area of \u200b\u200bthe figure bounded by curves y \u003d f (x) and y \u003d g (x) (F (x)< g (x)) и прямыми х= a , x = b , находится по формуле

3. If the curve is set by parametric equations x \u003d x (t), y \u003d y (t), then the area of \u200b\u200ba curvilinear trapezium bounded by this curve and straight x \u003d a, x \u003d b, is by the formula

4. Let s (x) be the area of \u200b\u200bthe body cross section of the plane, perpendicular to the axis OH, then the volume of the body part concluded between the perpendicular axes of the x \u003d a and x \u003d b axis is by the formula

5. Let the curvilinear trapezion, limited by the curve y \u003d f (x) and the straight y \u003d 0, x \u003d a and x \u003d b, rotates around the axis oh, then the volume of the rotation body is calculated by the formula

6. Let a curvilinear trapezoid, bounded by curve x \u003d g (y) and

straight x \u003d 0, y \u003d c and y \u003d d, rotates around the axis about y, then the volume of the rotational body is calculated by the formula

7. If the flat curve is attributed to the rectangular coordinate system and is set by the Y \u003d F (x) equation (or x \u003d f (y)), then the length of the arc is determined by the formula

1. Flat Figure area.

The area of \u200b\u200bthe curvilinear trapezium bounded by nonnegative function f (X), abscissa and direct x \u003d A., x \u003d B.is defined as S \u003d ∫ A B f x D x.

Square of curvilinear trapezium

Figure area limited function f (X)intersecting the abscissa axis is determined by the formula s \u003d σ i: f x ≥ 0 ∫ x i - 1 x i f x d x - σ i: f x< 0 ∫ x i - 1 x i | f x | d x , где x I. - zero functions. In other words, in order to calculate the area of \u200b\u200bthis figure, you need to split the segment zeros function f (X) On the part, integrate the function f. For each of the resulting gaps of the alternation, add separately integrals by segments on which the function f. Takes different signs, and subtract from the first second.

2. The area of \u200b\u200bthe curvilinear sector.

Curved sector area Consider the curve ρ = ρ (φ) in the polar coordinate system where ρ (φ) - continuous and non-negative on [α; β] function. Figure limited curve ρ (φ) and rays φ = α , φ = β is called the curvilinear sector. The area of \u200b\u200bthe curvilinear sector is s \u003d 1 2 ∫ α β ρ 2 φ d φ.

3. Scope of rotation.

Volume volume of rotation

Let the body formed by rotation around the axis of the Ox of a curvilinear trapezion bounded by continuous on the segment Function f (X). Its volume is expressed by the formula V \u003d π ∫ A B f 2 x D x.

To the problem of finding the volume of the body on the cross-sectional area

Let the body are concluded between the planes x \u003d A. and x \u003d B., and its cross-section area with a plane passing through the point x.- continuous on the segment function σ (x). Then its volume is V \u003d ∫ A B Σ x D x.

4. The length of the arc curve.

Let the curve R → T \u003d X T, Y T, Z T, then the length of its site limited by the values t \u003d α. and t \u003d β. It is expressed by the formula s \u003d ∫ α β x 't 2 + y' t 2 + z 't 2 dt.

The length of the arc flat curve in particular, the length of the flat curve defined on the coordinate plane Oxy equation y \u003d f (x), a ≤ x ≤ B, It is expressed by the formula S \u003d ∫ A B 1 + F 'x 2 DX.

5. Surface area of \u200b\u200brotation.

Surface surface area Suppose the surface is set by rotation relative to the OX axis of the function graphics y \u003d f (x), a ≤ x ≤ B, and function f. It has a continuous derivative on this segment. Then the area of \u200b\u200bthe rotation surface is determined by the formula π \u003d 2 π ∫ a b f x 1 + f 'x 2 d x.

The area of \u200b\u200bthe curvilinear trapezium, limited from the top of the function schedule y \u003d f (x)left and right - straight x \u003d A. and x \u003d B. respectively, from below - axis OX., calculated by the formula

The area of \u200b\u200bthe curvilinear trapezium, limited to the right schedule of the function x \u003d φ (y), top and bottom - straight y \u003d D. and y \u003d C. Accordingly, on the left - the axis Oy.:

The area of \u200b\u200bthe curvilinear figure limited from the top of the function graph y 2 \u003d F 2 (x), from below - the graph of the function y 1 \u003d F 1 (x)left and right - straight x \u003d A. and x \u003d B.:

The area of \u200b\u200bthe curvilinear figure limited to the left and right graphs of functions x 1 \u003d φ 1 (y) and x 2 \u003d φ 2 (y), top and bottom - straight y \u003d D. and y \u003d C. respectively:

Consider the case when a line that limits the curvilinear trapezium from above is set by parametric equations x \u003d φ 1 (t), y \u003d φ 2 (t)where α ≤ t ≤ β, φ 1 (α) \u003d a, φ 1 (β) \u003d b. These equations define some function y \u003d f (x) On the segment [ a, B.]. The area of \u200b\u200bthe curvilinear trapezium is calculated by the formula

Moving to the new variable x \u003d φ 1 (t), then dx \u003d φ "1 (t) dt, but y \u003d f (x) \u003d f (φ 1 (t)) \u003d φ 2 (t), therefore, \\ begin (displaymath)

Area in polar coordinates

Consider the curvilinear sector OAB, limited by a line given by the equation ρ=ρ(φ) in polar coordinates, two rays Oa. and OB.for which φ=α , φ=β .

Select break on elementary sectors Om k-1M k ( k \u003d 1, ..., n, M 0 \u003d A, M n \u003d b). Denote by Δφ K. The angle between the rays Om k-1 and Om K.forming the polar axis of the corner φ k-1 and φ K. respectively. Each of the elementary sectors Om k-1 m k Replace the circular sector with a radius ρ k \u003d ρ (φ "k)where φ "K. - The value of the corner φ From the interval [ φ k-1, φ k], and the central angle Δφ K.. The area of \u200b\u200bthe last sector is expressed by the formula .

expresses the area of \u200b\u200bthe "step" sector approximately replacing this sector OAB.

Sector Sector OAB called the limit of the "speed" sector at n → ∞. and λ \u003d max Δφ k → 0:

As T.

Arc Length Krivoy

Let on the segment [ a, B.] Differential function is set y \u003d f (x), whose schedule is an arc. Section [ a, B.] Thrumspace by n. pieces of points x 1, x 2, …, x N-1. These points will correspond to the point M 1., M 2., …, M n-1 Arcs, connect their broken line, which is called the broken, inscribed in the arc. Perimeter of this broken is denoted by s N., i.e

Definition. The length of the line of the line is called the limit of the perimeter inscribed into it broken when the number of links M k-1 m k It is unlimited, and the length of the greatest of them is striving for zero:

where λ is the length of the greatest link.

We will count the length of the arc from some of its point, for example, A.. Let in the point M (X, Y) The length of the arc is equal s., and at the point M "(X + Δ X, Y + ΔY) The length of the arc is equal s + ΔS.where, I\u003e ΔS - the length of the arc. From triangle MNM " We find the length of the chord :.

From geometrical considerations it follows that

that is, the infinitely small arc lines and the tightering of her chord is equivalent.

We transform the formula expressing the length of the chord:

Turning to the limit in this equality, we obtain a formula for a derivative function s \u003d S (x):

from which we find

This formula expresses an arc differential to a flat curve and has a simple geometrical meaning: expresses the theorem of Pythagora for an infinitely small triangle MTN. (dS \u003d MT., ).

Spatial curve arc differentials determined by the formula

Consider the arc of the spatial line given by parametric equations

where α ≤ t ≤ β, φ i (t) (i \u003d 1, 2, 3) - Differential argument functions t.T.

Integrating this equality by interval [ α, β ], we get a formula for calculating the length of this line arc

If the line lies in the plane OxyT. z \u003d 0. at all t∈ [α, β], so

In the case when the flat line is set by the equation y \u003d f (x) (a≤x≤b.), where f (x) - differentiable function, the last formula takes the form

Let the flat line be set by the equation ρ=ρ(φ) (α≤φ≤β ) In polar coordinates. In this case, we have parametric line equations x \u003d ρ (φ) cos φ, y \u003d ρ (φ) sin φwhere the polar angle is taken as a parameter φ . Insofar as

then the formula expresses the length of the arc line ρ=ρ(φ) (α≤φ≤β ) in polar coordinates, has the form

Body volume

Find the volume of the body if the area of \u200b\u200bany cross-section of this body is known perpendicular to some direction.

We divide this body to elementary layers with planes perpendicular to the axis OX. and defined equations x \u003d const. For any fixed x∈. Known Square S \u003d S (x) cross section of this body.

Elementary layer cut off with planes x \u003d x k-1, x \u003d x k (k \u003d 1, ..., n, x 0 \u003d a, x n \u003d b), replace the cylinder with a height Δx k \u003d x k -x k-1 and foundation area S (ξ k), ξ k ∈.

The volume of the specified elementary cylinder is expressed by the formula ΔV k \u003d e (ξ k) Δx k. Make up the amount of all such works

the integrated amount for this function S \u003d S (x) On the segment [ a, B.]. It expresses the volume of a stepped body consisting of elementary cylinders and approximately replacing this body.

The volume of this body is called the limit of the indicated stepped body when λ→0 where λ - the length of the greatest of elementary segments ΔX K.. Denote by V. the volume of this body, then by definition

On the other hand,

Consequently, the volume of the body according to the specified cross sections is calculated by the formula

If the body is formed by rotation around the axis OX. curvilinear trapezium limited from top of a continuous arc y \u003d f (x)where a≤x≤b.T. S (x) \u003d πF 2 (x) And the last formula takes the form:

Comment. The volume of the body obtained by the rotation of the curvilinear trapezium limited to the right schedule x \u003d φ (y) (c ≤ x ≤ D), around the axis Oy. Calculated by formula

Surface area of \u200b\u200brotation

Consider the surface obtained by the rotation of the line arc y \u003d f (x) (a≤x≤b.) Around the axis OX. (Suppose that the function y \u003d f (x) It has a continuous derivative). Fix value x∈., the argument of the function will give increment dX.which corresponds to the "elementary ring" obtained by the rotation of the elementary arc ΔL.. This is a "ring" by replacing the cylindrical ring - the side surface of the body formed by the rotation of the rectangle with the base equal to the Differential of the arc dL, and high h \u003d f (x). Cutting the last ring and turning it, we get a strip width dL and length 2πywhere y \u003d f (x).

Consequently, the differential surface area will express the formula

This formula expresses the surface area obtained by the rotation of the line arc y \u003d f (x) (a≤x≤b.) Around the axis OX..

Home\u003e Lecture

Lecture 18. Applications of a specific integral.

18.1. Calculation of plane patterns.

It is known that a certain integral on the segment is an area of \u200b\u200ba curvilinear trapezoid, limited by a graph of the function f (x). If the schedule is located below the axis oh, i.e. f (x)< 0, то площадь имеет знак “-“, если график расположен выше оси Ох, т.е. f(x) > 0, then the area has a "+" sign.

To find the total area, the formula is used.

The area of \u200b\u200bthe figure, bounded by some lines, can be found with certain integrals, if the equations of these lines are known.

Example. Find the area of \u200b\u200bthe figure limited to the lines y \u003d x, y \u003d x 2, x \u003d 2.

The desired area (shaded in the figure) can be found by the formula:

18.2. Finding the area of \u200b\u200bthe curvilinear sector.

To find the area of \u200b\u200bthe curvilinear sector, we introduce the polar coordinate system. The equation of the curve limiting the sector in this coordinate system has the form  \u003d f (), where  - the length of the radius - the vector connecting the pole with an arbitrary point of the curve, and  is the angle of inclination of this radius - vector to the polar axis.

The area of \u200b\u200bthe curvilinear sector can be found by the formula

18.3. Calculation of the length of the arc curve.

y y \u003d f (x)

s i y i

Length of a broken line that matches the arc can be found as
.

Then the length of the arc is equal
.

From geometrical considerations:

In the same time

Then you can show that

Those.

If the equation of the curve is set to parametrical, then taking into account the rules for calculating the derivative of parametrically specified, we get

where x \u003d  (t) and y \u003d  (t).

If set spatial curve, and x \u003d  (t), y \u003d  (t) and z \u003d z (t), then

If the curve is set in polar coordinatesT.

,  \u003d f ().

Example: Find the circumference length given by the equation x 2 + y 2 \u003d R 2.

1 way. Express the variable from the equation.

Find a derivative

Then s \u003d 2r. Received a well-known formula of the circumference length.

2 way. If you represent a given equation in the polar coordinate system, we obtain: R 2 COS 2  + R 2 SIN 2  \u003d R 2, i.e. function  \u003d f () \u003d R,
then

18.4. Calculation of volumes of bodies.

Calculation of body volume according to the well-known areas of its parallel sections.

Let there be a body of volume V. The area of \u200b\u200bany cross-section of the body q is known as a continuous function Q \u003d Q (X). We break the body on the "layers" by cross-sections passing through points x I splitting the segment. Because On the intermediate section of the partition, the function q (x) is continuous, then it takes the largest and smallest values \u200b\u200bon it. Denote them, respectively, M I and M i.

If we build cylinders with forming, parallel axes on these largest sections, then the volumes of these cylinders will be respectively equal to M I x I and M I X I here x i \u003d x i - x i -1.

By producing such constructions for all sections of the partition, we obtain cylinders whose volumes are equal, respectively
and
.

When striving for zero partitioning step , these amounts have a shared limit:

Thus, the volume of the body can be found by the formula:

The disadvantage of this formula is that in order to find the volume you need to know the function Q (x), which is very problematic for complex tel.

Example: Find the volume of the ball of radius R.

In the transverse sections of the ball, the circles of the variable radius are obtained. Depending on the current coordinates x, this radius is expressed by the formula
.

Then the function of cross sections has the form: Q (X) \u003d
.

Get a ball:

Example: Find the volume of arbitrary pyramid with a height of H and an area of \u200b\u200bS.

When crossing the pyramid with planes, perpendicular height, in the section we obtain figures similar to the base. The likeness ratio of these figures is equal to the X / H ratio, where x is the distance from the sequence plane to the top of the pyramid.

From geometry it is known that the ratio of areas of such figures is equal to the likeness of the likeness in the square, i.e.

From here we get the function of the areas of sections:

We find the volume of the pyramid:

18.5. Volume of rotation bodies.

Consider the curve specified by the equation y \u003d f (x). Suppose that the function f (x) is continuous on the segment. If the corresponding curvilinear trapezium with the bases a and b rotate around the axis oh, then we get the so-called body of rotation.

y \u003d f (x)

Because Each body cross section plane X \u003d const is a circle of radius
The volume of the body of rotation can be easily found according to the formula obtained above:

18.6. Surface surface of the body of rotation.

M I B.

Definition: Surface surface area The AV curve around this axis is called the limit to which the area of \u200b\u200bthe spinning surfaces of the broken, inscribed in the AV curve, with the desire to zero the largest lengths of the links of these broken.

We break an arc of AV on N parts by points m 0, m 1, m 2, ..., m n. The coordinates of the vertices of the resulting broken are coordinates x i and y i. When the broken breaks around the axis, we obtain the surface consisting of the side surfaces of truncated cones, the area of \u200b\u200bwhich is equal to p i. This area can be found by the formula:

Here s i is the length of each chord.

Apply Lagrange Theorem (see Lagrange Theorem) To relation
.