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Fig. 2.1

Transferring to the repeated integral we get the final formula for calculation of a triple integral

 

b

2 ( x)

f2

( x; y)

f (x; y; z)dz

 

f (x; y; z)dV = dx

dy

 

 

(2.2)

 

 

 

 

 

 

V

a

1 ( x)

f1 ( x; y )

 

 

Remarks.

1.If the region V is complex, then it should be splitted by finite number of correct regions to which the formula (2.2) is applied.

2.The order of integrating in the formula (2.2) can be another.

Example 2.1. Calculate the triple integral

xdxdydz V

where the region

V is bounded by the surfaces = 0, = 0, = 0, + = 2, = 5. Solution. We construct given surfaces in the Oxyz coordinate space (fig.

2.2).

Fig. 2.2

31

The projection of the region V onto xy-plane is the rectangle. So, the region V can be given by the system of inequalities:

0 ≤ ≤ 2, { 0 ≤ ≤ 5,

0 ≤ ≤ 2 − .

So, we have

 

 

 

2

 

 

5

2x

 

 

 

2

 

 

5

 

2 x

 

2

 

 

 

 

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

xdxdydz = xdx dy dz =

xdx z

0

 

dy = x(2 x)dx dy =

V

 

 

0

 

 

0

0

 

 

 

0

 

 

0

 

 

 

 

0

 

 

 

 

0

2

 

 

 

 

 

5

2

 

 

 

5

 

 

 

 

 

 

 

2

 

 

 

 

 

 

= x(2 x)dx dy = y

 

0

x(2

x)dx

= 5 (2x x2 )dx =

 

0

 

 

 

 

 

0

0

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

x

2

 

x

3

2

 

 

 

 

 

x

3

2

 

 

 

 

2

 

 

 

0

 

4

= 5 2

 

 

 

 

 

=5 x

2

 

 

 

 

 

=5 2

2

 

0

2

 

= .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

3 0

 

 

 

 

 

3 0

 

 

 

 

3

 

 

 

3

3

2.3.The Calculation of the Triple Integral in the Cylindrical Coordinates and in the Spherical Coordinates

The calculation of the triple integral, as and the double integral is often carried out by the substitution method that by the variable conversion.

Let the substitution = ( ; ; ), = ( ; ; ), = ( ; ; ) be made. If this functions in any region V* of a space Ouvw have the continuous partial derivatives and different from a zero determinant

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

( ; ; ) = |

 

 

 

 

 

 

 

 

 

 

|,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and the function ( ; ; ) is continuous in the region V then the following formula is valid

f (x; y; z)dV =

V

= f ( (u;v; w); (u;v; w); (u;v; w)) I (u;v; w) dudvdw. (2.3)

V

Here ( ; ; ) is the determinant of Jacoby or the transformation Jacobian.

32

For the calculation of a triple integral so called cylindrical coordinates are often applied.

The position of a point M(x;y;z) in space Oxyz can be determined by three numbers r, φ, z where r is the length of a radius-vector projection onto Ox axis, z – applicata of a point M (fig. 2.3). These three numbers ( ; ; ) are called the cylindrical coordinates of a point M.

The cartesian and the cylindrical coordinates are connected by the

equations

 

= ∙ cos , = ∙ sin , = .

(2.4)

Here ≥ 0, [0; 2 ], .

 

Fig.2.3

The right parts of equations (2.4) are continuously differentiable functions. The Jacobian of the coordinate’s transformation has the view

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

|

 

 

 

 

|

cos

− sin

0

=

 

 

 

 

 

 

= |sin

cos

0| = ≥ 0.

 

 

 

 

 

 

|

 

 

 

 

|

0

0

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

So, the formula (2.3) has the view

f (x; y; z)dV = f (r cos ; r sin ; z) rdrd dz.

V

V

(2.5)

So, the calculation of the triple integral in cylindrical coordinates is led to the integration with respect to r, to φ and to z like in cartesian coordinates.

Remark. The cylindrical coordinates are convenient in the cases when the region of integrating is formed by the cylindric surface.

33

Example 2.2. Calculate the triple integral

 

x

 

2

+

y

2

dxdydz

 

where

V the region V is bounded by the surfaces = 1, 2 + 2 = 4, = 2 + 2 +2.

Solution. We construct given surfaces in the Oxyz coordinate space (fig.

2.4).

Fig. 2.4

From the fig. 2.4 clear that a given integral is convenient to calculate in the cylindrical coordinates expressed by the formulas (2.4).

 

 

 

 

x

2

+ y

2

dxdydz =

 

 

r cos

2

+

r sin

 

2

rdrd dz =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

 

 

 

 

 

 

 

 

 

 

 

 

 

V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

2

 

 

2

 

2

 

 

2+r2

 

 

 

 

2

 

 

2

 

2+r

2

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

r drd dz =

d

r dr

 

dz =

 

d

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

z

 

 

 

 

 

r dr =

 

 

 

 

V

 

 

 

 

 

 

 

 

 

0

 

 

 

0

 

 

 

 

 

 

1

 

 

 

 

0

 

 

0

 

 

 

1

 

 

 

 

 

 

 

 

 

2

 

 

2

(

 

 

 

 

 

 

)

 

 

 

 

2

 

 

 

2

(

 

 

)

 

 

 

 

2

 

 

 

2

(

 

 

 

)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

d

 

 

2 + r2 1 r2dr =

 

 

d

 

 

1

+ r2

 

 

r2dr

=

 

d

 

 

 

r2

+ r4

 

dr =

 

0

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

0

 

r

 

2

 

 

 

 

 

 

0

 

 

 

0

 

 

 

0

 

0

 

2

 

2

 

 

 

 

 

 

 

 

 

 

 

 

r

3

 

 

 

 

 

 

 

 

2

 

 

 

2

 

 

 

 

 

 

 

 

 

(r

2

+ r

4

)dr =2

 

 

 

+

 

 

 

 

=2

 

 

+

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

5 0

 

 

 

 

 

 

3

 

 

 

5

 

 

3

 

 

5

 

 

 

 

 

 

 

 

 

=

2

 

8

+

32

= 2

 

40 + 96

=

272

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

5

 

 

15

 

 

15

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The spherical coordinates of a point M(x;y;z) in space Oxyz are called the three numbers , , where is the length of a radius-vector of a point M, is an angle between the radius-vector projection onto Oxy plane and Ox

34

axis, is an angle between the radius-vector of a point M and Oz axis (fig. 2.5). These three numbers ( , , ) are called the spherical coordinates of a point M.

Fig. 2.5

The cartesian and the spherical coordinates are connected by the

equations

 

= cos sin , = sin sin , = cos

(2.6)

Here ≥ 0, [0; 2 ], [0; ].

 

The right parts of equations (2.6) are continuously

differentiable

functions. The Jacobian of the coordinate’s transformation has the view

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

|

 

 

 

|

cos sin

− sin sin

cos cos

=

 

 

 

 

 

 

 

= |sin sin

cos sin

sin cos |

 

 

 

 

 

 

 

|

 

 

 

|

cos

 

0

− sin

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= sin sin |

sin sin

sin cos

| +

 

 

 

 

cos

 

− sin

 

 

 

 

+ cos sin |

cos sin

cos cos

| =

 

 

 

 

 

cos

 

− sin

= sin sin (− sin sin2 − sin cos2 ) +

+ cos sin (− cos sin2 − cos cos2 ) =

= − 2 sin2 sin ∙ 1 − 2 cos2 sin ∙ 1 = − 2 sin

So, the formula (2.3) has the view

35

f (x; y; z)dV =

 

V

 

 

= − f ( cos sin ; sin sin ; cos )

 

(2.7)

2

sin d d d .

 

 

 

V

*

 

 

 

 

 

Remark. The spherical coordinates are convenient in the cases when the region of integrating is sphere or its part. An equation for the sphere has the

view 2 + 2 + 2 = 2 where = .

 

 

 

Example

2.3.

Calculate

the

triple

integral

 

x

2

+ y

2

+ z

2

V

 

 

 

 

 

 

3

dxdydz

 

where the region V is bounded by the sphere

x2 + y2 + z2 = 4 and the plane = 0 ( ≥ 0).

Solution. We find a given integral in the spherical coordinates using the formulas (2.6).

 

x

2

+ y

2

+ z

2

3

dxdydz =

3

 

2

sin d d d =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

V

 

 

 

 

 

 

 

 

 

 

 

V

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

2

 

 

d =

 

 

(

cos )

 

6

=

 

 

 

 

 

 

 

 

 

 

 

 

 

= d sin d

5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

0

 

 

 

0

 

 

 

 

0

 

 

 

 

 

 

0

 

6

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

6

0

6

 

 

 

 

64

 

64

 

= ( 0) (cos + cos 0)

 

 

= 2

=

.

 

 

6

 

 

6

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2.3.The Applications of the Triple Integral

Let’s consider some examples of the triple integral applications.

The Body Volume

The volume of the region V in the cartesian coordinates is expressed by the formula

V = dV

V

or

V = dxdydz. V

(2.8)

The formula (2.8) in the cylindrical coordinates has the view

 

V = rdrd dz.

(2.9)

V

36

The formula (1.28) in the spherical coordinates has the view

V =

2

sin d d d .

 

 

V

 

 

(2.10)

The Mass of a Body

If the volume density is given then the body mass is found by the triple integral

m = (x; y; z)dxdydz, V

(2.11)

where ( ; ; ) is the volumetric density of mass distribution at a point

M(x;y;z).

The Static Moments and the Center of Mass Coordinates

of a Body

The static moments of a body relatively to Oxy, Oxz and Oyz coordinate planes respectively can be calculated by the formulas

Sxy = z (x; y; z)dxdydz,

V

Sxz = y (x; y; z)dxdydz, V

(2.12)

Syz = x (x; y; z)dxdydz.

V

The static moments and the center of mass coordinates of a body are connected by equations

x =

Syz

,

y

 

= Sxz

and z

 

= Sxy ,

 

c

c

c

m

 

m

 

m

 

 

 

 

 

The Moments of Inertia of a Body

(2.13)

The moments of inertia relatively to Oxy, Oxz and Oyz coordinate planes can be found by the formulas

37

Ixy

= z

2

(x; y; z)dV ,

 

 

 

 

 

 

Ixz

 

V

 

 

(x; y; z)dV ,

= y

2

 

 

 

 

 

 

 

 

 

V

 

 

 

I

yz

=

 

x2

(x; y; z)dV .

 

 

 

 

 

 

 

V

(2.14)

The moments of inertia relatively to the formulas

Ix

= ( y

2

+ z

2

)

 

 

 

 

I y

V

 

+ z

 

)

= (x

2

2

 

 

 

 

Iz

V

 

+ y

 

)

= (x

2

2

 

 

 

 

 

V

 

 

 

 

Ox, Oy, Oz axes can be found by

(x; y; z)dV ,

 

(x; y; z)dV ,

(2.15)

 

(x; y; z)dV .

 

Example 2.4. Find the volume, the mass, the static moment relatively to Oyz plane, the center of mass abscissa of a body bounded by the surfaces

= 2 + 2, = 4 if the density is ( ; ; ) = .

Solution. We construct given surfaces in the Oxyz coordinate space (fig.

2.6).

Fig. 2.6

We find the volume of a body by the formulas (2.8) and (2.9):

38

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

2

 

 

4

 

2

 

 

 

 

 

 

)rdr

V = dxdydz = rdrd dx =

d rdr dx = 2 (4 r

2

V

 

 

 

V

 

 

 

 

 

0

 

 

 

 

0

 

 

r

2

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

r

2

 

 

 

r

4

 

 

2

 

 

 

 

r

4

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 2 (4r r

3

)dr = 2 4

 

 

 

 

 

 

 

= 2 2r

2

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

2

 

 

 

4

 

 

0

 

 

 

 

4

 

 

0

 

 

 

 

 

2

4

 

 

 

0

4

 

 

2

 

 

 

 

(8 4 0 + 0) = 8 .

 

 

 

 

 

 

 

 

 

 

 

= 2 2 2

 

 

2 0

 

+

 

 

 

 

 

= 2

2

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

 

 

 

4

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

We find the mass of a body by the formula (2.11) taking into account the density ( ; ; ) = :

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

2

 

 

4

 

 

 

 

m = xdxdydz = x rdrd dx = d rdr

xdx =

 

 

 

V

 

 

 

 

 

 

 

V

 

 

 

 

 

 

 

 

0

 

0

 

 

r

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

x

2

4

 

 

 

 

 

 

1

2

 

 

 

)

 

 

 

 

2

 

 

 

 

 

 

 

)

 

 

 

 

 

2

 

 

 

 

 

 

 

 

(

 

 

 

 

 

 

 

(

 

 

 

 

 

 

= 2

 

 

2

r

 

 

rdr = 2

 

2 0

42 r2

 

rdr

=

 

16r r3

dr =

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r

2

 

r

4

2

 

 

 

 

 

 

2

4

 

 

 

 

0

4

2

 

 

 

= 16

 

 

 

 

 

=

8 2

2

 

 

 

8 0

2

+

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

4

0

 

 

 

 

 

 

4

 

 

 

 

4

0

 

 

= (32 4 0) = 28 .

The static moment of a body relatively to Oyz plane we find by the formula (2.12) taking into account the density ( ; ; ) = :

Syz = x x dxdydz = x

 

 

 

2

 

2

4

 

dx =

rdrd dx = d rdr x

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

2

 

 

 

V

 

 

 

 

V

 

 

 

0

 

0

r

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= 2

2

x3

4 rdr =

2 2

43 r6

 

rdr =

2

2

64r r7

 

dr =

 

 

 

(

)

 

(

 

 

r

2

 

 

 

 

3

 

 

 

)

 

0

3

 

 

3 0

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

39

 

 

 

 

 

 

 

 

 

2

r

 

 

r

 

2

 

2

 

 

2

 

 

 

0

 

 

64

 

2

 

8

 

 

=

32 2

 

8

32 0

 

8

=

 

 

 

 

 

 

2

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

2

 

8 0

 

3

 

 

8

 

 

 

8

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

(128 32 0) = 64 .

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The center of mass abscissa and the static moment of a body relatively to Oyz plane are connected by the first formula (2.13)

xc = Smyz = 6428 = 167 = 2 72 .

Analogically can be found the moments of inertia of a given body.

3.THE INDIVIDUAL TASKS

The students are recommended to solve the following tasks with respect to the variant suggested by a teacher:

1.Change the order of integrating.

2.Calculate the double integral.

3.Calculate the double integral in the polar coordinates.

4.Calculate the triple integral.

5.Find an area (mass, volume) of a flat plate, bounded by given lines.

6.Find the moment of inertia of a homogeneous body relatively to a given axis.

Variant 1

22

.. ∫ ∫ ( ; ) ;

0

. .

(2 + );

: = 2;

= 2;

 

 

 

 

40

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