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Let’s find each integral separately. To find the first integral we make the following substitution

					y = 2 sin dy = 2 cos d
	= arcsin				y						= arcsin 0 = 0;											= arcsin1 =
	= arcsin										= arcsin 0 = 0;									2		= arcsin1 =
					2				1											2						2
					2																					2

	1	2										2							2
	1					2													2
	I =			4 − y			dy =						4 − 2sin									2 cos d =
		0										0

2													2									2	1+ cos(2 )
=	4(1−sin		2	) 2 cos d = 4 cos												2	d = 4						1+ cos(2 )				d
=	4(1−sin			) 2 cos d = 4 cos													d = 4										d

0													0									0			2
0													0									0

		2												+				1 sin(2 )
		2
	= 2		1+ cos(2 )							)	d = 2													2	=
		(								)								2						2
		0																2						0
		0																						0
				= 2					+			1	sin(2		)	−		0			= .
				= 2				2	+			2	sin(2	2	)	−		0			= .
								2				2		2

To lead the first integral to the table view we apply the basic trigonometrical identity, the formula of half argument for cosine and the properties of the integral.

Let’s find the second integral using the properties of an integral of a single variable function, the table of integrals and the Newton-Leibniz formula.

I		=	2	4 − y2 dy =	4 y −	y3	2		= 4 2 −	23	− 4 0 +	03	=
	2		(
				)	3			0	3		3
			0

= 8 −	8	=	24 −8	=	16	.
	3		3		3

To calculate the third integral, we make the following substitution

																	t = 4 − y									2		,
																										2
	dt = (						4 − y			2				dy = −2 ydy															ydy = −											1		dt,
	dt = (						4 − y				)			dy = −2 ydy															ydy = −											2		dt,
																																								2
				t = 4 − y									,				y		= 0 t											= 4 − 0 = 4
				t = 4 − y								2	,				1										1			= 4 − 2							= 0.
												2					1										1
															y			= 2 t																	2
																																			2
																	2										2
At result we get
																																																	1		+1	4
																																																	1
	2								0							dt								1		0									1		4			1				1			t		2
I3	= y 4 − y					2	dy =						t							= −										tdt =							t				dt =								2
I3	= y 4 − y						dy =						t							= −										tdt =							t			2	dt =


	0								4							−2								2		4									2		0							2			1			+1		0
	0								4																	4											0										2			+1


						3		4
						3		4
		1		t		2				1				2							4				1																	8
	=					2		=											t	3				=					(		4		3	−			0		3	)= .
	=							=											t					=					(		4			−			0			)= .

		2			3			0		2				3							0				3																	3
					2			0					16									8
					2
Finally, we get				= ,						=						, =							. Let’s turn to an initial integral
Finally, we get				= ,						=						, =							. Let’s turn to an initial integral
			1					2					3				3				3
	V =																													1							2						3
	V =				(10 − x − y)dxdy = 40																								I			− 2 I						−			4 I			=
			D
	= 40 − 2							16	− 4						8		= 40 −												32			−		32		= 40 −									64			.
	= 40 − 2							3	− 4						3		= 40 −													3		−		3		= 40 −												.
								3							3															3				3												3

The Area of a Flat Figure

If we suppose that ( ; ) = 1 in the formula (1.5) then the cylindric body turns to the straight cylinder with the height = 1. As known the volume of such cylinder is equal to an area S of the base D. So, we get the formula for calculation of the domain area

S = dxdy	(1.16)
D

or in polar coordinates

S = rdrd

(1.17)

Example 1.4. Find an area of a figure bounded by the lines = 2 − ,

2 = 4 + 4.

Solution. We construct given lines in the Oxy coordinate plane (fig. 1.13). So, the integrating domain is the curved trapesoid.

Fig. 4.13

Taking into account the formula (1.17) and the limits of integrating performed on a figure 1.13 we have

4 x+4

2−x

S = dxdy = dx

dy +

dy =

−1

−

4 x+4

−

4 x+4

4x+4

2−x

dx + y

dx =

−1

−

4x+4

−

4x+4

= 2 0

dx + 8 (2 − x +

)dx = 2 0

4x + 4

4x + 4dx +

−1

								8										8													1		2		3
																			(2 − x)dx												1		2		3
						+			4x + 4dx +										(2 − x)dx							=			2I + I					+ I			.
								0										0
The first two terms have the same indefinite integral but different
integrating limits. Let’s find the indefinite integral at first.
																					t = 4x + 4																		1
I =																																							1
I =		4x + 4dx =											(4x																					1				=				tdt =
									dt		=		(4x			+ 4)					dx = 4 dx dx =													4	dt				4
																																		4
						1	+1
		1			t	2	+1				1			2										1
	=	1			t				+ C =		1			2				t	3	+ C =				1		t		3			+ C = t = 4x + 4 =



		4		1		+1					4			3										6
				2		+1
				2
														=	1					4x + 4				3	+ C.
														=	6					4x + 4					+ C.
															6
Let’s put the integrating limits
	1		1									0				1			(																						)
	1	=					4x + 4			3				=					(		4 0 + 4							3			−	4 (−1)				+ 4				3	)	=
	I									3																		3												3
	I		6													6
			6									−1				6
												−1
													=		1		(8 − 0) =							8	=			4		.

															6									6				3
					1									8			1		(																				)=
		I2	=		1				4x + 4			3			=		1					4 8 +				4			3		−	4 0 +				4		3
		I2	=		6				4x + 4					0	=		6					4 8 +				4					−	4 0 +				4
					6												6

										=		1		(216 −							8)	=	208				=		104			.
										=		6		(216 −							8)	=		6			=				3	.
												6												6							3
Let’s find the third integral
8														x2		8										82										02
=	(2 − x)dx = 2x −																			=		2 8 −									−	2	0		−					= −16.
=	(2 − x)dx = 2x −													2						=		2 8 −					2				−	2	0		−		2			= −16.
0														2				0									2										2
																		0
																					24

Let’s turn to the initial integral

4 x+4

2−x

S =

dxdy =

dy +

dy = 2I1 + I2 + I3 =

−1

− 4 x+4

= 2

104

−16 =

The Mass of a Flat Figure

According to the physical meaning of the double integral the mass of the flat plate is found by the formula (1.7).

The Static Moments and the Center of Mass Coordinates

of a Flat Figure

The static moments of a figure D relatively to Ox and respectively can be calculated by the formulas

Sx	= y (x; y)dxdy	and Sy	= x (x; y)dxdy.
	D		D

Oy axes

(1.18)

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

x =	Sy	and y		= Sx .
			c
c	m			m

The Moments of Inertia of a Flat Figure

(1.19)

The moment of inertia of a material point with mass m relatively to axis l is the product of the material point mass m by the square of the distance d between a point and an axis that is = ∙ 2. The moments of inertia relatively to axes Ox and Oy can be found by the formulas

M x	= y	2	(x; y)dxdy
		2
	D

and M	y	=	x2	(x; y)dxdy. (1.20)

The moment of inertia relatively to an origin is found by the formula

0 = + .

Example 1.5. Find the mass, the static moment relatively to Ox axis, the center of mass ordinate, the moment of inertia relatively to Oy axis of a flat plate bounded by the lines = 0, = √ and + = 2 if its density is =.

Solution. We construct given lines in the Oxy coordinate plane (fig. 1.14).

Fig. 5.14

To find the mass of the flat plate we use the formula (1.7) with respect the physical meaning of the double integral taking into account the density= and the integrating limits: = √ 1 = 2; + = 2 2 = 2 − ; 1 = 0; 2 = 1.

																		1				2− y							1			2 − y
																		1				2− y							1
m = (x; y)dxdy = ydxdy = ydy																									dx = x								y	2	ydy =
		D									D							0				y		2					0				y
		D									D							0				y							0
	1						)				1										)								y	2		y	3		y		4	1
=	(		− y − y2				)			=	(		2 y − y2 − y						3		)			=						−				−					=
=		2	− y − y2					ydy		=			2 y − y2 − y						3			dy		=			2			−				−				0	=
	0										0																		2			3			4			0
				2		y3			y4		1			2		13		14					2			03				04					1			1
	= y				−		−				0	=1			−		−			− 0					+			+			=1−					−
	= y				−		−					=1			−		−			− 0					+			+			=1−					−
						3			4							3 4											3 4								3 4
												=		12 − 4 − 3					=			5			.
												=				12			=			12			.
																12						12
																26

The static moment of a given figure relatively to Ox axis we calculate by the formula (1.18)

		Sx																												1			2− y				dx =
		Sx	= y (x; y)dxdy = y y dxdy = y																													dy					dx =
																															2
					D													D												0				y		2
					D													D												0				y
	1		2 − y									1	(							)								1												)
					2		y2dy ==						(	2 − y − y2						)				2dy =				(		2 y2		− y3 − y4								)
=		x		y			y2dy ==							2 − y − y2								y		2dy =						2 y2		− y3 − y4									dy =
	0			y								0																0
			y	3		y	4		y	5		1					y				y					y		1					1				1				1
				3			4			5								3					4				5						3					4			5
= 2			3		−	4		−	5			0	=		2		3		−	4				−		5			0		= 2		3		−		4		−		5	−
			3			4			5			0					3			4						5			0				3				4				5
			−2			03		+	04		+	05		=	2	−		1	−		1		=		40 −15 −12								=			13		.
			−2			3		+	4		+	5		=	3	−		4	−	5			=					60					=			60		.
						3			4			5			3			4		5								60								60

The center of mass ordinate is found by the formula (1.19).

						13
y		=	S	x	=	60	=	13	12	=	13	= 0.52.
				x		60
	c		m			5		60	5		25
	c

						12

The moment of inertia we find by the formula (1.20).

M y	= x		2	(x; y)dxdy = x						2	y dxdy = x				2	y dxdy =
			2							2					2
	D								D				D
1	2− y			1	x	3	2− y					1	1				3	)ydy =
1				1		3	2− y						1				3
= ydy		x2dx =							ydy	=			((2 − y)3	− (y2 )
= ydy		x2dx =			3		y	2	ydy	=		3	((2 − y)3	− (y2 )
0	y2			0	3		y					3	0

=1 1 (8 −12 y + 6 y2 − y3 − y6 )ydy =

3 0

=1 1 (8y −12 y2 + 6 y3 − y4 − y7 ) ydy =

3 0

			1				y	2			y	3				y	4			y	8		1
								2				3					4				8
		=	3		8		2		−12		3		+ 6			4		−		8				0		=
			3				2				3					4				8				0
		=		1			y2 − 4 y3						+	3	y4 −				y8				1		=
					4																		0
				3	4									2					8
				3										2					8
1								3			8						1										3		1
1	2				3	+		3	4	−	1					=	1			4		− 4 +					3	−	1	=
3	4 1	− 4 1				+		2	1	−	8		− 0			=	3			4		− 4 +					2	−	8	=
3								2			8						3										2		8
		=		1	(1.5 − 0.125) =									1.375					0.458.
		=		3	(1.5 − 0.125) =											3			0.458.
				3												3

The double integral can be also applied for the calculation of a figure surface.

2.THE TRIPLE INTEGRAL

2.1. The Basic Concepts and Definitions

The triple integral is the generalization of the integral to the case of a function of three variables. The theory of the triple integral is analogical to the theory of the double integral.

Let a continuous function = ( ; ; ) be given in the closed region V of a coordinate space Oxyz. Let’s split the region V by n elementary parts( = 1, , ). We denote the volume of such region by ∆ . Let’s choose an arbitrary point ( ; ; ) in each region . We compose the integral

sum

n
	f (x ; y ; z ) V
	i	i i	i
i=1

for a function = ( ; ; ) with respect to the

region V.

If the limit of an integral sum exists when unlimited magnification of n such that each elementary region shrinks to a point (that is the maximal diameter tends to zero (max → 0)) then it is called the triple integral of a function = ( ; ; ) with respect to the region V. We denote it as

f (x; y; z)dxdydz (or f (x; y; z)dV ).

V V

So, according to the definition we have

			n	i
				i
	f (x; y; z)dxdydz = lim			f (x; y; z) V	=		f (x; y; z)dV
V	n→		i=1			V
V	d	→0	i=1			V
	i

(1.21)

Here = is an element of the volume.

Theorem 2.1. The condition for existence of a triple integral. If the function = ( ; ; ) is continuous in the closed domain V then the limit of the integral sum (1.21) when n tends to the infinity and the maximal diameter tends to zero (max → 0) exists and does not depend on the way

of split of the region V by parts and the choice of the points ( ; ; ) in them.

The triple has the same properties as the double integral:

1.

c f (x; y; z)dV = c f (x; y; z)dV , c −const.

2.

( f (x; y; z) g(x; y; z))dV = f (x; y; z)dV g(x; y; z)dV .

3.
3.
	V

f (x; y; z)dV = f (x; y; z)dV f (x; y; z)dV ,
V	V
1	2

if = 1 2 and intersection of the regions is their bounded.

4.
4.
	V

f (x; y; z)dV 0

if ( ; ; ) ≥ 0 in the region V. If in the

integrating

f (x; y;

region

z)dV g(x; y;

f (x; y; z) g(x; y; z)

z)dV .

then

5. dV = V because in the case when ( ; ; ) = 1 any integral sum

has the view

n	i
	i
	V	=V
i=1

and equals to the volume of a body.

6. The estimation of a triple integral:

mV f (x; y; z)dV MV ,

where m is the smallest value of a function and M is the largest value of a function in the region V.

The

theorem about

the

average

value.

the

function

= ( ; ; )

is continuous

in the

closed

domain

then

a point

0(0; 0; 0) exists in this region such that

f (x; y; z)dV =

) V.

f (x

; y

; z

2.2.The Calculation of the Triple Integral in the Cartesian

Coordinates

The calculation of a triple integral is reduced to the sequential calculation of three definite integrals.

Let the region of integrating V be a body bounded by the surface =1( ; ) from the bottom and the surface = 2( ; ) on the top. The functions both 1( ; ) and 2( ; ) ( 1( ; ) ≤ 2( ; )) are continuous in the closed domain D which is the projection of a body onto Oxy plane. We suppose that the region V is correct in the direction of an axis Oz. It means that any straight line parallel to Oz intersects the bounded of the region no more than in two points. So, for any continuous in the region V function ( ; ; ) the formula is valid

				f	2	( x; y )

	f (x; y; z)dV	=					f (x; y; z)dz dS
					1
V			D	f		( x; y)

(2.1)

At first, we calculate the inside integral with respect z when x and y are constant within the change of z. The bottom limit of integral is the applicate of a point A which is the entrance point of a straight line parallel to Oz into the region V that is 1( ; ); the top limit is the applicate of a point B which is the exit point of a straight line from the region V that is 2( ; ). The result of calculation of this integral is the function of two variables x and y.

Let us first assume that the domain D is a curved trapezoid bounded by the straight lines = and = and the curves = 1( ) and = 2( ) which are continuous and such that 1( ) ≤ 2( ) for all [ ; ] (fig. 2.1). Such domain is called the correct relatively to the direction of Oy axis. Any straight line parallel to Oy axis intersects the domain bounded no more than in two points. Let’s construct the section of the cylindric body by the plane perpendicular to Ox axis ( = , [ ; ]).

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