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Now according to the method of parallel sections, the found volume of the cylindric body can be found in the following way

			(				)
	b		b			( x)
V =		S(x)dx =		2			f (x; y)dy dx.
					( x)
	a		a	1
	a		a

From the other hand the value of the double integral from nonnegative function is equal to the volume of the cylindric body. So,

	b	( x)	f (x; y)dy)dx.
		( x)
V = f (x; y)dxdy = ( 12( x)
D	a

This equality is usually written in view

	b			( x)
f (x; y)dxdy = dx				( x)
f (x; y)dxdy = dx		2
		2
			( x)
D	a	1
D	a

f (x; y)dy.

(1.8)

The formula (1.8) is the way of the double integral calculation in the Cartesian coordinates. The right part of an equation (1.8) is the two-fold (or iterated) integral of a function ( ; ) with respect to the domain D. An

integral

	2	( x)
	( x)
1

f (x; y)dy

is the inside integral. To calculate the two-fold

integral we find the inside integral supposing that x is a constant at first and after that we find the outside integral that is we integrate the result of the first integrating with respect to x in the limits from a until b.

If the domain D is bound by the straight lines = and = ( < ), the curves = 1( ) and = 2( ) with 1( ) ≤ 2( ) for all [ ; ], that is the domain is correct relatively to the direction of Ox axis then by dissecting a body by the plane = analogically we get

f (x; y)dx.

(1.9)

Here we suppose that y is constant.

Example 1.1. Reduce the double integral

f (x, y)dxdy D

to iterated

integrals if the domain D is bounded by the lines = ,	= 1, =	2
			.
		9

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

1.7).

Fig. 1.7

We got the curved trapezoid AOB. There are two variants to perform the double integral as the iterated integral. The first one is when the outside integral is with respect to y. In this case the limits of the inside integral are

		2
= 1	= , =	2	2 = 3√ and the limits of the outside integral
= 1	= , =	9	2 = 3√ and the limits of the outside integral
		9
are 0 ≤ ≤ 1. So, we get
			1	3	y
	f (x, y)dxdy = dy				f (x, y)dx.
	D		0	y

We can change the order of integrating. We will find the outside integral with respect to x. In this case the domain D contains two domains D1 and D2:

f (x, y)dxdy =

f (x, y)dxdy + f (x, y)dxdy.

The domain D1 is bounded by the lines =

and = for

0 ≤ ≤

1. The domain D2 is lying between the lines =

= 1 for 1 ≤ ≤

and

3. Thus,

f (x, y)dxdy = 1 dx x

f (x, y)dy + 3 dx 1

f (x, y)dy.

Example 1.2. Find the double integral ( + ) with respect to

the domain D bound by the straight lines 2 + = 1;

− = 2; = 0

in the cartesian coordinates.

Solution. We construct given lines in the Oxy coordinate plane (fig. 1.8). So, the integrating domain is the triangle ABC.

Fig. 1.8

To define the integrating limits with respect to Oy axis we express y from the given equations for the straight lines

− = 2 1 = 2 − 2 + = 1 2 = 1 − 2

To define the integrating limits, we solve a system

− = 2; {2 + = 1. 3 = 3 = 1; = −1.

The left limit 1 = 0; 2 = 1. So, we get an integral

		1	1−2				1	1−2
		1					1
( + ) = ∫ ∫				( + ) = ∫ ∫					+
		0	2−				0	2−
			2−					2−
	1−2		1	1−2			1	1−2
1			1				1
+ ∫ ∫		= ∫		∫	+ ∫			∫ =
0	2−		0	2−			0	2−
	2−			2−				2−
	1	1 − 2		1		2	1 − 2
		1 − 2				2	1 − 2
	= ∫ ( \|		) ∙ + ∫ (				\|	) =
	= ∫ ( \|		) ∙ + ∫ (			2	\|	) =
	0	2 −		0		2	2 −
1				1

= ∫(1 − 2 − (2 − )) ∙ + 12 ∫((1 − 2 )2 − (2 − )2) =

We got two integrals with respect to x. Let’s solve them separately. To lead the first integral to the table view we bring the similar terms, open the brackets and apply the properties of an integral of a single variable function. After that we use the table of integrals and the Newton-Leibniz formula.

1 1

	1	= ∫(1 − 2 − 2 + ) ∙ = ∫(−1 − ) ∙ =
		0					0
1					1				1	1
− ∫(1 + ) ∙ = − ∫( + 2) = − ∫ − ∫ 2 =
0					0				0	0
	2	1	3	1		1	(12 − 02) −	1	(13	− 03) = −	1		1
= −		\| −		\| = −								−		.
= −	2	\| −	3	\| = −		2		3			2	−	3	.
	2	0	3	0		2		3			2		3

1 = − 56.

To lead the second integral to the table view we apply the abbreviated multiplication formulas. There are two variants here. We can use the difference of squares formula or the formula of the difference square. Further we open the brackets, bring the similar terms and apply the properties of an integral of a single variable function. After that we use the table of integrals and the Newton-Leibniz formula. Let’s consider the first variant.

2 = ∫((1 − 2 )2 − (2 − )2) =

= ∫((1 − 2 − (2 − )) ∙ (1 − 2 + (2 − )) =

= ∫((1 − 2 − 2 + ) ∙ (1 − 2 + 2 − ) =

= ∫((−1 − ) ∙ (3 − 3 )) = −3 ∫(1 + ) ∙ (1 − ) =

1			3			1					1
= −3 ∫(1 − 2) = −3 ( −			3		) \|	1		= −3 (1 −			1	− 0) = −2.
= −3 ∫(1 − 2) = −3 ( −			3		) \|	0		= −3 (1 −			3	− 0) = −2.
0			3								3
0
Finally, we get
( + ) = 1 +	1					5			1	∙ (−2)			5
		2		= −				+			= −1			.
	2	2		= −			6	+	2		= −1		6	.

Remarks.

1.The formulas (1.7) and (1.8) is valid and for the case when ( ; ) < 0, ( ; ) .

2.If the domain D is correct in both directions, then the double integral can be calculated both applying the formula (1.7) and (1.8).

3.If the domain D is not correct either in the direction of Ox axis nor the direction of Oy axis then to reduce a double integral to the repeated, we should split the initial domain by parts correct in the direction of Ox axis or Oy axis.

4.It’s useful to remember that the outside limits in the two-fold integral are constant and the inside limits are as a rule variable.

1.6.The Calculation of the Double Integral in the Polar

Coordinates

To simplify the calculation of the double integral the substitution method is often applied that is new variables are introduced under the double integral sign.

Let’s define the transformation of the independent variables x and y as

= ( ; ) and = ( ; )

(1.10)

If the functions (1.9) in any domain D* of a plane Ouv have the continuous partial derivatives and different from a zero determinant

( ; ) = \|	\|,	(1.11)

and the function ( ; ) is continuous in the domain D then the following formula is valid

f (x; y)dxdy = f ( (u; v); (u; v)) I (u; v) dudv.		(1.12)
D	D*
	15

An equation (1.11) is formula of the variable’s substitution in the double integral. The functional determinant (1.10) is called the determinant of Jacobi or Jacobian (Carl Gustav Jacob Jacobi is a German mathematician).

We consider the partial case of the variable substitution which is often applied to calculate the double integral namely the substitution of the cartesian coordinates x and y by the polar coordinates r and .

As u and v we take the polar coordinates r and . The cartesian and the polar coordinates are connected by the equations

= ∙ cos , = ∙ sin

(1.13)

The right parts of equations (1.12) are continuously differentiable functions. The Jacobian of the coordinate’s transformation has the view

		( ∙ cos )	( ∙ cos )

= \|	\| = \|			\| = \|cos	− sin \| = .
		( ∙ sin )	( ∙ sin )	sin	cos

So, the formula (1.11) has the view

f (x; y)dxdy = f (r cos ; r sin ) r drd ,
D	*
	D

(1.14)

where D* – is the domain in the polar coordinates corresponding to the domain D in the cartesian coordinates. To calculate the double integral in polar coordinates the same methods are applied.

Let the domain D* shown on a figure 1.9 be given. It is bounded by the beams = and = ( < ) and the curves = 1( ) and = 2( )

( 1( ) ≤ 2( )).

Fig. 1.9

The domain D* is correct so the right part of (1.13) can be written in view

	r	( )
	2
f (r cos ; r sin ) r drd = d
	r ( )
	1

(r cos ; r sin ) r

dr.

(1.15)

The inside integral of (1.15) is found supposing that is constant.

Remarks.

1.The transfer to the polar coordinates is useful when integral function has the view ( 2 + 2). The domain D circle, ring or part of them.

2.In practice, the transition to polar coordinates is carried out by the

substitution = ∙ cos , = ∙ sin , = ; the equations for the lines bounding the domain D are also reduced to the polar coordinates. The domain D is not transforming to the domain D*. We just overlap the polar coordinate system on the cartesian system to define new limits of integrating with respect toand .

Example 1.3. Find the double integral with respect to the

√ 2+ 2

domain D bounded by the curve = cos 2 and the circle 2 + 2 = 0.25. Solution. We construct given lines in the coordinate plane and get the

integrating domain (shaded area on a fig. 1.10).

	Fig. 2.10
From the	circle equation 2 + 2 = 0.25 follows that the radius
2 = 0.25	= 0.5. So, the variable is changing from = 0.5 until
	17

= cos 2 . It is evident from the fig. 1.10 that − 6 ≤ ≤ 6. At result the initial integral can be performed in the polar coordinates.

			= cos
			= sin
			= { 2 + 2 = 2 } =															=					=
	√
	√	2 + 2												√ 2
			=
					cos 2
			6		cos 2							6			cos 2
															cos 2
		= = ∫ ∫								= 2		∫ ( \|						)
				0.5														0.5
			−6									−6
As we see on a fig. 1.10 the integrating domain is symmetric relatively
to Ox axis. That’s why the integral with the limits −																		≤ ≤			is equal to
to Ox axis. That’s why the integral with the limits −																6		≤ ≤		6	is equal to
																6				6
two integrals with the limits 0 ≤ ≤											.
two integrals with the limits 0 ≤ ≤									6		.
									6

6
6							1					1
2 ∫(cos 2 − 0.5) = 2 (							1	sin 2 −				1	) \| 6				= sin			−		=
2 ∫(cos 2 − 0.5) = 2 (								sin 2 −				2	) \| 6				= sin		3	−	6	=
0				2								2							3		6
0														0
														0

				√3
			=			−				≈ 0.342.
			=	2		−		6		≈ 0.342.

1.7. The Applications of the Double Integral

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

The Body Volume

As shown earlier with respect to the geometrical meaning of the double integral the volume of the cylindric body is found by the formula (1.5).

Example 1.3. Find a volume of a body bounded by the cylinder 2 +2 = 4 and the planes = 0, = 10 − − . (40pi)

Solution. We construct given surfaces in the coordinate space and get a body shown on a fig. 1.11

Fig. 3.11

According to the geometrical meaning of the double integral

V = f (x; y)dxdy.

So, we get

V = (10 − x − y)dxdy.

To perform an initial integral as iterated we consider the projection of a body onto Oxy plane (fig. 1.12).

Fig. 1.12

As we see from a figure 1.12 the integrating limits with respect to y are −2 ≤≤ 2. To find the integrating limits with respect to x we reduce an equation

2 + 2 = 4:

2 + 2 = 4 = ±√4 − 2

= ±√4 − 2 1 = −√4 − 2.

		= √4 − 2
	2
	2		4−y	2
	2		4−y
V = (10 − x − y)dxdy = dy						(10 − x − y)dx.
D	−2	−	4−y		2
		−	4−y

From the figure 1.12 it’s evident that the integrating domain is symmetric. So, it’s enough to find the double integral with respect to the quarter of the domain:

										2						4−y	2
										2						4−y
	V = (10 − x − y)dxdy = dy																			(10 − x − y)dx =
		D									−2				−	4−y		2
															−	4−y
2	4− y		2							2												2						2
	4− y																			x							4 − y
= 4 dy				(10 − x − y)dx = 4								(10x −								x			− yx)				4 − y		dy =
= 4 dy				(10 − x − y)dx = 4								(10x −											− yx)						dy =
																					2					0
0		0								0																0
														2		2
		2							(		4 − y				)	2

					4 − y	2		−								− y								4 − y	2
	= 4 10				4 − y			−		2						− y								4 − y			dy =
		0								2
		0

		2					2										2
				4 − y2 dy − 2			(			− y2		)														2 dy =
	40			4 − y2 dy − 2					4	− y2		dy − 4								y			4 − y			2 dy =
		0					0										0
					= 40 I					− 2 I			2	− 4 I					3		.
									1				2						3
										20

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