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Национальный авиационный университет

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Higher Mathematics. Part 3

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Change the order of integration. Firstly draw the domain of integration.

y2 +3

3. ∫ dx ∫ f (x, y)dy.

4. ∫ dy ∫ f (x, y)dx.

3− x

1− x2

π / 4

cos x

5. ∫ dx ∫

f (x, y)dy.

6. ∫ dx ∫ f (x, y)dy.

−1

sin x

8−2x

7. ∫ dx ∫ f (x, y)dy + ∫ dx ∫ f (x, y)dy.

−1

Calculate the double integral

8. ∫ dx∫ (x2 + y2 )dy.

9. ∫ dx∫

(x + y)

Evaluate the double integrals ∫∫ f (x, y)dxdy.

10. ∫∫ (x3 + 4y)dxdy, over the domain D bounded by the lines

x = 0,

x = 1 ,

y = 1 .

y = 0 and

11. ∫∫

dxdy, over the domain D bounded by the lines x = 2,

y = x and

xy = 1.

the hyperbola

12. ∫∫ ydxdy,

over the domain D bounded by the lines y = 0,

x + y = 2

y =

x .

and the parabola

Evaluate the following integrals using polar coordinates system.

13. ∫∫

25 − x2 − y2 dxdy,

over

the domain

D bounded by

the

circle

x2 + y2 = 9 .

14. ∫∫ (x2 + y2 )2 dxdy, if the domain D is the annulus 1 ≤ x2 + y2 ≤ 4.

15. ∫∫

4 − x2 − y2 dxdy,

over a domain D bounded by the circle

x2 + y2 = 4

and lines y = x ,

y =

3x ( x > 0, y > 0 ).

16. ∫∫ ydxdy,

over a domain

D bounded

by the

circles

x2 + y2

= 4x,

x2 + y2 = 8x and lines

y =

3x , x = 3y.

121

Calculate the area bounded by the lines.

17.	x2	= 4y + 4,	x2	= −2 y + 4.
18.	x2	+ y2 = 4,	y2	= 3x (x ≥ 0).

Using the geometrical interpretation of the double integral, calculate the volume of a cylindrical body bounded by the following surfaces

19. x2 + y2 + z = 4, z = 0.

20.

z = x2 + y2 , x = 0,

x = 1,

y = 0,

y = 2,

z = 0.

21.

z = 0,

y + z = 2,

y = x2 .

22.

Calculate the volume of a body bounded by the cylinder

x2 + y2 = Rx

and sphere x2 + y2 + z2

= R2 .

23.

Find

the mass

plate D bounded

the

curves

x2 + y2 = 4,

x2 + y2

= 16 ( x ≥ 0, y ≥ 0 ),

the

density

in each

point

(x, y) D equals

γ(x, y) =

x2 + y2

Find the centre of mass of the homogeneous figures bounded by the following lines

	24.	y = x2 + 1,		x − y + 3 = 0.
	25.	x2 + y2 = 4, x = y,						y =	3x ( x ≥ 0,		y ≥ 0 ).
									Answers
	4 x					1		4 y	4		4	1	x+5
	1. ∫dx∫ f (x, y)dy			or		∫dy ∫ f (x, y)dx + ∫dy∫ f (x, y)dx.						2. ∫ dx ∫ f (x, y)dy +
	0 x					0		y	3		y	−1 x+ 2
4	6		4			4 y−2			6 y−2			3	3
4	6					4 y−2			6 y−2			3	3
+∫dx ∫ f (x, y)dy				or		∫dy ∫ f (x, y)dx + ∫dy ∫ f (x, y)dx.						3. ∫dy ∫ f (x, y)dx +
1 x+ 2						1		−1	4 y−5			0 3− y
4	3					3		x	4		x	7	2
+∫dy∫ f (x, y)dx.					4. ∫dx ∫ f (x, y)dy + ∫dx ∫ f (x, y)dy + ∫dx ∫ f (x, y)dy.
3	0					0		0	3		x−3	4	x−3
		1− y2			1
1		1− y2		2				arcsin x	1		arccos x	0	4
5. ∫dy		∫ f (x, y)dx. 6. ∫					dy	∫ f (x, y)dx + ∫ dy ∫ f (x, y)dx. 7. ∫ dy∫ f (x, y)dx +
0	− 1− y2			0				0		1	0	−1 0
0	− 1− y2			0				0	2		0
									2
122

	4−	y
4		2					10									9			9						5			122
+∫dy	∫		f (x,	y)dx.	8.		10		.		9.	0,5.	10.			9	.	11.	9	. 12.					5	.	13.	122			π.		14.	21π.
+∫dy	∫		f (x,	y)dx.	8.		3		.		9.	0,5.	10.			4	.	11.	4	. 12.					12	.	13.			3	π.		14.	21π.
0	y
15. 12(		3 − 1).		16.	56	. 17. 8. 18. (4π +								3) / 3 . 19. 8π.						20.		16	.	21.			32	2	. 22.			2(3π − 4)			R3.
					3			8					8								3						15						9
23. 6.	24. (0,5; 2,6). 25.							8				− 2);	8		2 − 1)
23. 6.	24. (0,5; 2,6). 25.									(	3	− 2);		(	2 − 1)			.
								π					π

Micromodule 5

SELF-TEST ASSIGNMENTS

5.1. Change the order of integration. Firstly, draw the domain of integration.

	2	x+1
5.1.1. а)	∫ dx		∫	f (x, y)dy;
	−1	x2 −1
	2	x+ 2
5.1.2. а)	∫ dx ∫			f (x, y)dy;
	0	x
	1	2x+1
5.1.3. а)	∫ dx	∫		f (x, y)dy;
	0	−1
	2	x2 +1
5.1.4. а)	∫ dx		∫	f (x, y)dy;
	−1		0
	4	3
5.1.5. а)	∫ dx ∫		f (x, y)dy;
	0	x
	1	2− x2
5.1.6. а)	∫ dx	∫		f (x, y)dy;
	0		x
	1	x+ 2
5.1.7. а)	∫ dx ∫			f (x, y)dy;
	0	− x
	2		x+ 2
5.1.8. а)	∫ dx		∫	f (x, y)dy;
	−2		x
	1	4− x2
5.1.9. а)	∫ dx	∫		f (x, y)dy;
	0	1


4				0
b) ∫ dx						∫		f (x, y)dy .
2	−2−					4x− x2
1	5+			9−8x− x2
b) ∫ dx						∫			f (x, y)dy .
0				0
3				6
b) ∫ dx						∫			f (x, y)dy .
−1		3+			3+ 2x− x2
4		16− x2
b) ∫ dx ∫							f (x, y)dy .
2			− x
3		4− 3+ 2x− x2
b) ∫ dx						∫			f (x, y)dy .
−1				0
4			x
b) ∫ dx ∫							f (x, y)dy.
2			4x− x2
4				x−2
b) ∫ dx						∫		f (x, y)dy .

0−2− 4x− x2

b)	∫ dx		∫	f (x, y)dy .
	0 −		9−8x− x2
	0	−1+ −6x− x2
b)	∫ dx		∫	f (x, y)dy .
	−6		−1

123

1	0
5.1.10. а) ∫ dx	∫ f (x, y)dy;
−1	x2 −3

12− x

5.1.11.а) ∫ dx ∫ f (x, y)dy;

	−2				x2
	1		3x
5.1.12. а)	∫ dx ∫					f (x, y)dy;
	0		x2
	1	2 x
5.1.13. а)	∫ dx ∫					f (x, y)dy;
	0				x
	4				x
5.1.14. а)	∫ dx ∫					f (x, y)dy;
	0		− x
	1				− x2
5.1.15. а)	∫ dx ∫ f (x, y)dy;
	−3			2x−3
	2			x+ 2
5.1.16. а)	∫ dx ∫					f (x, y)dy;
	−1				x2
	3		x
5.1.17. а)	∫ dx∫ f (x, y)dy;
	1		1
			x
	1		2− y
5.1.18. а)	∫ dy ∫					f (x, y)dx;
	0			y2
	2		2 y
5.1.19. а)	∫ dy ∫					f (x, y)dx;
	1			2
				y
	1	1+				x
5.1.20. а)	∫ dx ∫					f (x, y)dy;
	0			− x2
	2	2x
5.1.21. а)	∫ dx ∫					f (x, y)dy;

12− x2

5.1.22.а) ∫ dx ∫ f (x, y)dy;

−1 x2

124

0	3+ −2x− x2
b) ∫ dx		∫			f (x, y)dy .
−2		0
5	6x− x2 −5
b) ∫ dx		∫		f (x, y)dy .
1		−2
1		x
b) ∫ dx		∫		f (x, y)dy .
0	−	10x− x2
6	2+	6x− x2
b) ∫ dx		∫			f (x, y)dy .
0		2
2	2+	2x− x2
b) ∫ dx		∫			f (x, y)dy .
0		x
2	36− x2
b) ∫ dx ∫			f (x, y)dy .
0		0
−1 1+ 6x+ x2 −5
b) ∫ dx		∫			f (x, y)dy .
−3		0
5	1+	10x− x2
b) ∫ dx		∫			f (x, y)dy .
0		0
2		4
b) ∫ dx		∫		f (x, y)dy .
1	1−	2x− x2
0	3− −2x− x2
b) ∫ dx		∫			f (x, y)dy .
−1		0
4		0
b) ∫ dx		∫			f (x, y)dy .
0	−	9+8x− x2
5		1
b) ∫ dx		∫			f (x, y)dy .
3	−	6x− x2 −5
2	2−	2x− x2
b) ∫ dx		∫			f (x, y)dy .
0		0

1x2

5.1.23.а) ∫ dx ∫ f (x, y)dy;

−1	x−1
2	3x−2
5.1.24. а) ∫ dx	∫ f (x, y)dy;
1	1

1	y2 +1
5.1.25. а) ∫ dy	∫ f (x, y)dx;

0− y2

1y+1

5.1.26. а)	∫ dy ∫			f (x, y)dx;
	0	− y
	2		2x
5.1.27. а)	∫ dx		∫	f (x, y)dy;
	0	x2 −1
	0	x2 + 2
5.1.28. а)	∫ dx		∫	f (x, y)dy;
	−1		1
	4	2	x
5.1.29. а)	∫ dx ∫			f (x, y)dy;
	1		x

5.1.30.а) ∫ dx ∫ f (x, y)dy;

0			x+ 4
b) ∫ dx ∫				f (x, y)dy .
−4	−		16− x2
−3			2
b) ∫ dx			∫	f (x, y)dy .
−5	−		6x+ x2 −5
3	3+		3+ 2x− x2
b) ∫ dx			∫	f (x, y)dy .
1			0
3	2+		6x− x2 −5
b) ∫ dx			∫	f (x, y)dy .
1			0
−1			3
b) ∫ dx			∫	f (x, y)dy .
−2	2−		−2x− x2
0	1+		−6x− x2
b) ∫ dx			∫	f (x, y)dy .
−3			0
5			0
b) ∫ dx			∫	f (x, y)dy .
1	−	10x− x2
0			2
b) ∫ dx			∫	f (x, y)dy .
−1	−		9+8x− x2

5.2. Calculate the double integral ∫∫ f (x, y)dxdy over the domain D.

5.2.1. ∫∫ (x + 2 y + 1)dxdy , over the domain D bounded by the lines: y = x2 ,

y = 1 , x = 0 , y = 0 .

5.2.2. ∫∫ (x2 + y2 )dxdy , over the domain D bounded by the lines: y = x − 1,

y = 3 − 2x , x = 0 .

5.2.3. ∫∫ (9 − y2 )dxdy , over the domain D bounded by the lines: y = 4 − 2x ,

y = 2x , y = 0 .

5.2.4. ∫∫ (x − y + 3)dxdy , over the domain D bounded by the lines: y = x ,

y = (x − 2)2 , y = 0 .

125

5.2.5. ∫∫(2x + 1)dxdy , over the domain D bounded by the lines: x = 4 , y2 = x.

5.2.6. ∫∫(x − 4y)dxdy , over the domain D bounded by the lines: y = x − 4 ,

	D
y = x ,	y = 0 .
5.2.7.	∫∫ (x2 − 2y)dxdy , over the domain D bounded by the lines:		x = 1 ,
	D	= 0 .
x = 2 , y = x , y		= 0 .
5.2.8.	∫∫ (x2 + 4 y)dxdy, over the domain D bounded by the lines:		x = 2 ,
	D
y = 2x , 2y = x .
5.2.9.	∫∫ (2x − 3y2 )dxdy , over the domain D bounded by the lines:		y = 2 ,
	D
y = 2x , 2y = x .

5.2.10. ∫∫ (x + 2 y −1)dxdy , over the domain D bounded by the lines: x = 1 ,

x = 0 , y = x2 , x + y = 0 .

5.2.11. ∫∫ 2xydxdy , over the domain D bounded by the lines: y = 2 ,

y = 1− x , y = x − 1.

5.2.12. ∫∫ (2y − x + 2)dxdy , over the domain D bounded by the lines: y = 1,

y= 2 − x2 .

5.2.13.∫∫ ( y − 2x + 6)dxdy , over the domain D bounded by the lines: y = 0 ,

y = x , y = 2 − x .

5.2.14. ∫∫(3x2 − y2 )dxdy, over the domain D bounded by the lines: y = x + 1, y = 1, y = xD, y = 0 .

5.2.15. ∫∫ (2x + 3)dxdy , over the domain D bounded by the lines: x = 0 ,

y= 1− x2 , y = −1− x .

5.2.16.∫∫ (x + 3y2 )dxdy , over the domain D bounded by the lines: y = x + 1,

x= −1 , y =Dx , x = 1 .

5.2.17.∫∫ (2x + y + 2)dxdy , over the domain D bounded by the lines: x = 0 ,

y = 0 , y = x2 + 1 , x = 1 .

126

5.2.18. ∫∫ ( x + 2y)dxdy, over the domain D bounded by the lines: y = x +1,

x = 0 , y = 0 , x = 1 .

5.2.19. ∫∫ (x − y )dxdy , over the domain D bounded by the lines: x = 2 ,

y = 1 , y = 0 , y = − x .

5.2.20. ∫∫ (2xy − 1)dxdy , over the domain D bounded by the lines: y = x + 1,

y = 1 , y = 0 , x = 1 .

5.2.21. ∫∫ (4x − y + 1)dxdy , over the domain D bounded by the lines:

y = − x2 , x = 0 , y = 1 , x = −1 .

5.2.22. ∫∫ (3 x + 4y)dxdy , over the domain D bounded by the lines: y = −2 ,

y = x , y = − x .

5.2.23. ∫∫ (x − 2y + 5)dxdy , over the domain D bounded by the lines: y = 0 ,

y = 1 , y = − x , x = 2 .

5.2.24. ∫∫ (x + 1) ydxdy , over the domain D bounded by the lines: y = 0,

y = x, y = x − 1, x = 2 .

5.2.25. ∫∫ (3x + y + 2)dxdy , over the domain D bounded by the lines: x = 1 ,

y = 2x , y = − x .

5.2.26. ∫∫ (x + 3y2 + 1)dxdy , over the domain D bounded by the lines: y = 3 ,

y = x2 −1 .

5.2.27. ∫∫ (x2 + 2xy + 2)dxdy , over the domain D bounded by the

lines: y = 0 , y = 2 − x , y = 1, x = 0 .

5.2.28. ∫∫ (4x − 2y + 1)dxdy , over the domain D bounded by the lines: y = 0 ,

y = x2 , x = −2 .
5.2.29. ∫∫ (xy + 1)dxdy , over the domain D bounded by the lines:	y =	1	,
	y =	x	,
D		x

y = 1, y = 2 , x = −1 .

127

5.2.30. ∫∫ (x + 4y + 2)dxdy , over the domain D bounded by the lines:

	1		D
y =		,	y = 2 , y = x .
	x

5.3. Calculate the area of a domain D bounded by the following lines. Evaluate the following integrals using polar coordinates system.

5.3.1.	x2 + y2 + 6x ≤ 0,			y ≥	3x,	y ≤ 0.
5.3.2.	x2	+ y2	+ 4 y ≤ 0,	y ≤	3x,	x ≤ 0.
5.3.3.	x2	+ y2	− 6x ≤ 0,	y ≤	3x,	y ≥ − x.
5.3.4.	x2	+ y2	− 4y ≤ 0,	y ≥	3x,	x ≥ 0.


5.3.5. y ≥ 0,			y ≤		3x,		x2 + y2		≤ 10x .
5.3.6. x2 + y2			≤ 8y,			3y ≥ x,			y ≤		3x.
5.3.7. y ≥ − x,				y ≤ −		3x,		x2 + y2 − 4y ≤ 0.
5.3.8. x2 + y2 + 10x ≤ 0,								3y ≤ x,			y ≥ x.
5.3.9. y ≥ 0,			y ≤ x,			x2 + y2 − 6x ≤ 0.
5.3.10. x2		+ y2		≤ 2y,		y ≤ x,			3y ≥ x.
5.3.11.	3y ≥ x,				y ≤ − x,			x2 + y2 + 4x ≤ 0.
5.3.12.	x2	+ y2		+ 10x ≤ 0,				y ≥ x,		y ≤ 0.
5.3.13.	y ≥ 0,			y ≤	3x,		x2 + y2			≤ 4x.
5.3.14.	x2	+ y2		≤ 2y,		y ≤		3x,		y ≤ − x.
5.3.15.	x2	+ y2		+ 4x ≤ 0,				3y ≥ − x,			y ≤ − 3x.
5.3.16.	x2	+ y2		+ 16y ≤ 0,				y ≤	3x.
5.3.17.	y ≤ x,			3y ≥ x,			y2 ≤ x(6 − x).
5.3.18.	x ≥ 0,			y ≥	3x,		x2 + y2			≤ 8y.
5.3.19.	x2	+ y2 + 4x ≤ 0,						3y ≤ − x,			y ≥ 0.
5.3.20.	x2	+ y2 + 10 y ≤ 0,						y ≥ x.
5.3.21.	3y ≥ − x,				y ≤ 0,			y2 ≤ x(8 − x).

5.3.22.3y ≤ − x, x2 + y2 ≤ 8y.

5.3.23.	x2	+ y2	+ 12 y ≤ 0,	y ≥ 3x,	3y ≤ x.
5.3.24.	x2	+ y2	+ 2 y ≤ 0,	y ≤ − x, y ≤	3x.

128

5.3.25.	y ≥ 3x, x2 + y2			≤ 6x.
5.3.26.	y ≤ − x,		3y ≥ − x,		x2 + y2 ≤ 8y.
5.3.27.	x2	+ y2 + 4x ≥ 0,		y ≤ 0,		3y ≥ x.
5.3.28.	y ≥ − x,		y ≤ − 3x,		x2 + y2 ≤ 8y.
5.3.29.	x2	+ y2	+ 4x ≤ 0,		3y ≤ x,	y ≥ x.
5.3.30.	x2	+ y2	+ 16 y ≤ 0,		3y ≤ − x, y ≤ 3x.

5.4. Using the geometrical sense of the double integral, calculate the volume of a cylindrical body bounded by the following surfaces.

5.4.1. x =

y = 4,

x = 0,

z = 0,

x + z = 6.

5.4.2. z = 4 − x2 ,

x + y − 2 = 0, x = 0, y = 0,

z = 0.

5.4.3. y = x2 ,

y = 0,

x = 2,

z = 4 − x2 ,

z = 0.

5.4.4. x + y = 2,

x = 0, z = x2 + y2 ,

x =

z = 0.

5.4.5. z = 1+ x2 ,

x + y − 2 = 0, x = 0,

y = 0, z = 0.

5.4.6. y = x2 ,

y = 0,

x + y = 2,

z = 0, z + y = 3.

5.4.7. z = x2 ,

x + y − 2 = 0,

y = 0, z = 0.

5.4.8. y = x,

y = 2x,

y = 2,

z = 6 − y2 ,

z = 0.

5.4.9. y =

x = 4,

x + 2 y = 0, z − x = 2, z = 0.

5.4.10. z = 1+ x2 ,

x = 2,

x + y − 4 = 0, y = 0,

z = 0.

5.4.11. x = y,

2y + x = 0,

x = 2, z = x2 + y2 ,

z = 0.

5.4.12. y = 2x,

y + 2x = 4,

y = 0,

z = x2 ,

z = 0.

5.4.13. y = x2 ,

y = 1,

z = y2 ,

z = 0.

5.4.14. y = 2x,

x = 2,

y = 1,

z = y2 ,

z = 0.

5.4.15. x − 2y = 0,

z = 4 − x2 , y = 0,

z = 0.

5.4.16. z = y2 ,

x + y = 6,

y = x, z = 0.

5.4.17. z = 4 − x2 ,

x + y − 4 = 0,

y = x,

x = 0,

z = 0.

5.4.18. z = 4 + y2 ,

x + y − 4 = 0,

x = 0,

y = 0,

z = 0.

5.4.19. z = 4 − y2 ,

x + y − 4 = 0,

y = x,

y = 0, z = 0.

5.4.20. z = y2 ,

x + y − 6 = 0, y = x, z = 0.

5.4.21. y =

y + x = 0, x = 1, z = x2 + y2 ,

z = 0.

5.4.22. y =

x = 0,

y = 2,

z = x2 + 1, z = 0.

129


5.4.23.	y = x2 ,	y = 4, x = 1, z = 2 + x,					z = 0.
5.4.24.	y = x + 1,	x = 1,		y = 0,		z = 1+ y2 , z = 0.
5.4.25.	y = x, y = x + 1,			x = 0,		x = 2,	z = x2 + y2 , z = 0.
5.4.26.	z = 2 + x2 ,		x + y − 1 = 0,			x = 0,	x − y = 1, z = 0.
5.4.27.	y = x2 , y + x = 0, x = 1,					z = 0,	z + x = 3.
5.4.28.	y = 4 − x2 ,		y = 0,		z = y, z = 0.
5.4.29.	x = 1− y2 ,		z =	x,	z = 0.
5.4.30.	x = 2, z =		y,	y = x, z = 0.

Micromodule 6

BASIC THEORETICAL INFORMATION.

TRIPLE INTEGRALS

Triple integrals. Principal concepts and definitions. Conditions of the existence and properties. Cylindrical and Spherical coordinates. Change of variable in double integral. Application of double integrals.

Key words: triple integral — подвійний інтеграл; integral sum — інтегральна сума; iterated (repeated) integral — повторний інтеграл; rectangular parallelepiped — прямокутний паралелепіпед; nonhomogeneous material plate —

неоднорідна матеріальна пластина, mass of a planar figure — маса плоскої фігури; moment of inertia — момент інерції.

Literature: [3, chapter 2, section 2.3], [9, chapter 10, § 2], [15, chapter 12, section 12.2], [16, chapter 14, § 11—14], [17, chapter 2, § 8].

6.1. Principal Definitions. Condition for the Existence of a Triple Integral

A triple integral is an analogue of a double integral and is introduced for a function of three variables.

Suppose we are given a material body occupying a three-dimensional region

Ωfilled with a matter. The density at each point of it is know too. Suppose

μ= μ(P) = μ(x, y, z) .

The task is required to find the mass m of the body.

We will break Ω into nonoverlaping cubable (i.e., having a volume) parts Ωi , i = 1, 2, 3… n having volumes Vi , respectively.

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