
математика_2семестр
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Вариант 2.
2x
ZZ
1: |
dx f(x; y) dy: |
11 x
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ZZ |
18x2y2 + 32x3y3 |
dxdy; D : x = 1; y = x3 |
; y = p3 |
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3: x = y2 2y; y = x: |
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y = 5p |
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x; y = |
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Вариант 3.
3
2y+3
ZZ
1: |
dy f(x; y) dx: |
02y2
ZZ p
2: 12xy + 9x2y2 dxdy; D : x = 1; y = x; y = x2:
D
3: y2 = 4x + 4; y = 2 x:
p
4: x2 + y2 = 2; y = x; z = 0; y = 0; z = 15x:
Вариант 4.
p
425 x2
ZZ
1: |
dx |
f(x; y) dy: |
03x
4
2: |
ZZ |
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8xy + 9x2y2 |
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dxdy; D : x = 1; y = x2 |
; y = p3 |
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3: 3y2 = 25x; 5x2 = 9y: |
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x + y = 2; y = p |
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x; z = 0; z = 12y: |
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Вариант 5.
1x+3
ZZ
1: |
dx f(x; y) dy: |
02x2
11

2: |
ZZ |
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18x2y2 + 32x3y3 |
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dxdy; D : x = 1; y = x2 |
; y = p3 |
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3: y = x2; 4y = x2; y = 4: |
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x = 20p2y; x = 5p2y; z = 0; z + y = |
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Вариант 6.
5y
04
ZZ
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dy |
f(x; y) dx: |
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9+y2 |
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2: |
27x2y2 |
+ 48x3y3 |
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dxdy; |
D : x = 1; y = x3; y = p |
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3: xy = 4; y = x; x = 4: |
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Вариант 7.
0x+3
ZZ
1: |
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dx |
f(x; y) dy: |
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2x2 |
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2: |
ZZ |
4xy + 3x2y2 |
dxdy; D : x = 1; y = x2 |
; y = p |
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3: x = 4 y2; x + 2y 4 = 0:
4: x2 + y2 = 2; x = py; x = 0; z = 0; z = 30y:
Вариант 8.
1x2+1
ZZ
1: |
dx |
f(x; y) dy: |
01
2: |
ZZ |
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8xy + 18x2y2 |
dxdy; D : x = 1; y = x2; y = p3 |
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12

3: y = x2; 4y = x2; x = 2; x = 2:
4: x + y = 2; x = py; z = 0; z = 125x :
Вариант 9.
13 y
ZZ
1: |
dy f(x; y) dx: |
02y2
2: |
ZZ |
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5xy + 9x2y2 |
dxdy; |
D : x = 1; y = x3; y = px: |
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3: x + 4 = y2; x + 3y = 0: |
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4: y = 17 2x; y = 2 2x; z = 0; x + z = |
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Вариант 10. p
425 y2
ZZ
1: |
dy |
f(x; y) dx: |
03y
4
2: |
ZZ |
6xy + 24x3y3 |
dxdy; D : x = |
1; y = x2; y = p |
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3: y = x2; y = 6 x; y = 0: |
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5p |
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5x |
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y = |
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Вариант 11.
03 x
ZZ
1: |
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dx f(x; y) dy: |
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2x2 |
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2: |
ZZ 4xy + 16x3y3 dxdy; D : x = 1; y = x3; y = p3 |
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x: |
D
3: y = 4 x2; y = x + 2:
13

4: x2 + y2 = 8; y = p2x; y = 0; z = 0; z = 1511x:
Вариант 12. p
49+y2
ZZ
1: |
dy |
f(x; y) dx: |
05y
4
2:
3:
4:
ZZ |
6x2y2 + 3 x4y4 |
dxdy; D : x = 1; y = x2; y = px: |
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D |
25 |
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y = 6 x; y = x5: p
x + y = 4; y = 2x; z = 0; z = 3y:
Вариант 13.
12 x
ZZ
1: |
dx f(x; y) dy: |
0x
2: |
ZZ |
xy 4x3y3 |
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dxdy; D : x = 1; y = x3; y = p |
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3: y = x3; y = 4x: |
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4: |
x = 6 |
; x = 18 |
; z = 0; z = |
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5p |
y |
5y |
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3 + p |
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Вариант 14.
2x+2
ZZ
1: |
dx f(x; y) dy: |
0x2
2: |
ZZ 3x2y2 + 3 x4y4 |
dxdy; D : x = 1; y = x3; y = p3 |
x: |
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D |
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3: y2 = 9x; y = x + 2: |
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x = 19p |
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4: |
2y; x = 4p2y; z = 0; y + z = 2: |
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14

Вариант 15.
1x2
ZZ
1: |
dx f(x; y) dy: |
0x2
2: |
ZZ |
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44xy + 16x3y3 |
dxdy; D : x = 1; y = x2 |
; y = p3 |
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3: y2 = x + 1; y2 = 9 x: |
30y |
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4: |
x2 + y2 = 8; x = p2y; x = 0; z = 0; z = |
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Вариант 16.
14 x2
ZZ
1: |
dx |
f(x; y) dy: |
02x+1
2: |
ZZ y cos xy dxdy; D : x = 1; x = 2; y = 2 |
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D |
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3: xy = 4; x + y 5 = 0: |
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x + y = 4; x = p |
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3x |
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2y; z = 0; z = |
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Вариант 17.
1x
ZZ
1: |
dx f(x; y) dy: |
0x
2: |
ZZ y2 sin 2 |
dxdy; D : x = 0; y = 2 |
; y = |
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xy |
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D |
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x y + 1 = 0; y = cos x; y = 0: |
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p |
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4: y = 6 3x; y = 3x; z = 0; x + z = 3:
p
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Вариант 18.
p
22 x
ZZ
1: |
dx f(x; y) dy: |
0x2
4
15

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xy |
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2: |
y2e 4 dxdy; |
D : x = 0; y = 2; y = x: |
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3: y = 2x x2; y = x: |
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5p |
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5 (3 + p |
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4: y = |
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Вариант 19.
p
325 x2
ZZ
1: |
dx |
f(x; y) dy: |
00
2: ZZ |
4ye2xydxdy; D : x = 1; x = 2 |
; y = ln 3; y = ln 4: |
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3: y = p |
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x; xy = 1; x = 2; y = 0: |
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4: x2 + y2 = 18; y = p |
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5x |
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3x; y = 0; z = 0; z = |
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Вариант 20.
p
425 x2
ZZ
1: |
dx |
f(x; y) dy: |
00
2: |
ZZ |
4y2 sin xy dxdy; D : x = 0; y = x; y = r |
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D |
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3: y = x2 + 4; 2x + y 4 = 0: |
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x + y = 6; y = p |
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4: |
3x; z = 0; z = 4y: |
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Вариант 21.
12 y
ZZ
1: |
dy f(x; y) dx: |
00
2: |
ZZ y sin xy dxdy; D : x = 1; x = 2; y = 2 |
; y = : |
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D
3: y = x2 + 2; x + y = 4:
16

pp
4: x = 7 3y; x = 2 3y; z = 0; x + z = 3:
Вариант 22.
p
4x
ZZ
1: |
dx f(x; y) dy: |
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ZZ y2e 2 dxdy; D : x = 0; y = x; y = p |
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3: y = x2 + 8; y = x2: |
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z = x2 + y2; y = x2; z = 0; y = 1: |
Вариант 23.
23 x
ZZ
1: |
dx f(x; y) dy: |
00
2: |
ZZ |
2y cos 2xy dxdy; D : x = 1; x = 2; y = 4 |
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3: |
y = 2 x + 1; y = 4 2x; y = 0: |
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4: y = x2; z + y = 2; x = 0; z = 0:
Вариант 24.
3
4x
ZZ
1: |
dx f(x; y) dy: |
0x2
2: |
ZZ |
8ye4xydxdy; D : x = 4 |
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2; y = ln 3; y = ln 4: |
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3: y = (x + 1)2; y2 = x + 1: |
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y + z = 1; x = y2 + 1; x = 0; y = 0; z = 0: |
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Вариант 25.
1p
2y
ZZ
1: |
dx f(x; y) dy: |
0y
2: |
ZZ |
3y2 sin 2 dxdy; |
D : x = 0; y = |
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D
3: y = (x 2)3; y = 4x 8:
p
4: z = 1 y; x2 = y; z = 0:
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Контрольная работа 7
Содержание контрольной работы 7
Задание 1
Вычислите криволинейный интеграл первого рода по данной линии.
Задание 2
Вычислите работу силы ~
F (x; y) при перемещении вдоль линии L от точ-
ки A до точки B.
Задание 3
Вычислите поверхностный интеграл первого рода по поверхности S, где S часть плоскости , отсеч¼нная координатными плоскостями.
Задание 4
Вычислите поверхностный интеграл второго рода по поверхности S, где S часть плоскости , отсеч¼нная координатными плоскостями, в направлении нормали, образующей острый угол с осью Oz. .
Указание.
Перед решением задач контрольной работы рекомендуется ознакомиться со следующими методическими указаниями:
1.Фаттахова М.В., Купчиненко М.Б. Криволинейные интегралы. Решение типовых задач: Методические указания. СПб.: СПбГТИ(ТУ),- 2008.- 32 c.
Условия задач контрольной работы 7
Вариант 1.
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L : x = t cos t; y = t sin t; z = t; |
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4; 0); B(0; 2): |
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AB; A( |
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F = x 2y |
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отрезок прямой |
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ZZ (2x + 3y + 2z)dS; : x + 3y + z = 3: |
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4: |
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3xdydz + (y + z)dxdz + (x z)dxdy; |
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Вариант 2. |
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L : x2 + y2 = 4: |
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4; 0); B(0; 2): |
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AB; A( |
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отрезок прямой |
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ZZ (2 + y 7x + 9z)dS; : 2x y 2z = 2: |
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4: |
ZZ (3x 1)dydz + (y x + z)dxdz + 4zdxdy; |
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Вариант 3. |
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1: |
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2: |
F~ |
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+ 2y ~{ + y2 |
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L : 2 |
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2 |
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3: |
ZZ (6x + y + 4z)dS; |
: 3x + 3y + z = 3: |
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4: |
ZZ xdydz + (x + z)dxdz + (y + z)dxdy; |
: 3x + 3y + z = 3: |
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S |
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Вариант 4. |
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Z |
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4p3 |
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3p |
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1: |
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dl; |
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L |
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L отрезок прямой AB; A( 1; 0); B(0; 1): |
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2: |
~ |
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L : x |
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+ y |
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= 4 (y > |
0) ; A(2; 0); B( 2; 0): |
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F = (x + y)~{ + 2x~{; |
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20