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14.28. ∫(x2 + y3 )dx +(x x − y)dy; L : y = x4 ,0 ≤ x ≤1.
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14.29. ∫ |
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14.30. ∫ xe−3y dx + yex4 dy; L : y = x2 ,0 ≤ x ≤1.
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Задание 15. Вычислить криволинейный интеграл второго рода ∫Р(x, y, z)dx +Q(x, y, z)dy + R(x, y, z)dz .
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15.1. ∫ydx + zdy + xdz; A(0 ;0;0); B(1;2;3);
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15.2. ∫(x + y)dx + (y + z)dy + (z + x)dz; A(0; 0;1); B(2;−2;3) .
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15.3. ∫e y dx + e2z dy + e3x dz; A(1 ;0;0); B(2;3;4) .
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15.4. ∫cos zdx +sin xdy + cos ydz; A(0 ;0;0); B(3;2;1) .
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15.5. |
∫(z − y)dx + (x − z)dy + (y − x)dz; A(0 ;1;0); B(1;2;3) . |
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15.6. |
∫ |
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15.7. ∫ yez2 dx + zex2 dy + xe y2 dz; A(0 ;0;0); B(2;3;5).
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15.8. |
∫ |
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dz; A(2;2;3); B(3;4;5). |
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15.9. |
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∫ |
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z + xdy + x + ydz; A(0 ;0;0); B(2;3;7 ) . |
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15.10. ∫ y3dx + z3dy + x3dz; A(0 ;0;1); B(2;4;7 ) .
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15.11. ∫ ydx + z2dy + x3dz; a =1;b = 2 ;c = 3.
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15.12. ∫
ydx + 3
zdy + xdz; a =1;b = 4 ;c = 8 .
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15.13. ∫ xe y dx + xez dy + ex3 dz; a =b = c =1.
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15.14. ∫ zdx + x2dy + 
ydz; a = c =1;b = 4.
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15.15. ∫ y2dx + zdy + x3dz; a =1;b = 3 ;c = 2 .
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15.16. ∫(z − y)dx + (x − z)dy + (y − x)dz; a =b = 2 ;c = 3; .
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15.17. ∫(z2 − y2 )dx + (x2 − z2 )dy + (y2 − x2 )dz
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a=1;b = −1 ;c = 2; .
15.18.∫ yzdx + zxdy + xydz; a = 2;b =1 ;c = −3; .
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15.19. ∫(y + z)dx + (z + x)dy + (x + y)dz; a =1;b = c = 2 .
L
15.20. ∫(x2 + z2 )dx + (z2 + x2 )dy + (x2 + y2 )dz
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15.21. ∫ ydx + zdy + xydz; R =1;h = 2 . |
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15.22. ∫ |
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15.23. ∫ xydx + |
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15.24. ∫ y2dx + x2dy +(x + y)dz; R = 3;h = 2 . |
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15.25. ∫ |
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15.26. ∫(x2 + y2 )dx + (x2 + y2 )dy +(2x − y)dz; R =1;h = 4 .
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15.27. ∫(x + y)dx + (y − x)dy + x2 ydz; R =2;h = 5 .
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15.28. ∫ xdx
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15.29. ∫(y −
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+ xy dy + xy2dz; R = 3;h = 2 .
2x)dx + (y − x)dy + x3dz; R =1;h = 2 .
15.30. ∫ |
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Задание 16. Вычислить криволинейный интеграл по замкнутому контуру
а) непосредственно, б) по формуле Грина.
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∫P(x, y)dx +Q(x, y)dy |
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1 |
∫(1− x2 )dx + 2xydy |
y = x, y = 0, x = 3 |
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2 |
∫(1+ y 2 )dx +(x + y)dy |
y = x, y = 2, x = 0 |
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∫(x2 +2xy)dx +(2xy + y2 )dy |
y = x +1, |
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y = 0, x = 0, x = 3 |
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∫ y2dx +(x + y)2 dy |
y = 2 − x, x = 2, y = 2 |
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∫(x + y)2 dx +(x2 − y2 )dy |
y = 2 − x, x = 0, y = 0 |
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∫ yx2dx + y2dy |
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∫2(y − x)dx + (x + y)dy |
y = 4x − x2 , y = 0 |
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8 |
∫2ydx + (y − x)dy |
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∫(xy + x)dx +(x − y)dy |
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10 |
∫x2dx +(x + y2 )dy |
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∫ y |
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x2 + y2 = 9 (I четверть), |
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∫ y |
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∫(2x + y2 )dx +(x + y)2 dy |
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14 |
∫ y2dx − x2dy |
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15 |
∫(x − y2 )dx +8xydy |
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∫− x2 ydx + xy2dy |
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x2 + y2 = 4 |
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17 |
∫(xy − 2y)dx + x2dy |
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18 |
∫(x + y)dx +3x2dy |
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x2 + y2 = R2 |
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19 |
∫(3x2 y + y)dx +(x3 −2y)dy |
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20 |
∫ x2 ydx + x3dy |
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y = x3 , x = 0, y = 8 |
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21 |
∫(2x2 +3y)dx +(y3 +2x)dy |
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y = 2 − 8 , y = 0 |
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∫(x2 + y2 )dx + |
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y = x, y = 2 − x, x = 0 |
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+ (3x2 + 2xy − y2 )dy |
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∫(2xy − y)dx +(x2 +2x)dy |
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24 |
∫(xy −5y2 )dx + |
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∫(ex sin y − y)dx + |
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π |
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cos y −1)dy |
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∫(3x2 +5y)dx +(2y3 − x)dy |
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∫(x −5y2 )dx + y3dy |
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y = |
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∫(3x2 y + y2 )dx +(x3 + y2 )dy |
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∫2ydx + (5x2 + 2x)dy |
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∫(x |
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Задание |
17. |
Вычислить |
поток |
векторного поля |
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а = аxi + аy j + аz k |
через часть плоскости S , |
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расположенную |
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в первом октанте (нормаль образует острый угол с осью OZ ). |
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а = аxi + аy j + аz k |
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a = x i + y j + z k |
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x + y + z = 2 |
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x = 2y + 2z = 2 |
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3 |
a = (2х+ у) i + у j + 2z k |
2x + 2y + z − 2 = 0 |
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a = (5 − 2х) i + x(x + у) j + xz k |
x + y + 2z = 2 |
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a = (2 + х) i + у j + z k |
6x +3y + 2z −6 = 0 |
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a = 3хi + 2( у − х) j + 2z k |
2x + y + z − 2 = 0 |
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a = x i − y j + (z + 4y) k |
2x + 6y +3z = 6 |
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a = 3хi +3у j +3z k |
3x + 2y + 6z −6 = 0 |
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a = x i − (y + 4x) j − z k |
2x + y + 2z − 2 = 0 |
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3x +3y + 2z = 6 |
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a = (3х −8) i +3у j +3z k |
3x + 2y + 2z −6 = 0 |
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a = (х +3y) i − 2 j + z k |
x +3y +3z −3 = 0 |
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a = (х + 2у) i + 2z k |
x + 2y + 2z = 2 |
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a = (x − 2y) i +3y j + (z −7) k |
x + y + z = 2 |
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a = 2хi −3(2y − x) j +3(y + z) k |
3x + y +3z −3 = 0 |
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a = (х −5) i + у j + z k |
3x +3y + z −3 = 0 |
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a = 2x i +5y j + (5z +3х) k |
x + y + z = 3 |
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a = 2(х+ у) i + (у −3) j + 2z k |
2x + 4y + z = 4 |
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a = (2х −3) i + 2у j + 2z k |
2x + y + 4z − 4 = 0 |
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a = (2х + у) i −5 j + 2z k |
4x + 2y + z − 4 = 0 |
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21 |
a = (4 − х) i + (2 − у) j − z k |
x + 2y + 2z = 4 |
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a = −20 i + y j + (z + х) k |
x + 2y + z − 4 = 0 |
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a = −хi + (7 − у) j − z k |
3x + 2y + z −6 = 0 |
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a = (6 + х) i −(3 − у) j + (z + 2) k |
x + 2y +3z = 6 |
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a = (2х −7) i + 2у j + 2z k |
2x + y +3z −6 = 0 |
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a = 8x i + (7 y −3х) j + 7(z − 4) k |
6x + 2y + z = 6 |
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a = хi + (у −9) j + z k |
2x + y + 6z −6 = 0 |
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a = (х + 2) i − у j + (z + 4у) k |
2x + 2y + z − 4 = 0 |
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a = хi + 4у j + (z −9у) k |
3x +3y + z = 6 |
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a = 2хi −(10 − у) j + (z − х) k |
2x + y + 2z −6 = 0 |
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