Example 5. Calculate the circulation of the vector field F = xzi + yx j + zyk along the line of intersection of the plane x + y + z = 3 with coordinate planes by direct calculation and by Stokes’ formula.
Solution. The first method (direct calculation).
C = ∫ xzdx + xydy + zydz = I1 + I2 + I3
L
where, I1 , I2 , I3 are integrals along line segments AB, BC and CA accordingly. Let us evaluate the integral I1 .
On moving along the segment AB from the point A to the point B the variable x changes from 3 to 0, y = 3 − x and
z = 0 (Fig. 9.11). Then
0
I1 = ∫ xzdx + yxdy + zydz =∫(3 − x)xd(3 − x) =
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Symmetry of the problem relative to the variables x, y and z implies that I2 = I3 = I1 = 4,5, therefore C = 3 4, 5 = 13, 5.
The second method. We can calculate the circulation of the vector field by Stokes’ formula. To this end we consider the upside of the plane σ (triangle ABC) and a positive direction ABCA corresponding to this side. We have
P = xz, Q = yx, R = zy, |
∂P |
= 0, |
∂Q |
= y, |
∂R |
= z, |
∂Q |
= 0, |
∂P |
= x, |
∂R |
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Putting these values in Stokes’ formula (9.14) and using the formula (9.12), we obtain
C = ∫ xzdx + xydy + zydz = ∫∫ ydxdy + zdydz + xdzdx
Let us express the last surface integral by double integrals over domains which are projections of the triangle ABC onto the coordinate planes:
C = ∫∫ |
ydxdy + ∫∫ |
zdydz + ∫∫ xdzdx, |
ABO |
BCO |
AOC |
where ABO, BCO, AOC are projections of the triangle ordinate planes Oxy, Oxz, Oyz. We have
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∫∫ |
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By analogy ∫∫ zdydz = |
∫∫ xdzdx = 4,5. Thus, C = 3 4,5 = 13,5. |
BCO |
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Example 6. Calculate a circulation of the vector F = yi + x j + k along a circle x2 + y2 = 1, z = 0 in the positive direction.
Solution. The first method (direct calculation). We can write down the equation of the given circle L in a parametric form:
x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π.
So as
P = − y = − sin t, |
Q = x = cos t, R = 1, |
dx = − sin tdt, |
dy = cos tdt, dz = 0, |
By definition of circulation (see a formula (9.12)) we obtain
C = ∫ (− y)dx + xdy + dz = |
2∫π ((− sin t )(− sin t ) + cos t cost )dt = |
L |
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2π |
2π |
= ∫ (sin2 t + cos2 t)dt = ∫ dt = 2π. |
0 |
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The second method (by Stokes’ formula). By formulas (9.12) and (9.14) we have (D : x2 + y2 ≤ 1) :
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C = ∫∫(1− (−1))dxdy = 2∫∫ dxdy = 2 ∫ dϕ∫ρdρ = 2π. |
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Example 7. Prove that the vector field |
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F = (3x2 y2 + 2xz3 )i + (2 yx3 − z2 ) j + (3x2 z2 − 2yz + 4z3 )k |
is potential, and find its potential. |
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Solution. To prove that the field is potential it |
is enough to prove that |
rot F = 0. We have |
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rot F = |
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3x2 y2 + 2xz3 |
2 yx3 − z2 |
3x2 z2 − 2yz + 4z3 |
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Therefore, the field of vector F is potential. |
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We can define a potential u(x, |
y, z) by the formula (9.16), taking the origin |
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u(x, y, z) = ∫ |
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+∫ 2yx3dy +∫(3x2 z2 − 2yz + 4z3 )dz = y2 x3 +x2 z3 − yz2 + z4 + C. |
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Micromodule 9 |
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CLASS AND HOME ASSIGNMENTS |
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1. Find the surfaces of level of the scalar field u = |
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2. Calculate a derivative of the scalar field u = x2 y + 2 y2 z − z2 x + 1 at the |
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M0 (0; 2; 1) in the direction of the point M1 (2; 4; 0) . |
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3. Calculate a derivative of |
the scalar field |
u = z |
x2 + y2 |
at the point |
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M (3; 4; 1) |
in direction of the vector forming the angles |
α = 45° |
and β = 60° |
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with the coordinate axes Ox and Оу respectively. |
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4. Calculate a derivative of the scalar field u = |
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at the point |
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M (6; − 2; 3) |
in direction of its gradient calculated at this point. |
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5. Calculate a derivative of |
the field u = ln (x2 + y2 − z) |
at the point |
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M (1; |
3; 3) |
of the circle x = 2 cos t, y = 2 sin t, z = 3 in the positive direction |
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of this circle. |
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6. Find the gradient of the scalar field u = 2x3 y − 3yz2 + zx − 2y + 3 at the |
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7. The scalar field u = arctg |
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calculated at the points M1 (1; 0) |
and M2 (−1; 1) . |
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8. Find points where a gradient of the scalar field u = ln(x − y−1) is equal to i + j.
9.Find vector lines of the field F = (z − y)i + (x − z) j + ( y − z)k.
10.Find a vector line of the field F = i − y2 j + z2 k , passing through the point M (1; 2; 3).
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11. Calculate the flux of the vector field F = (x + 3y)i + ( y − z) j + (z + x)k
across the upper part of circle, that is cut out by a cone x2 + y2 = z2 from the plane z = 1 .
12. Calculate the flux of the vector field F = xi + zk through the lateral surface of the circular cylinder x2 + y2 = 4 bounded by the planes z = 0 and z = 1 .
13. Calculate the flux of the vector field F = (x + y)i + yzj + (x − z)k
through the closed surface bounded by the paraboloid z = 1− x2 − y2 and the plane z = 0.
14. |
Calculate the flux of the vector field |
F = x2 i + y2 j + z2 k through the |
closed surface bounded by the hemisphere z = |
1− x2 − y2 and the plane z = 0. |
15. |
Calculate the flux of the vector field F = (3x − 2)i − 5yj + (2z + 1)k |
through |
the outer side of the cone |
x2 + y2 |
= z2 |
( 0 ≤ z ≤ 3 ), using Gauss– |
Ostrogradsky formula after completing the given surface to the closed one. |
Calculate the circulation of the vector field |
F along a closed contour L. |
16. |
F = zi − xj + yk , L is a circle |
z = x2 + y2 − 10, z = −1 . |
17. |
F = xyi + yzj + xzk , L is an ellipse x2 + y2 |
= 1, x + y + z = 1. |
18. F = y2 i + z2 j + x2 k , L is a closed contour ABCA, formed by intersection of the plane x + y + z = 3 with coordinate planes.
Calculate the circulation of the vector field F along a closed contour L using: a) direct integration; b) Stokes’ formula.
19.F = x2 y3i + j + zk , L is a circle x2 + y2 = R2 , z = 0 .
20.F = zi − yk , L is an ellipse x2 + y2 = 4, x + 2z = 5 .
21. F = x2 i + y2 j + z2 k , L is a circle x2 + y2 + z2 = 10, z = 1 .
22. Find the divergence and rotation of the vector field of a radious vector r = xi + yj + k .
Find the rotation of the vector field F at the point M (−1; 2; 0).
23.F = (x2 + y2 )i + ( y2 + z2 ) j + (z2 + x2 )k .
24.F = (2x2 y − z)i + (3xz2 + 1) j + (x2 + y2 + z2 )k .
25.Prove that the vector field
F = (4x3 z + y2 − 3)i + (2xy + z2 + 1) j + (x4 + 2yz)k
is potential, and find its potential.
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Answers
1. z = C x2 + y2 — are cones whose vertex is the origin of coordinates; Oz — is a symmetry axis.
2. 2. 3. |
(29 + 3 |
2) /10 ; (3 2 − 21) /10 . 4. 1. 5. 4 |
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grad u = −2i |
− 12 j + k. |
7. |
3π |
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8. (2; 1), |
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13. π . |
(0; –1). |
9. |
x + y + z = C , x2 + y2 + z2 = C . 10. |
x + |
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4π. |
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15. −36π. 16. −9π. |
17. −π. 18. –27. 19. |
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23. 2 j − 4k. 24. |
2i − 5 j − 8k. |
25. u = x4 y + xy2 + yz2 − 3x + y + C. |
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Micromodule 9
SELF-TEST ASSIGNMENTS
9.1. Calculate the flux of the vector F(x, y, z) across the outer surface of the pyramid
bounded by coordinate planes, and the given sloping plane, using GaussOstrogradsky formula.
№ |
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Equation of plane α |
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9.1.1 |
F = xyi + (x4 − y) j + (z − 2y)k |
2x + y + 2z = 2 |
9.1.2 |
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= xi + (2 − z) j + (z2 + 2x)k |
2x − 6 y + 3z = 6 |
F |
9.1.3 |
F = xzi + (2x3 + y) j + (z − 3y)k |
2x + 2 y − z = 2 |
9.1.4 |
F = (z − x)i + (z + y2 ) j + (x2 + 2 y)k |
3x − 6 y + 2z = 6 |
9.1.5 |
F = (x + 2)i + (z − 2 y) j + z2 k |
2x + 2 y − z = 2 |
9.1.6 |
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= x2 i + (2 − z) j + (z + 2x)k |
3x − 2y + 6z = 6 |
F |
9.1.7 |
F = ( yx + 1)i + (2z − x) j + zk |
2x + y − 4z = 4 |
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9.1.8 |
F = (x + z)i + (xy − 2) j + (3x + y)k |
3x + 2 y − 6z = 6 |
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9.1.9 |
F = (2x + yz)i + (2 y − x) j + z2 k |
6x − 4 y + 3z = 12 |
9.1.10 |
F = (x2 − 1)i + (x − zy) j + (x + yz)k |
6x + 3y + 4z = 12 |
9.1.11 |
F = (xy + 1)i + (2z − 3y) j + 3zk |
4x − 6 y + 3z = 12 |
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9.1.12 |
F = (x + 2 yz)i + (3y − x) j + ( y − 2z)k |
6x − 4y − 3z = 12 |
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9.1.13 |
F = (xz − 3)i + (x3 + 2 y) j + (z − 3y)k |
10x − 4y + 5z = 20 |
9.1.14 |
F = (z − x)i + (2 + y2 ) j + (x + 2 y)k |
12x − 20y + 15z = 60 |
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№ |
F(x, y, z) |
Equation of plane α |
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9.1.15 |
F = (xy − 2)i + (z − 2y) j + 2zk |
3x + 4y + 6z = 12 |
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9.1.16 |
F = (x − y)i + (4 − zy) j + (z + 2x)k |
15x − 10 y + 6z = 30 |
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9.1.17 |
F = (x2 + 1)i + (2z − y) j + (z + 2)k |
10x − 4y − 5z = 20 |
9.1.18 |
F = (x2 + z)i + xyj + (3x + y)k |
6x − 4y − 3z = 12 |
9.1.19 |
F = (2x + yz)i + (3y − x) j + (z − y)k |
4x − 6 y + 3z = 12 |
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9.1.20 |
F = xi + (x − z) j + (x + yz)k |
20x + 12y − 15z = 60 |
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9.1.21 |
F = (2x − y)i + (1− 2zy) j + z2 k |
15x − 6 y + 10z = 30 |
9.1.22 |
F = (x + y)i + (1+ y) j + (z + 4x)k |
6x + 4y − 3z = 12 |
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9.1.23 |
F = x2 i + (2x + y) j + (z − 3y)k |
10x + 5y + 4z = 20 |
9.1.24 |
F = (z − x)i + (z + 3y2 ) j + (x + 2 y)k |
20x + 4y − 5z = 20 |
9.1.25 |
F = (x + y)i + (z − 2 y) j + zxk |
2x − 4y − z = 4 |
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9.1.26 |
F = xyi + (2 − zy) j + ( y + 2x)k |
5x − 2y − 10z = 10 |
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9.1.27 |
F = xi + (2z − x) j + zyk |
2x − 4y + z = 4 |
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9.1.28 |
F = (x2 + z)i + xj + (3x + y)k |
15x + 3y + 5z = 15 |
9.1.29 |
F = (2x + yz)i + (2 y − x) j + zyk |
2x − 3y − 6z = 6 |
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9.1.30 |
F = xyi + (x − zy) j + (x + yz)k |
15x + 10y − 6z = 30 |
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9.2. Calculate the circulation of the vector field F along the line of intersection of the plane α with coordinate planes, using the direct calculation and Stokes’ formula (direction of integration is counter-clockwise, if to look from the origin of coordinates).
9.2.1. |
F = (x + y)i + x2 yj + zy)k , |
α : |
x + y + z = 2 . |
9.2.2. |
F = xzi + ( y − z) j + z2 k , |
α : |
x − y + z = 1 . |
9.2.3. |
F = xzi + (x + y) j + (z + y)k , |
α : |
x + 2 y + z = 2 . |
9.2.4. |
F = (x − y)i + (z + y) j + 4k , |
α : |
x − 3y + z = 3 . |
9.2.5. |
F = xyi + zyj + (z2 + x2 )k , |
α : |
x + y − 5z = 5 . |
9.2.6. |
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= (x + 2)i + ( y − 2z) j + (z + x)k , |
α : |
2x + 3y + z = 6 . |
F |
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9.2.7.F = (x + y)i + ( y − z) j + (z + 2x)k ,
9.2.8.F = xzi + xyj + yk ,
9.2.9.F = (x + 3z)i + 2 j + (z2 + 1)k ,
9.2.10.F = (2x + y)i + zyj + xzk ,
9.2.11.F = (x + y)i + zyj + 2k ,
9.2.12.F = (x + 1)i + ( y + z) j + zk ,
9.2.13.F = xzi + (2y + 1) j + (z − y)k ,
9.2.14.F = zi + yj + xk ,
9.2.15.F = (x + y)i + (z + y) j + (x + z)k ,
Calculate the circulation of the vector field the direct calculation and Stokes’ formula
9.2.16.F = (x2 − y2 )i + 4 j + (z + x)k ,
9.2.17.F = (x2 + y2 )i + zyj + 3k ,
9.2.18.F = (x + z)i + 4 j + (z2 − x)k ,
9.2.19.F = 2xyi + ( y2 + 2z) j − 2k ,
9.2.20.F = xi + yj + (z2 + y2 )k ,
9.2.21.F = yi + xj + (z2 − y2 )k ,
9.2.22.F = (x + y)i + (z + y) j + (z + x)k ,
9.2.23.F = x2 i + y2 j + zk ,
9.2.24.F = (z + x)i + zj + yk ,
9.2.25.F = (x2 + y)i + ( y2 + z) j + k ,
9.2.26.F = xyi + zyj + yzk ,
9.2.27.F = xi + zxj + (z + y)k ,
9.2.28.F = (x2 + z2 )i + 3 j + zk ,
9.2.29.F = (x − y)i + ( y − z) j + (z − x)k ,
9.2.30.F = 2i + y2 xj + zk ,
α: 2x + y + z = 2 .
α: 3x + y − z = 3 .
α: 6x + 2 y + 3z = 6 .
α: x − y − z = 1 .
α: x + y + 4z = 4 .
α: x + 2 y + 3z = 6 .
α: 2x + 4 y + z = 8 .
α: 5x + y + z = 5 .
α: 3x + 2y + 2z = 6 .
F along the closed line L, using
L: x2 + y2 = 4, z = 2.
L: z = x2 + y2 , z = 4.
L: x2 + y2 + z2 = 10, x = 1. L: x2 + z2 = 1, y = 0.
L: y = x2 + z2 , z = 1.
L: x2 + y2 + z2 = 5, z = 1.
L: z2 + y2 = 9, z = 0.
L: x = z2 + y2 , x = 9.
L: x2 + z2 = 1, x + y + z = 1.
L: x2 + y2 = 1, z = 4.
L: z = x2 + y2 − 3, z = 1. L: x2 + y2 = 4, x + z = 0.
L: x2 + z2 = 9, y = 2.
L: x = y2 + z2 + 2, z = 6. L: x2 + z2 = 4, y + z = 0.
9.3. Prove the vector field to be potential and find its potential.
9.3.1.F = (3x2 + y2 )i + (2xy + z) j + ( y + 3z2 )k.
9.3.2.F = (2x + y + z)i + (x + 2 y + z) j + (x + y + 2z)k.
9.3.3.F = (4x3 + yz)i + (4 y3 + xz) j + (xy + 4z3 )k.
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9.3.4.F = (3x2 + y2 z3 )i + (−3y2 + 2xyz3 ) j + 3z2 (xy2 − 1)k.
9.3.5.F = ( y2 z2 + 2xy)i + (2xyz2 + x2 ) j + 2xy2 z2 k.
9.3.6.F = (2xyz3 − 1)i + (x2 z3 + 2) j + (3x2 yz2 + 3)k.
9.3.7.F = (x + y + 2z)i + (x + y + 3z) j + (2x + 3y + z)k.
9.3.8.F = (2x + 2)i + (2 y − 3) j + (2z + 4)k.
9.3.9.F = (x + y − z)i + (x − y + z) j + (z + y − x)k.
9.3.10.F = (2x − z3 )i + 2( y + z) j + (2 y − 3xz2 )k.
9.3.11.F = (4x3 y2 z − 1)i + 2x4 yzj + (x4 y2 + 2z)k.
9.3.12.F = (2xy4 z4 + y)i + (4x2 y3 z4 + x) j + (4x2 y4 z3 − 1)k.
9.3.13.F = (3x2 + 3z + 2)i + (3y2 − 1) j + 3(x + z2 )k.
2 y2 z + yz2 )i + (2x3 yz + xz2 ) j + (x3 y2 +
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9.3.16.F = (x + 2 y − z)i + (2x − y + 3z) j + (z + 3y − x)k.
9.3.17.F = (3z2 + y2 x3 )i + (−3y2 + 2zyx3 ) j + 3x2 (zy2 − 1)k.
9.3.18.F = (2xyz3 − 1)i + ( y2 z3 + 2) j + 3( y2 xz2 + 1)k.
9.3.19.F = ( y2 z2 + 2xz)i + (2xzy2 + x2 ) j + 2xz2 y2 k.
9.3.20.F = (2 y − z3 )i + 2(x + z) j + (2x − 3yz2 )k.
9.3.21.F = (4z3 y2 x −1)i + 2z4 yxj + (z4 y2 + 2x)k.
9.3.22.F = (2 yx4 z4 + x)i + (4 y2 x3 z4 + y) j + (4 y2 x4 z3 − 1)k.
9.3.23.F = (3x2 + 3y + 2)i + (3z2 − 1) j + 3(x + y2 )k.
9.3.24.F = (3z2 y2 x + yx2 )i + (2z3 yx + zx2 ) j + (z3 y2 + 2zy)k.
9.3.25.F = (2xz4 − 4)i + 2yzj + (4x2 z3 + y2 )k.
9.3.26.F = (2xz + y3 )i + 3xy2 j + (x2 − 3z2 )k.
9.3.27.F = (2xy4 z − y2 )i + (4x2 y3 z − 2xy) j + (x2 y4 + 2)k.
9.3.28.F = (ln y + zx−1 )i + (ln z + xy−1 ) j + (ln x + yz−1 )k.
9.3.29.F = (2e2x y − ez )i + (e2x + ey z2 ) j + (2zey − xez )k.
9.3.30.F = (sin y − 2z cos 2x)i + (x cos y + cos z) j − ( y sin z + sin 2x)k.
228
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1.Бахвалов Н. С., Жидков Н. П., Кобельков Г. М. Численные методы. —
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матика: Підручник. У 2 ч. Ч. 1: Лінійна і векторна алгебра. Аналітична геометрія. Вступ до математичного аналізу. Диференціальне і інтегральне числення. — К.: Техніка, 2000. — 592 с.
7.Овчинников П. П. Вища математика: Підручник. У 2 ч. — Ч. 2: Диференціальні рівняння. Операційне числення. Ряди та їх застосування. Стійкість за Ляпуновим. Рівняння математичної фізики. Оптимізація і керування. Теорія ймовірностей. Числові методи. — К.: Техніка, 2000. — 792 с.
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М.: Наука, 1985. — Т. 2. — 456 с.
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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Module 1. SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . |
. 4 |
Micromodule 1. |
Number series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . |
4 |
Micromodule 2. |
Functional series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . |
35 |
Micromodule 3. |
Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . |
69 |
Micromodule 4. |
Fourier integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . |
87 |
Module 2. MULTIPLE INTEGRALS. LINE AND SURFACE
INTEGRALS. FIELD THEORY . . . . . . . . . . . . . . . . . . . . . . 98
Micromodule 5. Double integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Micromodule 6. Triple integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Micromodule 7. Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Micromodule 8. Surface integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Micromodule 9. Field theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
Literature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
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