Кузнецов Л. А / Векторный анализ. Кузнецов. Вариант 28

8 _ 04 _ 28 _1

a = (x + xz2 )i + yj +(z − zx2 )k,

S : x2 + y2

+ z2 =9, P : z = 0

(z ≥ 0).

П = Пн + Пбок

Пбок

= П − Пн

G G

G

∂a

∂ay

∂a

П = ∫∫

(a, n)dS = ∫∫∫div a dx dy dz = ∫∫∫

x

+

+

z dx dy dz =

S

V

V ∂x

AntiGTU

∂y

∂z

= ∫∫∫(1 + z2 +1 +1 − x2 )dx dy dz = ∫∫∫(3 + z2

− x2 )dx dy dz =

V

V

x = r cosθ cosϕ

2π

π

/ 2

3

=

y = r cosθ sinϕ

= ∫dϕ

∫ dθ∫r2 cosθ (3 + r2 s n2 θ −r2 cos2 θ cos2 ϕ) dr =

z = r sinθ

0

0

0

2π

π / 2

r

5

3

= ∫dϕ

∫

dθ

r3 cosθ +

(sin2 θ cosθ −cos3 θ cos2 ϕ)

| =

0

0

5

0

2π

π / 2

243

(sin

2

θ cosθ −(1 −sin

2

θ) cosθ cos

2

ϕ)

= ∫dϕ

∫

27 cosθ +

5

dθ =

0

0

2π

θ +

243

sin3 θ

−sinθ cos

2

ϕ +

sin3 θ

cos

2

π / 2

=

= ∫dϕ

27sin

5

3

3

ϕ

|

0

0

=

2π

81

27ϕ

−

81

sin 2ϕ

2π

= 54π

∫0

27 −

5

cos 2ϕ dϕ =

5

2

|

G

G

G

G

0

G

nн = (0;0;1); a

nн

= z − zx

2 ;(a

nн ) |z=0 = 0

Пн = ∫∫(aG nG

н )dS = ∫∫0 dS =

0

с

S

S

Пбок

= П − Пн

= 54π −0 = 54π

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8 _ 05 _ 28 _1

a = 2xi +3yj + 4zk

P : 2x +3y + z =1

nG ={2;3;1};

nG

=

nx2

+ ny2 + nz2

=

14

AntiGTU

0

n

ny

2

3

1

cosα =

Gx

=

;cos β =

G

=

;cos

γ =

Gz

=

14

14

14

n

n

n

dS = 1 + zx′ 2 + z′y

2 dx dy = 1 + −2 2 + −3 2 dx dy = 14 dx dy

GG

cosα + ay cos β + az cosγ )dS =

П = ∫∫andS = ∫∫(ax

S

S

= ∫∫

2

3

1

= ∫∫(2ax +3ay + az )dx dy =

ax

+ ay

+ az

14 dx dy

14

14

4

Dxy

Dxy

= ∫∫(2 2x +3 3y + 4(1 −2x −3y))dx dy = ∫∫(4 −4x −3y)dx dy =

Dxy

Dxy

1/ 2

(1−2 x) / 3

1/ 2

/ 2)

(1−2 x) / 3

= ∫ dx ∫

(4 −4x −3y)dy =

∫

dx (4 y −4xy −

3y2

|

=

0

0

0

0

=

1/ 2

2

−

10x

+

7

2x3

−

5x

2

7x

1/ 2

=

1

∫

2x

3

dx =

3

3

+

6

|

4

6

0

Скачано

с

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0.0

0.1

0.2

0.3

Скачано

с

AntiGTU

0.4

0.5

.

ru

2

2

2

S : z

= 9(x

+ y

)

z = 4

ntiGTA U

Т.к. поверхностьзамкнутая, товоспользуемсяформулой

.

Остроградского− Гаусса

∂a

∂ay

∂a

П = w∫∫a n dσ = ∫∫∫div a dx dy dz = ∫∫∫

x

+

+

z

dx dy dz =

∂x

∂y

∂z

σ

V

V

x = r cosϕ

= 42∫π dϕ

4∫/ 3 r dr ∫4

= ∫∫∫(−1 −1 +6)dx dy dz = 4∫∫∫dx dy dz =

dz

V

V

y = r sinϕ

0

0

3r

2π

4 / 3

2π

−r3 )

|

= 128

2π

256 π

= 4 ∫dϕ ∫ (4r −3r2 ) dr = 4 ∫dϕ(2r2

∫dϕ =

4 / 3

0

0

0

0

27

0

27

Скачано

с

0.0

AntiGTU

0.5

1.0

.

ru

8G

_ 09 _ 28G_1 G

G

a

= 2xy +i + 2xy j

+ z2 k

x

+ y

+ z

=

2

S

:

2

2

2

= 0(z ≥ 0)

z

nAtiGTU

Т.кповерхнлстьзамкнутая, товоспользуемсяформулойОстроградского

.

G G

G

∂a

∂ay

∂a

П = w∫∫(a; n)dS = ∫∫∫div a dx dy dz = ∫∫∫

x

+

+

z

dx dy dz =

∂x

∂y

∂z

S

V

V

= ∫∫∫(2x + 2 y + 2z )dx dy dz = 2∫∫∫(x + y + z )dx dy dz =

x = r cosθ cosϕ

y = r cosθ sinϕ

=

V

V

z = r sinθ

2π

π / 2

4 2

= 2 ∫dϕ ∫

cosθ dθ ∫r3 (sinθ +cosθ cosϕ +cosθ s

ϕ)dr =

0

0

0

2π

π / 2

r

4

4 2

= 2 ∫dϕ ∫

(sinθ cosθ +cos2 θ(cosϕ +sinϕ)) dθ

|

=

4

0

0

0

=

2∫π dϕ

π∫/ 2

(sinθ cosθ +1 +cos 2θ

(cosϕ +sinϕ)) dθ =

0

0

2

=

2π

sin2 θ

θ

+

sin 2θ

ϕ)

π / 2

=

2π

1

+

π(cosϕ +sinϕ)

dϕ

2

+

2

4

(cosϕ +sin

|

∫0

2

4

dϕ =

∫0

0

=

ϕ

+

π(sin−cos

ϕ)

2π

2

4

| =π

с

0

Скачано

2

1

0

Скачано

с

AntiGTU

1

2

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8 _10 _ 28

F = −xi + yj,

L : x2 +

y2

=1

(x ≥ 0, y ≥ 0),

9

M (1, 0), N (0,3).

A = ∫(Fx dx+ Fy dy) =

x = cos t

= π∫/ 2

L

y = 3sin t

0

π / 2

π / 2

= ∫ 10sin t cos t dt =

5sin2 t |

= 5

0

0

Скачано

с

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