Кузнецов Л. А / Векторный анализ. Кузнецов. Вариант 8

8 _ 01_ 08

u = arctg ( y

x) + xz,

S : x2 + y2

- 2z = 10,

M (2, 2, -1).

F (x, y, z)

= x2 + y2 - 2z -10

AntiGTU

UUR

¶F

¶F

¶F

.

нормальк поверхности S : N

= grad F (x, y, z) =

¶x

;

¶y

;

¶z

¶F = 2x;

¶F = 2 y; ¶F = -2z

¶x

¶y

¶z

¶F (M ) = 4;

¶F (M ) = 4; ¶F (M ) = 2

¶x

¶y

¶z

UUR

UUR

= 6 cosα = cos β = 2 / 3;cos γ = 1/ 3

N (M ) = (4; 4; 2);

N (M )

¶U =

1

× - y

+ z; ¶U =

1

×

1

;

¶U = x

1 + y2

/ x2

+ y2 / x2

¶x

x2

¶y

1

x

¶z

¶U (M ) = -5 ;

¶U (M ) =

1

; ¶U (M ) = 2

¶x

4

¶y

4

¶z

¶U

¶U

¶U

¶U

¶

-5 2 1 2

1

UUR

= ¶x

cosα + ¶y

cos β +

¶z

cos γ

UUR

|M =

×

+

× 3 + 2

×

= 0

4

3

4

3

¶N

¶N

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8 _ 02 _ 08

V =

6

-

6

+

2

U =

yz2

1

1

1

,

,

M

,

,

.

2x

2

2

2 y 3z

x

3

¶V =

-

6

AntiGTU

¶x

2x2

¶V =

6

2 y2

¶y

¶V =

-2

3z2

¶z

¶U = - yz2

¶x

x2

¶U =

z2

x

¶y

¶U =

2 yz

¶z

x

gradV = {¶V ;

¶V ;

¶V }; gradU

= {

¶U ;

¶U ;

¶

}

¶z

¶x

¶y

¶z

¶x

¶y

gradV (M ) = {-

6;

6; -2};

gradV (M )

= 4

gradU (M ) = {-

2

;

2

;

2

};

gradU (M )

=

4

3

3

3

с

3

2

2

4

6

×

+

6 ×

-

gradV (M ) × gradU (M )

cosα =

=

3

3

3

= 0

gradV (M )

×

gradU (M )

4 × 4 / 3

α = arccos(cosα )

= π

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.

ru

dx

=

dy

=

dz

dy = 0

y = C1

+ C

2z

0 3x

3x dx = 2z dz

3x2

Скачано

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линий поля a

y = C1

AntiGTU

= 2z2

3x2

− 2z2

= C

.

ru

8 _ 05 _ 08 _1

a = yj + 3zk

.

P :

x 2 + y + z = 1

R

= {1/ 2;1;1};

R

=

nx2

+ ny2

+ nz2

=

3

AntiGTU

n

n

n

ny

2

n

1

2

2

cosα =

=

; cos β =

=

;cos γ

=

=

Rx

R

Rz

3

3

3

n

n

n

2 + z¢ 2

3

dS = 1 + z¢

dx dy = 1 + -1/ 2 2 + -1 2 dx dy =

dx dy

x

y

2

R R

П = ∫∫ andS = ∫∫(ax cosα + ay cos β + az cos γ )dS =

S

S

2

2

3

2 y

3

= ∫∫ y

+ 3z

dx dy = ∫∫

+ 2(1

- x / 2 - y)

dx dy =

3

2

3

2

D

3

D

xy

xy

=

3

D∫∫

2 y

3

D∫∫

(2

- x - 4 y / 3) dx dy =

2

3

2

+ 2 - x - 2 y dx dy =

xy

xy

3

2

(2− x ) / 2

3

2

(2− x ) / 2

=

∫ dx ∫

(2 - x - 4 y / 3) dy =

∫ dx (2 y - xy - 4 y2 / 6)

|

=

2

0

0

2

0

0

3

2

4

4x

x

2

3

2x

2

x

3

2

3

8

=

∫

-

+

dx

=

4x

-

+

|

=

×

= 4

2

0

3 3

3

2 3

3

9 0

2 9 3

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