Кузнецов Л. А / Векторный анализ. Кузнецов. Вариант 23

∂U

=

∂U cosα +

∂U cos β + ∂U cosγ

∂I

∂x

∂y

∂z

I

=

1 +1 =

2 cosα = 0;cos β =1/

∂U

=

3x2

∂x

∂U

=

1

2 y =

y

∂y

2 y2 + z2

y2 + z2

∂U

=

1

2z =

z

∂z

2 y2 + z2

y2 + z2

∂U

|

= 3; ∂U

| = −3 ;

∂U

|

= 4

∂x

M

∂y

M

5

∂z

M

5

Скачано

с

nAtiGTU

2;cos

γ = −1/

2 .

γ = −3

1

−

4

1 = −7

5

5

2

2

5

2

V =

3

x

2

+3y

2

−2z

2

,

2

3

, M

1

,

3

2

U = x

yz

2,

3

2

.

∂V = 3x2

AntiGTU

∂x

∂V

= 6 y

∂y

∂V

= −4z

∂z

∂U

= 2xyz3

∂x

∂U

= x2 z3

∂y

∂U

= 3x2 yz2

∂z

{∂V

∂V ;

∂V

{∂U

∂U

∂

gradV =

;

};gradU =

;

;

}

∂z

∂y

∂z

∂x

∂y

∂x

gradV (M ) ={6; 2; −2

6};

gradV (M )

= 8

gradU (M ) ={

6,3

, 6};

gradU (M )

= 4

6

cosα =

gradV (M )

gradU (M )

=

6

2 +6

6 −12 6

= 0

gradV (M )

gradU (M )

с

8 4

6

α = arccos(0) =π / 2

Скачано

.

ru

8 _ 04 _ 23 _1

aG

= (x + z)Gi +( y + z)Gj +(z − x − y)kG

S : x2 + y2 + z2 = 4; P : z = 0(z ≥ 0)

П = Пн + Пбок

Пбок = П −−Пн

AntiGTU

G G

G

∂a

∂ay

∂a

П = ∫∫(a, n)dS = ∫∫∫div a dx dy dz

= ∫∫∫

x

+

+

z

dx dy dz =

∂x

∂y

∂z

S

V

V

=

∫∫∫

(1 +1 +1)dx dy dz = 3 dx dy dz = 3 1 4 πR3 = 3 1 4 π 23 =16π

∫∫∫

2

3

2

3

G

V

G G

V

G

G

nн

=

(0;0; −1); a nн

= x + y − z;(a nн ) |z=0 = x

− y

Пн

G

G

∫∫(x − y)dS =

x = r cosϕ

2π

2

= ∫∫(a

nн )dS =

y = r sinϕ

=

∫dϕ∫r2 (cosϕ −s nϕ) dr =

S

S

0

0

2π

2

2π

2

2

= ∫(cosϕ −sinϕ)dϕ∫r2 dr =

(sinϕ +cosϕ) |

∫r2 dr = 0 ∫r2 dr = 0

0

0

0

0

0

Пбок

= П − Пн

=16π −0 =16π

Скачано

с

.

ru

8 _ 05 _ 23 _1

aG = xiG

+ 2 yGj +5zkG

.

P : x + 2 y + z / 2 =1

GG

cos β +cz cosγ )dS

П = ∫∫(an)dS = ∫∫(ax cosα + ay

S

S

AntiGTU

x + 2 y + z / 2 =1 2x + 4 y + z = 2 nG ={2;4;1}

G

=

21

cosα = 2 /

21;cos β = 4 /

21;cosγ =1/

21

dS =

1 +

zx′

2 +

z′y

2

dx dy =

1 + −2 2 +

−4 2 dx dy =

21dx dy

1 1/ 2−x / 2

2

4

1

П =

21∫dx

∫

x +

2 y +

5(2 −2x −4 y)

dy

=

21

21

21

0

0

1

1/ 2−x

/ 2

1

(1−x) / 2

1

= ∫dx

∫

(10 −8x −12 y)dy = ∫dx (10 y −8xy −6 y2 )

|

=

∫(5x2 / 2 −6x +7 / 2)dx =

0

0

0

0

0

5x

3

+ 7x

1

=

4

=

−3x2

|

6

2

0

3

Скачано

с

Скачано

с

0.8

AntiGTU

1.0

.

ru

8G

_ 08G_ 23 G

G

a

= xi −2 y j +

3zk

S : x

+ y

= z

2

2

2

z = 2x

AntiGTU

Т.кповерхностьзамкнутая, товоспользуемсяформулой

.

Остроградского− Гаусса

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

= ∫∫∫(1 −2 +3)dx dy dz = 2∫∫∫dx dy dz =

x = r cosϕ

y = r sin ϕ

=

V

V

z = z

2r cosϕ

π

/ 2

2 cos ϕ

π / 2

2 cos ϕ

r (

2r cosϕ −r2 ) dr =

= 2 ∫

dϕ

∫

r dr

∫

dz = 2 ∫ dϕ

∫

−π / 2

0

r2

−π / 2

0

π

/ 2

2 cos ϕ

(2r2 cosϕ −r3 ) dr = 2

π

/ 2

2r

3

cosϕ

r

4

2 cos ϕ

= 2

∫

dϕ

∫

∫

dϕ

−

|

=

3

−π / 2

0

−π / 2

4

0

= 2

π∫/ 2

dϕ

4 cos4 ϕ

= 8 π∫/ 2

cos4 ϕ dϕ

= 2 π∫/ 2

(1 +cos 2ϕ)2

dϕ =

−π / 2

3

3 −π / 2

3 −π / 2

=

2

π / 2

(1 + 2 cos 2ϕ +cos

2

2ϕ) dϕ =

2 π

/ 2

+ 2 cos 2ϕ +

1 +cos 4ϕ

dϕ =

3

∫

∫

1

2

−π / 2

3 −π / 2

=

2

3ϕ

+sin 2ϕ +

sin 4ϕ

π / 2

=π

с

3

2

8

|

−π / 2

Скачано

8JG_10 _ 23

G

G

F = ( y2 − y)i

+(2xy + x) j

L : x2 + y2 = 3( y ≥ 0)

.

M (3;0); N (−3;0)

nAtiGTU

x = 3cos t

A = ∫(Fx

dx+ Fy dy)=

dx = −3sin t dt

= 3sin t

= 3cos t dt

=

L

y

dy

π

((9sin2 t −3sin t )(−3sin t dt) +(2 3cos t 3sin t +3cos t)(3cos t dt))=

= ∫

0

π

(−9sin3 t +3sin2 t +18cos2 t sin t +3cos2 t )dt =

= 3∫

0

= 3

π

2

π

1 −cos 2t

π

2

t

π

1 +cos 2t

=

−9∫sin

t sin t dt +3∫

2

dt −18∫cos

(−sin t dt) +3∫

2

0

0

0

0

π

2

t

sin 2t π

cos3 t

π

t

s

2t π

= 3

9∫(1

−cos

t) (−sin t dt) +3

−

4

|

−

18

3

| +3

+

4

|

=

0

2

0

0

2

0

= 3

π

π

2

t

(−sin t dt) +

3π

+12

+

3π

9 cos

π

cos3

π

+3π +12

9∫−sin t dt −9∫cos

2

2

= 3

| −9

3

|

0

0

0

0

= 3(−18 +6 +3π +12)= 9π

Скачано

с

8 _11_ 23 _1

aG

= 7ziG− xGj + yzkG

x = 6cos t

Г :

y = 6sin t

AntiGTU

z =1/ 3

dx = −6sin t dt;dy = 6cos t dt;dz = 0

Ц = ∫(ax dx+ ay dy+ az dz)

2π

7

1

=

= ∫

3

(−6sin t dt) −6cos t 6cos t dt

Г

0

=

2∫π (7 (−2sin t dt) −36cos2 t dt )= −2∫π (14sin t +18(1 +cos 2 ))dt =

0

0

= −

sin 2t 2π

= −36π

−14cos t +18

t +

2

|

0

0.6

0

с

0.4

0.2

0