Задачи по векторному анализу. Михайлов В.К - 143 стр.

UptoLike

Рубрика: 

143
m
ax bx a b x+++ =
12
1
2
1
22
2
222
3
2
(()).
5.27.
I
m
Rh
Rh
R
=
+
+
4
30 0
030
002
22
22
2
(/)
(/).
5.28.
I
mR mR
xxx=
++ =
22
1
2
2
2
3
2
5
700
070
002
5
772 1;( ).
5.29.
ρ
ρρ
jEE==
Λ ;,
0
0
1
ñëåäîâàòåëüíî,
ρ
j
A
м
=−
0
2
5
10
6
2
=−
0
2
5
2
A
мм
;
≈===22 3 6 9
0
123
;,,;
αλλλ
{} {} {}
ρρ ρ
=
=−
=−
ee e
12 3
1
3
122
1
3
21 2
1
3
22 1,, , ,, , ,, .
5.30.
ρρ
DE==
≈===
;;,,;
εε ε α ε ε ε
00
0
12 3
0
2
3
34 1 4 7
       m 2 2
          ( a x1 + b 2 x22 + ( a 2 + b 2 ) x32 ) = 1.
       12


           ( R2 + h 2 / 3)      0         0 
         m                    2   2           
5.27. I =        0         ( R + h / 3)   0 .
         4
                 0              0       2 R2 

               7 0 0
          mR2                     mR2
5.28. I =        0 7 0;               ( 7 x12 + 7 x22 + 2 x32 ) = 1.
           5                      5
               0 0 2

                  0
      ρ    ρ ρ  
5.29. j = ΛE; E = 0 , ñëåäîâàòåëüíî,
                  
                 1

           0               0 
       ρ         6 A             A 
       j = − 2 ⋅ 10  2  = − 2         ;
                  м            мм 2 
           5             5 


       α ≈ 22 0 ; λ1 = 3, λ 2 = 6, λ 3 = 9;


       ρ 1             ρ 1              ρ 1
       e1′ = {1,2 ,2}, e2′ = {2 ,1,−2}, e3′ = {− 2 ,2 ,−1}.
            3               3                3

                       0
      ρ       ρ
          ∃ 0 E = ε 0  2; α ≈ 34 0 ; ε 1 = 1, ε 2 = 4 , ε 3 = 7;
5.30. D = εε
                       
                       3


                             143