Поверхностные интегралы. Скопина М.А. - 23 стр.

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N
X
l=1
g
f
x
l
dx
l
β β
0
+ (1)
d
N
X
l=1
f
g
x
l
β dx
l
β
0
=
Ã
N
X
l=1
f
x
l
dx
l
β
!
(g β
0
) + (1)
d
(f β)
Ã
N
X
l=1
g
x
l
dx
l
β
0
!
=
ω
0
+ (1)
d
ω
0
.
ω
C
(2)
d
2
ω := d() = 0.
ω 0
ω = f
d
2
ω =
N
X
l=1
N
X
k=1
2
f
x
l
x
k
dx
l
dx
k
=
X
k<l
2
f
x
l
x
k
dx
l
dx
k
X
k>l
2
f
x
l
x
k
dx
k
dx
l
= 0.
ω = fβ
d
2
ω = d(df β) = d
2
f β (df ).
= 0 d
2
ω = 0
U R
N
V R
M
Φ C
(1)
U V V d
ω =
X
i∈I
d
f
i
dy
i
1
. . . dy
i
d
.