Поверхностные интегралы. Скопина М.А. - 37 стр.

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СПбГУ | Санкт-Петербург

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Рубрика: 

(ϕ
∗
k
)
−1
(e
j
) =
e
e
j−1
j = k + 1, . . . , d (ϕ
∗
k
−1
)
0
= A
T
k
A
T
k
k
B = (ϕ
∗
k
−1
)
0
· ϕ
0
· (ϕ
∗
n
)
0
= (ϕ
∗
k
−1
◦ ϕ ◦ ϕ
∗
n
)
0
A k n
1
A k det A = (−1)
n+k
det B
k = 0 ϕ(e
n
) = 0
ϕ(0) = e
r
ϕ(x) = e
r
+ A · x n A
−e
r
e
j
− e
r
ψ(x) = e
1
+ A
0
· x A
0
A
1 r
(ϕ
∗
0
−1
)
0
A
0
−1
det ψ
0
= (−1)(−1)
n+1
det(ϕ
∗
0
−1
◦ ψ ◦ ϕ
∗
n
)
0
.
ϕ(γ
n
) = 0 ψ(γ
n
) = 0
(ψ
−1
◦ ϕ)(γ
n
) = γ
n
det(ψ
−1
◦ ϕ)
0
= det(ϕ
∗
n
−1
◦ (ψ
−1
◦ ϕ) ◦ ϕ
∗
n
)
0
=
det((ψ ◦ ϕ
∗
n
)
−1
◦ (ϕ ◦ ϕ
∗
n
))
0
.
ϕ ψ
ϕ ◦ ϕ
∗
n
ψ ◦ ϕ
∗
n
n = 0 ϕ
−1
sign det ϕ
0
= sign (det(ϕ
−1
)
0
)
−1
=
(−1)
k
sign (det(ϕ
∗
0
−1
◦ ϕ
−1
◦ ϕ
∗
k
)
0
)
−1
=
(−1)
k
sign det(ϕ
∗
k
−1
◦ ϕ ◦ ϕ
∗
0
)
0
.♦

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