Интегральное исчисление функции одной переменной. - 43 стр.

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§2. ïÐÒÅÄÅÌÅÎÎÙÊ ÉÎÔÅÇÒÁÌ, ÏÓÎÏ×ÎÙÅ Ó×ÏÊÓÔ×Á. . . 43
×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ:
S =
β
Z
α
y(t) · x
0
(t) dt.
ðÒÉÍÅÒ 12. ÷ÙÞÉÓÌÉÔØ ÐÌÏÝÁÄØ ÏÂÌÁÓÔÉ, ÏÇÒÁÎÉÞÅÎÎÏÊ ÜÌÌÉÐÓÏÍ x =
= 3 cos t, y = 2 sin t.
÷ÙÞÉÓÌÉÍ ÐÌÏÝÁÄØ ×ÅÒÈÎÅÊ ÐÏÌÏ×ÉÎÙ É ÕÄ×ÏÉÍ. úÄÅÓØ x [3, 3], ÐÏ-
ÜÔÏÍÕ t ÉÚÍÅÎÑÅÔÓÑ ÏÔ π ÄÏ 0,
S = 2 ·
0
Z
π
2 sin t(3 sin t) dt = 12
π
Z
0
sin
2
t dt =
= 12
π
Z
0
1 cos 2t
2
dt = 12
t
2
sin 2t
4
π
0
= 6π.
ðÒÉÍÅÒ 13. ÷ÙÞÉÓÌÉÔØ ÐÌÏÝÁÄØ ÆÉÇÕÒÙ, ÏÇÒÁÎÉÞÅÎÎÏÊ ÃÉËÌÏÉÄÏÊ x =
= t sin t, y = 1 cos t, t [0, 2π].
S =
2π
Z
0
(1 cos t)
2
dt =
3
2
t 2 sin t +
1
4
sin 2t
2π
0
= 3π.
II.3. ðÌÏÝÁÄØ ËÒÉ×ÏÌÉÎÅÊÎÏÇÏ ÓÅËÔÏÒÁ × ÐÏÌÑÒÎÙÈ ËÏÏÒÄÉÎÁÔÁÈ r = r(ϕ),
α 6 ϕ 6 β, ×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ:
S =
1
2
β
Z
α
(r(ϕ))
2
dϕ.
ðÒÉÍÅÒ 14. îÁÊÔÉ ÐÌÏÝÁÄØ ËÁÒÄÉÏÉÄÙ r = cos ϕ + 1, ϕ [0, 2π].
S =
1
2
2π
Z
0
(cos ϕ + 1)
2
=
1
2
3
2
ϕ +
1
4
sin 2ϕ = 2 sin ϕ
2π
0
=
3π
2
.
ðÒÉÍÅÒ 15. îÁÊÔÉ ÐÌÏÝÁÄØ ÌÅÍÎÉÓËÁÔÙ r
2
= 2 cos 2ϕ.
äÌÑ ×ÙÞÉÓÌÅÎÉÑ ÏÂÝÅÊ ÐÌÏÝÁÄÉ ÄÏÓÔÁÔÏÞÎÏ ÕÄ×ÏÉÔØ ÐÌÏÝÁÄØ ÐÒÁ×ÏÇÏ
Ï×ÁÌÁ, ËÏÔÏÒÏÍÕ ÏÔ×ÅÞÁÅÔ ÉÚÍÅÎÅÎÉÅ ÕÇÌÁ
π
4
6 ϕ 6
π
4
.
S = 2 ·
1
2
· 2
π/4
Z
π/4
cos 2ϕ = sin 2ϕ
π/4
π/4
= 1 (1) = 2.
§2. ïÐÒÅÄÅÌÅÎÎÙÊ ÉÎÔÅÇÒÁÌ, ÏÓÎÏ×ÎÙÅ Ó×ÏÊÓÔ×Á. . .                                                    43

×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ:
                                         Zβ
                                    S=           y(t) · x0 (t) dt.
                                         α
   ðÒÉÍÅÒ 12. ÷ÙÞÉÓÌÉÔØ ÐÌÏÝÁÄØ ÏÂÌÁÓÔÉ, ÏÇÒÁÎÉÞÅÎÎÏÊ ÜÌÌÉÐÓÏÍ x =
= 3 cos t, y = 2 sin t.
   ÷ÙÞÉÓÌÉÍ ÐÌÏÝÁÄØ ×ÅÒÈÎÅÊ ÐÏÌÏ×ÉÎÙ É ÕÄ×ÏÉÍ. úÄÅÓØ x ∈ [−3, 3], ÐÏ-
ÜÔÏÍÕ t ÉÚÍÅÎÑÅÔÓÑ ÏÔ π ÄÏ 0,
         Z0                               Zπ
 S =2·        2 sin t(−3 sin t) dt = 12              sin2 t dt =
         π                                  0
                                             Zπ                                            π
                                                     1 − cos 2t           t sin 2t
                                     = 12                       dt = 12     −                    = 6π.
                                                         2                2   4              0
                                             0
   ðÒÉÍÅÒ 13. ÷ÙÞÉÓÌÉÔØ ÐÌÏÝÁÄØ ÆÉÇÕÒÙ, ÏÇÒÁÎÉÞÅÎÎÏÊ ÃÉËÌÏÉÄÏÊ x =
= t − sin t, y = 1 − cos t, t ∈ [0, 2π].
                Z2π                                                        2π
                                     3              1
             S = (1 − cos t)2 dt =     t − 2 sin t + sin 2t                       = 3π.
                                     2              4                        0
                   0

  II.3. ðÌÏÝÁÄØ ËÒÉ×ÏÌÉÎÅÊÎÏÇÏ ÓÅËÔÏÒÁ × ÐÏÌÑÒÎÙÈ ËÏÏÒÄÉÎÁÔÁÈ r = r(ϕ),
α 6 ϕ 6 β, ×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ:
                                                 Zβ
                                       1
                                    S=                (r(ϕ))2 dϕ.
                                       2
                                                 α

  ðÒÉÍÅÒ 14. îÁÊÔÉ ÐÌÏÝÁÄØ ËÁÒÄÉÏÉÄÙ r = cos ϕ + 1, ϕ ∈ [0, 2π].
               Z2π                                                               2π
         1                          1   3    1                                              3π
      S=          (cos ϕ + 1)2 dϕ =       ϕ + sin 2ϕ = 2 sin ϕ                          =      .
         2                          2 2      4                                     0         2
               0

   ðÒÉÍÅÒ 15. îÁÊÔÉ ÐÌÏÝÁÄØ ÌÅÍÎÉÓËÁÔÙ r 2 = 2 cos 2ϕ.
   äÌÑ ×ÙÞÉÓÌÅÎÉÑ ÏÂÝÅÊ ÐÌÏÝÁÄÉ ÄÏÓÔÁÔÏÞÎÏ ÕÄ×ÏÉÔØ ÐÌÏÝÁÄØ ÐÒÁ×ÏÇÏ
Ï×ÁÌÁ, ËÏÔÏÒÏÍÕ ÏÔ×ÅÞÁÅÔ ÉÚÍÅÎÅÎÉÅ ÕÇÌÁ − π4 6 ϕ 6 π4 .
                            Zπ/4                               π/4
                   1
              S =2· ·2             cos 2ϕ dϕ = sin 2ϕ                 = 1 − (−1) = 2.
                   2                                           −π/4
                          −π/4