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Silindrik jismning xajmini hisoblash
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ikki karrali integrallar
- Bu sahifa navigatsiya:
- silindrik jismning xajmi
- 4. Og’irlik markazining koordinatlarini hisoblash.
- 5. Inersiya momentlarini hisoblash.
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2. Silindrik jismning xajmini hisoblash. Yuqoridan ( )
, = sirt, quyidan 0 = z tekislik, yon tomondan to’g’ri silindrik sirt bilan hamda XOY tekislikda D sohani hosil qiladigan silindrik jismning xajmi
( ) ∫∫ = D dxdy y x f V ,
integral bilan xisoblanadi. 2-misol. 2 1
y + = , 0 , 5 , 3 = = = z y x z sirtlar bilan chegaralangan I oktantdagi jismning hajmini hisoblang. Yechish. Hajmi hisoblanishi kerak bo’lgan jism yuqoridan x z 3 = tekislik, yondan 2 1
y + = parabolik silindr, 5 = y tekislik bilan chegaralangan. Shunday kilib, [ ] ( ) . . 12 12 24 4 2 2 2 3 4 2 4 3 4 3 ) 1 ( 5 3 3 3 4 2 2 0 2 0 2 0 4 2 3 5 1 2 0 2 2 bir кub x x dx x x dx x x dy xdx xdxdy V D x = − = − ⋅ = = − = − = + − = = = ∫∫ ∫ ∫ ∫ ∫ +
bo’ladi. 3. S
tatik momentlarini hisoblash. Plastinka har bir nuqtasidagi zichlik funksiyasi ( )
y x, γ bo’lsa, uning massasi ( )
∫∫ =
dxdy y x m , γ integral bilan hisoblanadi. Plastinkaning
o’qlarga nisbatan statik momentlari.
( ) ∫ ∫
= D x dxdy y x y M , γ , ( )
∫ ∫ =
y dxdy y x x M , γ formulalar bilan hisoblanadi. 4. Og’irlik markazining koordinatlarini hisoblash. Plastinka birjinsli, ya’ni nt cos
= γ bo’lganda uning og’irlik markazining koordinatlari
S xdxdy S M x х
∫∫ =
S ydxdy S M y D x c ∫∫ = =
formulalar yordamida topiladi, bu yerda S ,
D sohaning yuzi. 5. Inersiya momentlarini hisoblash. Plastinkaning ОХ va ОУ o’qlariga nisbatan
( )
∫∫ =
x dxdy y x y J , 2 γ ,
( ) ∫∫ = D y dxdy y x x J , 2 γ
formulalar bilan, koordinatlar boshiga nisbatan inersiya momenti ( ) ( ) ∫∫ + = + = D y x J J dxdy y x y x J , 2 2 0 γ formula bilan aniqlanadi. 3-misol. 4 2 , 4 4 2 2 + − = + = x y x y chiziqlar bilan chegaralangan figuraning og’irlik markazining koordinatlarini toping. Yechish. Chiziqlar OX o’qiga nisbatan simmetrik bo’lganligi uchun 0 =
y
x ni topamiz: =
− = − − − = = = ∫∫ ∫ ∫ ∫ ∫ − −
y dy y y dy dy dxdy S D y y 2 0 2 4 4 4 2 0 2 0 2 2 2 2 2 4 3 3 2 4 4 2 4 2 2
8 12 8 2 6 12 6 2 0 3 = − = − = y y ( ) ( ) ∫∫ ∫ ∫ ∫ − − = − − − = = =
y y c dy y y xdxdy dy xdxdy x 2 0 2 4 4 4 2 0 2 2 2 2 2 2 16 4 4 4 8 1 2 8 1 8 1
( ) = + − = ∫ dy y y 4 2 0 2 16 3 2 3 3 8 1 5 2 80 3 2 3 8 1 2 0 5 3 = + − = y y y .
Demak ) 0 ; 5 2 ( C .
1. Ikki karrali integralning ta’rifini toping. A) ta’rif. ( ) i n i i i n S y x f S ∆ = ∑ = 1 , integral yig’indining, D sohaning qismlarga bo’linish usuliga,
qismda
) , ( i i i y x P nuqtaning tanlanishiga bog’liq bo’lmagan 0 → λ dagi(
λ qism sohalar diametrlarining eng kattasi) limiti mavjud bo’lsa, bu limitga ) , ( y x f funksiyaning D sohadagi ikki karrali integrali deyiladi va ( ) ∫∫ D ds y x f ,
simvol bilan belgilanadi В ) ta’rif. ( )
n i i i n S y x f S ∆ = ∑ = 1 , integral yig’indining, D sohaning qismlarga bo’linish usuliga,
qismda
) , ( i i i y x P nuqtaning tanlanishiga bog’liq bo’lmagan
0 → dagi limiti mavjud bo’lsa, bu limitga ) , ( y x f funksiyaning D
sohadagi ikki karrali integrali deyiladi va ( )
D ds y x f ,
simvol bilan belgilanadi D) ta’rif. ( )
n i i i n S y x f S ∆ = ∑ = 1 , integral yig’indining, 0 →
dagi( λ qism sohalar diametrlarining eng kattasi) limiti mavjud bo’lsa, bunga limitga ) , ( y x f funksiyaning D sohadagi ikki karrali integrali deyiladi va ( )
D ds y x f ,
simvol bilan belgilanadi E) ta’rif. ( )
n i i i n S y x f S ∆ = ∑ = 1 , integral yig’indining, D sohaning qismlarga bo’linish usuliga,
qismda
) , ( i i i y x P nuqtaning tanlanishiga bog’liq bo’lmagan 0 → λ dagi(
λ qism sohalar diametrlarining eng kattasi) limiti mavjud bo’lmasa, bu limitga ) , ( y x f funksiyaning D sohadagi ikki karrali integrali
deyiladi va ( ) ∫∫ D ds y x f ,
simvol bilan belgilanadi 2.
D
soha ) ( ), ( 2 1 x y y x y y = = funksiyalar grafklari hamda
b x va a x = = to’g’ri chiziqlar bilan chegaralangan bo’lsa, ya’ni ( )
( ) ≤ ≤ ≤ ≤
y y x y b x a 2 1 tengsizliklar bilan aniqlangan bo’lsa, ikki karrali integral qanday hisoblanadi? A) (
( ) ( ) ( ) ( ) ( ) ( )
∫∫ ∫ ∫ ∫ ∫ = = D b a x y x y b a x y x y dy y x f dx dx dy y x f ds y x f 2 1 2 1 , , ,
formula yordamida В ) ( ) ( )
( ) ( )
( ) ( )
( ) ∫∫ ∫ ∫ ∫ ∫ = = D d s y x y x d s y x y x dx y x f dy dy dx y x f dxdy y x f 2 1 2 1 , , ,
formula yordamida D) ( ) ( ) ( ) ( )
( ) ( ) ( ) ∫∫ ∫ ∫ ∫ ∫ = = D a a x y x y a a x y x y dx y x f dx dx dy y x f ds y x f 2 1 2 1 , , ,
formula yordamida E) ( ) ( ) ( ) ( )
( ) ( ) ( ) ∫∫ ∫ ∫ ∫ ∫ = = D b a x y x y b a x y x y dx y x f dy dy dx y x f ds y x f 2 1 2 1 , , ,
formula yordamida
3. D soha
( ) ( )
≤ ≤ ≤ ≤ y x x y x d y с 2 1
tengsizliklar bilan aniqlangan bo’lsa , ikki karrali integral qanday hisoblanadi? A) ( )
( ) ( )
( ) ( )
( ) ( )
∫∫ ∫ ∫ ∫ ∫ = =
d с
x y x d с
x y x dx y x f dy dy dx y x f dxdy y x f 2 1 2 1 , , ,
formula yordamida В ) ( ) ( ) ( )
( ) ( )
( ) ( )
∫∫ ∫ ∫ ∫ ∫ = =
d s y x y x d s y x y x d у
x f d х
х
у
x f dxdy y x f 2 1 2 1 , , ,
formula yordamida D) ( ) ( ) ( )
( ) ( )
( ) ( )
∫∫ ∫ ∫ ∫ ∫ = =
d s y x y x d s y x y x d у
x f d х
dx y x f dxdy y x f 2 1 2 1 , , ,
formula yordamida E)
( ) ( )
( ) ( )
∫∫ ∫ ∫ = D с с y x y x dх y x f dу dxdy y x f 2 1 , ,
formula yordamida II darajali testlar
4. ∫∫ D ydxdy x ln integralni D soha:
4 0 ≤ ≤ x ,
y ≤ ≤ 1 to’g’ri to’rtburchak bo’lganda hisoblang. A) 16
В ) 8
D) 4 E) 0
5.
( ) ∫∫ − D dxdy y x integralni 1 2
2 : 2 − = − = x y x y D , chiziqlar bilan chegaralangan soha bo’lganda hisoblang. A) 4
15 4
В ) 4
D) 15 4 E) 0
III darajali testlar
6. 4 2 , 4 4 2 2 + − = + = x y x y chiziqlar bilan chegaralangan figuraning og’irlik markazining koordinatlarini toping. A)
) 0 ; 8 (
B)
) 0 ; 2 (
D)
) 0 ; 5 2 ( C
E) ) 0 ; 5 (
2 1
y + = , 0 , 5 , 3 = = = z y x z sirtlar bilan chegaralangan I oktantdagi jismning hajmini hisoblang. A)
. . 24 bir êub V =
B) . . 36 bir êub V =
D) . . 8 bir êub V =
E) . . 12 bir êub V =
35.6-ilova Mustaqil bajarish uchun topshiriqlar
1. (
dxdy y xy x D ∫∫ + − 2 3 2 integralni 2 ,
0 : 2 = = = y y x x D chiziqlar bilan chegaralangan soha bo’lganda hisoblang. 2.
0 , 2 2 = + − =
x y y x chiziqlar bilan chegaralangan yuzani ikki karrali integral yordamida hisoblang. 3.
4 4 , 2 2 + = − = x y x y chiziqlar bilan chegaralangan yuzani hisoblang. 4. 4
0 , 0 , 0 , 8 2 2 = + + = = = = +
y x z y x y x
sirtlar bilan chegaralangan jism hajmini hisoblang. 5. 0
0 , 4 2 , 2 2 = = = + + = z y z y x y x sirtlar bilan chegaralangan silindrik jismning hajmini hisoblang. 6.
2 , 1 , 2 , 2 2 = = = = x x x y x y chiziqlar bilan chegaralangan yuzaning og’irlik markazini toping. 7.
y x x y = = 2 2 , parabolalar bilan chegaralangan yuzaning og’irlik markazini toping.
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