Sadrzaj za M2 -...
Transcript of Sadrzaj za M2 -...
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1. Izra unaj krivolinijski integral od ta ke A(1,0) do ta ke B(0,2). ydyxdxxyIL
2)1(
a) po pravoj 2x+y=2 b) duž parabole 44 2yxc) duž elipse x=cost ; y= 2sint
Rješenja:
a) Skicirajmo datu pravu (uputa vidi sliku desno).
2x+y=2y = 2 - 2x dy = -2dx
111212
23
64
4)1264(
)44()122(
)2)(22(1)22(
)1(
0
1
0
1
20
1
30
1
42
0
1
3
320
1
2
20
1
2
1
xxxxdxxxx
dxxxdxxx
dxxxdxxx
ydyxdxxyIL
b) Skicirajmo parabolu (uputa: vidi sliku iznad).
dyydxyxyx24
1442
2
dyyyydyyyy
ydyyydyyyy
ydyydyyyyydyxdxxyIL
162282
1621
21
4
41
21
41)1(
532
0
42
422
0
3
222
0
22
2
209
.1517
1545203012103
342
54
32
412
68
816
4032
9664
43
68409623
228162
0
22
0
32
0
42
0
52
0
62
0
2345 yyyyydyyyyyy
c) Skicirajmo elipsu (uputa: vidi sliku sa prethodne stranice).
x = cost y = 2sint
dy = 2costdt
20
sincos
:3
t
tytx
L
34
322
3121
414
3sin2cos
4cos4
3sin
3cossin
cossin
4cos
4sin
sincos
sincos
cossin2sinsincos4
)cossin2sinsincos4(sincos4)sincossin2(
cos2sin2cos)sin()1sin2(cos)1(
20
320
20
4
332
4433
2
0
2
0
22
0
3
22
0
332
0
2
2
0
22
3
ttt
ctcuduudutdt
uttdtt
ctcuduudutdtdutdt
uttdtt
tdtttdttdtt
dtttttttdttdtttt
tdttttdtttydyxdxxyIL
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1. Izra unati površinu figure koja je ograni ena krivom:
a) elipsom , x acost y b sint ;b) petljom Dekartovim listom .3 3 3 0x y axy
Rješenja:a)
Slika 1: elipsa Koristit emo sljede u formulu:
1
1 ,2 C
P xdy y∮ dx
gdje je (vidi sliku 1)
1
0 2
x acostC y b sin
tt
Izra unajmo izvode od x i y: dx a sint dtdy bcost dt
Uvrstimo u formulu:
1
2
0
2 22 2 2 2
0 0
2
0
1 1 ( ) 2 2
1 1 ( ) 2 2
21 1 1 1 2 002 2 2
C
P xdy ydx acost bcost b sint a sint dt
abcos t ab sin t dt ab cos t sin t dt
ab dt ab t ab ab
∮
Kona no rješenje: .P ab
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b)
Slika 2: Dekartov list
Da bismo koristili formulu
1
1 ,2 C
P xdy y∮ dx
moramo pre i na parametarsku jedna inu krive uzevši:
, yy tx tx
idimo da polarni radijus OM (vidi sliku 2), gdje je O(0,0) i M(x,y), opisuje cijelu petlju
te na dobiveni rezultat unijeti i smjenu
Vkrive kada t ide od 0 do + .
Uvrstimo smjenu y tx u x3 3 3 0y axy yxt
pa emo imati: 33
3 3 2
3 3 2
2
3
3
3
3 0
1 3 0 / :
(1 ) 3 0
1 3 0
1 3
31
x tx ax tx2x t tax x
x t taxx
x t ta
x t ta
taxt
3
31taxt
3
3
313
1
y tat ty tat t
229
Pa dalje ra unamo izvod za x: 3 2
3 2
3 2
3 2
3
3 2
3 1 3 (3 )
(1 )1 (3 )3
(1 )1 23(1 )
a t ta tdx dt
tt t tdx a dtt
tdx a dtt
te i za y : 3 2 2
3 2
3 2
3 2
3 3
3 2
3
3 2
6 1 3 (3 )
(1 )
2 1 (3 )3
(1 )2 2 33
(1 )23
(1 )
at t at tdy dt
t
t t tdy at dt
tt tdy at dtt
tdy at dtt
omnožimo izvode sa dx i dy sa y i x, redom P3
3 23
32 2
33
2 3
3 3 2
32 2
3 3
3ta 23(1 )1
291
3 1 2 3 1 (1 )
1 29(1 )
tx dy at dttt
t a ty dx a dtt t
tx dy a t dtt
ty dx a t dtt
ad uvrstimo dobijene rezultate: S3 3
2 2 2 23 3 33
0
3 3 2 32 2 2
3 33 30 0
1 1 2 1 2 9 92 2 (11
1 2 1 2 9 1 92 21 1
C
t tP xdy y dx a t a ttt
t t a ta t dt t dtt t
)dt∮
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32 2 2
122 23
20 12
2 2 22 1
11
119 3 3
2 21 13
3 3 1 3 0 12 2 1 2
t uua t adt t dt du duu ut
t dt du
a a au du u a2
1
32
Prema tome 23
2aP .
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1. Izra unati površinski integral:
a) , gdje jedszyxI )346( oblast ravni x+2y+3z = 6, u prvom oktantu;
b) dsxazyKW
)( 22 , gdje je W površina cilindra , koja se nalazi
izme u ravni z = 0 i z = h.
222 ayx
Rješenja:
a) Skicirajmo oblast (vidi sliku desno)
x+2y+3z = 6/:6
1236zyx
segmentni oblik jedna ine ravni
yxz 2631
2y-x-63z6 3z2yx
dxdyyz
xzyxzyxfdszyxf
D
22
1)),(,,(),,(
3231
yzxz
314
914
94
9111
22
yz
xz
Projekcija na xOy ravan izgleda: Nacrtati projekciju (uputa:vidi xOy ravan sa slike iznad).
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yxyxyx
yx
yxyyyy
xxxx
266183
618336
6030
606
12
1
12
1
yxy
D260
30:
1454)63459(142212
103
142
)2110(142)126606(314
1236412)42436(25
314
)26(6)26(2)26(25
314
6225
314625
314
6253142646
314)346(
3
0
3
0
23
0
3
3
0
23
0
2
3
0
22
3
0
2
26
0
26
0
26
0
23
0
3
0
26
0
yyy
dyyydyyy
dyyyyyy
dyyyyy
dyxxyxdxyxdy
dxdyyxdxdyyxyxdszyxI
yyyy
DD
b) dsxazyKW
)( 22 z = 0 i z = h 222 ayx
Skicirajmo oblast W (vidi sliku na sljede oj stranici)
dxdyzy
xyzzxyxfdszyxf
W D
22
1)),,(,(),,(
253
0
22
21
zy
xax
xy
KKK222
222
xayayx
22
22
22
xay
ixay
xay
hzaxa
D0
:
2222
22
2211
xaadxdzdxdz
xaxdxdz
xaxds
ha
a
h a
aD
DW
zdzxa
dxadzdxadxdzxa
za
xaadxdzxazxadsxazyK
002222
22
2222221
22
24
24
24
coscos
24
sin1cos
24
sincos
24
2
2
cossin
222
222
2
22
2
2
22
2
2
22
2
2
2
22
2
2
222
22
22
2
ahhathaha
dtahhattdtahha
tatdtaahha
taatdtaahha
tay
tax
tdtadxtax
xadxahhaa
a
a
254
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22
22
22
0
xaadxdyds
zy
xax
xy
xay
222sincos
2
2
2
cossin
22
22
2
2
2
222
2
222
2
22
2
0
2
0222222
22
2222222
ahtahdtahtaa
tdtaah
tax
tax
tdtadxtax
xadxahz
xadxazdz
xadxa
xaadxdzz
xaadxdzxazxadsxazyK
a
a
hh a
a
a
aD
DW
haahahhaahahhaK 4422
4 2222
2
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(Zadaci su skinuti sa stranice: \pf.unze.ba\nabokovZa uocene greske pisati na [email protected])
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1. Na i protok (fluks) vektorskog polja 2 2 2 1p xi y j x y k kroz elipsoidu 2 2 2
2 2 2 1x y za b c
.
Rješenje:
Slika 1: elipsoid
Kako je S zatvorena površ možemo primijeniti formulu Gauss - Ostrogradski
yx z
S
vv vpn ds div p dxdydz dxdydzx y z
Imamo: 2 2 2, , , , 1x y zp v v v x y x y
1, 2 , 0yx zvv vyx y z
Oblast je ograni ena elipsoidom (vidi sliku 1) 2 2 2
2 2 2 1x y za b c
1 2 (*)y dxdydz
Uvedimo sferne koordinate:
329
2
'
0 10
0 2
x ra sin cosy rb sin sinz rccosdxdy dz r sin abcdr d d
r
2
'1 2
2
0 0 01
2
0 01
2
0 0
* (1 2rb sin sin ) r sin abc dr d d
abc r dr sin d 1 2rb sin sin d
abc r dr (2 2rb sin ( cos 2 cos 0)sin d
abc r dr (2 2rb sin 1 1 )sin d
12
0 01
2
0 01
2
01
2
01
2
0
abc r dr 2 sin d
2 abc r dr sin d
2 abc ( cos cos 0)r dr
2 abc ( ( 1) 1)r dr
4 abc r dr
1 4 4 abc (1 0)3 3
abc
Prema tome 43
abc .
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Pismeni dio ispita iz Matematike II, 03.02.2011.
GRUPA A 1. Na i površinu figure koja je ograni ena linijama 02 , 2y x x y .2. Na i ekstreme funkcije 3 23 15 12z x xy x y.3. Na i zapreminu tijela ograni enog ravnima
1, 3, 1, 5, 2 1 0, 0.x x y y x y z z
4. Izra unati krivolinijski integral 2 2 22 ,Ic
z x y z ds ako je c kriva
2 2cos , cos , sin , 0, .2 2r rx t y t z r t t
GRUPA B 1. Izra unati površinu rotacionog tijela koje se dobije rotacijom parabole 2 4y x
od ta ke 0x do ta ke 2.x2. Na i uslovne ekstreme funkcije y6 4 3z x uz uslov 2 2 1.x y3. Na i zapreminu tijela ograni enog ravnima
1, 2, 2, 2, 4 3 2 0, 0.x x y y x y z z
4. Izra unati krivolinijski integral 2 2 24 ,Ic
y x y z ds ako je c kriva
6 3sin , cos , sin , 0, .3 3 2a ax t y a t z t t
Pismeni dio ispita iz Matematike II, 18.02.2011
GRUPA A
1. Izra unati integrale: 3 2
3 21 2
1 0
1 ,1 sin cos
dxI x x dx I .x x
2. Izmjeniti poredak integracije u integralu 22 2
1 2
,x x
x
I dx f x y dy .
3. Izra unati površinski integral 2 1S
P z d ,S S je dio sfere
u prvom oktantu. 2 2 2 4x y z
4. Izra unati integral 2
2x
e
0
1 x
I dxxe
pomo u diferenciranja po parametru
ako je 1.
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GRUPA B
1. Izra unati integrale: 4 4
1 20 0
sin, .sin cos1 2 1
dx xI I dxx xx
2. Izmjeniti poredak integracije u integralu 21 2
0
,x
x
I dx f x y dy .
3. Izra unati površinski integral 2 4S
x dS , gdje je (S) omota površi
2 2 2
,0 3.4 4 9x y z z
4. Izra unati pomo u diferenciranja po parametru integral
22 2 2
0
ln sin cos , 0.I x x dx
Pismeni dio ispita iz Matematike II, 23.06.2011.
GRUPA A
1. Izra unati dužinu luka krive 1ln1
x
x
eye
od ta ke sa apscisom 1x do ta ke sa
apscisom 2.x2. Izra unati pomo u dvostrukog integrala zapreminu tijela kojeg ograni avaju
površi22 2 2 2 2 2 22 0, ,x y az x y a x y a 0.
3. Izra unati pomo u Greenove formule krivolinijski integral 2 2 2 2ln ,
c
I x y dx y xy x x y dy ako je c kontura koja
ograni ava oblast .2 2 2, 2, 0y x x y4. Izra unati površinski integral 2 ,
S
I x y z dS ako je S polulopta
2 2 2 9, 0.x y z z
GRUPA B
1. Izra unati dužinu luka krive 2
2 2ln ( 0)ay a aa x
od ta ke 0,0A do ta ke
4, ln .2 3aB a
2. Izra unati pomo u dvostrukog integrala zapreminu tijela kojeg ograni avaju površi 2 2 2 2 24 ( 0), 1 1 , 4 0x y x x y x z x z i ravan 0.z
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3. Izra unati krivolinijski integral ,c
xds ako je c lemniskata
22 2 2 2 2 , 0x y a x y a .
4. Izra unati površinski integral ,S
I x y z dS ako je S polulopta
2 2 2 4, 0.x y z z
Pismeni dio ispita iz Matematike II, 08.07.2011.Grupa A
1. Odrediti jedna inu tangentne ravni na površ 2 2 2
2 2 2 1x y za b c
, koja je normalna na
pravoj1 2 3x y z .
2. Izra unati integral po glatkom luku koji spaja ta ke A i B
2 211
AB
y x x xydx dy dzy z z y z
, (1,1,1), (1, 2,3), , , : 0, 0, 0A B AB x y z x y z
3. Izra unati zapreminu onog dijela lopte 2 2 2 1x y z koji se nalazi unutar cilindra .2 2( 1)x y 1
)x4. Dato je vektorsko polje 2( 2 ,1 ,xA e z xy x e z
L
. Pokazati da je polje A potencijalno i odrediti mu potencijal. Izra unati integral A dr , gdje je L duž PQ, P (0, 1, -1), Q (2, 3,
0), orijentisana od P prema Q.
Grupa B 1. Dokazati da proizvoljna tangentna ravan površi S: obrazuje sa
koordinatnim ravnima tetraedar stalne zapremine
3( 0,konstanta)xyz a a
392
V a .
2. Izra unati integral po glatkom luku koji spaja ta ke A i B
2( )AB
zxdy xydz yzdxx yz
(7, 2,3),A (5,3,1),B xzy
.
3. Izra unati zapreminu tijela koje je ograni eno površima .2 2 2 2 2, 2 , ,x y y x y y z y z 0
4. Dato je vektorsko polje 2 2 2 2 2 2(2 ( ) , 2 ( ) , 2 ( ) )A x y z yz y z x xz z x y xy . Pokazati da je polje A potencijalno i odrediti mu potencijal. Izra unati fluks vektorskog polja Akroz spoljnu stranu polusfere 2 2 2: 2 0, 0x y z z y
Pismeni dio ispita iz Matematike II, 15.09.2011.
GRUPA A 2 3 3 12x y 01. Izra unati površinu figure koju odre uju prava i dio elipse
6 u prvom kvadrantu.2 24 9 3x y333
222 8y
5102
.dxI ydyx
2. Promij ra unati dvostruki integral eniti poredak integracije i iz
3. Izra unati površinu dijela površi 2 2 1 0x y z koji se nala a ni zi iznad r v0.z
4. Dati su krivolinijski integrali 1 2
1 2 2 2, ,x2 2
c c
dy ydx xdy ydxI Ix y x y
gdje je 1c duž
,AB 1,2 , 1,4 ,A B orjentisan ki B, a 2c je parabola
koja prolazi kroz ta ke
a od ta ke A prema ta
1, 2 , 1, 4A B i 1 11, .2 4
C Dokazati da je 1 2I I i
A B
izra unati taj broj.
volinijskog etverougla ome enog parabolama
GRUP
1. Izra unati površinu kri2
2 2 2, , 2 , 3 .x3
y x y y x y x
2. Promijeniti poredak integracije i izra unati dvostruki integral 2 2
20 0
.lna a ya x x a
3. Izra unati površinu dij s 2
dxI ydy
x a
ela fere 2 2 2x y z a koji se nalazi u unutrašnjosti
cilindra2 2
2 2 1, 0 .x y b a
ti krivolinijsk tegra
a b
4. Izra una i in l 31 arctg ,y
c
yI dx y e dyx x
ako je c pozitivno
ene isje kom kružnog prstena orjentisana kontura oblasti odre2 21 4, 0 .x y y x
eni dio ispita izPism Matematike II, 23.09.2011.
RUPA A 1. Izra unati površiG
2 3 3 12 0x y i dionu figure koju odre uju prava elipse 6 u prvom kvadrantu.
2. Promij ra unati dvostruki integral
24 9x y2 3
eniti poredak integracije i iz
222 8y
5102
.dxI ydyx
3. Izra unati površinu dijela površi 2 2 1 0x y z koji se nala a ni 0.zzi iznad r v
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1 2
1 22 2 2 2, ,c c
xdy ydx xdy ydxI I4. Dati su krivolinijski integrali x y x y
gdje je 1c duž ,AB
1, 2 , 1, 4A B i
1, 2 , 1, 4 ,A B
prolazi kroz ta ke
orjentisana od ta ke A prema ta ki B, a je parabola koja 2c1 11, .
2 4C Dokazati da je 1 2I I i izra
taj broj.
RUPA B i površinu krivolinijskog etverougla ome enog parabolama
unati
G1. Izra unat
22 2 2, , 2 , 3 .x
3y x y y x y x
2. Promijeniti poredak integracije i izra unati dvostruki integral 2 2
lna a ya
20 0
.x x a dx
I ydyx a
3. Izra unati površinu dijela sfere 22 2 2x y z a koji se nalazi u unutrašnjosti
cilindra2 2
1, 0 .x y b a2 2a b
4. Izra unati krivolinijski integral 31 arctg ,y
c
yI dx y e dyx x
ako je c pozitivno
ene isje kom kružnog prstena orjentisana kontura oblasti odre2 21 4, 0 .x y y x
Pismeni dio ispita iz Matematike II, oktobar 2011.
1. Odrediti jedna inu tangentne ravni na površ 2 2 2
2 2 2 1x y za b c
, koja je normalna na
pravoj1 2 3x y z .
2. Promijeniti poredak integracije i izra unati dvostruki integral 2 2
lna a ya
20 0
.x x a dx
I ydyx a
3. Izra unati krivolinijski integral ,c
xds ako je c lemniskata
4. Dato je vektorsko polje
, 0x y a x y a22 2 2 2 2
2 2
.2 2 2 2(2 ( ) , 2 ( ) , 2 ( ) )A x y z yz y z x xz z x y xy .
tencijalno i odrediti mu potencijal. Izra unati fluks Pokazati da je polje A povektorskog polja A kroz spoljnu stranu polusfere 2 2 2: 2 0, 0x y z z y .
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