sveska iz matematike ii akademska 2011 2012 v3 1dio
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1 dio matematika 2TRANSCRIPT
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Univerzitet u Zenici Mašinski fakultet Akademska 2011/2012.
Sadržaj sveske iz predmeta Matematika 2Odsjeci: Inžinjerski dizajn proizvoda, Inžinjerska ekologija, Menadžment proizvodnim
tehnologijama, Održavanje
Sedmica broj 1 i 2 (Odre eni integrali)
• Tablica integrala. 3 • Izra unavanje odre enih integrala pomo u neodre enih. 4 • Osobine odre enih integrala. 5 • Zamjena promjenjivih u odre enom integralu. 6• Nepravi (nesvojstveni) integral. 13• Primjena odre enog integrala: I Izra unavanje površine ravne figure. 14
• II Zapremina rotacionog tijela. 27• III Dužina luka krive. 29 • IV Izra unavanje površine dijelova zakrivljenih ploha (Komplanacija obrtne pov.). 31
Sedmica broj 3 i 4 (Funkcije dvije promjenjive)
• Funkcije dvije promjenjive. Definiciono podru je f-je dvije promjenjive. 33 • Tablica izvoda. 37 • Parcijalni izvodi. 38 • Totalni i parcijalni diferencijal. 40 • Parcijalni izvodi i diferencijl višeg reda. 42 • Parcijalni izvodi i diferencijl složenih funkcija. 45 • Parcijalni izvodi višeg reda složenih funkcija. 51 • Ekstremi funkcija dvije promjenjive 55
• Uslovni ekstremi funkcija dvije promjenjive 67• Jedna ina tangentne ravni i jedna ina normale na površ 75
Sedmica broj 5, 6 i 7 (Višestruki integrali)
• Dvojni (dvostruki) integrali 83• Smjena promjenjivih u dvojnim integralima 107
• Trojni (trostruki) integrali 131• Ra unanje trostrukih integrala uvo enjem cilindri nihi sfernih koordinata 141
• Primjena dvojnog integrala 151• Primjena trostrukog integrala 167
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Sedmica broj 8, 9 i 10(Krivoliniski integrali)
• Krivoliniski integrali prve vrste (po luku) 187• Primjena krivoliniski integrali prve vrste: Ra unanje površine cilindri ne površi 197
• Krivoliniski integrali druge vrste (po koordinatama) 201• Green-Gausova formula 215
• Primjena krivoliniski integrali druge vrste: Ra unanje površine ravne figure 225 • Nezavisnost krivoliniskog integrala od vrste konture. Odre ivanje primitivnih f-ja. 235
Sedmica broj 11 i 12 (Površinski integrali)
• Površinski integrali I vrste 247• Površinski integrali II vrste 263
• Primjena površinskog integrala 275• Stoksova formula 279• Formula Gaus-Ostrogradskog 287
Sedmica broj 13 (Integrali ovisni o parametru)
• Diferenciranje svojstvenog integrala ovisnog o parametru 293• Diferenciranje nesvojstvenog integrala ovisnog o parametru 297
Sedmica broj 14 i 15 (Vektorska teorija polja)
• Skalarno polje. Gradijent skalarnog polja. Vektorsko polje. Rotor i divergencija vektorskog polja. 303
• Cirkulacija i fluks vektorskog polja. 319
Dodatak(Ispitni rokovi)
• Svi ispitni rokovi iz 2011 godine 331
Zbirke zadataka za dodatno usavršavanje i napredovanje:• Berman: Zbirka zadataka iz Matemati ke analize, Nau na knjiga, 1978 • Peri , Tomi , Kara i : Zbirka riješenih zadataka iz Matematike II, Svjetlost, 1987 • Uš umli , Mili i : Zbirka zadataka iz Matematike II, Nau na knjiga,• Ferenci, Ungar, omi , Cvijetanovi , Uzelac: Zbirka zadataka iz Matematike za studente Tehni kih fakulteta, Nau na knjiga, 1983 • Ze i , Huskanovi , Alajbegovi : Matematika 1 za tehni ke fakultete, MF, 2009
(sveska je skinuta sa stranice pf.unze.ba\nabokov Za sve uo ene greške pisati na [email protected])
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(Zadaci su skinuti sa stranice: \pf.unze.ba\nabokovZa uocene greske pisati na [email protected])
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1. Izra unaj trostruki integral . 2
0
11
1
)4(2
dzzdydxIx
Rješenje:
340
3410)
322(10)
3(10)1(10
10)28(2
4)4(
1
1
31
1
1
1
2
11
1
1 1
1
2
0
2
0
1
1
22
0
111
1 2222
xxdxx
dxydydxdyzzdxdzzdydxIxxxx
2. Izra unaj trostruki integralG yxdxdydz
1, gdje je G ograni ena ravnima :
a) x+y+z=1, x=0, y=0, z=0;
b) x=0, x=1, y=2, y=5, z=2, z=4.
Rješenja:
a)G yxdxdydz
1 x=0, y=0, z=0
Skicirajmo oblast G (vidi sliku desno).
x+y+z=1 1111zyx
x= 0 je yOz ravan y= 0 je xOz ravan z= 0 je xOy ravan
Odredimo projekciju oblasti na xOy ravan: Nacrtati sliku (uputa: pogledati xoy ravan sa slike desno).
x+y+z=1z= 0
135
x+y=1z= 1-x-y
yxzxy
x
101010
Sa slike projekcije odredimo granice:
21
211
2
)1())1(1
1(
11
11
1
0
21
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1
0
1
0
1
0
1
0
1
0
1
0
xx
dxxdxydydxdyyxyx
dx
dyzyx
dxdzyx
dydxyx
dxdydz
xxx
yx xyx
x
G
b)G yxdxdydz
1 x=0, x=1, y=2, y=5, z=2, z=4.
Skicirajmo oblast G (vidi sliku).
dxxdxx
dxxxttdtdx
xtyxty
dtdytyx
yxdydx
yxdyzdxdz
yxdydx
x
x
x
x1
0
1
0
1
0
4
1
1
0
1
0
4
1
1
0
5
2
4
2
1
0
5
2
4
2
1
0
5
2
1ln24ln2
1ln4ln2ln22
4512
1
12
11
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11ln
1
1ln1ln
44ln
4
4ln4ln
xxdxxx
xvxdxdu
dvdxxudxx
xxdxxx
xvxdxdu
dvdxxudxx
54ln10
45ln
45ln2
45
225
ln2
2ln245ln8
25ln21ln24ln8
25ln2
122
44222ln5ln2
121ln2
424ln2
1054
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
xx
xdxdx
xdxdx
xxdxxx
xxdxxx
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