Ministerstvo obrazovani Rossisko Federacii Sankt-Peterburgski gosudarstvenny universitet nizkotemperaturnyh i piwevyh tehnologi KRATNYE INTEGRALY metodiqeskie ukazani k vypolneni rasqetnogo zadani dl studentov 2-go kursa vseh specialьnoste Cankt-Peterburg 2003
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Ministerstvo obrazovani Rossisko Federacii
Sankt-Peterburgski gosudarstvenny universitet
nizkotemperaturnyh i piwevyh tehnologi
KRATNYE INTEGRALY
metodiqeskie ukazani
k vypolneni rasqetnogo zadani
dl studentov 2-go kursa vseh specialьnoste
Cankt-Peterburg 2003
UDK 518(07.07)
Kurova L.V., Podolьski V.A. Kratnye integraly. Metod. ukaza-ni k vypolneni rasqetnogo zadani dl studentov 2-go kursa vsehspec. – SPb.: SPbGUNiPT, 2003. – 31 s.
Priveden pordk vypolneni rasqetnogo zadani, obrazec vypolneni tipovogo rasqeta
i nabor tipovyh zadaq.
RecenzentKand. fiz.-mat. nauk, doc. G.V. Karpuhin
Odobreny k izdani Sovetom fakulьteta holodilьno tehniki
c© Sankt-Peteburgski gosudarstvenny universitet
nizkotemperaturnyh i piwevyh tehnologi, 2003
Ukazani k vypolnenitipovogo rasqeta
Obwie trebovani k vypolneni zadani
Vypolnenie individualьnogo tipovogo rasqeta po teme ”Kratnyeintegraly” imeet celь:
• zakrepitь teoretiqeskie znani u studentov po prosluxannomuna lekcih materialu;
• razvitь umenie samostotelьno rabotatь nad predmetom;
• nauqitь studentov priemam vyqisleni dvonyh i tronyh in-tegralov, pravilьno, vybira ishod iz uslovi konkretno za-daqi, neobhodimu sistemu koordinat.
Obъem zadani tipovogo rasqeta, sostowego iz konkretnyh pri-merov, opredelets prepodavatelem individualьno dl kaжdogo stu-denta.
Zadani vypolnts studentami v otdelьno tetradi.Vypolnennoe zadanie student dolжen sdatь prepodavatel na pro-
verku, a potom zawititь ego. Pri zawite student obzan pravilьnootvetitь na lbo teoretiqeski vopros, pereqenь kotoryh privo-dits niжe.
Pri vypolnenii zadani rekomenduets:
• uslovie zadaqi zapisatь polnostь;
• rexenie izloжitь podrobno, t.e. so vsemi preobrazovanimi irasqetami, s ukazaniem ispolьzovannyh formul ili teorem;
• risunki k zadaqam vypolnitь qetko, razborqivo karandaxom.
Pri rexenii primerov tipovogo rasqeta i pri podgotovke k egozawite rekomenduets ispolьzovatь sleduwu literaturu:
1. Bermant A.F., Aramanoviq I.G. Kratki kurs matematiqeskogoanaliza. M.: Nauka, 1973.
3
2. Piskunov N.S. Differencialьnoe i integralьnoe isqislenie.M.: Nauka, 1970.
3. Bugrov .S., Nikolьski S.M. Differencialьnoe i integralь-noe isqislenie. M.: Nauka, 1980.
4. Bugrov .S., Nikolьski S.M. Differencialьnye uravneni.Kratnye integraly. Rdy. Funkcii kompleksnogo peremennogo.M.: Nauka, 1981.
5. Myxkis A.D. Lekcii po vysxe matematike. M.: Nauka, 1973.
6. Danko P.E., Popov A.G., Koжevnikova T.. Vysxa matematikav upraжnenih i zadaqah. Q.1–2. M.: Vysx. xk., 1980.
7. Zaporoжec G.I. Rukovodstvo k rexeni zadaq po matematiqe-skomu analizu. M.: Vysx. xk., 1966.
8. Berman G.N. Sbornik zadaq po kursu matematiqeskogo analiza.M.: Nauka, 1969.
9. Demidoviq B.P. Sbornik zadaq i upraжneni po matematiqe-skomu analizu. M.: Nauka, 1969.
10. Ilьin V.A., Poznk З.G. Osnovy matematiqeskogo analiza. M.:Nauka, 1979. Q.I–II.
11. Budak B.M., Fomin S.V. Kratnye integraly i rdy. M.: Nauka,1965.
12. Ter-Krikorov A.M., Xabunin M.I. Kurs matematiqeskogo ana-liza. M.: Nauka, 1988.
Teoretiqeskie voprosy
1. Obъem cilindriqeskogo tela.
2. Opredelenie dvonogo integrala.
3. Svostva dvonogo integrala.
4. Vyqislenie dvonogo integrala svedeniem ego k povtornomu vdekartovo sisteme koordinat.
4
5. Zamena peremennyh v dvonom integrale. Dvono integral vpolrno sisteme koordinat.
6. Priloжeni dvonogo integrala:
a) vyqislenie plowadi plosko figury;
b) vyqislenie massy plosko figury;
c) vyqislenie momentov inercii plosko figury.
7. Opredelenie tronogo integrala.
8. Vyqislenie tronogo integrala v dekartovo sisteme koordi-nat.
9. Zamena peremennyh v tronom integrale.
10. Trono integral v cilindriqesko sisteme koordinat.
11. Trono integral v sferiqesko sisteme koordinat.
12. Priloжeni tronogo integrala:
a) vyqislenie obъema tela;
b) vyqislenie massy tela;
c) vyqislenie momentov inercii tela.
Obrazec vypolneni tipovogo rasqeta
Primer 1. Izmenitь pordok integrirovani v povtornom integra-le
−√
3∫
−2
dx
√4−x2∫
0
f(x, y)dy +
0∫
−√
3
dx
2−√
4−x2∫
0
f(x, y)dy.
Reqь idet o summe dvuh dvonyh integralov po oblastm D1 i D2,zadannyh sootvetstvenno:
D1 :
x = −2, x = −√
3,
y = 0, y =√
4 − x2i D2 :
x = −√
3, x = 0,
y = 0, y = 2 −√
4 − x2.
5
Po suwestvu, my moжem rassmatrivatь зtu summu kak odin dvo-no integral po oblasti D, kotora vlets obъedineniem oblaste
D1 i D2:
D :
y = 0,
y =√
4 − x2 , y = 2 −√
4 − x2 .Pri зtom okruжnosti x2 + y2 = 4 ix2 +(y− 2)2 = 4 peresekats v toqkeA(−
√3, 1) (ris. 1), koordinaty ko-
toro vlts rexeniem sistemy
uravneni
x2 + y2 = 4,x2 + (y − 2)2 = 4.Ris.1
−√
3
Poзtomu, izmeniv pordok integrirovani, my moжem zapisatь dvo-no integral po oblasti D v vide odnogo povtornogo integrala:
1∫
0
dy
√4y−y2
∫
−√
4−y2
f(x, y)dx.
Primer 2. Vyqislitь dvono integral∫
D
∫
(54x2y2 + 150x4y4)dxdy
po oblasti D : x = 1, y = x3, y = −√x (ris. 2).
Ris.2
Vybira pordok integrirovani sle-duwim obrazom: vnutrenni integral– po peremenno y, vnexni – po peremen-no x, poluqim
∫
D
∫
(54x2y2 + 150x4y4)dxdy =
1∫
0
dx
x3
∫
−√x
(54x2y2 + 150x4y4)dy ==
1∫
0
dx(
18x2y3 + 30x4y5)∣
∣
∣
x3
−√x
= 11.
6
Primer 3. Vyqislitь plowadь figury D :
x2 + y2 = 12,
y2 =√
6x.
Kak izvestno, plowadь plosko figury D daets dvonym integra-lom S =
∫
D
∫
dxdy. V naxem sluqae oblastь D ograniqena okruжnostь
x2 + y2 = 12 i parabolo y2 =√
6x, kotorye peresekats v toqkahA(
√6,√
6) i B(√
6,−√
6). Koordinaty toqek vlts rexenimi si-
stemy uravneni
x2 + y2 = 12,
y2 =√
6x.Uqityva simmetriqnostь oblasti D (ris. 3)
otnositelьno osi Ox, moжno vyqislitь integralpo polovine oblasti i rezulьtat umnoжitь na 2:
S = 2
∫
D1
∫
dxdy = 2
√6
∫
0
dy
√12−y2
∫
y2/√
6
dx =
Ris. 3
√6
−√
6
= 2
√6
∫
0
dy(
√
12 − y2 − y2
√6
)
= 3π + 2.
Primer 4. Vyqislitь plowadь figury D, gde oblastь zadana sle-
duwim obrazom:
x2 + y2 − 4y = 0,x2 + y2 − 8y = 0,
y =√
3x, x = 0.Figura D ograniqena dvum okruжnostmi
x2 + (y − 2)2 = 4 i x2 + (y − 4)2 = 16 radiusami2 i 4 sootvetstvenno, a takжe dvum prmymiy =
√3x i x = 0 (ris.4). Peredem k polrnym
koordinatam
x = r cos ϕ,y = r sinϕ, (x2 + y2 = r2). Ris. 4
ϕ
Pri зtom oblastь D otobrazits v oblastь G :
r = 4 sinϕ,r = 8 sinϕ,
tg ϕ = 1/√
3, cos ϕ = 0
7
ili G :
r = 4 sinϕ,r = 8 sinϕ,ϕ = π/6, ϕ = π/2
, a зlement plowadi dxdy =⇒ rdrdϕ.
Poзtomu
S =
∫
D
∫
dxdy =
∫
G
∫
rdrdϕ =
π/2∫
π/6
dϕ
8 sin ϕ∫
4 sin ϕ
rdr = 24
π/2∫
π/6
sin2 ϕdϕ = 4π + 3√
3.
Primer 5. Vyqislitь plowadь figury D : (x2 + y2)2 = a2(x2 − 3y2).Preжde vsego otmetim, qto x2 − 3y2 ≥ 0, tak kak neotricatelьna le-va qastь ravenstva, a sama figura simmetriqna otnositelьno oseOx i Oy. Peredem k polrnym koordinatam, pri зtom poluqim
r2 = a2(cos2 ϕ − 3 sin2 ϕ) t. e. r = a√
cos2 ϕ − 3 sin2 ϕ = a√
2 cos 2ϕ − 1 . Izskazannogo vyxe sleduet, qto cos2 ϕ − 3 sin2 ϕ ≥ 0 ili tg2 ϕ ≤ 1/3 =⇒−1
√3 ≤ tg ϕ ≤ 1
√3 =⇒ πk − π/6 ≤ ϕ ≤ π/6 + πk (k = 0,±1,±2, . . .). V
naxem sluqae dl k sleduet ostavitь znaqeni k = 0 i k = 1. Takim
obrazom, oblastь D =⇒ G :
r = a√
2 cos 2ϕ − 1 ,−π/6 ≤ ϕ ≤ π/6, 5/6π ≤ ϕ ≤ 7/6π.
Dalee integriruem po qetverto qasti oblasti G i poluqennyrezulьtat umnoжaem na 4:
S = 4
π/6∫
0
dϕ
a√
2 cos 2ϕ−1∫
0
rdr = 2a2
π/6∫
0
(2 cos 2ϕ − 1)dϕ = a2(√
3 − π
2
)
.
Primer 6. Vyqislitь trono integral
∫∫
Ω
∫
x2z dxdydz po oblasti
Ω :
y = 3x, y = 0, x = 2,z = xy, z = 0.
Ishodny integral zapixem v vide
∫∫
Ω
∫
x2z dxdydz =
∫
D
∫
dxdy
xz∫
0
x2z dz ,
8
gde oblastь D – proekci tela Ω na plos-kostь xOy, t. e. D : y = 3x, y = 0, x = 2 (ris. 5).Dalьnexie vyqisleni dostatoqno prosty:
∫
D
∫
dxdy
xz∫
0
x2z dz =
∫
D
∫
dxdy x2 z2
2
∣
∣
∣
∣
xy
0
=
=1
2
2∫
0
x4 dx
3x∫
0
y2 dy = 144
.
Ris.5
Primer 7. Vyqislitь obъem tela Ω :
x = 17√
2y, x = 2√
2y,z = 0, z + y = 1/2.
Telo Ω ograniqeno cilindriqeskimi poverh-nostmi x = 17
√2y i x = 17
√2y i ploskostmi
z = 0 i z + y = 1/2 (otmetim, qto x ≥ 0). Egoproekci na xOy predstavlet sobo oblastьD : x = 17
√2y, x = 2
√2y, y = 1/2 (ris. 6). Poзtomu
V =
∫∫
Ω
∫
dxdydz =
∫
D
∫
dxdy
1/2−y∫
0
dz =
∫
D
∫(
1
2− y
)
dxdy =
1/2∫
0
(
1
2− y
)
dy
17√
2y∫
2√
2y
dx = 1 .
Ris.6
Primer 8. Vyqislitь obъem tela Ω :
x2 + y2 + 2x = 0,
z =25
4− y2, z = 0.
Dannoe telo predstavlet sobo cilindr, uravnenie kotorogomoжno zapisatь v vide (x + 1)2 + y2 = 1, ograniqenny snizu plos-kostь xOy, a sverhu cilindriqesko poverhnostь s obrazuwe,parallelьno osi Ox. Lboe seqenie зto poverhnosti ploskostь,perpendikulrno osi Ox, vlets parabolo z = 25/4 − y2. Obъemtela raven:
9
V =
∫∫
Ω
∫
dxdydz =
∫
D
∫
dxdy
25/4−y2
∫
0
dz =
∫
D
∫(
25
4− y2
)
dxdy,
gde oblastь D predstavlet sobo krug (x + 1)2 + y2 ≤ 1, s centrom vtoqke C(−1, 0) i radiusom 1 (ris. 7).
Vyqisl dvono integral, peredem kpolrnym koordinatam, pri зtom oblastь Dotobrazits v G : r = −2 cos ϕ, π/2 ≤ ϕ ≤ 3π/2.Vvidu simmetrii zadaqi integrirovatь moж-no po polovine oblasti i rezulьtat udvoitь.
Takim obrazom,
V = 2
∫
G1
∫(
25
4− y2
)
dϕ rdr = 2
π∫
π/2
dϕ
−2 cos ϕ∫
0
rdr =
ϕ
Ris.7
=
π∫
π/2
cos2 ϕ (25−8 sin2 ϕ cos2 ϕ) dϕ =1
2
π∫
π/2
(1+cos 2ϕ)(25−2 sin2 2ϕ) dϕ = 6π .
Zameqanie. V зto zadaqe moжno bylo srazu pereti k cilindri-qeskim koordinatam: x = r cos ϕ, y = r sinϕ, z = z. Togda
Ω =⇒ Λ :
r + 2 cos ϕ = 0,π
2≤ ϕ ≤ 3π
2, z =
25
4− y2,
dxdydz =⇒ dϕ rdrdz .
V rezulьtate poluqim tako жe integral:
V =
∫∫
Λ
∫
dϕ rdrdz = 2
∫
G1
∫
dϕ rdr
25/4−y2
∫
0
dz, gde G1 :
r = −2 cos ϕ,π
2≤ ϕ ≤ π.
Primer 9. Vyqislitь obъem tela Ω :
z =√
36 − x2 − y2,
z =
√
x2 + y2
3.
Telo predstavlet sobo xarovo segment – qastь konusa 3z2 = x2 + y2,
10
nahodwus nad ploskostь xOy, na-krytu sverhu sferiqesko poverhno-stь x2 + y2 + z2 = 36 s centrom v naqa-le koordinat i radiusom 6 (ris. 8.) Takkak ono obladaet cilindriqesko simme-trie (osь Oz vlets osь simmetriibeskoneqnogo pordka), udobno srazu жepereti k cilindriqeskim koordinatam:
x = r cos ϕ,y = r sinϕ,z = z;
3√
3
Ris.8
Ω =⇒ Λ :
z = +√
36 − r2,
z =r√3;
dxdydz =⇒ dϕ rdrdz.
Poluqim:
V =
∫∫
Λ
∫
dϕ rdrdz =
∫
D
∫
dϕ rdr
√36−r2∫
r/√
3
dz =
∫
D
∫(
√
36 − r2 − r√3
)
dϕ rdr.
Oblastь D :
r = 3√
3,0 ≤ ϕ ≤ 2π
estь proekci Ω na ploskostь xOy i
predstavlet sobo krug s centrom v naqale koordinat, radius koto-
rogo r = 3√
3 vlets rexeniem sistemy uravneni
z =√
36 − r2,
z =r√3;
V =
∫
D
∫(
√
36 − r2 − r√3
)
dϕ rdr
2π∫
0
dϕ
3√
3∫
0
(
√
36 − r2 − r√3
)
rdr = 72π.
Primer 10. Vyqislitь obъem tela Ω :
36 ≤ x2 + y2 + z2 ≤ 144,
−√
x2 + y2
3≤ z ≤ −
√
x2 + y2
15,
0 ≤ y ≤ −√
3x.
11
Naxe telo zaklqeno meжdu koniqeskimi x2+y2 = 3z2 i x2 + y2 = 15z2
i koncentriqeskimi sferiqeskimi x2 + y2 + z2 = 36 i x2 + y2 + z2 = 144poverhnostmi i rassekaets ploskostmi y = 0 i y = −
√3x. Raspolo-
жeno ono pod ploskostь xOy (z ≤ 0). Seqeni tela koordinatnymiploskostmi xOz i xOy pokazany na ris. 9 i 10.
Ris. 9
θ
z=−|x|/√
15
z=−|x|/√
3
Ris. 10
ϕy=−
√3x
Peredem k sferiqeskim koordinatam po formulam:
x = r sin θ cos ϕ,y = r sin θ sinϕ,z = r cos θ;
dxdydz =⇒ dϕ sin θdθ r2dr.
Iz uslovi zadaqi (i risunkov) vidno, qto x ≤ 0 i z ≤ 0, a зtooznaqaet, qto cos ϕ ≤ 0 i cos θ ≤ 0, pri зtom sin θ ≥ 0, tak kak ugolθ ∈ [0, π]. Poзtomu oblastь Ω otobraжaets v Λ:
Λ :
36 ≤ r2 ≤ 144,
−√
r2 sin2 θ
3≤ r cos θ ≤ −
√
r2 sin2 θ
15,
0 ≤ r sin θ sinϕ ≤ −r√
3 sin θ cos ϕ
⇒
6 ≤ r ≤ 12,
−sin θ√3
≤ cos θ ≤ −sin θ√15
,
0 ≤ sinϕ ≤ −√
3 cos ϕ
Rassmotrim podrobnee neravenstvo 0 ≤ sinϕ ≤ −√
3 cos ϕ. Takkak cos ϕ ≤ 0, to dl ego rexeni udobno vvesti novu peremennuϕ = π − ϕ, togda ϕ = π − ϕ. Pri зtom sinϕ = sin(π − ϕ) = sin ϕ, acos ϕ = cos(π − ϕ) = − cos ϕ (cos ϕ ≥ 0). Razdeliv neravenstvo nacos ϕ, poluqim 0 ≤ tg ϕ ≤
√3. Iz mnoжestva ego rexeni vybira-
em 0 ≤ ϕ ≤ π/3 ili, vernuvxisь k peremenno ϕ, 2π/3 ≤ ϕ ≤ π(sm. ris. 10). Podobnym obrazom iz neravenstva, soderжawego ugol θ,poluqim π − arctg
√15 ≤ θ ≤ 2π/3. Teperь oblastь Λ zapixem v okon-
12
qatelьnom vide:
Λ :
6 ≤ r ≤ 12,
π − arctg√
15 ≤ θ ≤ 2π
3,
2π
3≤ ϕ ≤ π.
Obъem vyqislets prosto:
V =
∫∫
Λ
∫
dϕ sin θdθ r2dr =
π∫
2π/3
dϕ
2π/3∫
π−arctg√
15
sin θdθ
12∫
6
r2dr =
= 168π
(
1
2− cos(arctg
√15)
)
.
Preobrazuem rezulьtat, ispolьzu formulu cos α = 1/√
1 + tg2 α :
cos(arctg√
15) =1
√
1 + tg2(arctg√
15)=
1
4.
Okonqatelьno imeem: V = 42π.
13
Rasqetnye zadani
1. Izmenitь pordok integrirovani
V zadaqah 1.1–1.30 izmenitь pordok integrirovani v zadannyhpovtornyh integralah. Oblastь integrirovani izobrazitь na qer-teжe.
1.1.
−1∫
−2
dy
0∫
−√2+y
f(x, y) dx +
0∫
−1
dy
0∫
−√−y
f(x, y) dx .
1.2.
1∫
0
dy
0∫
−√y
f(x, y) dx +
√2
∫
1
dy
0∫
−√
2−y2
f(x, y) dx .
1.3.
1∫
0
dy
y∫
0
f(x, y) dx +
√2
∫
1
dy
√2−y2
∫
0
f(x, y) dx .
1.4.
1∫
0
dy
√y
∫
0
f(x, y) dx +
2∫
1
dy
√2−y
∫
0
f(x, y) dx .
1.5.
−1∫
−√
2
dx
0∫
−√
2−x2
f(x, y) dy +
0∫
−1
dx
0∫
x
f(x, y) dy .
1.6.
1/√
2∫
0
dy
arcsin y∫
0
f(x, y) dx +
1∫
1/√
2
dy
arccos y∫
0
f(x, y) dx .
1.7.
−1∫
−2
dy
√2+y
∫
0
f(x, y) dx +
0∫
−1
dy
√−y∫
0
f(x, y) dx .
14
1.8.
1∫
0
dy
0∫
−√y
f(x, y) dx +
e∫
1
dy
− ln y∫
−1
f(x, y) dx .
1.9.
−1∫
−√
2
dx
√2−x2∫
0
f(x, y) dy +
0∫
−1
dx
x2
∫
0
f(x, y) dy .
1.10.
−√
3∫
−2
dx
0∫
−√
4−x2
f(x, y) dy +
0∫
−√
3
dx
0∫
−2+√
4−x2
f(x, y) dy .
1.11.
1∫
0
dx
1∫
1−x2
f(x, y) dy +
e∫
1
dx
1∫
lnx
f(x, y) dy .
1.12.
1∫
0
dy
3√
y∫
0
f(x, y) dx +
2∫
1
dy
2−y∫
0
f(x, y) dx .
1.13.
π/4∫
0
dy
sin y∫
0
f(x, y) dx +
π/2∫
π/4
dy
cos y∫
0
f(x, y) dx .
1.14.
−1∫
−2
dx
0∫
−(2+x)
f(x, y) dy +
0∫
−1
dx
0∫
3√
x
f(x, y) dy .
1.15.
1∫
0
dy
√y
∫
0
f(x, y) dx +
e∫
1
dy
1∫
ln y
f(x, y) dx .
1.16.
1∫
0
dy
0∫
−√y
f(x, y) dx +
2∫
1
dy
0∫
−√2−y
f(x, y) dx .
1.17.
1∫
0
dy
0∫
−y
f(x, y) dx +
√2
∫
1
dy
0∫
−√
2−y2
f(x, y) dx .
15
1.18.
1∫
0
dy
y3
∫
0
f(x, y) dx +
2∫
1
dy
2−y∫
0
f(x, y) dx .
1.19.
√3
∫
0
dx
0∫
−2+√
4−x2
f(x, y) dy +
2∫
√3
dx
0∫
−√
4−x2
f(x, y) dy .
1.20.
−1∫
−2
dy
0∫
−(2+y3)
f(x, y) dx +
0∫
−1
dy
0∫
3√
y
f(x, y) dx .
1.21.
1∫
0
dy
y∫
0
f(x, y) dx +
e∫
1
dy
1∫
ln y
f(x, y) dx .
1.22.
1∫
0
dx
x2
∫
0
f(x, y) dy +
√2
∫
1
dx
√2−x2∫
0
f(x, y) dy .
1.23.
π/4∫
0
dx
sin x∫
0
f(x, y) dy +
π/2∫
π/4
dx
cos x∫
0
f(x, y) dy .
1.24.
−1∫
−√
2
dy
0∫
−√
2−y2
f(x, y) dx +
0∫
−1
dy
0∫
y
f(x, y) dx .
1.25.
1∫
0
dx
x3
∫
0
f(x, y) dy +
2∫
1
dx
2−x∫
0
f(x, y) dy .
1.26.
√3
∫
0
dx
2−√
4−x2∫
0
f(x, y) dy +
2∫
√3
dx
√4−x2∫
0
f(x, y) dy .
1.27.
1∫
0
dx
0∫
−√x
f(x, y) dy +
2∫
1
dx
0∫
−√
2−x
f(x, y) dy .
16
1.28.
1∫
0
dx
x∫
0
f(x, y) dy +
√2
∫
1
dx
√2−x2∫
0
f(x, y) dy .
1.29.
1∫
0
dy
√y
∫
0
f(x, y) dx +
√2
∫
1
dy
√2−y2
∫
0
f(x, y) dx .
1.30.
1∫
0
dx
√x
∫
0
f(x, y) dy +
2∫
1
dx
√2−x
∫
0
f(x, y) dy .
2. Vyqislitь dvonye integraly.
V zadaqah 2.1–2.30 vyqislitь dvonye integraly∫
D
∫
f(x, y) dxdy,
esli oblastь D zadana ukazannymi linimi.
2.1.
∫
D
∫
(12x2y2+16x3y3) dxdy,
D : x = 1, y = x2, y = −√x .
2.2.
∫
D
∫
(9x2y2 + 48x3y3) dxdy,
D : x = 1, y =√
x, y = −x2.
2.3.
∫
D
∫
(36x2y2−96x3y3) dxdy,
D : x = 1, y = 3√
x, y = −x3.
2.4.
∫
D
∫
(12x2y2+16x3y3) dxdy,
D : x = 1, y = x3, y = − 3√
x .
2.5.
∫
D
∫
(27x2y2+48x3y3) dxdy,
D : x = 1, y = x2, y = − 3√
x .
2.6.
∫
D
∫
(18x2y2+32x3y3) dxdy,
D : x = 1, y = 3√
x, y = −x2.
2.7.
∫
D
∫
(18x2y2+32x3y3) dxdy,
D : x = 1, y = x3, y = −√x .
2.8.
∫
D
∫
(27x2y2+48x3y3) dxdy,
D : x = 1, y =√
x, y = −x3.
2.9.
∫
D
∫
(4xy + 3x2y2) dxdy,
D : x = 1, y = x2, y = −√x .
2.10.
∫
D
∫
(12xy + 9x2y2) dxdy,
D : x = 1, y =√
x, y = −x2.
2.11.
∫
D
∫
(8xy + 9x2y2) dxdy,
D : x = 1, y = 3√
x, y = −x3.
2.12.
∫
D
∫
(24xy + 18x2y2) dxdy,
D : x = 1, y = x3, y = − 3√
x .
17
2.13.
∫
D
∫
(12xy + 27x2y2) dxdy,
D : x = 1, y = x2, y = − 3√
x .
2.14.
∫
D
∫
(8xy + 18x2y2) dxdy,
D : x = 1, y = 3√
x, y = −x2.
2.15.
∫
D
∫
(4
5xy +
9
11x2y2) dxdy,
D : x = 1, y = x3, y = −√x .
2.16.
∫
D
∫
(4
5xy + 9x2y2) dxdy,
D : x = 1, y =√
x, y = −x3.
2.17.
∫
D
∫
(24xy − 48x3y3) dxdy,
D : x = 1, y = x2, y = −√x .
2.18.
∫
D
∫
(6xy + 24x3y3) dxdy,
D : x = 1, y =√
x, y = −x2.
2.19.
∫
D
∫
(4xy + 16x3y3) dxdy,
D : x = 1, y = 3√
x, y = −x3.
2.20.
∫
D
∫
(4xy + 16x3y3) dxdy,
D : x = 1, y = x3, y = − 3√
x .
2.21.
∫
D
∫
(44xy + 16x3y3) dxdy,
D : x = 1, y = x2, y = − 3√
x .
2.22.
∫
D
∫
(4xy + 176x3y3) dxdy,
D : x = 1, y = 3√
x, y = −x2.
2.23.
∫
D
∫
(xy − 4x3y3) dxdy,
D : x = 1, y = x3, y = −√x .
2.24.
∫
D
∫
(4xy + 176x3y3) dxdy,
D : x = 1, y =√
x, y = −x3.
2.25.
∫
D
∫
(6x2y2 +25
3x4y4) dxdy,
D : x = 1, y = x2, y = −√x .
2.26.
∫
D
∫
(9x2y2+25x4y4) dxdy,
D : x = 1, y =√
x, y = −x2.
2.27.
∫
D
∫
(3x2y2 +50
3x4y4) dxdy,
D : x = 1, y = 3√
x, y = −x3.
2.28.
∫
D
∫
(9x2y2+25x4y4) dxdy,
D : x = 1, y = x3, y = − 3√
x .
2.29.
∫
D
∫
(xy − 9x5y5) dxdy,
D : x = 1, y = 3√
x, y = −x2.
2.30.
∫
D
∫
(54x2y2+150x4y4) dxdy,
D : x = 1, y = x3, y = −√x .
18
3. Vyqislitь plowadь plosko figury (I)
V zadaqah 3.1–3.30 vyqislitь s pomowь dvonogo integrala v de-kartovyh koordinatah plowadь plosko figury, ograniqenno uka-zannymi linimi.
3.1. y =3
x, y = 4ex, y = 3, y = 4 . 3.2. x =
√
36 − y2, x = 6 −√
36 − y2 .
3.3. x2 + y2 = 72, 6y = −x2 (y ≤ 0) . 3.4. x = 8 − y2, x = −2y .
3.5. y =3
x, y = 8ex, y = 3, y = 8 . 3.6. y =
√x
2, y =
1
2x, x = 16 .
3.7. x = 5 − y2, x = −4y . 3.8. x2 + y2 = 12, −√
6y = x2 (y ≤ 0) .
3.9. y =√
12 − x2, y = 2√
3−√
12 − x2, x = 0 (x ≥ 0) .
3.10. y =3
2
√x, y =
3
2x, x = 9 . 3.11. y = 11 − x2, y = −10x .
3.12. y = 20 − x2, y = −8x . 3.13. y = sinx, y = cos x, x ≥ 0 .
3.14. y =√
18 − x2, y = 3√
2 −√
18 − x2 .
3.15. y = 32 − x2, y = −4x . 3.16. y =2
x, y = 5ex, y = 2, y = 5 .
3.17. y = 3√
x, y =3
x, x = 9 . 3.18. x2 + y2 = 36, 3
√2y = x2 (y ≥ 0) .
3.19. y = 6 −√
36 − x2, y =√
36 − x2, x = 0 (x ≥ 0) .
3.20. y =25
4− x2, y = x − 5
2. 3.21. y =
√x, y =
1
x, x = 16 .
3.22. y =2
x, y = 7ex, y = 2, y = 7 . 3.23. x = 27 − y2, x = −6y .
3.24. x =√
72 − y2, 6x = y2, y = 0 (y ≥ 0) .
3.25. y =√
6 − x2, y =√
6 −√
6 − x2 . 3.26. y =3
2
√x, y =
3
2x, x = 4 .
3.27. y = sinx, y = cos x, x = 0 (x ≤ 0) .
3.28. y =1
x, y = 6ex, y = 1, y = 6 . 3.29. y = 3
√x, y =
3
x, x = 9 .
3.30. y =√
24 − x2, 2√
3y = x2, x = 0 (x ≥ 0) .
19
4. Vyqislitь plowadь plosko figury (II)
V zadaqah 4.1–4.30 nati plowadь figury, ograniqenno dannymilinimi, v polrnyh koordinatah.
4.1. a) y2 − 2y + x2 = 0,y2 − 4y + x2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)3 = a2x2y2.
4.2. a) x2 − 4x + y2 = 0,x2 − 8x + y2 = 0,y = 0, y = x/
√3;
b) (x2 + y2)3 = a2y4.
4.3. a) y2 − 6y + x2 = 0,y2 − 8y + x2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)2 = a2(x2 + 4y2).
4.4. a) x2 − 2x + y2 = 0,x2 − 4x + y2 = 0,y = 0, y = x;
b) (x2+y2)3 = a2y2(4x2+3y2).
4.5. a) y2 − 8y + x2 = 0,y2 − 10y + x2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)4 = a2x3y3.
4.6. a) x2 − 4x + y2 = 0,x2 − 8x + y2 = 0,y = 0, y = x;
b) (x2 + y2)2 = a2(3x2 + y2).
4.7. a) y2 − 4y + x2 = 0,y2 − 6y + x2 = 0,y = x, y = 0;
b) (x2 + y2)3 = a2(x2 − y2)2.
4.8. a) x2 − 2x + y2 = 0,x2 − 10x + y2 = 0,y = 0, y =
√3x;
b) (x2 + y2)3 = a2xy3.
4.9. a) y2 − 6y + x2 = 0,y2 − 10y + x2 = 0,y = x, x = 0;
b) (x2 + y2)3 = a2x3y.
4.10. a) x2 − 2x + y2 = 0,x2 − 4x + y2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)3 = a2(x4 + x2y2 + y4).
20
4.11. a) y2 − 2y + x2 = 0,y2 − 4y + x2 = 0,x = 0, y =
√3x;
b) (x2 + y2)3 = a2y2(x2 + 3y2).
4.12. a) x2 − 2x + y2 = 0,x2 − 6x + y2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)2 = a2(3x2 + y2).
4.13. a) y2 − 4y + x2 = 0,y2 − 6y + x2 = 0,y =
√3x, x = 0;
b) (x2 + y2)5 = a6xy3.
4.14. a) x2 − 2x + y2 = 0,x2 − 8x + y2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)7 = a8x4y2.
4.15. a) y2 − 2y + x2 = 0,y2 − 6y + x2 = 0,y = x/
√3, x = 0;
b) (x2 + y2)5 = a4x2y4.
4.16. a) x2 − 2x + y2 = 0,x2 − 4x + y2 = 0,y = x/
√3, y = 0;
b) (x2 + y2)5 = a4x4y2.
4.17. a) y2 − 2y + x2 = 0,y2 − 10y + x2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)2 = a2(x2 + 2y2).
4.18. a) x2 − 2x + y2 = 0,x2 − 6x + y2 = 0,y = 0, y = x/
√3;
b) (x2 + y2)7 = a8x2y4.
4.19. a) y2 − 4y + x2 = 0,y2 − 10y + x2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)2 = a2(x2 − y2).
4.20. a) x2 − 2x + y2 = 0,x2 − 6x + y2 = 0,y = x, y = 0;
b) (x2 + y2)5 = a6x3y.
4.21. a) y2 − 2y + x2 = 0,y2 − 4y + x2 = 0,y = x, x = 0;
b) (x2 + y2)2 = ax3.
4.22. a) x2 − 2y + y2 = 0,x2 − 4x + y2 = 0,y = 0, y =
√3x;
b) (x2 + y2)2 = a2(5x2 + 7y2).
21
4.23. a) y2 − 6y + x2 = 0,y2 − 8y + x2 = 0,y = x, x = 0;
b) (x2 + y2)2 = a2xy.
4.24. a) x2 − 4x + y2 = 0,x2 − 8x + y2 = 0,y = 0, y =
√3x;
b) (x2 + y2)2 = 2a2(x2 + 3y2).
4.25. a) y2 − 4y + x2 = 0,y2 − 8y + x2 = 0,y = x, x = 0;
b) (x2 + y2)2 = a2y3.
4.26. a) x2 − 4x + y2 = 0,x2 − 8x + y2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)2 = 4ay3.
4.27. a) y2 − 4y + x2 = 0,y2 − 8y + x2 = 0,x = 0, y =
√3x;
b) (x2 + y2)2 = a2(3x2 − y2).
4.28. a) x2 − 4x + y2 = 0,x2 − 6x + y2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)3 = a2x2(x2 +3y2).
4.29. a) y2 − 2y + x2 = 0,y2 − 10y + x2 = 0,y = x/
√3, x = 0;
b) (x2 + y2)3 = a2x4.
4.30. a) x2 − 6x + y2 = 0,x2 − 10x + y2 = 0,y = x/
√3, y =
√3x;
b) (x2 + y2)2 = a2(x2 − 3y2).
5. Vyqislitь trono integral
V zadaqah 5.1–5.30 vyqislitь tronye integraly∫∫
Ω
∫
f(x, y, z) dxdydz
po oblasti Ω, zadanno ukazannymi poverhnostmi.
5.1.
∫∫
Ω
∫
x dxdydz,
Ω :
y = 10x, y = 0, x = 1,z = xy, z = 0.
5.2.
∫∫
Ω
∫
dxdydz
(1 + x/3 + y/4 + z/8)4,
Ω :
x/3 + y/4 + z/8 = 1,x = 0, y = 0, z = 0.
22
5.3.
∫∫
Ω
∫
15(y2 + z2 dxdydz,
Ω :
z = x + y, x + y = 1,x = 0, y = 0, z = 0.
5.4.
∫∫
Ω
∫
(3x + 4y) dxdydz,
Ω :
y = x, y = 0, x = 1z = 5(x2 + y2), z = 0.
5.5.
∫∫
Ω
∫
(1 + 2x3) dxdydz,
Ω :
y = 9x, y = 0, x = 1,z =
√xy, z = 0.
5.6.
∫∫
Ω
∫
(27 + 54y3) dxdydz,
Ω :
y = x, y = 0, x = 1,z =
√xy, z = 0.
5.7.
∫∫
Ω
∫
y dxdydz,
Ω :
y = 15x, y = 0, x = 1,z = xy, z = 0.
5.8.
∫∫
Ω
∫
dxdydz
(1 + x/16 + y/8 + z/3)5,
Ω :
x/16 + y/8 + z/3 = 1,x = 0, y = 0, z = 0.
5.9.
∫∫
Ω
∫
(3x2 + y2) dxdydz,
Ω :
z = 10y, x = y = 1,x = 0, y = 0, z = 0.
5.10.
∫∫
Ω
∫
(15x + 30y) dxdydz,
Ω :
z = x2 + 3y2, z = 0,y = x, y = 0, x = 1.
5.11.
∫∫
Ω
∫
(4 + 8z3) dxdydz,
Ω :
y = x, y = 0, x = 1z =
√xy, z = 0.
5.12.
∫∫
Ω
∫
(1 + 2x3) dxdydz,
Ω :
y = 36x, y = 0, x = 1,z =
√xy, z = 0.
5.13.
∫∫
Ω
∫
21xz dxdydz,
Ω :
y = x, y = 0, x = 2,z = xy, z = 0.
5.14.
∫∫
Ω
∫
dxdydz
(1 + x/10 + y/8 + z/3)6,
Ω :
x/10 + y/8 + z/3 = 1,x = 0, y = 0, z = 0.
5.15.
∫∫
Ω
∫
(x2 + 3y2) dxdydz,
Ω :
z = 10x, x + y = 1,x = 0, y = 0, z = 0.
5.16.
∫∫
Ω
∫
(60y + 90z) dxdydz,
Ω :
y = x, y = 0, x = 1,z = x2 + y2, z = 0.
23
5.17.
∫∫
Ω
∫
(9 + 18z) dxdydz,
Ω :
y = 4x, y = 0, x = 1,z =
√xy, z = 0.
5.18.
∫∫
Ω
∫(
10
3x +
5
3
)
dxdydz,
Ω :
y = 9x, y = 0, x = 1,z =
√xy, z = 0.
5.19.
∫∫
Ω
∫
3y2 dxdydz,
Ω :
y = 2x, y = 0, x = 2,z = xy, z = 0.
5.20.
∫∫
Ω
∫
dxdydz
(1 + x/2 + y/4 + z/6)4,
Ω :
x/2 + y/4 + z/6 = 1,x = 0, y = 0, z = 0.
5.21.
∫∫
Ω
∫
x2 dxdydz,
Ω :
z = 10(x + 3y), x + y = 1,x = 0, y = 0, z = 0.
5.22.
∫∫
Ω
∫
(8y + 12z) dxdydz,
Ω :
y = x, y = 0, x = 1,z = 3x2 + 2y2, z = 0.
5.23.
∫∫
Ω
∫
63(1 + 2√
y) dxdydz,
Ω :
y = x, y = 0, x = 1,z =
√xy, z = 0.
5.24.
∫∫
Ω
∫
(x + y) dxdydz,
Ω :
y = x, y = 0, x = 1,z = 30x2 + 60y2, z = 0.
5.25.
∫∫
Ω
∫
xyz dxdydz,
Ω :
y = x, y = 0, x = 2,z = xy, z = 0.
5.26.
∫∫
Ω
∫
dxdydz
(1 + x/6 + y/4 + z/16)5,
Ω :
x/6 + y/4 + z/16 = 1,x = 0, y = 0, z = 0.
5.27.
∫∫
Ω
∫
y2 dxdydz,
Ω :
z = 10(3x + y), x + y = 1,x = 0, y = 0, z = 0.
5.28.
∫∫
Ω
∫(
5x +3z
2
)
dxdydz,
Ω :
y = x, y = 0, x = 1,z = x2 + 15y2, z = 0.
5.29.
∫∫
Ω
∫
(x2 + 4y2) dxdydz,
Ω :
z = 20(2x + y), x + y = 1,x = 0, y = 0, z = 0.
5.30.
∫∫
Ω
∫
dxdydz
(1 + x/8 + y/3 + z/5)6,
Ω :
x/8 + y/3 + z/5 = 1,x = 0, y = 0, z = 0.
24
6. Vyqislitь obъem tela (I)
V zadaqah 6.1–6.30 vyqislitь s pomowь tronogo integrala obъ-em tela, ograniqennogo ukazannymi poverhnostmi. Sdelatь qerteжdannogo tela i ego proekcii na ploskostь xOy.
6.1. y = 16√
2x, y =√
2x,z = 0, x + z = 2.
6.2. y = 5√
x, y = 5x/3,z = 0, z = 5
√x/3 + 5.
6.3. x2 + y2 = 2, y =√
x, y = 0,z = 0, z = 15x.
6.4. x + y = 2, y =√
x,z = 12y, z = 0.
6.5. x = 20√
2y, x = 5√
2y,z = 0, z + y = 1/2.
6.6. x = 5√
y/2, x = 5y/6,z = 0, z = 5(3 +
√y)/6.
6.7. x2 + y2 = 2, x =√
y, x = 0,z = 0, z = 30y.
6.8. x + y = 2, x =√
y,z = 12x/5, z = 0.
6.9. y = 17√
2x, y = 2√
2x,z = 0, x + z = 1/2.
6.10. y = 5√
x/3, y = 5x/9,z = 0, z = 5(3 +
√x)/9.
6.11. x2 + y2 = 8, y =√
2x, y = 0,z = 0, z = 15x/11.
6.12. x + y = 4, y =√
2x,z = 0, z = 3y.
6.13. x = 5√
y/6, x = 5y/18,z = 0, z = 5(3 +
√y)/18.
6.14. x = 19√
2y, x = 4√
2y,z + y = 2, z = 0.
6.15. x2 + y2 = 8, x =√
2y, x = 0,z = 0, z = 30y/11.
6.16. x + y = 4, x =√
2y,z = 0, z = 3x/5.
6.17. y = 6√
3x, y =√
3x,z = 0, x + y = 3.
6.18. y = 5√
x/6, y = 5x/18,z = 0, z = 5(3 +
√x)/18.
6.19. x2 +y2 = 18, y =√
3x, y = 0,z = 0, z = 5x/11.
6.20. x + y = 6, y =√
3x,z = 0, z = 4y.
6.21. x = 7√
3y, x = 2√
3y,z = 0, y + z = 3.
6.22. x = 5√
y/3, x = 5y/9,z = 0, z = 5(3 +
√y)/9.
6.23. x2 + y2 = 18, x =√
3y,x = 0, z = 0, z = 10y/11.
6.24. x + y = 6, x =√
3y,z = 4x/5, z = 0.
6.25. y =√
15x, y =√
15x,z = 0, z =
√15(1 +
√x).
6.26. x2 + y2 = 50, y =√
5x,z = 0, z = 3x/11, y = 0.
6.27. x + y = 8, y =√
4x,z = 0, z = 3y.
6.28. x = 16√
2y, x =√
2y,z + y = 2, z = 0.
25
6.29. x = 15√
y, x = 15y,z = 0, z = 15(1 +
√y).
6.30. x2 + y2 = 50, x =√
5y,z = 0, z = 6y/11, x = 0.
7. Vyqislitь obъem tela (II)
V zadaqah 7.1–7.30 vyqislitь s pomowь tronogo integrala obъ-em tela, ograniqennogo ukazannymi poverhnostmi, ispolьzu cilin-driqeskie koordinaty.
7.1. a) x2 + y2 = 2y,z = 0, z = 5/4 − x2;
b) z =√
9 − x2 − y2,9z/2 = x2 + y2.
7.2. a) x2 + y2 = y,x2 + y2 = 4y,z = 0, z =
√
x2 + y2;
b) z = 15√
x2 + y2/2,z = 17/2 − x2 − y2.
7.3. a) x2 + y2 = 8√
2x, z = 0,z = x2 + y2 − 64 (z ≥ 0);
b) z =√
4 − x2 − y2,
z =√
(x2 + y2)/255.
7.4. a) x2 + y2 + 4x = 0,z = 0, z = 8 − y2;
b) z =√
64 − x2 − y2,x2 + y2 = 60, z = 1(vnutri cilindra).
7.5. a) x2 + y2 = 6x,x2 + y2 = 9x,z = 0, z =
√
x2 + y2,y = 0 (y ≤ 0);
b) z =√
16/9 − x2 − y2,2z = x2 + y2.
7.6. a) x2 + y2 = 6√
2y, z = 0,z = x2 + y2 − 36 (z ≥ 0);
b) z = 3√
x2 + y2,z = 10 − x2 − y2.
7.7. a) x2 + y2 = 2y,z = 0, z = 9/4 − x2;
b) z =√
25 − x2 − y2,
z =√
(x2 + y2)/99.
7.8. a) x2 + y2 = 2y,x2 + y2 = 5y,z = 0, z =
√
x2 + y2;
b) z =√
100 − x2 − y2,z = 6, x2 + y2 = 51(vnutri cilindra).
7.9. a) x2 + y2 + 2√
2y = 0, z = 0,z = x2 + y2 − 4 (z ≥ 0);
b) z = 21√
x2 + y2/2,z = 23/2 − x2 − y2.
7.10. a) x2 + y2 = 4x,z = 0, z = 10 − y2;
b) z =√
16 − x2 − y2,6z = x2 + y2.
26
7.11. a) x2 + y2 = 7x,x2 + y2 = 10x,z = 0, z =
√
x2 + y2,y = 0 (y ≤ 0);
b) z =√
9 − x2 − y2,
z =√
(x2 + y2)/80.
7.12. a) x2 + y2 = 8√
2y, z = 0,z = x2 + y2 − 64 (z ≥ 0);
b) z =√
81 − x2 − y2,z = 5, x2 + y2 = 45(vnutri cilindra).
7.13. a) x2 + y2 = 2y,z = 0, z = 13/4 − x2;
b) z =√
1 − x2 − y2,3z/2 = x2 + y2.
7.14. a) x2 + y2 = 3y,x2 + y2 = 6y,z = 0, z =
√
x2 + y2;
b) z = 6√
x2 + y2,z = 16 − x2 − y2.
7.15. a) x2 + y2 = 6√
2x, z = 0,z = x2 + y2 − 36 (z ≥ 0);
b) z =√
36 − x2 − y2,
z =√
(x2 + y2)/63.
7.16. a) x2 + y2 = 2√
2y, z = 0,z = x2 + y2 − 4 (z ≥ 0);
b) z =√
64 − x2 − y2,z = 4, x2 + y2 = 39(vnutri cilindra).
7.17. a) x2 + y2 = 4x,z = 0, z = 12 − y2;
b) z =√
144 − x2 − y2,18z = x2 + y2.
7.18. a) x2 + y2 = 8x,x2 + y2 = 11x,z = 0, z =
√
x2 + y2,y = 0 (y ≤ 0);
b) z = 3√
x2 + y2/2,z = 5/2 − x2 − y2.
7.19. a) x2 + y2 = 4√
2x, z = 0,z = x2 + y2 − 16, (z ≥ 0);
b) z =√
9 − x2 − y2,
z =√
(x2 + y2)/35.
7.20. a) x2 + y2 = 4y,z = 0, z = 4 − x2;
b) z =√
49 − x2 − y2,z = 3, x2 + y2 = 33(vnutri cilindra).
7.21. a) x2 + y2 = 4y,x2 + y2 = 7y,z = 0, z =
√
x2 + y2;
b) z =√
36 − x2 − y2,9z = x2 + y2.
7.22. a) x2 + y2 = 4√
2y, z = 0,z = x2 + y2 − 16 (z ≥ 0);
b) z = 9√
x2 + y2,z = 22 − x2 − y2.
27
7.23. a) x2 + y2 + 2x = 0,z = 0, z = 17/4 − y2;
b) z =√
16 − x2 − y2,
z =√
(x2 + y2)/15.
7.24. a) x2 + y2 = 9x,x2 + y2 = 12x,z = 0, z =
√
x2 + y2
y = 0 (y ≥ 0);
b) z =√
36 − x2 − y2,z = 2, x2 + y2 = 27(vnutri cilindra).
7.25. a) x2 + y2 + 2√
2x = 0, z = 0,z = x2 + y2 − 4 (z ≥ 0);
b) z =√
4/9 − x2 − y2,z = x2 + y2.
7.26. a) x2 + y2 = 4y,z = 0, z = 6 − x2;
b) z = 12√
x2 + y2,z = 28 − x2 − y2.
7.27. a) x2 + y2 = 10x,x2 + y2 = 13x,z = 0, z =
√
x2 + y2,y = 0 (y ≥ 0);
b) z =√
9 − x2 − y2,
z =√
(x2 + y2)/8.
7.28. a) x2 + y2 = 2√
2x, z = 0,z = x2 + y2 − 4 (z ≥ 0);
b) z =√
25 − x2 − y2,z = 1, x2 + y2 = 21(vnutri cilindra).
7.29. a) x2 + y2 = 2x,z = 0, z = 21/4 − y2;
b) z =√
64 − x2 − y2,12z = x2 + y2.
7.30. a) x2 + y2 = 5y,x2 + y2 = 8y,z = 0, z =
√
x2 + y2;
b) z = 9√
x2 + y2/2,z = 11/2 − x2 − y2.
8. Vyqislitь obъem tela (III)
V zadaqah 8.1–8.30 vyqislitь s pomowь tronogo integrala obъemtela, zadannogo neravenstvami, ispolьzu sferiqeskie koordinaty.
8.1. 1 ≤ x2 + y2 + z2 ≤ 49, −√
x2 + y2
35≤ z ≤
√
x2 + y2
3, −x ≤ y ≤ 0.
8.2. 4 ≤ x2 + y2 + z2 ≤ 64,
√
x2 + y2
15≤ z ≤
√
x2 + y2
3, −
√3x ≤ y ≤ 0.
8.3. 16 ≤ x2 + y2 + z2 ≤ 100, z ≤√
x2 + y2
24, −
√3x ≤ y ≤ − x√
3.
28
8.4. 4 ≤ x2 + y2 + z2 ≤ 36, z ≥ −√
x2 + y2
63, 0 ≤ y ≤ − x√
3.
8.5. 1 ≤ x2 + y2 + z2 ≤ 36, z ≥√
x2 + y2
99, −
√3x ≤ y ≤
√3x.
8.6. 25 ≤ x2 + y2 + z2 ≤ 100, z ≤ −√
x2 + y2
99,√
3x ≤ y ≤ −√
3x.
8.7. 1 ≤ x2 + y2 + z2 ≤ 49, 0 ≤ z ≤√
x2 + y2
24, y ≤ − x√
3, y ≤ −
√3x.
8.8. 25 ≤ x2 + y2 + z2 ≤ 121, −√
x2 + y2
24≤ z ≤ 0, y ≥ − x√
3, y ≥ −
√3x.
8.9. 4 ≤ x2 + y2 + z2 ≤ 64, −√
x2 + y2
35≤ z ≤
√
x2 + y2
3, x ≤ y ≤ 0.
8.10. 16 ≤ x2 + y2 + z2 ≤ 100,
√
x2 + y2
15≤ z ≤
√
x2 + y2
3,√
3x ≤ y ≤ 0.
8.11. 16 ≤ x2 + y2 + z2 ≤ 100, z ≤√
x2 + y2
3, −
√3x ≤ y ≤ − x√
3.
8.12. 16 ≤ x2 + y2 + z2 ≤ 64, z ≥ −√
x2 + y2
63, − x√
3≤ y ≤ −
√3x.
8.13. 4 ≤ x2 + y2 + z2 ≤ 49, z ≥√
x2 + y2
99, y ≤ 0, y ≤
√3x.
8.14. 36 ≤ x2 + y2 + z2 ≤ 121, z ≤ −√
x2 + y2
99, y ≥ 0, y ≥
√3x.
8.15. 4 ≤ x2 + y2 + z2 ≤ 64, 0 ≤ z ≤√
x2 + y2
24, y ≤
√3x, y ≤ x√
3.
8.16. 36 ≤ x2 + y2 + z2 ≤ 144, −√
x2 + y2
24≤ z ≤ 0, y ≥
√3x, y ≥ x√
3.
8.17. 9 ≤ x2 + y2 + z2 ≤ 81, −√
x2 + y2
3≤ z ≤
√
x2 + y2
35, 0 ≤ y ≤ −x.
8.18. 36 ≤ x2+y2+z2 ≤ 144, −√
x2 + y2
3≤ z ≤ −
√
x2 + y2
15, 0 ≤ y ≤ −
√3x.
29
8.19. 36 ≤ x2 + y2 + z2 ≤ 144, z ≤√
x2 + y2
3,√
3x ≤ y ≤ x√3.
8.20. 36 ≤ x2 + y2 + z2 ≤ 100, z ≥ −√
x2 + y2
63,
x√3≤ y ≤
√3x.
8.21. 9 ≤ x2 + y2 + z2 ≤ 64, z ≥√
x2 + y2
99, y ≤ x√
3, y ≤ − x√
3.
8.22. 49 ≤ x2 + y2 + z2 ≤ 144, z ≤ −√
x2 + y2
99, y ≥ x√
3, y ≥ − x√
3.
8.23. 9 ≤ x2 + y2 + z2 ≤ 81, 0 ≤ z ≤√
x2 + y2
24, y ≤ 0, y ≤ x√
3.
8.24. 49 ≤ x2 + y2 + z2 ≤ 169, −√
x2 + y2
24≤ z ≤ 0, y ≥ 0, y ≥ x√
3.
8.25. 16 ≤ x2 + y2 + z2 ≤ 100, −√
x2 + y2
3z ≤
√
x2 + y2
35, 0 ≤ y ≤ x.
8.26. 64 ≤ x2 + y2 + z2 ≤ 196, −√
x2 + y2
3≤ z ≤ −
√
x2 + y2
15, 0 ≤ y ≤
√3x.
8.27. 64 ≤ x2 + y2 + z2 ≤ 196, z ≤√
x2 + y2
3, − x√
3≤ y ≤ 0.
8.28. 64 ≤ x2 + y2 + z2 ≤ 144, z ≥ −√
x2 + y2
63, 0 ≤ y ≤ x√
3.
8.29. 16 ≤ x2 + y2 + z2 ≤ 81, z ≥√
x2 + y2
99, y ≤ 0, y ≤ −
√3x.
8.30. 16 ≤ x2 + y2 + z2 ≤ 100, 0 ≤ z ≤√
x2 + y2
24, y ≤ 0, y ≤ − x√
3.
30
Soderжanie
Ukazani k vypolneni tipovogo rasqeta . . . . . . . . 3Obwie trebovani k vypolneni zadani . . . . . . . . . . 3Teoretiqeskie voprosy . . . . . . . . . . . . . . . . . . . . . 4Obrazec vypolneni tipovogo rasqeta . . . . . . . . . . . . 5
Rasqetnye zadani . . . . . . . . . . . . . . . . . . . . . . . 141. Izmenitь pordok integrirovani . . . . . . . . . . . . . 142. Vyqislitь dvonye integraly . . . . . . . . . . . . . . . 173. Vyqislitь plowadь plosko figury (I) . . . . . . . . . . 194. Vyqislitь plowadь plosko figury (II) . . . . . . . . . 205. Vyqislitь trono integral . . . . . . . . . . . . . . . . 226. Vyqislitь obъem tela (I) . . . . . . . . . . . . . . . . . . 257. Vyqislitь obъem tela (II) . . . . . . . . . . . . . . . . . . 268. Vyqislitь obъem tela (III) . . . . . . . . . . . . . . . . . 28
31
Kurova Ldmila VladimirovnaPodolьski Vladimir Aleksandroviq
KRATNYE INTEGRALYmetodiqeskie ukazani
k vypolneni rasqetnogo zadanidl studentov 2-go kursa vseh specialьnoste
Kontrolьnye zadani vypolnenyv redakcionno-izdatelьsko sisteme
LATEX 2E
Redaktor E.O. Trusova
Podpisano k peqati 2003. Format 60x84 1/16. Bumaga pisqa. Peqatь ofsetna.
Usl.peq.l. 1,86. Peq.l. 2,9. Uq.-izd.l. 1,81. Tiraж 350 зkz. Zakaz N0 S 66
SPbGUNiPT. 191002, Sankt-Peterburg, ul. Lomonosova, 9
IPC SPbGNiPT. 191002, Sankt-Peterburg, ul. Lomonosova, 9
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