Tich phan boi

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Cc v d mu v cc bi tp tch phn bi 2 v tch phn bi 3

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Bi 01 : TnhGii10 0.xxy dy dx 10xxdx dy xy )`===10x yx ydx dy xy dx2xyxx102

=dx ) x x (21103 2 =104 34x3x21

=241=Bi 02: TnhGii12 20( ) .yyx y dx dy + 12 20( )yyx y dx dy + ` ) 13203yyxxy dy = + ` )13/ 25/ 2 3043 3yy y dy| |= + |\ 145/ 2 7/ 201 2 2 4. . .3 5 7 3 4yy y

= +

353=Bi 03: TnhGii1 12 20 0.(1 ) (1 )dx dyx y + + y, cn ly tch phn u l hng s v trong hm ly tch phn ta cth tch ring cc bin x v y. Cho nn: + +10102 2) y 1 ( ) x 1 (dy dx

+|||

\|+= 102102y 1dyx 1dx] ]1 10 0. arctgx arctgy =4.4 =162=Bi04 : TnhGii21 12 20 01xdy dxx y++ + dxy ) x 1 (dy102x 1 y0 y2 2 )`+ ++ ==2112 2001.1 1xyarctg dxx x+

| |=|+ + \

122 201 101 1xarctg arctg dxx x | |+= | |+ +\ dx4x 11102

+=120ln 14x x

= + +

ln(1 2) ln14 = + ln (1 2)4= +Bi 5: TnhGii,Axy dx dy =Trong A l min gii hn bi trc Ox, x = 2a v x2= 4ay.Ady dx xy ===)`=a0 ya 2 xay 4 xdy dx xydy2y xa 2ay 2a02

==a02 2dy2ay 4 y a 43a3y2aya 2 dy ) y ay ( a 24a0a03 22=)` = =Bi 6: TnhGii( )Dxy x ydx dy +( )Dxy x ydx dy +dx3xy2y x10x2x3 2 2

+ =dx x31x21x65107 6 4||

\| =563= )`+ ===10x y2x y2 2dx dy ) xy y x (Trong D l min gii hn bi y = x2v y = x.Bi 07: TnhGiiNu R l min gii hn bi ng trn x2+y2= 1 trong gc phn t th nht, tnh2.1Rxydx dyy R2dy dxy 1xy = ==)`=a0 x2x 1 y0 y2dx dyy 1xydx y 1 x10 x2x 102=

=dx 0 1 ) x 1 ( 1 x102)` + =+ =10dx } 1 x { x =102dx ) x x (103 23x2x

=3121 =61=Bi 08 : Bng cch i th t tch phn, tnh:Gii10.xxxy dy dx , v 0, 1 y x y x x x = = = =Bng cch i th t tch phn, ta cn tnh tch phn theo bin x trc, ngha l tnh tch phn theo phng Ox.Ta c: min ly tch phn ng gii hn bi:QuansttheophngOx,tac minDlminutrongkhongy=0ny=1v ng vo l x = y2v ng ra x = y.Do ta c:dx dy xy10xx)` ====)`=1 y0 yy x2y xdy dx xydy y2xy2y1 y0 y2

===

=105 3dy2y2y106 412y8y

=12181 =241=Bi 09: Bng cch i th t tch phn, tnh:Gii210.yyxy dx dy Ta c:dy dx xy10y 2 xy x )` ==Min ly tch phn gii hn bi x2= y vx + y = 2 trong khong y = 0 n y = 1.TheophngOy,minDl minucng ng vo y = 0 nhng khng cng ng ra nn ta chia min D thnh 2 min R1v R2Trn min R1, trong khong x = 0 n x = 1, ng vo y =0vngray = x2. Do :1Rxy dx dy ===)`=10 x2x y0 ydx dy xydx2xy2x0102

==105dx x21121=Trn min R2, trong khong x = 1 n x = 2, ng vo y = 0 v ng ra y = 2 - x. Do :2Rxy dx dy dx dy xy2 x1 xx 2 y0 y == ==)`= ==)`=21 x212x 202dx2) x 2 ( xdx2xy1 2R R Rxy dx dy xy dxdy xy dx dy = + 245121+ =247=524=Bi 10: Xc nh cn ly tch phn theo 2 phng Ox v Oy ca:Gii( , ) .Df xydxdyD l cung trn nm trong on t3 n 1 ca na di ng trn (O;2)23, 1, 4 , 0 x x y x y = = = =Do :Theo phng Oy: Min ly tch phn gii hn bi:Theo phng Ox: Min ly tch phn l min u nhng khng cng ng vo v ng raTa chia min D thnh 3 min: ABEF, BECD v cung CD.Bi 11GiiTm din tch ca min gii hn bi x2= 4ay v y2= 4ax.t x2= 4ay (1)y2= 4ax (2)Gii (1) v (2) ta tm c ta giao im l (0, 0) v (4a, 4a)YcbtDS dx dy =dx dya 40 xax 2 ya 42xy ===)`= dx yax 2a 42xa 40 x

==dxa 4x0 ax 2a 402

=a 4032 / 3a 12xa1.32) ax ( 2

=a 12) a 4 () a 4 ( .3a 432 / 3 =3a 163a 322 2 =3a 162=t x = x(u,v) v y = y (u,v) tha: Khi : 1. f(x,y)=f(x(u,v),y(u,v))minDtrongmtphng Oxy s bin thnh min D trong mt phng Ouv. 2. Cng thc tch phn 2 lp s c thay i thnh:Cng thc i bin'( , ) ( , )DDf x ydxdy F u vJ dudv = ( , )( , )x xxyu vJy y u vu v = = ( )1 1; ( ; ) xy uvV d:,'3 22 2 2 23 31 1( ; ); ;DDf x y dxdyuu u uu uF dudv duF dvv v v v v v| | | |= = ||\ \ D gii hn bi: y = x2, y = 2x2, y = x,3 y x =Dgiihnbi1cpparabol v 1 cp ng thng.t: u = y/x ; v = y/x2. Ta c:2 23;u u ux y Jv v v= = =Trongtrnghpta cc. Ta c:Tch phn 2 lp trong ta cc21( )( )( ; )( cos , sin )( , )RRr rr rf xydxdyf r r rdrdf r rdr d = == == = ` ) 1.NuminlytchphnDgii hn bi 2 tia xut pht t cc:tip xc vi bin caminDtiAv B von ngcongAPB cphng trnh,onngcong AQB cphngtrnh:th (1) c tnh nh sau:Tch phn 2 lp trong ta cc1( ) r r =,( ) = = 2( ) r r =21( )( )( , )r rr rf r dr d = == = ` ) 2. Nu gc O nm trong min D v mi tia xut pht t O uct bin ca min HD ti 1 im c bn knh vec t l th:3.Trongta cc tchtchphn2lpthngtnhtch phn theo r trc.4. Ta ch i sang h ta cc khi:-Hm di du tch phn c cha , ng thi min D gii hn bi cc ng thng i qua O.- Min ly tch phn D l hnh trn, hnh trn lch, gii hn ca hai hnh trn, hoc ng cong c cha x2+ y2Tch phn 2 lp trong ta cc( ) r r =( ) 20 0( ; ) ( cos ; sin )rDf x y dxdy f r r rdr d = ` ) 2 2x y +1. Nu min D i xng qua Ox v f(x;y) = f(x;-y) th:(vi D1 l phn ca D ng vi y > 0)Nu min D i xng qua Ox v f(x;y) = -f(x;-y) th:2. Nu min D i xng qua Oy v f(x;y) = f(-x;y) th:(vi D2 l phn ca D ng vi x > 0)Nu min D i xng qua Oy v f(x;y) = -f(-x;y) th:3. Nu D i xng qua Ox, Oy v f(x;y)= f(-x;y)= f(x;-y)= f(-x;-y)1 s tnh cht:1( ; ) 2 ( ; )D Df x ydxdy f x ydxdy = ( ; ) 0Df xydxdy =2( ; ) 2 ( ; )D Df x ydxdy f x ydxdy = ( ; ) 0Df x y dxdy =*( ; ) 4 ( ; )D Df x ydxdy f x ydxdy = Kt qu quan trng:2 20 0(2 1)!! , ( 2 )(2 )!! 2sin cos(2 )!! , ( 2 1)(2 1)!!n nkn kkxdx xdxkn kk == == ++ Lu :- xc nh cn tch phn trong ta cc ta xt:-Tm iu kin ca r 0.0, 0 2 r Bi 11GiiTnh sin ,Dr drd Vi D gii hn bi r = a(1-cos), a>0, nm pha trn na mp(Oxy).sinDr drd (1 cos )0 0sinr arr dr d = = = = ` ) (1 cos )200sin2ard

=

220 0 0sin cos ( sin ) sin22ad d d = ` ) 2 30cos cos 2cos2 3 2a = + ` )3a 42=Bi 12GiiNu R l min gii hn bi na ng trn r = 2acosnm pha trn trc honh.CMR:2 32sin .3Rr drd a =ng trn r= 2acoshay r2= 2arcosC phng trnh trong h ta (Oxy) l:x2+ y2- 2ax = 0 ngtrntm(a,0),bn knh a v i qua gc ta .2 cos/2300sin .3ard =

=

/23 30sin8 cos3a d = / 23308cos .( sin )3ad = / 23 408 cos.3 4a

=

( )320 13a= 3a 23=2sinRr drd 2 cos /220 0sinarr dr d = = = ` ) D l min gii hn bi ng trn tm I(a;b) , bn knh R bt k.t:X = x a, Y = y bKhi : 2 21( ; ) ( ; )DX Yf xydxdy f X a Y bdXdY+ = + + 20 0( cos ; sin )af a r b r rdr d = + + ` ) Bi 13Gii2 2/ 22 20ln( )a yayx y dx dy+ Min ly tch phn c xc nh bi:cos sin4x y r r = = =2 2 2 2 2x a y x y a r a = + = =0 042ay Bng cch chuyn qua ta cc, tnh:2 2/ 22 20ln( )x a y y ay x yx y dx dy= == = + ` ) ( )/ 420 0ln . .arr r dr d == = = ` ) / 4220 0012 . ln 2. . ln4 2 4r aarrrd r rdr r r== ==| || | | | | || = = || | | |\ \ \ \ 22ln4 2aa a | |= |\ / 40 02 lnarr r dr d= = = ` ) Bi 14GiiS dng php bin i x + y = u v y = uv, tnh1 1/( )0 0.xy x ye dx dy+ /( ) y x yDTrong e dy+Min ly tch phn D b gii hn bi trc Ox, x + y = 1 v y thay i t y = 0 n y = 1.T php bin i ta c: x = u(1-v) v y = uvuu vu v 1vyuyvxuxJ = == Xc nh cn ca D:(i) Trc Ox (y = 0): suy ra uv = 0. Vy u = 0, v = 0Do 0 y 1 v x 0 nn u > 0 v v > 0(ii)Trc Oy, (x = 0): u(1-v) = 0 nn u = 0, v = 1, since x > 0Do x 0 nn u(1 - v) > 0u > 0 v v < 1(iii) ng thng x + y = 1 cho ta u = 1.Vy min D gii hn bi u = 0, u = 1, v = 0 v v = 1/ y x yRe dx dy+( )111 1 11/ 200 0 0 0 01.2uvu u vvuvu v vvv u u v uu e du dv udu e dv u e=== = ==== = = = = | || || |= = = `| | |\ ) \ \ ( 1)2e =Bi 15GiiTm din tch ca min giao gia 2 ng trn x2+ y2= a2v x2+ y2= 2ax.Ta c:) 1 ...( a r a y x2 2 2= = +2 22 2 cos ...(2) x y ax r a + = =Giao im ca 2 ng trn, ta c = /3Trong min R1, r thay i t 0 n r = a v thay i t 0 n /3Trong min R2, r thay i t 0 n r = 2acos v thay i t /3 n /2Vy = 2{S(R1) + S(R2)}2 cos/3 / 20 0 /3 02 2aar rrdr d r dr d = = = = = + ` ` ) ) /3 /22 20 0 /312. 2 4 cos2ad rdr a d | || |= + | |\ \ 222 33 2aa = Bi 16GiiD l min gii hn bi 2 ng trn r = 2asin v r = 2bsin,b > a, tnh2 2( ) .Dx y dxdy +2 2( )Dx y dxdy +2 sin/ 230 2 sin2r br ardr d == = = ` ) 2 sin/ 2402 sin2.4bard| |= |\ / 24 4 4 4 4 4 401 3.1 3(16 16 ) sin 8( ). . ( )2 4.2 2 2 2b a d b a b a = = = Th tch ca min gii hn bi 2 mt z = z1(x,y) = c1v z = z2(x,y) = c-2c xc nh bi:Th tchcamingiihntrongkg3chiu theo tch phn 2 lpdy dx | z z | V1 2R =Tch phn bi ba =R Rdz dy dx ) z , y , x ( dv ) z , y , x ( f =======b xa x) x (2y y) x (1y y) y , x (2z z) y , x (1z zdx dy dz ) z , y , x ( fCho x = rcos, y = rsin , z = zTch phn 3 lp trong ta tr( , , )( , , )x x xr zxyz y y ythen Jr z r zz z zr z = = cos sin 0sin cos 00 0 1rr r = =Bng cch s dng php i bin( , , ) ( , , ) ( , , )V V Vf x y z dV f x y z dxdydz F r z rdrd dz = = Cho x = rcossin , y = rsinsin, z = rcosTch phn 3 lp trong ta cu2( , , )th sin( , , )xyzJ rr = =2( , , ) sinVFr r drd d =( , , ) ( , , )V Vf xyzdV f xyzdxdydz = Bng cch s dng php i binNhn xt:Tnh th tch bng cch s dng tch phn 3 lpV VdV V dx dy dz = = 1.R VV dV rdrd dz = = 22. sinV RV dV r dr d d = = trong ta trtrong ta cuProblem20Solution. dx dy dz ) z y x ( Evaluateccbbaa2 2 2 + +Since all the limits are constants we can integrate in the given order. |||

\|+ +ccbbaa32 2dx dy3zz y x x

+ + =ccbb32 2dx dy3a 2) a 2 ( y ) a 2 ( xdx y3a 23ya 2 y ax 2bbcc3 32

+ + =dx3b a 43ba 4 x ab 4cc3 32

+ + =cc3 3 33bx a 4x3ab 43xab 4

+ + =3bc a 83c ab 83abc 83 3 3+ + =) c b a (3abc 82 2 2+ + =Problem21Solution. dz dy dx ) z y x ( Evaluate11z0z xz x ++ + ++ +11z0z xz xdz dx dy ) z y x ( +

+ + =11z xz xz02dz dx zy2yxy + =11z02dz dx ) z 2 xz 4 (dz x z 2 z x 2z0112 2

+ =0 z dz z 4 dz ) z 2 z 2 (11114 3113 3=

= = + = Problem22Solution.z y x 1dx dy dzEvaluate102x 102z2x 102 2 2 102x 102y2x 102 2 2 2dx dyz ) y x 1 (dz

=102x 102y2x 102 21dx dyy x 1zsin =102x 10dx dy2 =

=1022x 1010dx x 12dx y2

+ =x sin21x 12x21 282=Problem23SolutionFind by triple integration the volume of the sphere of radius a.Vdz dy dx == ==a0 x2x2a0 y2y2x2a0 zdx dy dz 8Changing to spherical coordinate systemdr dy dz = r2. sin dr d d) 1 (3a 43= == = = = = = =a0 r22 /02 /02 /02 /0a0 r2dr r d d sin 8d drd sin r 8 VProblem24Solutionegration int triple g sin u 1czbyaxellipsoid the of volume the Find222222= + + == ==a0 x2a2x1 b0 y2b2y2a2x1 c0 zdx dy dz 8 V Volume ==

=a0 x2b2y2a2x1 c02a /2x 1 b0 ydx dy z 8 ==)` =a0 x2b /2x21ab0 y2222dx dybyax1 c ==)`|||

\| =a0