01 Slide - Chuong 2 Tich Phan Boi

7/24/2019 01 Slide - Chuong 2 Tich Phan Boi

1/212

VI TCH PHN A2

Chng 2. TCH PHN BI

CBGD. L Hoi Nhn

Ngy 26 thng 7 nm 2015

CBGD. L Hoi Nhn () Chng 2. TCH PHN BI Ngy 26 thng 7 nm 2015 1 / 115
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Mc lc

1 Tch phn hai lp

nh nghaCch tnh tng qutTch phn trn hnh ch nhtTch phn trn hnh thang loi 1Tch phn trn hnh thang loi 2i bin tng qutTch phn trong ta cc

2 Tch phn ba lpnh ngha

Cch tnh tng quti bin tng qutTch phn trong ta trTch phn trong ta cu



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nh ngha

Cho hm s z = f(x, y) xcnh trn minD.

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nh ngha

Phn hoch minD

thnh n

min con, din tch ca mimin con l Si vi i =1, 2, ..., n.

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nh ngha

Trn mi min conSita chnty im Mi(xi, yi) v lp

tng

In =n

i=1

f(xi, yi)Si.

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nh ngha

Trn mi min conSita chnty im Mi(xi, yi) v lp

tng

In =n

i=1

f(xi, yi)Si.

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nh ngha

Cho n sao cho max Si 0.

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nh ngha

Cho n sao cho max Si 0. Nulimn In

= I tn ti hu hn, khng ph

thuc vo cch chia min Dv cch chnMi th:

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nh ngha




Ta ni hmfkh tch trn DvI ltch phn caf trnD.

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nh ngha




Ta ni hmfkh tch trn DvI ltch phn caf trnD.

K hiu:

I =D

f(x, y)dA=D

f(x, y)dxdy

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Din tch hnh phng

Cho min phngDtrong mt phng Oxy. Khi din tch ca minDc tnh theo tch phn hai lp

S=D

dx dy.

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Th tch vt th

Cho vt th c dng hnh tr, vi mt y pha di c phng trnhz=z1(x, y), mt y pha trn c phng trnh z=z2(x, y)vDl hnhchiu ca vt th trn mt phng Oxy. Khi th tch ca vt th ctnh theo tch phn hai lp

V =D

[z2(x, y) z1(x, y)] dx dy.



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Din tch mawjt cong

Cho mt congSc phng trnhz=z(x, y)xc nh trn minD(hnhchiu caStrn mt phngOxy). Din tch mtS l

S=D

1+z2x +z2ydxdy

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Cch tnh tch phn hai lp

Chuyn tch phn hai lp vtch phn xc nhtheo th bin ly tchphn khc nhau.



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Chuyn tch phn hai lp vtch phn xc nhtheo th bin ly tchphn khc nhau.

Ta xt ba trng hp ca min D: Hnh ch nht, Hnh thang loi 1v hnh thang loi 2.

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Biu din hnh hc minD

. Xc nh r phng trnh cc ng cong

v ta cc nh ca Dtrn hnh v .Nhn nh r minDl hnh thang loi 1, loi 2 hay hnh ch nht.Xc nh cc chn ca bin xvy. Lu : C t nht mt bin s ccc chn lhng s.

Cc biu din ngHnh thang loi 1 Hnh thang loi 2 Hnh ch nht

0 x 2x+1 y 2x+1

y2 x y0 y 1

1 x 40 y 3

Biu dinsai: x y x+12y

2

x 2y2

+1 Tt c cc cn u cha

bin!!!

p dng cng thc tng ng vi hnh dng ca minD. Tnh cctch phn thu c t phi sang tri.


C h h h h h i l


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Biu din hnh hc minD




0 x 2x+1 y 2x+1

y2 x y0 y 1

1 x 40 y 3


2

x 2y2

+1 Tt c cc cn u cha

bin!!!



C h h h h h i l
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Biu din hnh hc minD


v ta cc nh ca Dtrn hnh v .Nhn nh r minDl hnh thang loi 1, loi 2 hay hnh ch nht.Xc nh cc chn ca bin xvy.Lu : C t nht mt bin s ccc chn lhng s.


0 x 2x+1 y 2x+1

y2 x y0 y 1

1 x 40 y 3


2

x 2y2

+1 Tt c cc cn u cha

bin!!!



C h h h h h i l
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Biu din hnh hc minD




0 x 2x+1 y 2x+1

y2 x y0 y 1

1 x 40 y 3


2

x 2y2

+1 Tt c cc cn u cha

bin!!!



T h h t h h h ht
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Tch phn trn hnh ch nht

Hnh ch nht gii hn bi ccng thngx=a,x=b,y=cvy=d.


T h h t h h h ht
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Mt imM

(x, y

) Dc tnhcht:


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Mt imM(x, y) D

c tnhcht:a x bvc y d.


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Mt imM(x, y) D

c tnhcht:a x bvc y d.

D

f(x, y)dxdy =

ba

dc

f(x, y)dydx=

dc

ba

f(x, y)dxdy.


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TnhI = D

(4

x

y)dxdyviD

l min 0

x

1, 1

y

2.


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TnhI = D

(4

x

y)dxdyviD

l min 0

x

1, 1

y

2.

p dng cng thc (2.4) v th hai, ta c

I =

10

dx

21

(4 x y)dy=1

0

52 x

dx=2.


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TnhI = D

(4

x

y)dxdyviD

l min 0

x

1, 1

y

2.

p dng cng thc (2.4) v th hai, ta c

I =

10

dx

21

(4 x y)dy=1

0

52 x

dx=2.

p dng cng thc (2.4) v th ba, ta c

I =

21

dy

10

(4 x y)dx=2

1

72 y

dy=2.


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TnhI =D

xln ydxdyviDl min 0 x 4, 1 y e.


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TnhI =D

xln ydxdyviDl min 0 x 4, 1 y e.

Nhn xt rng min ly tch phn l hnh ch nht v hm ly tchphn c dng tch ca hai hm mt bin.p dng cng thc (2.5) trang 52 ta c,

I =

4

0

xdx

e

1

ln ydy=8


Bi tp Tch phn trn hnh ch nht I
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Bi tp Tch phn trn hnh ch nht I

T bi tp1-10,tnh cc tch phn trn minDc cho

1D

(6y2

2x)dxdy D : 0 x 1; 0 y 2

2

D

x

y2dxdy D : 0 x 4; 1 y 2

3

D

xycos ydxdy D : 1 x 1; 0 y

4

D

ysin(x+y)dxdy D : x 0; 0 y

5

D

exydxdy D : 0 x ln 2; 0 y ln 2


Bi tp Tch phn trn hnh ch nht II
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Bi tp Tch phn trn hnh ch nht II

6

D

xyexy2

dxdy D : 0 x 2; 0 y 1

7

D

xy3

x2 +1dxdy D : 0 x 1; 0 y 2

8 D

y

x2

y2

+1

dxdy D

: 0

x

1; 0

y

1

9

D

1xy

dxdy D : 1 x 2; 1 y 2

10D

ycos xydxdy D : 0 x ; 0 y 1

11 Tnh th tch ca min b chn pha trn bi paraboloidz=x2 +y2,pha di bi hnh vungD : 1 x 1; 1 y 1.


Bi tp Tch phn trn hnh ch nht III
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Bi tp Tch phn trn hnh ch nht III

12 Tnh th tch ca min b chn pha trn bi paraboloidz=16

x2

y2, pha di bi hnh vung

D:0

x

2; 0

y

2.

13 Tnh th tch ca min b chn pha trn bi mt phngz=2 x y, pha di bi hnh vungD :0 x 1; 0 y 1.

14 Tnh th tch ca min b chn pha trn bi mt phng z=y2

, pha

di bi hnh ch nhtD

:0

x

4; 0

y

2.15 Tnh th tch ca min b chn pha trn bi mt cong

z=2 sin xcos y, pha di bi hnh ch nhtD :0 x

2; 0 y

4.

16 Tnh th tch ca min b chn pha trn bi mt cong z=4

y2,pha di bi hnh ch nhtD :0 x 1; 0 y 2.


Tch phn trn hnh thang loi 1
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p g

Hnh thang loi 1 l hnh thangcong gii hn bi cc ng:

y=1

(x)y=2(x)x=ax=bvi 1(x) 2(x).


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p g


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p g

Nu imM(x, y) Dth a x b1(x) y 2(x) .


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p g

Nu imM(x, y) Dth a x b1(x) y 2(x) .

Cng thc tch phn lp

Df(x, y)dxdy=

b

a

2(x)

1(x)

f(x, y)dydx


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TnhI =D

(4x+2)dxdy viDl min gii hn bi cc ngy=x,

y=x2,x=1 vx=2.


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TnhI =D


y=x2,x=1 vx=2.


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TnhI =D


y=x2,x=1 vx=2.

Ta c th biu din minDnhsau: 1 x 2 vx y x2.


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TnhI =D


y=x2,x=1 vx=2.

Ta c th biu din minDnhsau: 1 x 2 vx y x2.p dng cng thc (2.2) tac

I = 223 .


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TnhI =D

(x3 +xy)dxdyviDl min gii hn bi cc ng y=x2

vy=

x.


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TnhI =D

(x3 +xy)dxdyviDl min gii hn bi cc ng y=x2

vy=

x.


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TnhI =D

(x3

+xy)dxdyviDl min gii hn bi cc ng y=x2

vy=

x.

Ta c minD: 0 x 1,x2 y x.


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TnhI =D

(x3

+xy)dxdyviDl min gii hn bi cc ng y=x2

vy=

x.

Ta c minD: 0 x 1,x2 y x.p dng cng thc (2.2) tac

I = 536.


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TnhI =D

2xdxdyviDl min gii hn bi cc ng y=x2,

x+y=2 vy=0.


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TnhI =D

2xdxdyviDl min gii hn bi cc ng y=x2,

x+y=2 vy=0.


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TnhI =D

2xdxdyviDl min gii hn bi cc ng y=x2

,

x+y=2 vy=0.

MinD

gm 2 min conD1

:0 x 1; 0 y x2 vD2:1 x 2; 0 y 2 x.p dng tnh cht 3 trang 49 v cngthc (2.2) ta c

I =12

+34

=11

6 .


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Hnh thang loi 2 l hnh thangcong gii hn bi cc ng:x=1(y),x=2(y),y=c,y=dvi 1(y) 2(y).


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Nu imM(x, y) Dth(x, y)tha

c y d1(y)

x

2(y) .


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Nu imM(x, y) Dth(x, y)tha

c y d1(y)

x

2(y) .

D

f(x, y)dxdy=

d

c

2(y)

1(y)

f(x, y)dxdy


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Tnh tch phnD

(x y)dxdy, trong ,Dl min gii hn bi cc

ngy= 1,y=1, x=y2 vy=x+1.


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Tnh tch phnD

(x y)dxdy, trong ,Dl min gii hn bi cc

ngy= 1,y=1, x=y2 vy=x+1.


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Hy tnh tch phn kp sau:

2

0

dx

4x2

0

x.e2y

4 ydy.


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2

0

dx

4x2

0

x.e2y

4 ydy.

Vit li minDdi dng bt ng thc.


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2

0

dx

4x2

0

x.e2y

4 ydy.


Biu din hnh hc minD.


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2

0

dx

4x2

0

x.e2y

4 ydy.


Biu din hnh hc minD.Xc nh cn ca tch phn theo hnh thang loi 2.


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2

0

dx

4x2

0

x.e2y

4 ydy.


Biu din hnh hc minD.Xc nh cn ca tch phn theo hnh thang loi 2.Tnh tch phn v trnh by bi gii.


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2

0

dx

4x2

0

x.e2y

4 ydy.

Ta c, minD: 0 x 2 v0 y 4 x2.


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2

0

dx

4x2

0

x.e2y

4 ydy.

Ta c, minD: 0 x 2 v0 y 4 x2.MinDgii hn bi cc ngx=0, x=2, y=0 vy=4 x2.


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2

0

dx

4x2

0

x.e2y

4 ydy.

Ta c, minD: 0 x 2 v0 y 4 x2.MinDgii hn bi cc ngx=0, x=2, y=0 vy=4 x2.


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2

0

dx

4x2

0

x.e2y

4 ydy.

Ta c, minD: 0 x 2 v0 y 4 x2.MinDgii hn bi cc ngx=0, x=2, y=0 vy=4 x2.Suy raD: 0 y 4 v0 x 4 y.


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2

0

dx

4x2

0

x.e2y

4 ydy.

Ta c, minD: 0 x 2 v0 y 4 x2.MinDgii hn bi cc ngx=0, x=2, y=0 vy=4 x2.Suy raD: 0 y 4 v0 x 4 y.

Ta cI = e8

14 .


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Tnh D

2xdxdy viDl min gii hn bi cc ng y=x2,x+y=2

vy=0.


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Tnh D


vy=0.


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Tnh D


vy=0.

Ta c minD: 0 y 1 vy x 2 y.

p dng cng thc (2.3) ta c

I =116 .


Tch phn trn min bt k
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Tnh tch phn

D

dxdy viDl min gii hn bi cc ng y=x2,

y=2x2,x=y2 vx=3y2.


Tch phn trn min bt k
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Tnh tch phn

D

dxdy viDl min gii hn bi cc ng y=x2,

y=2x2,x=y2 vx=3y2.


Bi tp Tch phn trn hnh thang I

Trong cc bi tp 1 8 hy biu din hnh hc ca cc min ly tch phn
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Trong cc bi tp1-8, hy biu din hnh hc ca cc min ly tch phn1 0

x

3; 0

y

2x

2 1 x 2; x 1 y x23 2 y 2; y2 x 44 0 y 1; y x 2y5 0

x

1; ex

y

e6 1 x e2; 0 y ln x7 0 y 1; 0 x arcsin y8 0 y 8;1

4y x y 13


Bi tp Tch phn trn hnh thang II

Trong cc bi tp 9 18 biu din tch phn

dxdy trn min D theo
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Trong cc bi tp9-18,biu din tch phn

D

dxdytrn minDtheo

hai th t khc nhau9 y=x3, y =8, x=0

10 y=2x, y=0, x=311 y=3x, y=x2

12 y=ex, y=1, x=213 y=

x, y=0, x=9

14 y=tan x, x=0, y=115 y=ex, y=1, x=ln 316 y=0, x=0, y=1, y =ln x17 y=3 2x, y=x, x=018 y=x2, y =x+2


Bi tp Tch phn trn hnh thang III

Trong cc bi tp19-24,hy biu din hnh hc min ly tch phn ca
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g p , y y pcc tch phn lp sau y. Sau , tnh gi tr ca chng.

19

0

dx

x0

xsin ydy

20

0

dx

sin x

0

ydy

21

ln 81

dy

ln y0

ex+ydx

22

21

dy

y2y

dx


Bi tp Tch phn trn hnh thang IV1 y2
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23

0

dy

0

3y3exydx

24

41

dx

x

0

32

e y

x

Trong cc bi tp25-28,tnh tch phn ca hm ftrn min c cho25 f(x, y) =

xy

trn min phng nm trong gc phn t th nht v b

chn bi cc ng thng y=x, y =2x, x=1, x=226 f(x, y) =x2 +y2 trn min tam gic vi cc nh (0, 0), (1, 0)v

(0, 1)27 f(u, v) =v utrn min tam gic ct t gc phn t th nht

ca mt phnguvbi ng thngu+v=1


Bi tp Tch phn trn hnh thang V28 f(s, t) =es ln ttrn min nm trong gc phn t th nht ca mt
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( , ) g g pphngstv nm pha trn ng cong s=ln tvi tt=1 nt=2

Mi bi tp29-32l mt tch phn trn h ta Descartes. Hy biudin hnh hc min ly tch phn v tnh gi tr ca cc tch phn .

29

0

2

vv

2dpdvtrn mt phngpv

30

10

1s20

8tdtdstrn mt phngst

31

3

3

1

cosx0

3cos tdudttrn mt phngtu


Bi tp Tch phn trn hnh thang VI3

2 42u2
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32

0

1

4 2uv2

dv dutrn mt phnguv

33 Tm th tch ca vt th b chn trn bi paraboloidz=x2 +y2 vb chn di bi tam gic c ba cnh ta trn cc ng thngy=x, x=0 vx+y=2 ca mt phng xy.

34

Tm th tch ca vt th c gii hn pha trn bi mt tr z=x2

v gii hn pha di bi hnh phng c gii hn bi paraboly=2 x2 v ng thngy=xtrong mt phng xy.

35 Tm th tch ca vt th m mt y ca n l min phng thucmt phngxy, b chn bi cc ngy=4

x2 v ng thng

y=3x; pha trn ca vt th l mt phngz=x+4.36 Tm th tch ca vt th b chn bi gc phn tm th nht ca mt

phng ta , mt tr x2 +y2 =4 v mt phng z+y=3.


Bi tp Tch phn trn hnh thang VII37 Tm th tch ca vt th b chn bi gc phn tm th nht ca mt
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phng ta , mt phng x=3 v mt tr parabol z=4 y2.38 Tm th tch ca vt th c ct t gc phn tm th nht ca hta Oxyzbi mt z=4 x2 y2.39 Tm th tch ca ci nm ct t gc phn tm th nht ca h ta

Oxyzbi mt trz=12 3y2 v mt phngx+y=2.40

Tm th tch ca vt th ct t ci ct hnh vung|x| + |

y| 1 bimt phngz=0 v mt phng 3x+z=3.

41 Tm th tch ca vt th bit rng n b chn pha trc v pha sau

bi mt phngx=2 vx=1; hai bn bi mt tr y= 1x

; pha

trn v pha di bi hai mt phng z=x+1 vz=0.


Bi tp Tch phn trn hnh thang VIII42 Tm th tch vt th bit rng n b chn pha trc v pha sau bi

1
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x= 3

; hai bn bi mt tr y= 1cos x

; pha trn bi mt tr

z=1+y2 v pha di l mt phngOxy.43 Tnh din tch hnh phng gii hn bi cc ngx=4y y2 v

x+y=6.44 Tnh din tch hnh phng gii hn bi cc ng thngy=x,

x=2y,x+y=2 vx+3y=2.


i bin trong tch phn hai lp
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Gi s, h phng trnh
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Gi s, h phng trnh

x=x(u, v)y=y(u, v)

xc nh mt php bin i 1 - 1 t Dvo D;



Gi s, h phng trnh
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G s, p g t

x=x(u, v)y=y(u, v)

xc nh mt php bin i 1 - 1 t Dvo D;x, yl nhng hm co hm ring lin tc trnD

CBGD L Hoi Nhn () Chng 2 TCH PHN BI Ngy 26 thng 7 nm 2015 40 / 115


Gi s, h phng trnh
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, p g

x=x(u, v)y=y(u, v)

xc nh mt php bin i 1 - 1 t Dvo D;x, yl nhng hm co hm ring lin tc trnDv

J=(x, y)(u, v)

=

xu x

v

yu yv

=0.

Khi , ta c cng thc i bin tng qut

D

f(x, y)dxdy =D

f(x(u, v), y(u, v))|J|dudv.


i bin tng qut
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Tnh tch phnD

dxdy viDl min gii hn bi cc ng y=x2

,

y=2x2,x=y2 vx=3y2.


i bin tng qut
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Tnh tch phnD

dxdy viDl min gii hn bi cc ng y=x2

,

y=2x2,x=y2 vx=3y2.

tu=x2

y vv=

y2

x .

Ta c (u, v)(x, y)

=3 = J=13

.

Min

D:

1

2u

v

1

3v

1.

Suy raI = 19

.


H ta cc
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H ta cc
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H ta cc gm: im Ochotrc c gi l "cc" v tiaOPc gi l "trc cc".


H ta cc
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H ta cc gm: im Ochotrc c gi l "cc" v tiaOPc gi l "trc cc".

Ta cc ca imMthuc mtphng gm hai yu t: bn knhvectorr=OMv gc cc= (OP,

OM).


H ta cc
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Cho trc h trc Oxy, chn h ta cc (O,Ox)th imM(x, y)cta ccM(r, )vi

x=r. cosy=r. sin

.


H ta cc
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Cho trc h trc Oxy, chn h ta cc (O,Ox)th imM(x, y)cta ccM(r, )vi

x=r. cosy=r. sin

.

Nu chn (r, )tha 0 2(hay r ) vr>0 th h phngtrnh trn xc nh php bin i 1 - 1

viJ=r.


ng cong trong ta cc

ng cong trong ta cc c cho bi phng trnh
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F(r, ) =0 hayr=r().


ng cong trong ta cc

ng cong trong ta cc c cho bi phng trnh
http://find/


88/212

F(r, ) =0 hayr=r().

Hnh minh ha ca ng cong r= cos.


ng cong trong ta cc


89/212

Tia (Ot)hp vi tiaOxmt gc 0c phng trnh = 0 = constvir 0.


ng cong trong ta cc
http://find/


90/212

Tia (Ot)hp vi tiaOxmt gc 0c phng trnh = 0 = constvir 0.

CBGD L Hoi Nhn () Chng 2 TCH PHN BI Ng 26 thng 7 nm 2015 45 / 115

ng cong trong ta cc
http://find/


91/212

Phng trnh ca ng trn tm Obn knha:r=a vi 0 2.

CBGD L H i Nh () Ch 2 TCH PHN BI N 26 th 7 2015 46 / 115

ng cong trong ta cc
http://find/


92/212

Phng trnh ca ng trn tm Obn knha:r=a vi 0 2.


ng cong trong ta cc

Ph h I ( ) b k h
http://find/


93/212

Phng trnh ng trn tm I(a, 0), bn knha:r=2a cosvi

2

2 .


ng cong trong ta cc

Ph h I ( 0) b k h 2 i
http://find/


94/212

Phng trnh ng trn tm I(a, 0), bn knha:r=2a cosvi

2

2 .


ng cong trong ta cc
http://find/


95/212

Phng trnh ng trn tm I(0, b), bn knhb:r=2bsinvi0 .

CBGD L H i Nh () Ch 2 TCH PHN BI N 26 h 7 2015 48 / 115

ng cong trong ta cc
http://find/


96/212

Phng trnh ng trn tm I(0, b), bn knhb:r=2bsinvi0 .


ng cong trong ta ccng cong trong Phng trnh trong Min xc nh

h ta Oxy h ta cc
http://find/


97/212

Tiay=k.x =arctan k r 0(x.y 0)Tiay=k.x =arctan k r 0

(x.y 0)ng trn

(x a)2 +y2 =a2 r=2a cos 2 2(a>0)

ng trnx2 + (y b)2 =b2 r=2bsin 0

(b>0)ng trnx2 +y2 =a2 r =a 0 2

(a>0)


Cch xc nh cn ca tch phn trong ta cc I

1 Biu din hnh hc minD.V minDv k hiu phng trnh bi h O Ch h h
http://find/


98/212

cc bin ca n theo ta Oxy. Chuyn cc phng trnh ny vta cc.

2 Tm cn ca r.V tiaLt cc xuyn qua minDtheo chiu tngcar. nh du gi tr carm ti L ln lt i vo v i ra khiminD. y chnh l cn di v cn trn ca r. Thng thng, cccn ny ph thuc vo gc c to bi Lv chiu dng ca trcx.

3 Tm cn ca .Tm gi tr nh nht v gi tr ln nht ca mn bao ly minD. y l cc cn ca .


Tch phn trong ta cc
http://find/


99/212

Php bin i

x=r. cosy=r. sin

vi 0 2(hay r ) vr>0 bin minDtrong mt phng cc thnh min Dtrong mtphngOxyvi|J| =r.

C G () C 2 C 26 201 1 / 11

Tch phn trong ta cc
http://find/


100/212

Php bin i

x=r. cosy=r. sin

vi 0 2(hay r ) vr>0 bin minDtrong mt phng cc thnh min Dtrong mtphngOxyvi|J| =r.

D

f(x, y)dxdy=D

f(r. cos, r. sin).r.drd.

() /

Tch phn trong ta cc
http://find/


101/212

D l min kn bao gc ta gii hn bi ng cong knr=r()th min (D)l:0 2v 0 r r().

Tch phn trong ta cc
http://find/


102/212

D l min kn bao gc ta gii hn bi ng cong knr=r()th min (D)l:0 2v 0 r r().

D

f(x, y)dxdy =

2

0

d

r()

0

f(r. cos, r. sin).r.dr.


Tch phn trong ta cc

l min hnh trn tm O bn knh a th (D ) l 0 2 v
http://find/


103/212

Dl min hnh trn tm O, bn knh ath (D )l0

2v

0 r a.


Tch phn trong ta cc

http://find/


104/212


2v

0 r a.Ta c cng thc

D

f(x, y)dxdy=

20

d

a0



Tch phn trong ta cc

http://find/


105/212


2v

0 r a.Ta c cng thc

D

f(x, y)dxdy=

20

d

a0


Tnh tch phnI =D

ex2y2dxdytrong , Dl min hnh trn

x2 +y2 a2,a>0.


Tch phn trong ta cc
http://find/


106/212


Tch phn trong ta cc

MinDc gii
http://find/


107/212

hn bi cc ng=, = ,r=h1(),r=h2().


Tch phn trong ta cc

MinDc gii
http://find/


108/212

hn bi cc ng=, = ,r=h1(),r=h2().

Ta c cng thc

D

f(x, y)dxdy=

d

h2()

h1()



Tch phn trong ta cc

Tnh tch phn I = arctany

dxdy vi D l min 1 x2 + y2 9 v
http://find/


109/212

Tnh tch phnI D

arctanx

dxdyviDl min 1

x +y

9 v

x3 y x.3.


Tch phn trong ta cc

Tnh tch phn I = arctany

dxdy vi D l min 1 x2 + y2 9 v
http://find/


110/212

Tnh tch phnI D

arctan

x

dxdyviDl min 1

x +y

9 v

x3 y x.3.

x

0 1 2 3 4

y

0

1

2

3

4

MinD: 1 r 3 v

6

3.

Suy ra

I =

2

6 .


Tch phn trong ta cc

I x2 y2dxdy D
http://find/


111/212

Tnh tch phnI

=D

4

x2

y2dxdyvi

D lna trnca hnh

trn (x 1)2 +y2 1.


Tch phn trong ta cc

Tnh tch phn I 4 x2 y2dxdy vi D l ca hnh


112/212

Tnh tch phnI =D

4 x y dxdyviD lna trnca hnhtrn (x 1)2 +y2 1.


Tch phn trong ta cc

Tnh tch phn I = 4 x2 y2dxdy vi D l na trn ca hnh
http://find/


113/212

Tnh tch phnI = D

4 x y dxdyviD lna trnca hnhtrn (x 1)2 +y2 1.

MinD: 0

2 v0 r 2cos.Suy ra

I =8

3.

22

3 .


Tch phn trong ta cc

Tnh tch phnI = dxdyviDl min gii hn bi ng cong
http://find/


114/212

p D

y g g g

(x2 +y2)2 =2xy.


Tch phn trong ta cc

http://find/


115/212

p D

y g g g

(x2 +y2)2 =2xy.


Tch phn trong ta cc

http://find/


116/212

D

y g g g

(x2 +y2)2 =2xy.

Nhn xt: minDl min c tmi xngOv hm ly tch phn

chn nn ta ch cn tnh tch phntrn na minDthuc gc phnt th nht.

MinD: 0 2

v

0 r sin 2Suy raI =2I =1.


Bi tp tch phn trong ta cc ITrong cc bi tp1-14,chuyn cc tch phn trong ta Descartessang ta cc. Tnh gi tr ca chng.

11 x2
http://find/


117/212

1

1

0

dy dx

2

1

0

1y2

0

(x

2

+y

2

)dx dy

3

2

0

4y2

0

(x2 +y2)dx dy

4

aa

a2x2

a2x2

dy dx


Bi tp tch phn trong ta cc II

5

60

y0

x dx dy
http://find/


118/212

6

20

x0

y dy dx

7

3

1

x

1

dy dx

8

22

y

4y2

dx dy

9

01

0

1x2

2

1+

x2 +y2dy dx


Bi tp tch phn trong ta cc III

10

11

1x2

1 x2

2(1+x2 +y2)2

dx dy
http://find/


119/212

11

ln 20

(ln 2)2y20

e

x2+y2 dx dy

12

11

1y2

1y2

ln(x2 +y2 +1)dx dy

13

10

2x2x

(x+2y)dy dx


Bi tp tch phn trong ta cc IV

14

21

2xx20

1(x2 +y2)2

dy dx
http://find/


120/212

15 Tnh din tch hnh phng nm ngoi ng trn r=1 v nm trong

ng trnr= 2

3cos .

16 Tnh din tch hnh phng gii hn bi ng congx3 +y3 =a.x.y

v cc trc ta (phn nm trong gc phn t th nht).17 Tnh din tch ca phn mt Hyperbolic paraboloid z=x2 y2 nm

pha trong mt trx2 +y2 =a2.18 Tnh din tch ca phn mt phng z=2xnm pha trong mt

paraboloidz=x2

+y2

.19 Tnh th tch vt thVgii hn bi mt cu x2 + y2 + z2 =2 v mtnnz=

x2 +y2 (phn nm pha trong mt nn).


Bi tp tch phn trong ta cc V20 Tnh th tch vt thVgii hn bi mt nn z=

x2 +y2 v

paraboloidz=x2 +y2.
http://find/


121/212


nh ngha Tch phn ba lpCho hm sf(x, y, z)xc nh trn minVng, b chn trong khnggianxyz.
http://find/


122/212



Chia min thnhnmin con khng dm ln nhau c tn v th
http://find/


123/212

Vtch gi chung lv1,v2, . . . ,vn.



Chia min thnhnmin con khng dm ln nhau c tn v th
http://find/


124/212

Vtch gi chung lv1,v2, . . . ,vn.Trong mi min con vi vii=1, . . . ,nly im Mi(xi, yi, zi)v lptng tch phn

In=

n

k=1

f(

xi, y

i, z

i).

vi.


nh ngha Tch phn ba lpGidi l ng knh ca min vi.
http://find/


125/212


nh ngha Tch phn ba lpGidi l ng knh ca min vi. Chon sao chomax di 0.
http://find/


126/212


nh ngha Tch phn ba lpGidi l ng knh ca min vi. Chon sao chomax di 0. Nu tn ti gii hn lim

maxdi0In=Imt cch c lp

V i( i i i )
http://find/


127/212

vi cch chia minVv cch chn cc imMi(xi, yi, zi)ca minvi




h h V h h Mi( i i i )
http://find/


128/212

vi cch chia minVv cch chn cc imMi(xi, yi, zi)ca minvith ta ni




i h hi i V h h i Mi( i i i ) i
http://find/


129/212

vi cch chia minVv cch chn cc imMi(xi, yi, zi)ca minvith ta niHm sf(x, y, z)kh tch trn minV.




i h hi i V h h i Mi( i i i ) i h
http://find/


130/212

vi cch chia minVv cch chn cc imMi(xi, yi, zi)ca minvith ta niHm sf(x, y, z)kh tch trn minV.Il tch phn 3 lp ca hm f(x, y, z)trn minVv c k hiu l

I=

V

f(x, y, z)dV =

V

f(x, y, z)dxdydz.


Th tch vt th
http://find/


131/212

Mt vt thVtrong khng gian c th tch c tnh theo tch phn balp

V = V

dx dy dz.


khi lng vt th
http://find/


132/212

Mt vt thVtrong khng gian, c hm khi lng ring (cn c gi lhm mt khi lng) ti mi im l (x, y, z), c khi lng ctnh theo tch phn ba lp

m=V

(x, y, z)dx dy dz.


Vt th hnh tr m rng

Ta ni, minVl mt th hnh tr m rng nu n c gii hn bi
http://find/


133/212

Hai mt cong n ginz=z1(x, y)


Vt th hnh tr m rng

http://find/


134/212

Hai mt cong n ginz=z1(x, y)vz=z2(x, y)


Vt th hnh tr m rng

http://find/


135/212

Hai mt cong n ginz=z1(x, y)vz=z2(x, y)trong z1 z2.


Vt th hnh tr m rng
http://find/


136/212

Mt bn c gii hnbi mt tr c ng sinh

song song vi trcOz.


Vt th hnh tr m rng
http://find/


137/212

Mi imM(x, y, z) Vu c cao ztha

z1(x, y) z z2(x, y).


Vt th hnh tr m rng
http://find/


138/212

GiDl hnh chiu caV

trn mt phngOxy.


Tch phn ba lp trong h ta Descartes

1 Phc tho minV.Xc nh mt y pha trn v mt y pha di ca minV Suy ra
http://find/


139/212

Xc nhmt y pha trnvmt y pha dica min . Suy racn cazXc nh min hnh chiuDcaVtrn mt phngOxy.

2 Phc tho minD.Thc hin nh trong tch phn hai lp. Suy racn caxvy.

3 Chuyn v tch phn lp.Da vo cc kt qu trn, a tchphn

V

f(x, y, z)dx dy dzv tch phn lp.

4 Suy ra gi tr ca tch phn.Tnh cc tch phn xc nh thu c bc 3 v suy ra kt qu ca bi ton.



D l hnh thang loi 1th a x b
http://find/


140/212

V :

a x b(x) y (x)z1(x, y) z z2(x, y)



http://find/


141/212

V :

a x b(x) y (x)z1(x, y) z z2(x, y)

Cng thc:

V

f(x, y, z)dxdydz=

ba

(x)(x)

z2(x,y)z1(x,y)

f(x, y, z)dzdy dx



http://find/


142/212

V : (x) y (x)

z1(x, y) z z2(x, y)

Cng thc:

V

f(x, y, z)dxdydz=

ba

(x)(x)

z2(x,y)z1(x,y)

f(x, y, z)dzdydx



Tnh tch phnI =

V

ydxdydztrong Vl min gii hn bi cc mt

y=x2,y+z=1 vz=0.
http://find/


143/212

y , y



Tnh tch phnI =

V


y=x2,y+z=1 vz=0.
http://find/


144/212

y , y

MinVv hnh chiuDtrn mt phngOxy



Tnh tch phnI =

V


y=x2,y+z=1 vz=0.
http://find/


145/212

y y




Tnh tch phnI =

V


y=x2,y+z=1 vz=0.
http://find/


146/212




Tnh tch phnI =

V


y=x2,y+z=1 vz=0.
http://find/


147/212




D l hnh thang loi 2th

V c y d

( ) ( )
http://find/


148/212

V : (y) x (y)

z1(x, y) z z2(x, y)




V : c y d

(y) x (y )
http://find/


149/212

V : (y) x (y)

z1(x, y) z z2(x, y)

Cng thc:

V

f(x, y, z)dxdydz=

d

c

(y)

(y)

z2(x,y)

z1(x,y)

f(x, y, z)dzdx dy




V : c y d

(y) x (y )
http://find/


150/212

V : (y) x (y)

z1(x, y) z z2(x, y)

Cng thc:

V

f(x, y, z)dxdydz=

d

c

(y)

(y)

z2(x,y)

z1(x,y)

f(x, y, z)dzdxdy



Tnh tch phnI =

V

11 x ydxdydztrong Vl min gii hn

bi cc mtx+y+z=1, x=0, y=0 vz=0.
http://find/


151/212



Tnh tch phnI =

V


http://find/


152/212




Tnh tch phnI =

V


http://find/


153/212




Tnh tch phnI =

V


http://find/


154/212




Tnh tch phnI =

V


http://find/


155/212




Tnh tch phnI =

V


http://find/


156/212




Tnh tch phnI =

V


http://find/


157/212




Vl hnh hnh hp ch nhtth

V :a x b; c y d; g z h.
http://find/


158/212



Vl hnh hnh hp ch nhtth

V :a x b; c y d; g z h.
http://find/


159/212

Cng thc:

V

f(x, y, z)dxdydz=

ba

dc

hg

f(x, y, z)dz dy dx



1 Tnh tch phnI =

11 x ydxdydzvi minVtha
http://find/


160/212

V

1 x y0 x 1, 2 y 5 v 2 z 4.

2 Tnh tch phnI =V

xyzdxdydzvi minVtha 0 x 2,0 y 2v 0 z 1.



Vc dng

V : a

x

bc y dg z h

.
http://find/


161/212

g z h



Vc dng

V : a

x

bc y dg z h

.
http://find/


162/212

g z h

vf(x, y, z) =f1(x).f2(y).f3(z)



Vc dng

V : a

x

bc y dg z h

.
http://find/


163/212

g z h

vf(x, y, z) =f1(x).f2(y).f3(z)ta c cng thc:

V

f(x, y, z)dxdydz=

ba

f1(x)dx

dc

f2(y)dy

hg

f3(z)dz


Bi tp tch phn ba lp trong h ta Descartes I

Trong cc bi tp1-14,hy tnh cc tch phn lp c cho

1

1

0

1

0

1

0

(x2 +y2 +z2)dz dy dx

2 3y 8x2y2
http://find/


164/212

2

0

0

x2+3y2

dz dy dx

3

e1

e21

e31

1xyz

dx dy dz

4

1

0

33x0

33xy0

dz dy dx


Bi tp tch phn ba lp trong h ta Descartes II

5

60

10

32

ysin z dx dy dz

6

1 1 2(x+y+z) dy dx dz
http://find/


165/212

1

0

0

7

3

0

9x2

0

9x2

0

dz dy dx

8

20

4y2

4y2

2x+y0

dz dx dy


Bi tp tch phn ba lp trong h ta Descartes III

9

10

2x0

2xy0

dz dy dx

10

1 1x2 4x2yx dz dy dx
http://find/


166/212

0

0

3

11

0

0

0

cos (u+v+w) du dv dw, trong khng gian uvw

12

10

e

1

e1

s.es. ln r.(ln t)2

t dt dr ds, trong khng gianrst

13

40

ln 1cosx

0

2t

ex dx dt dv, trong khng giantvx


Bi tp tch phn ba lp trong h ta Descartes IV

14

70

20

4q20

qr+1

dp dq dr, trong khng gianpqr.

Trong cc bi tp15-19,tch cc tch phn ba lp c cho15

(x2z+y)dx dy dztrong Vl min c gii hn bi cc mt
http://find/


167/212

V

( + y) y g V g

x=0, x=1, y=1, y =3, z=0, z=2.

16V

2yex dx dy dztrong Vl min c cho bi0 y 1, 0 x y, 0 z x+y.

17 Vx2y2z2 dx dy dztrong Vl min c gii hn bi

z=y+1, y+z=1, x=0, x=1, z=0.


Bi tp tch phn ba lp trong h ta Descartes V

18

V

xy dx dy dztrong Vl gc phn tm th nht b chn bi

cc mt phng ta v na pha trn ca mt cu x2 +y2 +z2=4.

19V

y2 dx dy dztrong Vl khi t din nm trong gc phn tm
http://find/


168/212

V

th nht, c gii hn bi cc mt phng ta v mt phng2x+3y+z=6.

20 Tm th tch ca vt th nm trong gc phn tm th nht, b chnbi mt phngz=x,y x=2 v mt tr y=x2. Tm ta trng tm ca vt th .

21 Tnh khi lng ca mt hnh lp phng n v bit rng mt

khi lng ca n ti mi im1 bng khong cch t im n mt mt cho trc ca khi lpphng.


Bi tp tch phn ba lp trong h ta Descartes VI

2 bng bnh phng khong cch t im n mt nh cho trc cakhi lp phng.

22 Tm th tch v ta trng tm ca vt th, bit rng vt th b

chn trn bi mt trx2

+z

=4, b chn di bi mt phngx+z=2 v cc mt bn l hai mt phng y=0, y =3.23 Cho V l vt th b chn trn bi mt phng z = y b chn di bi
http://find/


169/212

23 ChoVl vt th b chn trn bi mt phng z=y, b chn di bimt phngxyv xung quanh cc mt phng x=0, x=1, y=1.Tm khi lng ca

Vbit rng mt ca n ti mi im bng

bnh phng khong cch t im n gc ta . Tm ta trng tm caV.

24 Tnh tch phn ca hm f(x, y, z) =x+y3 +ztrn hnh cu n vc tm l gc ta .

25 Tnh tch phn ca hm f(x, y, z) =a1.x+ a2.y+ a3.z+ a4trn hnhcu n v c tm l gc ta , trong a1, a2, a3, a4l cc hng s.


Bi tp tch phn ba lp trong h ta Descartes VII

26 Tnh tch phn ca hm f(x, y, z) =x2.y2 trn min b chn trn bimt try2 +z=4, b chn di bi mt phngy+z=2 v xungquanh bi cc mt phngx=0, x=2.

27

Tnh th tch ca vt th c gii hn bi cc mt paraboloidz=2 x2 y2 vz=x2 +y2.
http://find/


170/212


i bin tng qut

Gi s, h phng trnh

x=x(u, v,w)y=y(u, v,w)z=z(u, v,w)

xc nh mt php bin

i 1 - 1 tV voV;
http://find/


171/212


i bin tng qut

Gi s, h phng trnh


xc nh mt php bin

i 1 - 1 tV voV;x, y, zl nhng hm c o hm ring lin tctrnV
http://find/


172/212


i bin tng qut

Gi s, h phng trnh


xc nh mt php bin

i 1 - 1 tV voV;x, y, zl nhng hm c o hm ring lin tctrnVv
http://find/


173/212

J=

xu xv x

w

yu yv y

w

zu

zv

zw

=0

th


i bin tng qut

Gi s, h phng trnh


xc nh mt php bin

i 1 - 1 tV voV;x, y, zl nhng hm c o hm ring lin tctrnVv
http://find/


174/212

J=

xu xv x

w

yu yv y

w

zu

zv

zw

=0

th

Vf(x, y, z)dxdydz= V

f(x(.), y(.), z(.))

|J

|dudvdw


Tch phn trong ta tr

Ta tr ca imM(x, y, z)c dngM(r, , z), trong im(r, )l ta cc ca M(x, y, 0)trong mt phng Oxy.
http://find/


175/212


Tch phn trong ta tr


Lin h gia ta Oxyz

v ta tr th hin qua php bin i:
http://find/


176/212


Tch phn trong ta tr


Lin h gia ta Oxyzv ta tr th hin qua php bin i: x = rcos

i
http://find/


177/212

y = rsinz = z

Php bin i ny c JacobianJ=r.


Tch phn trong ta tr


Lin h gia ta Oxyzv ta tr th hin qua php bin i: x = rcos

i
http://find/


178/212

y = rsinz = z

Php bin i ny c JacobianJ=r.

Cng thc i bin

V

f(x, y, z)dxdydz=V

f(rcos , rsin , z).r.drddz.


Tch phn trong ta tr

Tnh

V

(x2 +y2)dxdydzbitVl min gii hn bi cc mt

x2 +y2 =1, x2 +y2 =4, y=0, y=x(phnx>0), z=0 vz=2.
http://find/


179/212


Tch phn trong ta tr

Tnh

V


x2 +y2 =1, x2 +y2 =4, y=0, y=x(phnx>0), z=0 vz=2.
http://find/


180/212


Tch phn trong ta tr

Tnh

V


x2 +y2 =1, x2 +y2 =4, y=0, y=x(phnx>0), z=0 vz=2.
http://find/


181/212


Tch phn trong ta tr

Tnh

V


x2 +y2 =1, x2 +y2 =4, y=0, y=x(phnx>0), z=0 vz=2.
http://find/


182/212


Tch phn trong ta tr

Tnh

V


x2 +y2 =1, x2 +y2 =4, y=0, y=x(phnx>0), z=0 vz=2.
http://find/


183/212


Tch phn trong ta tr

Tnh

V


x2 +y2 =1, x2 +y2 =4, y=0, y=x(phnx>0), z=0 vz=2.
http://find/


184/212


Tch phn trong ta tr

Tnh

V


x2 +y2 =1, x2 +y2 =4, y=0, y=x(phnx>0), z=0 vz=2.
http://find/


185/212


Tch phn trong ta tr

Tnh

V

zdxdydzbitVl min gii hn bi cc mtz=x2 +y2,

z=0, x2 +y2 =4.
http://find/


186/212

CBGD. L Hoi Nhn () Chng 2. TCH PHN BI Ngy 26 thng 7 nm 2015 100 /

115

Tch phn trong ta tr

Tnh

V


z=0, x2 +y2 =4.
http://find/


187/212


115

Tch phn trong ta tr

Tnh

V


z=0, x2 +y2 =4.


188/212


115

Tch phn trong ta tr

Tnh

V


z=0, x2 +y2 =4.


189/212


115

Tch phn trong ta tr

Tnh

V


z=0, x2 +y2 =4.
http://find/


190/212


115

Tch phn trong ta tr

Tnh

V


z=0, x2 +y2 =4.
http://find/


191/212


115

Tch phn trong ta tr

Tnh tch phn lpI =

2a

0

dx

2axx2

2axx2

dy

4a2x2y2

0

dz.
http://find/


192/212


115

Tch phn trong ta tr

Tnh tch phn lpI =

2a

0

dx

2axx2

2axx2

dy

4a2x2y2

0

dz.

Biu din min ly tch phnVtrong h ta Descartes. Xc nh
http://find/


193/212

cc mt gii hn caV.


115

Tch phn trong ta tr

Tnh tch phn lpI =

2a

0

dx

2axx2

2axx2dy

4a2x2y2

0

dz.

Biu din min ly tch phnVtrong h ta Descartes. Xc nh ii h V
http://find/


194/212

cc mt gii hn caV.

Chuyn tch phn lp thnh tch phn ba lp trnV.


115

Tch phn trong ta tr

Tnh tch phn lpI =

2a0

dx

2axx2

2axx2dy

4a2x2y2

0

dz.

Biu din min ly tch phnVtrong h ta Descartes. Xc nh ii h V
http://find/


195/212

cc mt gii hn caV.Chuyn tch phn lp thnh tch phn ba lp trn

V.

i bin sang ta tr.


115

Tch phn trong ta tr

Tnh tch phn lpI =

2a0

dx

2axx2

2axx2dy

4a2x2y2

0

dz.

Biu din min ly tch phnVtrong h ta Descartes. Xc nh t ii h V
http://find/


196/212

cc mt gii hn caV.Chuyn tch phn lp thnh tch phn ba lp trn

V.

i bin sang ta tr.

Tnh tch phn trong ta tr.


115

Bi tp Tch phn trong ta tr I

Bng cch i bin sang ta tr, tnh cc tch phn t1-13

1 I =V

xyz5dxdydztrong Vl phn chung ca hai hnh cu

x2

+y2

+z2 a

2

vx2

+y2

+ (z a)2 a

2

.2 I =

V

((x+y)2 z)dxdydztrong Vl min gii hn bi mt
http://find/


197/212

Vz=0 v (z 1)2 =x2 +y2.

3 I =V

x2 +y2dxdydztrong Vl min gii hn bi

z=x2 +y2 vz=1.

4 I = V

(x2 +y2)dxdydztrong V

l min gii hn bi mt

2z=x2 +y2 vz=2.


115

Bi tp Tch phn trong ta tr II

5 I =V

z

x2 +y2dxdydztrong Vl min gii hn bi cc mt

x2 +y2 =2x,y=0, z=0 vz=3.

6 I =V

(x2 y2)dxdydztrong Vl min gii hn bi cc mtx2 +y2 =2zvz=2.
http://find/


198/212

7 I = Vzx

2 +y2dxdydztrong

Vl min gii hn bi cc mt

y2 =3x x2,z=0 vz=2.8 I =

V

dxdydztrong Vl min c cho bi

0 x 1, 0 y 1 x2, 0 z

4 (x2 +y2).


115

Bi tp Tch phn trong ta tr III

9 I =V

z3dxdydztrong Vl min c cho bi

1 x 1, 0 y 1 x2,

x2 +y2 z 1.

10 I =V

1x2 +y2 dxdydztrong Vl min c cho bi0 x

9 y2, 0 y 3, 0 z

9 (x2 +y2).
http://find/


199/212

11 I = V

zdxdydztrong V

l min c cho bi

0 x 1, 0 y 1 x2, 0 z

1 x2 y2.12 I =

V

sin(x2 +y2)dxdydztrong Vl min c cho bi

0 x 1, 0 y 1 x2, 0 z 2.


115

Bi tp Tch phn trong ta tr IV

13 I =V

x2 +y2dxdydztrong Vl min c cho bi

1 x 1,1 x2 y 1 x2, x2 + y2 z 2 (x2 + y2).

14 Tnh th tch vt th b chn trn bi mt nn x2

+y2

=z2

, b chndi bi mt phngxyv xung quanh bi mt tr x2 +y2 =2ax.15 Tnh th tch vt th b chn trn bi mt paraboloid trn xoay

x2 + y2 = az b chn di bi mt phng xy v xung quanh bi mt
http://find/


200/212

x +y =az, b chn di bi mt phngxyv xung quanh bi mttrx2 +y2 =2ax.

16 Tnh th tch vt th b chn trn bi mt congz=a

x2 +y2, bchn di bi mt phngxyv xung quanh bi mt trx2 + y2 =ax.

17 Tnh th tch vt th b chn trn bi mt phng 2z=4+x, b chndi bi mt phngxyv xung quanh bi mt tr x2 +y2 =2x.

18 Tnh th tch vt th gii hn paraboloidz=x2 +y2 v mt phngz=x.


115

Bi tp Tch phn trong ta tr V

19 Tnh th tch vt th b chn trn bi mt cu x2 +y2 +z2 =25, bchn di bi mt phngz=

x2 +y2 +1.

20 Tnh th tch vt th b chn di bi mt z=

3(x2 +y2), b chntrn bi mt cu n vx2 +y2 +z2 =1.

21 Tnh th tch vt th b chn trn bi mt hypeboloidz2 =a2 +x2 +y2, b chn di bi mt nn z2 =2(x2 +y2).

22 Tnh th tch vt th b chn di bi mt phng xy, nm pha trong
http://find/


201/212

p g y , p gmt cux2 +y2 +z2 =9 v bn ngoi mt trx2 +y2 =1.

23 Tnh th tch vt th nm gia hai mt tr x2 +y2 =1 vx2 +y2 =4, b chn trn bi mt ellipsoidx2 +y2 +4z2 =36 v bchn di bi mt phngxy.

24 Tnh th tch ca vt th b chn di bi paraboloidz=x2 +y2, b

chn trn bi paraboloidz=x2

+y2

+1 v xung quanh bi mt trx2 +y2 =1.


115

Bi tp Tch phn trong ta tr VI

25 Tnh th tch ca vt th c ct ra t mt "bc tng dy"1 x2 +y2 2 bi mt nn z=

x2 +y2.

26 Tnh th tch ca vt th nm bn trong mt cu x2 +y2 +z2 =2v bn ngoi mt tr x2 +y2 =1.

27 Tnh th tch ca vt th c gii hn bi cc mt x2 +y2 =1,z=0 vy+z=4.

28 Tnh th tch ca vt th c gii hn bi cc mt x2 +y2 =4,
http://find/


202/212

g y ,z=0 vx+y+z=4.

29 Tnh th tch ca vt th c gii hn trn bi mt paraboloidz=5 x2 y2 v b chn di bi mt paraboloidz=4x2 +4y2.

30 Tnh th tch vt th b chn trn bi mt paraboloidz=9 x2 y2, b chn di bi mt phngz=0 v nm bn ngoi

mt trx2

+y2

=1.


115

Bi tp Tch phn trong ta tr VII

31 Tnh th tch vt th c ct ra t hnh trx2 + y2 1 bi mt cux2 +y2 +z2 =4.

32 Tnh th tch vt th b chn trn bi mt cux2 +y2 +z2 =2 v bchn di bi mt paraboloid z=x2 +y2.

33 Cho vt thVb chn trn bi mt paraboloidz=1 (x2 +y2)vb chn di bi mt phngxy.

1 Tnh th tch caV.
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203/212

2 Tnh khi lng caVbit rng mt khi lng ca n ti mi

im bng khong cch t im n mt phngxy.3 Tnh khi lng caVbit rng mt khi lng ca n ti miim bng bnh phng ca khong cch t im n gc ta .


115

Tch phn trong ta cu

Ta cu ca mt im Mtrong khng gian gm (r, ,)vir=OM,
http://find/


204/212


115

Tch phn trong ta cu

Ta cu ca mt im Mtrong khng gian gm (r, ,)vir=OM, = (

Oz,

OM),
http://find/


205/212


115

Tch phn trong ta cu


Oz,

OM), = (

Ox,

OM),Ml hnh chiu caM

trn mt phngOxy.
http://find/


206/212


115

Tch phn trong ta cu


Oz,

OM), = (

Ox,

OM),Ml hnh chiu caM

trn mt phngOxy.Lin h gia ta Oxyzv ta cu

x = rcos sin y = rsin sin z = rcos
http://find/


207/212


115

Tch phn trong ta cu


Oz,

OM), = (

Ox,

OM),Ml hnh chiu caM


http://find/


208/212

Php bin i ny c Jacobian:J=r2 sin


115

Tch phn trong ta cu


Oz,

OM), = (

Ox,

OM),Ml hnh chiu caM


http://find/


209/212

Php bin i ny c Jacobian:J=r2 sin Cng thc i bin

V

f(x, y, z)dxdydz

=V

f(rcos sin , rsin sin , rcos ).r2 sin dddr


115

Tch phn trong ta cu

1 Tnh tch phnI =V

zdxdydzviVl min gii hn bi mt cu

x2 +y2 +z2 =2 v mt nn z= x2 +y2 (phn pha trong mtnn).

2 Tnh tch phnI =V

(x2 +y2 +z2)dxdydzviVl min gii hn
http://find/


210/212

Vbi hai mt cux2 +y2 +z2 =a2 vx2 +y2 +z2 =b2 (a


211/212

0 y 1 x2, x2 +y2 z 2 (x

2 +y2).

3 I =V

(x2 +y2 +z2)dxdydztrong Vl min c cho bi

0 x

4 y2, 0 y 2,

x2 +y2 z

4 x2 y2.


115

Bi tp Tch phn trong ta cu II

4 I =V

z

x2 +y2 +z2dxdydztrong Vl min c cho bi

0 x

9 y2, 0 y 3, 0 z

9 (x2 +y2).

5 I =V

1x2 +y2 +z2dxdydztrong Vl min c cho bi

0 x 1, 0 y

1 x2, 0 z

1 x2 y2.
http://find/


212/212


115
http://find/

Documents

01 Slide - Chuong 2 Tich Phan Boi