01 Slide - Chuong 3 Tich Phan Duong Va Mat

7/24/2019 01 Slide - Chuong 3 Tich Phan Duong Va Mat

1/89

Chng 4

TCH PHN NG V TCH PHN MT

CBGD. L Hoi Nhn

Ngy 22 thng 10 nm 2015

CBGD. L Hoi Nhn () TCH PHN NG V TCH PHN MT

Ngy 22 thng 10 nm 2015 1 / 53
http://find/http://goback/


2/89

Mc lc

1 Tch phn ng loi 1nh nghaCch tnh

ng dng2 Trng vector

3 Tch phn ng loi 2nh ngha

Cch tnhCng thc Greennh l c bn

ng dng4 Tch phn mt loi 1

nh nghaCch tnh

ng dng5 Tch phn mt loi 2

nh nghaCch tnhCng thc Gauss - OstrogradskiCng thc Stokesng dng


Ngy 22 thng 10 nm 2015 2 / 53
http://goforward/http://find/http://goback/


3/89

nh ngha tch phn ng loi 1

nh ngha 1.1 ((5) trang 109)

Cho hm s f(x, y, z)xc nh trn cungLt A n B.Chia cung AB thnh n cung nh khng gim ln nhau bi cc imchia lin tip: A A0, A1, ...,An B. K hiu di cung Ai1Ai lsi.


Ngy 22 thng 10 nm 2015 3 / 53
http://find/


4/89




Trn mi cung Ai1Aichn im Mity v lp tng tch phnIn =

ni=1

f(Mi).si.

CBGD. L Hoi Nhn () TCH PHN NG V TCH PHN MTNgy 22 thng 10 nm 2015 3 / 53


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ni=1

f(Mi).si.

Cho n sao chomax si 0. Nu Inc gii hn hu hn Ikhng ph thuc v cc chia cung AB v cch chn Mith I cgi l tch phn ng loi 1 ca f trn cung AB.



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ni=1

f(Mi).si.

Cho n sao chomax si 0. Nu Inc gii hn hu hn Ikhng ph thuc v cc chia cung AB v cch chn Mith I cgi l tch phn ng loi 1 ca f trn cung AB.

K hiuI =L

f(x, y, z).ds.

http://find/


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Cch tnh tch phn ng loi 1

tnh tch phn ng loi 1 ta thc hin cc bc sau y:

Tham s ha ng congL.Vit phng trnh tham s ca ngcongLmt cch thch hp. Xc nh cn ca tham s v tnh viphn cungds.a tch phn ng loi 1 v tch phn xc nh. Thay cc ktqu gmx, y, ztrong phng trnh caL,ds, cn ca tham s trn vo mt trong cc cng thc (4.6) - (4.8) trang 111.

Tnh tch phn xc nh.Tnh tch phn xc nh thu c bn trnv suy ra p s.

http://find/


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Bng tm tt cch tnh tch phn ng loi 1

Phng trnh Vi phn cung Bin ly Cng thctham s ds= tch phn p dngy=g(x) 1+g2(x)dx x (4.8) x=x(t)y=y(t)

x2(t) +y2(t)dt t (4.7)

x=x(t)y=y(t)z=z(t)

x2(t) +y2(t) +z2(t)dt t (4.6)

http://find/


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V d v Tnh tch phn ng loi 1 I

1 I =L

xdsviLl phn paraboly=x2

2 tx=0 nx=2.

2 I =L

2xdsviLl phn paraboly=x2 t (0, 0)n (1, 1).

3

I =L x

2

dsviLl phn t th nht ca ng trnx2

+y2

=4.

4 I =L

(x 4

3 +y 4

3 )dsviLl cung phn t th nht ca ng astroid

x

2

3 +y

2

3 =a

2

3 .5 I =

L

x2dsviLl ng giao tuyn ca hai mt phngx y+z=0 vx+y+2z=0 t gc n im (3, 1,2).

http://find/


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V d v Tnh tch phn ng loi 1 II

6 I =L

2y2 +z2dsviLl ng giao tuyn ca mt cu

x2 +y2 +z2 =a2 v mt phngy=x. s: 2a2

http://find/


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di cung

Cng thc 1.1 (trang 110 dng 12 )L=

L

ds

.



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di cung

Cng thc 1.1 (trang 110 dng 12 )L=

L

ds

.

V d 1.11 Tnh di mt nhp ca ng Cycloid x=a.(t sint),

y=a.(1

cos t)vi t

[0, 2].2 Tnh di mt nhp ca ng l xo x=a. cos t, y=a. sin t,

z=b.t vi t [0, 2].

http://find/


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Khi lng, Moment v tm khi

Cng thc 1.2 (trang 113 v 114)

Khi lng cung.Mt cungLc khi lng ring ti mi im Ml (M)c khi lng lm=

L

(M)ds.

http://find/


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L

(M)ds.

Moment tnhca cungLi vi cc mt ta lMxy=L

z(M)ds; Mxz=L

y(M)ds; Myz=L

x(M)ds

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L

(M)ds.

Moment tnhca cungLi vi cc mt ta lMxy=L

z(M)ds; Mxz=L

y(M)ds; Myz=L

x(M)ds

Tm khica cungLlxc= Myzm ; yc=Mxz

m ; zc=

Mxym

V d 1.2

Tm khi lng v tm khi ca dy c dng ng inh c x=cos t,y=sin t, z=t vi0 t bit rng(x, y, z) =z .

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Ta trng tm ca cung ng cht

Cng thc 1.3 ((4.10) trang 114)Khi cungLng cht th ta trng tm c tnh theo cng thc:xc=

1L

L

xds; y c=1L

L

yds; z c=1L

L

zds

V d 1.31 Tm ta trng tm ca na trn ng trn tm O bn knh R.2

Tm ta trng tm ca ng inh ct(t) =a. cos ti +a sin tj +b.tk.


T
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Trng vector


Trng vectorxc nh trn min l mt hm vectorF(x, y, z)vi(x, y, z)

.


T
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Trng vector


Trng vectorxc nh trn min l mt hm vectorF(x, y, z)vi(x, y, z)

.

Ta c,

F(x, y, z) =Fx(x, y, z)

i +Fy(x, y, z)

j +Fz(x, y, z)

k


h h
http://find/


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ng cong tch phn

nh ngha.ng cong tch phn ca trng vector l ng congm vector tip tuyn ca n cng phng vi vector ca trng iqua im .


t h h
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ng cong tch phn

nh ngha.ng cong tch phn ca trng vector l ng congm vector tip tuyn ca n cng phng vi vector ca trng iqua im .

Cng thc 2.1 ((4.1) trang 107)ng cong tch phn ca trng vector

F tha mn h phng trnh vi

phn:dx

Fx=

dy

Fy=

dz

Fz


T b t
http://find/


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Trng bo ton

nh ngha 2.2 ((3) trang 107)Trng vector

F c gi l trng bo ton nu tn ti hm s

(x, y, z)sao cho

F(x, y, z) = (x, y, z) =

xi +

yj +

yk


Trng bo ton
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Trng bo ton



(x, y, z)sao cho

F(x, y, z) = (x, y, z) =

xi +

yj +

yk

Hm (x, y, z)c gi lhm th vca trng bo ton

F .


Trng bo ton
http://find/


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Trng bo ton



(x, y, z)sao cho

F(x, y, z) = (x, y, z) =

xi +

yj +

yk


F .

Mt mc ca (x, y, z)c gi lmt ng th.


Trng bo ton
http://find/


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Trng bo ton



(x, y, z)sao cho

F(x, y, z) = (x, y, z) =

xi +

yj +

yk


F .

Mt mc ca (x, y, z)c gi lmt ng th.

NuF l trng vector phng th ng mc ca hm th v cgi lng ng th.


http://find/


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Cho trng vectorF xc nh trn ng congL : r = r (t)vit [a, b].




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Chia cungLbi cc im chia lin tip A0,A1, ..., An. Ta k hiu,Ai1Ai=

ri, i=1, ..., n.


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ri, i=1, ..., n.


ni=1

F(Mi).

ri.


http://find/


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ri, i=1, ..., n.


ni=1

F(Mi).

ri.

Cho n

sao chomax

|

ri

| 0. Nu In c gii hn hu hn I,

khng ph thuc vo cch chia cung AB v cch chn Mi th I cgi l tch phn ng loi 2 trn cung AB.


http://find/


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ri, i=1, ..., n.


ni=1

F(Mi).

ri.

Cho n

sao chomax

|

ri

| 0. Nu In c gii hn hu hn I,

khng ph thuc vo cch chia cung AB v cch chn Mi th I cgi l tch phn ng loi 2 trn cung AB.

K hiu ca tch phn ng loi 2: I=L

F.dr .


http://find/


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Nur = (x, y, z)thdr = (dx, dy, dz)


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Nur = (x, y, z)thdr = (dx, dy, dz)v gi s rngF(x, y, z) = (P,Q,R)

thF.dr =Pdx+Qdy+Rdz.


http://find/


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g p g

Nur = (x, y, z)thdr = (dx, dy, dz)v gi s rngF(x, y, z) = (P,Q,R)

thF.dr =Pdx+Qdy+Rdz.

Do ngi ta cn vit tch phn ng loi 2 dng:

I = L

Pdx+Qdy+Rdz.


Cch tnh tch phn ng loi 2
http://find/


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p g

Ta tnh tch phn ng loi 2 bng cch chuyn v tch phn xc nhtheo cc bc sau:

1 Tham s ha ng congL.Vit phng trnh tham s caLmtcch thch hp v suy ra cc vi phndx,dy,dztheo vi phn ca

tham s. Xc nh cn ca tham s tng ng vi im u v imcui caL.2 Chuyn tch phn ng loi 2 v tch phn xc inh.Thayx,y,

ztrong phng trnh caL; cc vi phndx,dy,dzv hai cn catham s vo tch phn ng.

3 Tnh tch phn xc nh.Tnh tch phn xc nh thu c v suyra p s.


Bng tm tt cch tnh tch phn ng loi 2
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g p g

Phng trnh Vi phn ca Bin ly Cng thctham s cc bin s tch phn p dngy=g(x) dy=g(x)dx x (4.14) x=x(t)y=y(t)

dx=x(t)dtdy=y(t)dt t (4.13)

x=x(t)y=y(t)z=z(t)

dx=x(t)dtdy=y(t)dtdz=z(t)dt

t trang 118


V d v Tnh tch phn ng loi 2 I
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p g

1 Hy tnh tch phnI =

L

x2dx+y2dyvi

1 L :y=x,A(0, 0),B(1, 1), tch phn ly tAnB.2 L :y=x2, tch phn ly tAnB.

2 Hy tnh tch phn ng loi 2:I =

L

xdy ydxviLl ng

astroidx=a cos3 t,y=a sin3 tt imA(a, 0)nB(0, a).

3 Hy tnh tch phn ng loi 2:I =L

(x+y)dx+ (x y)dy trong

Ll ng elipx2

a2 +y2

b2 =1 chy ngc chiu kim ng h.


V d v Tnh tch phn ng loi 2 II
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4 Tnh tch phn ng loi haiI =

L

(y2 z2)dx+2yzdy x2dz

trong Ll ng cong c phng trnh tham s x=t,y=t2,z=t3 (0 t 1) ly theo chiu tng ca tham s.

5 Tm cng sn sinh bi lc

F = (x+y)i + (x z)j + (z y)k di

chuyn vt t imA(1, 0,

1)n imB(0,

2, 3)dc theo ng

thngAB.6 Tnh tch phnI =

L

ydx xdy tA(1, 0)nB(0,1)dc theo

1 on thngAB.2 ba gc phn t ca ng trn n v tmO(0, 0).


Cng thc Green
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nh l 3.1 ((3) trang 121)

Cho P v Q v cc hm s lin tc cng cc o hm ring trn minphngDlin thng, hu hn. Khi , ta c

L

Pdx+Qdy=

D

Qx P

y

dxdy

viLl bin ca minDv tch phn ng ly theo chiu dng.

1 Xc nhP(x, y)vQ(x, y)(ln lt l "h s" cadxvdy).2 Tnh hiu Q

x P

y.

3 Xc nh minD(Lu : c th s dng ta cc).4 Tnh tch phn hai lp trnD


V d v Cng thc Green I
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1 ChoI =

L

(1 x2)ydx+x(1+y2)dyviLl ng trn

x2 +y2 =R2 bng hai cch:1 Tnh trc tip.2 Dng cng thc Green.

2 Tnh tch phn

I =L

(ex sin y ky)dx+ (ex cos y k)dy

trong cc trng hp sau:1 Ll na pha trn ca ng trnx2 +y2 =ax, tch phn ly t im

A(a, 0)n gcO(0, 0).2 Ll cung paraboly=2x x2 pha trn trcOx.3 Ll cung ng congy=sin x+1 t imA(0, 1)nB(, 1).


V d v Cng thc Green II
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4 Ll cung paraboly=4 x2 t imA(1, 3)nB(2, 0)3 Tnh tch phn

I =L

xydx+x2y3dy

viLbin ca tam gic c cc nh c ta l (0, 0), (1, 0)v(1, 2).

4 Tnh tch phn

I =L

y3dx x3dy

viLbin ca hnh vnh khn tm Obn knh ln lt l 1 v 2.


V d v Cng thc Green III
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5 Tnh tch phn

I =L

ex((1 cos y)dx (y sin y)dy)

viLchy theo chu tuyn dng ca min phng 0 x v0

y

sin x6 Tnh tch phn

I =

x2+y2=R2

e(x2+y2).(cos2xydx+sin 2xydy).


nh l bn mnh tng ng
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nh l 3.2 (4 trang 125)

Cho P v Q v cc hm s lin tc cng cc o hm ring trn minphngDhu hn. Bn mnh sau tng ng nhau:

1 Tn ti hm (x, y)sao cho d(x, y) =Pdx+Qdy,(x, y) D.


nh l bn mnh tng ng
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nh l 3.2 (4 trang 125)



2

Qx =

Py,(x, y) D.


nh l bn mnh tng ng
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nh l 3.2 (4 trang 125)



2

Qx =

Py,(x, y) D.

3

L

Pdx+ Qdy=0vi mi ng cong knL nm hon ton trongD.


nh l bn mnh tng ng
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nh l 3.2 (4 trang 125)



2

Qx =

Py,(x, y) D.

3

L

Pdx+ Qdy=0vi mi ng cong knL nm hon ton trongD.

4AB

Pdx+Qdy khng ph thuc vo ng ly tch phn m ch ph

thuc vo A v B.


p dng nh l bn mnh tng ng tnh tchh l i 2 I
http://find/


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phn ng loi 2 I

1 Xc nh cc hm sP(x, y)vQ(x, y). Kim chng ng thcQx

=Py

.

2 S dng mnh 4 hoc mnh 1 tnh tch phn1 Mnh 4.Ta chn ng ly tch phn n gin. Ri tnh trc tip

tch phn trn ng cong va chn.


p dng nh l bn mnh tng ng tnh tchh l i 2 II
http://find/


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phn ng loi 2 II

2 Mnh 1.Tm hm (x, y)theo cng thc (4.20) v (4.21) trang127:

(x, y) =

xx0

P(x, y0)dx+

yy0

Q(x, y)dy+C

hoc

(x, y) =x

x0

P(x, y)dx+y

y0

Q(x0, y)dy+C

trong (x0, y0) D. Khi ,B

A

Pdx+Qdy=(B) (A).


V d v p dng nh l bn mnh tng ng I
http://find/


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Trong cc cu(1) - (3), hy tnh cc tch phn ng loi 2

1

(3,2)(1,1)

(x+2y)dx+ydy(x+y)2

.

2

(2,3)

(0,0)

(xy2 +y)dx+ (x2y+x)dy.

3

(0,)

(0,0)

(ex sin y+2x)dx+ (ex cos y+2y)dy.

4

(1,2)(0,0)

exy(1+xy)dx+x2exydy.


V d v p dng nh l bn mnh tng ng II
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5

(2,2)

(1,0)

2x+1y2

x2 dx+

2y

x

dy.

6 Tm cc hng sa, bsao cho tch phn

I = L

y(1 x2 +ay2)dx+x(1 y2 +bx2)dy

(1+x2

+y2

)2

khng ph thuc vo ng ly tch phn. Hy tnh tch phn trnvia, bva tm trong trng hpLc im u lO(0, 0)v imcuiA(1, 1).


Din tch hnh phng
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Cng thc 3.1 ((4.19) trang 123)

Mt minDhu hn c bin l ng cong knLc din tch c tnhbi cng thc

S(D) =1

2L xdy ydx


Din tch hnh phng
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Cng thc 3.1 ((4.19) trang 123)

Mt minDhu hn c bin l ng cong knLc din tch c tnhbi cng thc

S(D) =1

2L xdy ydx

V d 3.1

Dng tch phn ng loi 2 tm din tch hnh trn bn knh R.


Tnh cng ca lc bin thin
http://find/


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Cng thc 3.2 (trang 117 dng 5

)

Mt lcF =Pi +Qj +Rk l di chuyn cht im t A n B theoqu oLc cng c tnh theo cng thc

W = L

F dr =

LPdx+Qdy+Rdz


Tnh cng ca lc bin thin
http://find/


52/89

Cng thc 3.2 (trang 117 dng 5

)

Mt lcF =Pi +Qj +Rk l di chuyn cht im t A n B theoqu oLc cng c tnh theo cng thc

W = L

F dr =

LPdx+Qdy+Rdz

V d 3.2

Hy tnh cng ca lcF = yi +xj + zk l di chuyn cht im trncung l xo x=cos t, y=sin t, z=t t im ng vi t=0n im ngvi t=2.


Tch phn mt loi 1
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53/89

Cho hm sf(x, y, z)xc nh trn mt cong Strong khng gian.


Tch phn mt loi 1
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54/89


Chia mt congS thnhnmin con, k hiu cc min con ny v dintch ca n l Si.


Tch phn mt loi 1
http://find/


55/89



Trn mi min con ny ta chn im Mity v lp tng

In =

n

i=1 f(Mi)Si.


Tch phn mt loi 1
http://find/


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In =

n

i=1 f(Mi)Si.

Chon sao cho max Si 0. Nu limn =Ihu hn khng

ph thuc vo cch chia mt congS

, cch chn cc imMi thIc gi l tch phn mt loi 1 ca hmf(x, y, z)trn mt congS.


Tch phn mt loi 1
http://find/


57/89




In =

n

i=1 f(Mi)Si.

Chon sao cho max Si 0. Nu limn =Ihu hn khng

ph thuc vo cch chia mt congS

, cch chn cc imMi thIc gi l tch phn mt loi 1 ca hmf(x, y, z)trn mt congS.

K hiu lI =S

f(x, y, z)dS.


Cch tnh tch phn mt loi 1
http://find/


58/89

Nu mt congSc phng trnhz=z(x, y)vi (x, y) D(Dl hnhchiu caStrn mt phngOxy) th yu t din tch mt

dS=

1+z2x +z2ydxdy.

S

f(x, y, z)dS=

D

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

1+z2x +z2ydxdy


http://find/


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Xc nh phng trnh mt congSv a v dngz=z(x, y). SuyradS=

1+z2x +z2ydxdy.

Xc nh hnh chiuDcaStrn mt phngOxybng cch xcnh ng bin caS.Thayz=z(x, y)vdSva tnh c trn vo biu thc

S

f(x, y, z)dS.

Tnh tch phn hai lp thu c.


Tch phn mt loi 1
http://find/


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V d 4.1

Tnh tch phn I =S

(x2 +y2)dS viS lna pha trnca mt cu

x2 +y2 +z2 =a2.


Tch phn mt loi 1
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V d 4.1

Tnh tch phn I =S

(x2 +y2)dS viS lna pha trnca mt cu

x2 +y2 +z2 =a2.


Tch phn mt loi 1
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V d 4.1

Tnh tch phn I =S

(x2 +y2)dS viS lna pha trnca mt cux2 +y2 +z2 =a2.


Tch phn mt loi 1
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V d 4.2

Tnh tch phn I =S

(x2

+y2

)dS viS lmt binca min

V :

x2 +y2 z 1. s:

22

+

2


Tch phn mt loi 1
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V d 4.2

Tnh tch phn I =S

(x2

+y2

)dS viS lmt binca min

V :

x2 +y2 z 1. s:

22

+

2


ng dng ca tch phn mt loi 1
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Din tch ca mt congSc tnh theo cng thc

S=S

1dS.

V d 4.3Tnh din tch ca phn mt paraboloid x2 +y2 z=0 nm pha dimt phngz=4. s:

(17

17 1)6

V d 4.4Tnh din tch ca phn mt phngx+2y+2z=5 c gii hn bihai mt trx=y2 vx=2 y2.


http://find/


66/89

Mt congSchm mt khi lngti im(x, y, z)l (x, y, z)thkhi lng ca mt l

m=S

(x, y, z)dS

V d 4.5Tnh khi lng ca paraboloid trn xoayz= 1

2(x2 +y2)vi0 z 1

bit rng mt khi lng ca n l (x, y, z) =z. s: 2(63+1)

15

V d 4.6Tnh khi lng ca na mt cu x2 +y2 +z2 =a2 viz 0bit rngmt khi lng ca n l (x, y, z) = z

a. s: a2


http://find/


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Bi tp 4.1

Tnh din tch ca phn mt paraboloid x2

+y2

z=0 trong cctrng hp sau:1 Phn pha di mt phngz=2.2 Phn nm gia hai mt phngz=2 vz=6.

Bi tp 4.2

Tnh din tch ca chm cux2 +y2 +z2 =2,z 0nm pha trongmt trx2 +y2 =1. s:2(22)

Bi tp 4.3

Tnh din tch ca ellip c ct t mt phng z=cx,cl hng s,bi mt trx2 +y2 =1.


http://find/


68/89

Bi tp 4.4

Tnh din tch ca phn mt x2

2z=0 nm pha trn tam gic cgii hn bi cc ng thngx= 3,y=0 vy=xca mt phngOxy.

Bi tp 4.5

Tnh din tch ca phn mt x2 2y 2z=0 nm pha trn tam gicc gii hn bi cc ng thng x=2,y=0 vy=3xca mtphngOxy.

Bi tp 4.6Tnh din tch ca chm cux2 +y2 +z2 =2 c ct bi mt nnz=

x2 +y2.


Tch phn mt loi 2
http://find/


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nh ngha 5.1 (Mt nh hng)

Mt nh hng l mt cong c trang b mt trng vector php tuynn v, bin thin lin tc trn n.


Tch phn mt loi 2
http://find/


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nh ngha 5.2 (Tch phn mt loi 2 - nh ngha 10 trang 135)

Tch phn mt loi 2 c dng:

I =S

Pdydz+Qdzdx+Rdxdy

CBGD L Hoi Nhn () TCH PHN NG V TCH PHN MTNgy 22 thng 10 nm 2015 41 / 53

http://find/


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Cng thc 5.1 ((4.33) trang 136)Xt mt congSc phng trnh z=z(x, y)vi(x, y) D( viDlhnh chiu caStrn mt phng Oxy). Khi ,

S Pdydz+Qdzdx+Rdxdy

= D

P.z

x

+Q.

z

y

+R

dxdy

vi du "+" tng ng tch phn ly theo pha trn ca

Sv du "

"

tng ng tch phn ly theo pha di caS.



V d 5 1
http://find/


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V d 5.1

Tnh I =S

xdydz+dzdx+xz2

dxdy viSl phn tm th nht camt cu x2 +y2 +z2 =1. Tch phn ly theo pha trn caS.



V d 5 1
http://find/


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V d 5.1

Tnh I =S

xdydz+dzdx+xz2




V d 5 1
http://find/


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V d 5.1

Tnh I =S

xdydz+dzdx+xz2



Tch phn mt loi 2

V d 5.2


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V d 5.2

Tnh I =S

(y z)dydz+ (z x)dzdx+ (x y)dxdy viSl bin cahnh nn gii hn bi z=

x2 +y2 v z=1. Tch phn ly theo pha

ngoi caS.


Tch phn mt loi 2

V d 5.2
http://find/


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V d 5.2

Tnh I =S

(y z)dydz+ (z x)dzdx+ (x y)dxdy viSl bin cahnh nn gii hn bi z=

x2 +y2 v z=1. Tch phn ly theo pha

ngoi caS.


Tch phn mt loi 2

V d 5.2
http://find/


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Tnh I =S (y z)dydz+ (z x)dzdx+ (x y)dxdy viSl bin ca

hnh nn gii hn bi z=

x2 +y2 v z=1. Tch phn ly theo phangoi caS.


nh l Gauss - Ostrogradski
http://find/


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nh l 5.1 ((8) trang 140)Nu P, Q, R l cc hm s lin tc cng cc o hm ring ca chngtrn min hu hnVth

S Pdydz

+Qdzdx+Rdxdy

=

V

Px

+Qy

+Rz

dxdydz

trong Sl bin ca minVv tch phn ly theo pha ngoi caS.


Cng thc Gauss - Ostrogradski

V d 5.3
http://find/


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Tnh tch phn I =S xydydz+yzdzdx+zxdxdy viSl bin ca hnh

chp gii hn bi cc mt x=0, y=0, z=0v x+y+z=1. Tchphn ly theo pha ngoi caS



V d 5.3
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Tnh tch phn I =S xydydz+yzdzdx+zxdxdy viSl bin ca hnh

chp gii hn bi cc mt x=0, y=0, z=0v x+y+z=1. Tchphn ly theo pha ngoi caS


http://find/


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V d 5.4

Tnh tch phn I = Sxdydz+ydzdx+zdxdy vi

Sl mt cu tm O,

bn knh a. Tch phn ly theo pha ngoi caS.



V d 5.5
http://find/


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V d 5.5

Tnh tch phn I =S

x3

dydz+y3

dzdx+z3

dxdy viSl na trn camt cu x2 +y2 +z2 =a2, tch phn ly theo pha trn caS.



V d 5.5
http://find/


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5 5

Tnh tch phn I =S

x3

dydz+y3

dzdx+z3

dxdy viSl na trn camt cu x2 +y2 +z2 =a2, tch phn ly theo pha trn caS.

CBGD L H i Nh () TCH PHN NG V TCH PHN MTN 22 th 10 2015 48 / 53

http://find/


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V d 5.6

Tnh tch phn I = Sx2dydz+y2dzdx+z2dxdy vi

Sl pha ngoi ca

bin hnh lp phng0 x a,0 y a,0 z a.


nh l Stokes
http://find/


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nh l 5.2 ((7) trang 139)Nu P, Q, R l cc hm s lin tc cng cc o hm ring ca chngtrn mt cong hu hnSth

L Pdx+Qdy+Rdz=

S

Ry Q

z

dydz+

Pz R

x

dzdx+

Qx P

y

dxdy

trong

Ll bin ca mt cong

S, hng ly tch phn mt v chiu ly

tch phn ng tun th quy tc vn nt chai.


nh l Stokes

V d 5 7
http://find/


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V d 5.7

Tnh tch phn I =L

(y+z)dx+ (z+x)dy+ (x+y)dz viLl giao

tuyn ca mt cu x2 +y2 +z2 =a2 v mt phng x+y+z=0. Tchphn ly ngc chiu kim ng h nhn t chiu dng ca trc Oz.


nh l Stokes

V d 5 7
http://find/


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V d 5.7

Tnh tch phn I =L

(y+z)dx+ (z+x)dy+ (x+y)dz viLl giao

tuyn ca mt cu x2 +y2 +z2 =a2 v mt phng x+y+z=0. Tchphn ly ngc chiu kim ng h nhn t chiu dng ca trc Oz.


Thng lng ca trng vector
http://find/


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Cng thc 5.2 ((4.31) trang 135)

Thng lng ca trng vector

F =P.i +Q.

j +R.

k i qua mt nh

hngS

l

=S

Pdydz+Qdzdx+Rdxdy

CBGD L H i Nh () TCH PHN NG V TCH PHN MTN 22 h 10 2015 52 / 53
http://find/


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HT CHNG 4

C G () C G C 22 10 201 3 / 3
http://find/

Documents

01 Slide - Chuong 3 Tich Phan Duong Va Mat