高數(shù)課件同濟(jì)版上下冊(cè)

1第二節(jié)

對(duì)坐標(biāo)的曲線積分curvilinear

integral問題的提出對(duì)坐標(biāo)的曲線積分的概念coordinates對(duì)坐標(biāo)的曲線積分的計(jì)算兩類曲線積分之間的關(guān)系小結(jié)思考題作業(yè)第十章曲線積分與曲面積分一、問題的提出實(shí)例變力沿曲線所作的功L

:

A

fi

B常力沿直線所作的功W

=

F

ABF

(

x,

y)

=

P(

x,

y)i

+

Q(

x,

y)

j分割A(yù)

=M0

,M1

(x1

,y1

),,Mn-1

(xn-1

,yn-1

),Mn

=BMi

-1

Mi

=

(Dxi

)i

+

(Dyi

)

j

對(duì)坐標(biāo)的曲線積分

OxyM2Mn-1M1A

=

M0B

=

MnMi

-1

L2DxiiDyF

(xi

,hi

)Mii

=1n?

[P(xi

,hi

)

Dxi

+

Q(xi

,hi

)

Dyi

]nii

=1ni求和

W

=

DW取F

(xi

,hi

)

=

P(xi

,hi

)i

+

Q(xi

,hi

)

j取近似

DWi

?

F

(xi

,hi

)

Mi

-1

Miii

i

i

i

i即

DW

?P(x

,h

)Dx

+

Q(x

,h

)Dy近似值精確值取極限

W

=

lim[P(xi

,hi)

Dxi

+

Q(xi

,hi

)

Dyi

]lfi

0i

=1

對(duì)坐標(biāo)的曲線積分

Mi

-1

Mi

=

(Dxi

)i

+

(Dyi

)

jF

(xi

,hi

)

B

=

MOxyA=

M02Mn-1M1MnMi

-1

LDxii3DyMi二、對(duì)坐標(biāo)的曲線積分的概念1.

定義設(shè)L為xOy面內(nèi)從點(diǎn)A到點(diǎn)B的一條有向光滑

曲線弧,函數(shù)P(x,y),Q(x,y)在L上有界.用L上的點(diǎn):M1

(

x1

,

y1

),Mn-1

(

xn-1

,

yn-1

)把L分成n個(gè)有向小弧段Mi-1

Mi

(i

=1,2,,n;M0

=

A,

Mn

=

B).設(shè)Dxi

=

xi

-

xi

-1

,

Dyi

=

yi

-

yi

-1

,點(diǎn)(xi

,hi

)為M

i

-1

Mi

上任意取定的點(diǎn).

對(duì)坐標(biāo)的曲線積分

4n如果當(dāng)各小段長(zhǎng)度的最大值l

fi

0時(shí),

P(xi

,hi

)Dxi的極限總存在,則稱此極限為函數(shù)i

=1P(x,y)在有向曲線弧L上對(duì)坐標(biāo)x的曲線積分,或稱第二類曲線積分.記作L

P(x,y)dx,即nL

P(

x,

y)dx

=

lim

P(xi,hi

)Dxilfi

0

i

=1n類似地定義L

Q(x,y)dy

=lim

Q(xi

,hi

)Dyilfi

0

i

=1稱Q(x,y)在有向曲線弧L上對(duì)坐標(biāo)y的曲線積分.積分弧段 被積函數(shù)

對(duì)坐標(biāo)的曲線積分

5式存在條件當(dāng)P(x,y),Q(x,y)在光滑曲線弧L上連續(xù),第二類曲線積分存在.組合形式L

P(

x,

y)dx

+

L

Q(

x,

y)dy其中F

=

(

P

,Q

),=L

P(x,y)dx

+Q(x,y)dy

=L

F

ds

“點(diǎn)積”形ds

=

(dx,dy).

對(duì)坐標(biāo)的曲線積分

64.

物理意義⌒

=AB⌒

Pdx

+

Qdy⌒

變力F

=

P(

x,

y)i

+

Q(

x,

y)

j

沿AB所作的功WABW

=

⌒

F

ds

對(duì)坐標(biāo)的曲線積分

ds

=

(dx,dy)75.

推廣空間有向曲線弧Γ,ii

i

inP(x

,h

,z

)DxGlfi

0

i

=1P(

x,

y,

z)dx

=

limG

Pdx

+

Qdy

+

Rdzii

i

inQ(x

,h

,z

)DyGlfi

0

i

=1Q(

x,

y,

z)dy

=

limnR(x

,h

,z

)Dz

i

i

i

ilfi

0

i

=1R(

x,

y,

z)dz

=

limG

對(duì)坐標(biāo)的曲線積分

86.

性質(zhì)(1)如果把L分成L1和L2

,則(2)設(shè)L是有向曲線弧,-L是與L方向相反的P(

x,

y)dx

+

Q(

x,

y)dyLL1L2對(duì)坐標(biāo)的曲線積分與曲線的方向有關(guān).Pdx

+

Qdy

=LPdx

+

Qdy

+1L有向曲線弧,則L2-

L=

-L

P(

x,

y)dx

+

Q(

x,

y)dy

對(duì)坐標(biāo)的曲線積分

-

LLOPdx

+

QdyxyO9xy思想是

化為定積分計(jì)算.對(duì)坐標(biāo)的曲線積分與曲線的方向有關(guān).因此下限應(yīng)是起點(diǎn)的坐標(biāo),

上限是終點(diǎn)的坐標(biāo).三、對(duì)坐標(biāo)的曲線積分的計(jì)算

對(duì)坐標(biāo)的曲線積分

10

y

=y

(t

)由a變到b時(shí),點(diǎn)M

(x,y)從L的起點(diǎn)A沿L運(yùn)動(dòng)到終點(diǎn)B,j

(t

),y

(t

)在以a及b為端點(diǎn)的閉區(qū)間上具連續(xù),L的參數(shù)方程為

x

=j

(t

),當(dāng)參數(shù)t單調(diào)地定理設(shè)P(x,y),Q(x,y)在曲線弧L上有定義且=LP(

x,

y)dx

+

Q(

x,

y)dyy

(t

)]P[j

(t

),j

(t

)dt

+

Q[j

(t

),y

(t

)]y

(t

)dt2

(t

)?0,則有一階連續(xù)導(dǎo)數(shù),且j

2

(t

)+y曲線積分L

P(x,y)dx

+Q(x,y)dy存在,且ab=

b{P[j

(t

),y

(t

)]j￠(t

)

+

Q[j

(t

),y

(t

)]y

￠(t

)}dtLP(

x,

y)dx

+

Q(

x,

y)dy

對(duì)坐標(biāo)的曲線積分

11特殊情形(1)L

:y

=y(x)

x起點(diǎn)為a,終點(diǎn)為bL

P(

x,

y)dx

+

Q(

x,

y)dyL

P(

x,

y)dx

+

Q(

x,

y)dy(2)L

:x

=x(y)

y起點(diǎn)為c,終點(diǎn)為d則ba￠{P[

x,

y(

x)]

+

Q[

x,

y(

x)]

y

(

x)}dx={P[

x(

y),

y]x

(

y)

+

Q[

x(

y),

y]}dydc￠=則對(duì)坐標(biāo)的曲線積分,12L

:

y

=y

(t

)

x

=

j

(t

)a=

{P[j

(t

),y

(t

)]j￠(t

)

+

Q[j

(t

),y

(t

)]y

￠(t

)}dtL

P(

x,

y)dx

+

Q(

x,

y)dybz

=

w

(t

)G

:

y

=y

(t

),

x

=

j

(t

)G

P(

x,

y,

z)dx

+

Q(

x,

y,

z)dy

+

R(

x,

y,

z)dz(3)推廣t起點(diǎn)a

,終點(diǎn)b{P[j

(t

),y

(t

),w

(t

)]j

￠(t

)=ba+

Q[j

(t

),y

(t

),w

(t

)]y

(t

)+

R[j

(t

),y

(t

),w

(t

)]w

(t

)}dt

對(duì)坐標(biāo)的曲線積分

13例計(jì)算L2xydx,其中L為拋物線

y

=

x上B(1,1)y2

=

x從A(1,-1)到B(1,1)的一段弧.解(1)化為對(duì)x的定積分y

=

–

xxydx

=Lx(-

x

)dx=10=

232x

dx

=45AOxydx

+⌒OBxydx⌒+x

xdx1010

對(duì)坐標(biāo)的曲線積分

O14xA(1,-1)y(2)化為對(duì)y的定積分x

=

y21-1y2dx

=2

ydy,y從-1到1=

45L

xy

d

x

=y

2

ydy

對(duì)坐標(biāo)的曲線積分

計(jì)算L2xydx,其中L為拋物線

y

=

x上從A(1,-1)到B(1,1)的一段弧.OyxA(1,-1)15B(1,1)y2

=

x其中Γ是由點(diǎn)A(1,1,1)到點(diǎn)B(2,3,4)的直線段.1

2

3解直線AB的方程為x

-1

=y

-1

=z

-1化成參數(shù)式方程為x

=1

+t,y

=1

+2t,z

=1

+3tA點(diǎn)對(duì)應(yīng)t

=0,B點(diǎn)對(duì)應(yīng)t

=1,于是G

xdx

+

ydy

+

(

x

+

y

-

1)dz(6

+

14t

)dt

=

13=10例計(jì)算G

xdx

+ydy

+(x

+y

-1)dz

對(duì)坐標(biāo)的曲線積分

16例L2計(jì)算L是上半圓周y

=a

2

-x

2

,反時(shí)針方向;L是x軸上由點(diǎn)A(a,0)到點(diǎn)B(-a,0)的線段.解(1)中L的參數(shù)方程為x

dx

+(y

-x)dy,其中x

=

a

cos

t,

y

=

a

sin

tA點(diǎn)對(duì)應(yīng)

t

=

0,

B點(diǎn)對(duì)應(yīng)

t

=

p

.B(-a,0)

OA(a,0)

x原式==

-

2

a3

-

p

a23

2

對(duì)坐標(biāo)的曲線積分

y17y原式=ax

dx2(2)L的方程為y

=0,-ax從a到-a.3=

-

2

a3A(a,0x)B(-a,0)

O

對(duì)坐標(biāo)的曲線積分

18L2計(jì)算(2)L是x軸上由點(diǎn)A(a,0)到點(diǎn)B(-a,0)的線段.x

dx

+(y

-x)dy,其中

x2

+

y2

+

z2

.r

=

xi

+

yj

+

zk

,

r

=|

r

|=

對(duì)坐標(biāo)的曲線積分

例位于原點(diǎn)(0,0,0)處的電荷q產(chǎn)生的靜電場(chǎng)中,一單位正電荷沿光滑曲線Γ:x

=

x(t

),

y

=

y(t

),

z

=

z(t

),

a

￡

t

￡

b從點(diǎn)A移到點(diǎn)B,設(shè)A對(duì)應(yīng)t

=a

,B對(duì)應(yīng)根據(jù)庫(kù)倫定律,位于點(diǎn)M處的單位正電荷受到求電場(chǎng)所作的功W.的電場(chǎng)力r

,即解設(shè)點(diǎn)M(x,y,z)的向徑OM

=q

F

=

r

3

rAB19W

=

⌒

F

ds因此所求的功為=

qr

(

b

)r

(a

)2drrq

G3(

x2

+

y2

+

z2

)2a3(

x2

+

y2

+

z2

)2b

(

xx

+

yy

+

zz

)dt=

q

1

1

r(b

)

r(a

)=

q

-其中r(a

),r(b

)分別是點(diǎn)A和B到原點(diǎn)的距離.ds

=

(dx,dy,dz

)1(

x2

+

y2

+

z2

)22

x2

+

y2

+

z2(

xx

+

yy

+

zz

)dt=

對(duì)坐標(biāo)的曲線積分

r

r

=

xi

+

yj

+

zkrq

3F

=W

=此G

例F表d明s

=,靜G電r場(chǎng)3

r電場(chǎng)ds力r作=功|

r只|=與x正2

電+荷y2

運(yùn)+z220動(dòng)的起點(diǎn)和終點(diǎn)的位置有關(guān),而與運(yùn)動(dòng)的路徑無(wú)關(guān).=凡q是具xd有x這+種yd特y

+性zd的z

力場(chǎng)d,r稱=保2(守xd力x

場(chǎng)+y.dy

+zdz)補(bǔ)充在分析問題和算題時(shí)常用的對(duì)稱性質(zhì)對(duì)坐標(biāo)的曲線積分,當(dāng)平面曲線L是分段光滑的,

關(guān)于

x

軸對(duì)稱,

L在上半平面部分與下半平面部分的走向相反時(shí),則P(x,

y)為y的偶函數(shù)

P(x,

y)為y的奇函數(shù)1P(

x,

y)dxL其中L1是曲線L的上半平面的部分.類似地,

對(duì)L

Q(x,y)dy的討論也有相應(yīng)的結(jié)論.

對(duì)坐標(biāo)的曲線積分

21L

P(

x,

y)dx

=

2

0ABCDA

|

xy

|

+1例計(jì)算dx

+dy,其中ABCDA為|

x

|

+|

y

|=1,解

法一

直接化為定積分計(jì)算,

-

x

+

y

=

1ABdx

+

dy=+xy

+

1

BC

-

xy

+

1DA

-

xy

+

1+xy

+

1dx

+

dy+CDBC

CD

DAdx

+

dy取逆時(shí)針方向.由曲線積分的性質(zhì).則+

+

+

A(1,0)B(0,1)x

+

y=

1C(-1,0)-

x

-

y

=

1x

-

y

=

1D(0,-1)dx

+

dy

對(duì)坐標(biāo)的曲線積分

=

0

dx

-

dx

+

-1

dx

+

dx

+

0

dx

-

dx1

x(1

-

x)

+

1

0

-

x(1

+

x)

+

1

-1

x(-1

-

x)

+

10

1

dx

+

dx

0+

0

x(1

-

x)

+

1=ABCDA-10=

2dx-

x(1

+

x)

+

11+

20

x(1

-

x)

+

1dxx

=

-t

0=

-10dx

+

dx-

x(1

+

x)

+

11+

0

x(1

-

x)

+

1dx

+

dxO22xyABCDA

|

xy

|

+1dx+ABCDA

|

xy

|

+1ABCDA

|

xy

|

+1dx法二利用對(duì)稱性質(zhì),將原式分成兩部分,即原式=對(duì)曲線關(guān)于x軸對(duì)稱,L在上半部分的走向與L在下半部分的走向相反,-

x

-

y

=

1dyx

+

y

=

1A(1,0)B(0,1)-

x

+

y

=

1C(-1,0)D(0,-1)=

0ABCDA

|

xy

|

+1被積函數(shù)為y的偶函數(shù).

dx

對(duì)坐標(biāo)的曲線積分

ABCDA

|

xy

|

+1dx

+

dy計(jì)算O23xx

-

y

=

1yABCDA

|

xy

|

+1dy對(duì)曲線關(guān)于y軸對(duì)稱,L在右半部分的走向與L在左半部分的走向相反,被積函數(shù)為x的偶函數(shù).=

0ABCDA

|

xy

|

+1dy|

xy

|

+1ABCDAdx

+

dy

=

0所以,

對(duì)坐標(biāo)的曲線積分

ABCDA|

xy

|

+1dx

+

dy計(jì)算ABCDAdx|

xy

|

+1

=

0-

x

-

y

=

1-

x

+

y

=

1C(-1,0)D(0,-1)B(0,1)x

+

y

=

1A(1,0)O24xx

-

y

=

1yL

Pdx

+

Qdy

=

L

(

P

cosa

+

Q

cos

b

)dscosa

=

j

(t

)

, cos

b

=j

￠2

(t

)

+y

￠2

(t

)j

￠2

(t

)

+y

￠2

(t

)y

(t

)

y

=y

(t

)L上點(diǎn)(x,y)處的切線向量的方向角為a

,b

,則

對(duì)坐標(biāo)的曲線積分

dx

=

j

(t

)dt,

dy

=y

(t

)dt,ds

=

j￠2

(t

)

+y

￠2

(t

)dt四、兩類曲線積分之間的關(guān)系設(shè)有向有向曲平線弧面L曲的線切弧向?yàn)榱縇為：

x

=j

(t

),25G

Pdx

+

Qdy

+

RdzGA

dr

=GAt

ds可用向量表示A

=

(

P,

Q,

R)t

=

(cosa

,

cos

b

,

cosg)dr

=t

ds

=(dx,Gd上y,d點(diǎn)z()x