曲線曲面積分第十一章

第十一章線面積分(習(xí)題課)題組一：曲線積分的計(jì)算題xds

,1.

求I

=L其中L

為曲線x

=y2從點(diǎn)(0,0)到點(diǎn)(1,1)的一段.解:xyoA(1,1)120I

=y

1+

4

y

dy12=

1

(5 5

-1)計(jì)算其中L為圓周提示:利用極坐標(biāo),ds

=原式

=

LOa

xyrr

2

+

r￠2

dq

=

a

dqax

dst說明:若用參數(shù)方程計(jì)算,則x2

+

y

2

d

td

s

=2.

計(jì)算4343LI

=2(x

+

y

)ds,2

2其中

L

為星形線

x

3

+

y

3

=

a

3

(a

>

0).yxo2

2

2解:

L

:

x

3

+

y

3

=

a

3

(a

>

0)x

=

a

cos3

q(0

￡q

￡

2p

)L

:

y

=

a

sin3

q4343I

=(x

+

y

)ds0Lp2=

444(a

cos3

q)3+

(a

sin3

q)3dq

dq(

dx

)2

+

(

dy

)2

dq734

40p2(cos

q

+sin

q)

|

sinq

cosq

|

dq=12a7=

4a

33.

計(jì)算其中L為雙紐線(

a

>

0

)(x2

+

y2

)

2

=

a2

(x2

-

y2

)解:在極坐標(biāo)系下它在第一象限部分為cos

2q

(0

￡q

￡

p

4

)L1

:

r

=

a利用對稱性

,

得40pr

sinq=

4r2

(q

)

+

r￠2

(q

)

dq420pa

sinq

dq=

4yox機(jī)動目錄上頁下頁返回結(jié)束4.

計(jì)算其中G為球面x2

+y2解:22r

=9

-(|

-1

|)2=

2I

=

9

ds

=

9

·

2p

·

2

=18p2

G

22+z

2

=9

與平面x

+z

=1的交線.G

的半徑為則機(jī)動目錄上頁下頁返回結(jié)束5.

計(jì)算

I

=

L

(x

-

y)dx

+

(

y

-

x)dy,

其中

L

是沿逆2

2時(shí)針方向進(jìn)行的以原點(diǎn)為中心,

a

為半徑的上半圓周.解:yxoaBA補(bǔ)線段如圖.BA所以

(

x2

-

y)dx

+

(

y2

-

x)dyLBAI

=

22-

x)dy-

(x

-

y)dx

+

(

yBAD=ax2

dx-a0dxdy

-323a=

-D6.

計(jì)算I

=

L

(12xy

+

e

)dx

-(cos

y

-

xe

)dy,y

y到點(diǎn)O(0,0),解:yxoB(2,0)A(-1,1)其中曲線L

由點(diǎn)A(-1,1)沿曲線y

=x2再沿直線y

=0

到點(diǎn)B(2,0)

的路徑.

補(bǔ)線段

BC

,

CA

如圖.C(2,1)于是I

=BCCA-

-

LBCCAD=10y(-12x)dxdy

-2(12x-1(2e

-

cos

y)dy

-+

e)dx120-

y=

-

dy12xdx+20

+

e

+sin1=

-1

+

e

+sin17.

求I

=

L

(3x

-

3x y

+1)dx

-(x

-

y)dy,5

2

3其中L

是r

=1+cosq自點(diǎn)A(0,2)

到點(diǎn)B(p,0)

的一段弧.解:yoA(2,0)

xB(0,

0)注意題目所給A

,B兩點(diǎn)坐標(biāo)為極坐標(biāo).因?yàn)?x?Q

=

-3x2

=

?P?y所以積分與路徑無關(guān).于是I

=5

23(3x

-

3x y

+1)dx-(x

-

y)dyAB052=(3x

+1)dx=

-34

ydx

-(x

-1)dy

,8.

求I

=(x

-1)2

+

y2

L其中:(1)

L

為圓周x2

+y2

-2

y

=0的正向;(2)

L

為橢圓

4x2

+

y2

-8x

=

0

的正向;解:(1)L

:

x2

+

y2

-

2

y

=

0L

:

x2

+

(

y

-1)2

=1所以點(diǎn)(1,0)不在L

所圍區(qū)域D

內(nèi),于是I

=(D?Q

?P?x

?yD0dxdy-

)dxdy

==

0.(2)L

:

4x2

+

y2

-8x

=

024y2L

:

(x

-1)

+=1所以點(diǎn)(1,0)在L

所圍區(qū)域D

內(nèi),補(bǔ)曲線dL

:

(x

-1)2

+

y2

=

d2

,于是I

=d

ydx

-(

x

-1)dy(

x

-1)2

+

y2LLd

ydx

-(x

-1)dy=

0

-(x

-1)2

+

y2LdL

:

x

=1+d

cosq

(q

:

2p

fi

0)

y

=

d

sinq02pdq=

-=

-2p.d

ydx

-(x

-1)dy-(x

-1)2

+

y2Lyy2x(x2

+

y2

)l

x(x2

+

y2

)ldx

-

dy9.

確定參數(shù)l

的值,使得在不經(jīng)過直線y

=0

的區(qū)域上是某個(gè)函數(shù)u

(x,y)的全微分,并求出u

(x,y).解:

?Q=?x222-2x(x2

+

y2

)l-1(x

+

y

+

lx

)?P

=?yy2y2x(x2

+

y2

)l-1(2ly2

-

x2

-

y2

)由?x?Q

=

?P?y得2l

=

-

1取積分路徑如圖.yxo(0,y)(0,1)(x,y)u(x,

y)

=(0,1)(

x,

y

)12122

222dyyy--x(

x2

+

y2

)x

(

x

+

y

)dx

-22

2dxy-1x

x(x

+

y

)=

0yx2

+

y2=+

C2L上任意一點(diǎn)的線密度為r(q)=sec

q

,0

￡

q

￡

p

),求:(1)曲線段的弧長;(2)曲線段的重心;解:L

:

r

=

a(1+

cosq)a2

(1+

cosq)2

+

a2

sin2

qdq22a

1+

cosqdq

=

2a

|

cos

q

|

dq題組二：線積分的應(yīng)用題和證明題1.

已知曲線

L

的極坐標(biāo)方程為

r

=

a(1

+

cosq)

(a

>

0,L20pqcos

dq(1)

s

=

ds

=

2a=

4a.ds

=

r2

+

r￠2

dq

==(2)

m

=L220r(q)ds

=p

q

qsec 2a

cos

dq

=

2ap.yLxr(q)dsm

=0p=a(1+

cosq)

cosq

sec

q

2a

cos

q

dq2

20p=a(1+

cosq)

cosq

2adq=

pa2

.xLyr(q)dsm

=0p=a(1+

cosq)

sinq

2adq2=

4amya所以

x

=

=

,m

2mx2aa

2ay

=

=

.

故重心坐標(biāo)為G(

,

).m

p

2

p3.

設(shè)曲線積分L

F

(x,y)(ydx

+xdy)與積分路徑無關(guān),且方程F(x,y)=0

所確定的隱函數(shù)的圖形過點(diǎn)(1,2)(其中F(x,y)是可微函數(shù)),求由F(x,y)=0

所確定的曲線.解:設(shè)

P

=

F

(x,

y)

y,

Q

=

F

(x,

y)x,則?P

=

?Q?y

?xF

+

yFy

=

F

+

xFxyF

xFx

=

y又F

(x,

y)

=

0dy

=

-

Fxdx

Fydy

=

-

yxdy

+

ydx

=

0d

(xy)

=

0dx

xxy

=

C又知曲線過(1,2)C

=

2.故所求曲線為xy

=2.4.

設(shè)f

(x)在(–￥

,+￥)上具有一階連續(xù)導(dǎo)數(shù),且f

(1)=1,對于任何閉曲線

L都有

L

4x ydx

+

xf

(x)dy

=

0,求

f

(x).3解:設(shè)

P

=

4x3

y,

Q

=

xf

(x).由?P

=?Q?y

?x4x3

=[xf

(x)]x4

+

C

=

xf

(x)f

(1)

=1C

=

0故所求函數(shù)為f

(x)

=

x3

.6.

設(shè)

L

是圓周(x

-1)2

+(

y

-1)2

=1,取逆時(shí)針方向,又Lf

(x)xf(

y)dy

-

y

dx

?

2p.f

(x)為正值連續(xù)函數(shù),證明:解:yo1

x1左=應(yīng)用格林公式有

1

Df

(x)[

f

(

y)

+

]dxdy積分區(qū)域D具有輪換對稱性D

f

(

y)dxdy

=

D

f

(x)dxdyD]dxdy

1

f

(x)=[

f

(x)

+Df

(x)

1

dxdyf

(x)?

2=

2D

dxdy

=

2p.題組三：曲面積分的計(jì)算與應(yīng)用S1.

計(jì)算I

=1+4zdS,其中是z

=x2

+y2

上z

￡1

部分.解:xozy1dS

=

1

+

z

2

+

z

2

dxdy,x

yD

:

x2

+

y2

￡1.xyI

=

Dxy1+

4(x2

+

y2

)1+

4(x2

+

y2

)dxdy12002pdq=(1+

4r

)rdr=

3p.xoy2.

設(shè)

:

x2

+

y2

+

z

2

=

a2f

(x,

y,

z)

d

S

.計(jì)算I

=a2

-x2

-y

2

的2解:錐面z

=x2

+y2

與上半球面z

=交線為設(shè)1為上半球面夾于錐面間的部分,它在xoy

面上的z

1x

yD投影域?yàn)?/p>

Dxy

={(x,

y)

x2

+

y2

￡

1

a2

},

則I

=

(x

+

y

)

d

S2

211I

=

(x2

+

y

2

)

d

S=

Dx

y(x2

+

y

2

)2

aar

20120

r

dra2

-

r

2=2pdq6=

1

p

a4

(8

-

5

2)aa2

-

x2

-

y

2d

x

d

yxoy1z

x

yD接2.l3.

證明:若是任一分片光滑的閉曲面,

為一固定方向,S和l

的夾角.證明:設(shè)n

=(cosa

,cos

b,cosg),(常向量)則Scosq

d

S

=0cos(n,l

)

d

SS=

S

(cosa

cosa

￠+

cos

b

cos

b￠+

cosg

cosg￠d

SGauss公式=

0.則

cosqdS

=0,其中q

是上任一點(diǎn)的外法線n32S4.

計(jì)算I

=(x2

+

y2

+

z2

)x

cosa

+y

cos

b

+z

cosgdS,

其中是球面x2

+y2

+z2

=R2

,a

,b,g

是其外法線的方向角.解:33d

x

d

y

d

zRW=

1Gauss公式5.

用三種方法計(jì)算I

=

xdydz

+ydzdx

+(z2

-2z)dxdy,S其中是介于z

=1

與z

=2

之間的錐面z

=x2

+

y2xoyz21部分的上側(cè).解:

方法1:Gauss公式補(bǔ)S1

:z

=1上側(cè)，S2

:z

=2下側(cè).SS1

S2

S1

S2I

=

-

-

=

-W

2zdxdydz

-x2

+

y2

￡1

(-1)dxdy

-(-x2

+

y2

￡4

0dxdy)21zD2zdzdxdy=

-

+p2212zpz

dz=

-.213p=

-+p方法2:

利用第一類曲面積分轉(zhuǎn)化xo接5.-1zy21D

:1

￡

x2

+

y2

￡

4.xyS

:

z

=

x2

+

y2

,cosa

dS

=

dydzcos

bdS

=

dzdxcosgdS

=

dxdycosgdydz

=

cosa

dxdycosgdzdx

=

cos

b

dxdy2xy-zxcosa

=1

+

z

2

+

z2xy-zycos

b

=1

+

z

2

+

z21xycosg

=1

+

z

2

+

zcosa

=

-zxcosgcosacosg

=

-zydydz

=

-zxdxdyydzdx

=

-z

dxdy接5.-2S

:

z

=x2

+

y2

,D

:1

￡

x2

+

y2

￡

4.xyxozy21SI

=

xdydz

+

ydzdx

+(z2

-

2z)dxdydydz

=

-zxdxdydzdx

=

-zydxdy=

x(-zxdxdy)

+

y(-zydxdy)

+(z

-

2z)dxdy2S=(-xz

-

yz

+

z2

-

2z)dxdyx

ySDxy(x2

+

y2

-

3

x2

+

y2

)dxdy=

22012pdq=132p(r

-

3r)rdr

=

-接5.-3方法3:

直接計(jì)算xozy21由對稱性可知S

S

xdydz

=

ydzdx取1S

:

x

=

z2

-

y2

,Dyz

:1

￡

z

￡

2,-z

￡

y

￡

z.SxdydzS1

xdydz

=

2Dyzz2

-

y2

dydz=

-212

zdz-z=

-2z2

-

y2

dy73p.=

-取

S

:

z

=

x2

+

y2

,D

:1

￡

x2

+

y2

￡

4.2S(z

-

2z)dxdy

=Dxyxy(x2

+

y2

-

2x2

+

y2

)dxdy22012pdq=(r

-

2r)rdr)p=

(15

-

282

3所以I

=3-

2

·

7

p

+

1528

13(

-

)p

=

-

p.2

3

232Sxdydz

+

ydzdx

+

zdxdy

,6.

計(jì)算I

=(x2

+

y2

+

z2

)其中為球面

x2

+

y2

+

z2

=1

的外側(cè).為任一不經(jīng)過原點(diǎn)的閉曲面的外側(cè).解

(1)

:

I

=

S

xdydz

+

ydzdx

+

zdxdy

=

3

dxdydz

=

4p.W(2)

:(a)曲面不包圍原點(diǎn)(0,0,0).利用Gauss公式得:W?x

?y

?zI

=

(

?P

+

?Q

+

?R

)dxdydz

=

0.(b)曲面包圍原點(diǎn)(0,0,0).2

2

2

2做一足夠小的球面

Se

:

x

+

y

+

z

=

e

,

并取其內(nèi)側(cè),則在S

Se上可用高斯公式.接6.于是32I

=

()

xdydz

+

ydzdx

+

zdxdy-(

x2

+

y2

+

z2

)

SSe

SeW?x

?y

?z=

(

?P

+

?Q

+

?R

)dxdydz3eSe-

1xdydz

+

ydzdx

+

zdxdy33dxdydze=

0

-

(-1)

1W

e=

4p.S7.

計(jì)算I

=

x(8

y

+1)dydz

+2(1

-y2

)dzdx

-4

yzdxdy,z

=

y

-1其中是由曲線x

=

0(1

￡

y

￡

3)繞y

軸旋轉(zhuǎn)一周而得的曲面,它的法向量與y

軸正向的夾角大于2p

.解:旋轉(zhuǎn)曲面為S

:y

-1

=x2

+z2

,xzD

:

x2

+

z2

￡

2.S補(bǔ)0S32

2S0

:y

=3,(x

+z

￡

2),

取右側(cè).則S0)x(8

y

+1)dydz

+

2(1

-

y2

)dzdx

-

4

yzdxdyI

=

(-SS0W=S02(1

-

y2

)dzdx1dxdydz

-31Dydy=Dxz16dzdxdxdz

+31=p

(

y

-1)dy+32p=

34pyzo

1x8.

求在圓柱面

x2

+

y2

=

ay(a

>

0)