高數(shù)a491對(duì)弧長(zhǎng)曲線積分

積分學(xué)積分學(xué)定積分二重積分三重積分曲線積分

第9引例:曲線形構(gòu)件的質(zhì) 弧段為AB其線密度為采用

,zk

MkMk-1 M

nknG是空間中一條有限長(zhǎng)的光滑曲線,G上的一個(gè)有界函數(shù),G的任意分割和對(duì)nlim

f

)Dsk

Gf(x,y,z)lfi0k

MkG上對(duì)弧長(zhǎng)的曲線積分或第一類曲線積分.稱為G稱為MGr(xyz

Mk-G如果L是xoy面上的曲線弧 f(x,y)ds= lfi0kLLf(xyds在L上f(x, 否ds0,dx可能為負(fù)(1)G[f(x,y,z)–g(x,y,z)=Gf(x,y,z)ds–Gg(x,y,z)（k為常數(shù)1212

Gf(x,y,z)

f(x,y,z)ds+

f(x,y,z)G由 f(x,y,z)￡g(x,y,

則Lf(xyz)ds￡Lg(x,y 求曲線積

f(x,y)ds=

f[j(t),y(t)]j2(t)

2(t)dL證:

nnlfi0k點(diǎn)(xk,hk)對(duì)應(yīng)參數(shù)為j2j22 kj2j2)+y2=n則nlfi0k

Dtkj2j22nlfi0k(1)Dsk0Dtk0,因此積分限必須滿足ab(2)ds=(dx)2+(d =j2(t)+y2(t)d

d 2a=f(x,y(x)) a如果方程為極坐標(biāo)形式Lrr(qa￡q￡b),b a

r2(q)+r2(q)推廣G:x=f(t),y=y(t),z=w(t)(a￡t￡b則Gf(x,yj2(t)+y2(t)+w2=bfj2(t)+y2(t)+w2a

d例1.計(jì)算其中L是拋物B(1,1).

Oyy=yy=L例1.計(jì)算其中L是拋物B(1,1).解:L:y= (0￡x￡1)

Oyy=Lyy=L111+4x21+4x232 1320 05=1(55

1)例2.計(jì)算半徑為R,中心角為 稱軸的轉(zhuǎn)動(dòng)慣量I(設(shè)線密度m=1).LL o例2.計(jì)算半徑為R,中心角為 稱軸的轉(zhuǎn)動(dòng)慣量I(設(shè)線密度m=1).L 解:建立坐標(biāo)系如圖,L I=

y2oL:x=Ry=R

(-a￡q￡a=aR2

sin2q(-Rsinq)2+(Rcosq)2

3qsin2q=R-asinqdq=2R2

例3.計(jì)算曲線積分

其中G為螺旋例3.線

解:G(x

+z2)=

2p[a2+k2t2]da2a2+ka2+ka2+k3

(3a2+4p2k2例4.Craq0qp所圍區(qū)域的邊界

y= r= py= a例4.Craq0qp所圍區(qū)域的邊界

y= r= p提示

y= a 若L1與L2關(guān)于x軸對(duì)

L=L1+f(x-y)f(x,y時(shí)Lf(x,y)ds1f(x-y)f(x,y時(shí)Lf(x,y)ds1

f(x,若L1L關(guān)于y2f(-x,y)f(x,y)時(shí)Lf(x,y)ds1f(-xy)f(x,y)時(shí)Lf(x,y)ds1

f(x,若L1與L2f(-x-y)f(x,y)時(shí)Lf(x,y)ds1f(-x-y)f(x,y時(shí)Lf(x,y)ds1

f(x,例5.計(jì)算其中L為雙紐(x2+y2)2=a2(x2-y2 (a>0)yyox例5.計(jì) (x2+y2)2=a2(x2-y2解:在極坐標(biāo)系下L1:r= cos2q(0￡q￡p4

(a>0)y ,

f(x,y)ds

f(-x,r2(q)r2(q)+r2(qp

4r 04a2cosqdqy3o例6.L:x2y2=y3o (2xy+3x2+4y2)L例6.L:x2

=1周長(zhǎng)為a, (2xy+3xL

+4y2)提示L2xyds 12L(43)ds=12Lds上下上下

L2xyds= 2xy

y3y3o= +2x(-)例7.計(jì)算其中G為球面x2+y22z29與平面xz=1的交線2例7.計(jì)算其中G為球面x2+y22z29與平面xz=1的交線2 1(x-1)2+1y2=1

,2x+z=12x

G y=2z=12則

0ds

(-2sinq)2

+( 2sinq)2dq= I

2

例8.計(jì)算

例8.計(jì)算

解:由對(duì)稱性可 x2G

y2dsG

z2\

x2ds=3

(x

a2ds=1a22p3 3思考例8中G改為

X2+Y2+Z2=解:Yy+1,G

X+Y+Z== =G(X+1)23

+2GXds在原點(diǎn),故X0+2在原點(diǎn),故X0例9.有一半圓弧

y(x,- R例9.有一半圓弧

解:d

=kmdsR2

(x,2dFy=kmds - R2R

pqcosqdq=2kqsinq+cosqR

pqsinqdq=2k[-qcosq+sinqR F(4k,2kp 例10.設(shè)均勻螺旋形彈簧L的方程 例10.設(shè)均勻螺旋形彈簧L的方程 解:ρ(常數(shù)(1)

(x2+y2)rds

2pa2 da2+ka2+ka2+ka2+ka2+k(2LmLra2+k =

a2+kcosta2+k0=ar=kr

+k 0+k 00

sintdt=tdt=2p2kra2+k00kp第二 上 結(jié)定義Lf(x,ydsGf(x,y,z)Gaf(x,y,z)+bg(x,y,z)bLg(xyz)ds(ab為常數(shù)1Gf(x,y,z)ds=1

f(x,y,z)ds+

f(x,y,z)2則Lf(xyz)ds￡Lg(x,y2則Lf(xyz)ds￡Lg(x,y

(G由G1,G2組成Lf(x,y)ds

a

2(t)+y2(t)d2bLf(x,y)ds=b

f(x,y(x)) Lf(x,r2q=bf(r(q)r2