數(shù)學(xué)分析簡明教程答案

第二十一章曲線積分與曲面積§1一型曲線積分與曲面積（線性性）L

f(x,y,z)ds,

g(x,y,z)dsk1k2Lk1f(xyzk2g(xyz)dsLk1f(x,y,z)k2g(x,y,z)dsk1Lf(x,y,z)dsk2Lg(x,y,z)ds 1dsl,其中l(wèi)L的長度L(可加性)LLLLL

f(x,yz)ds存在 12Lf(xyz)ds12

f(x,y,z)ds1Lf(x,y,z)ds1

f(x,y,z)ds+

2f(x,y,z)ds2單調(diào)性

f(xyz)ds

g(x,yz)ds均存在，且在L上的每一點(diǎn)p都有fpgpLfp)dsLgp)ds

fp)ds存在，則

fpdsLf(p)ds

f(p)（中值定理）LfpLp0LLfp)dsfp0)l lL的長度Sfp)dsgp)ds （線性性）設(shè)kk是實(shí)常數(shù)，則kfpkgp)ds存在, Sk1f(p)k2g(p)dsk1f(p)dsk2g(p)ds 1dss,sS的面積S(可加性)SS1S2SS1S2S1S2fp)ds存在Sfp)dsfp)ds f(p)ds=f(p)ds+f( (單調(diào)性)Spfpgpf(p)dsg(p)ds 5

fpdsfp)ds

f(p)ds （中值定理）fpSp0S,fp)dsfp0)ssSS

(x2y2ds,L是以(0,02,00,1為頂點(diǎn)的三角形LLL1L2L2:y

0y0xL:y1x

0x所以(x2y2ds

(x2y2)ds(x2y2)ds

(x2y2 1

5 2 x25y2dy

x2dxx21 5353

0 2

x2y2ds,Lx2y2ax

(aL的參數(shù)方程為

xaacos;yasin

0 則xasin,yacos

ds

x'2y'2da x2yx2y

222所 ds 2

d2

2a2

cos

d

d2a2

Lxyzds,其中L為螺線xacostyasintzbt(0ab

0t2解：xasintyacostzds

x'2y'2z'2dt

a2b2所以

xyzds2a2btcostsint0

a2b2aa2ba2b2

a2b2a2b2

(x2y2z2ds,L與(3)相同La2b2解 (x2y2za2b2L

0 0

(xL

y3)ds,Lx3y3a3解：L1的參數(shù)方程為:xacost,yasin 0 x'3acos2tsint,y'3asin2tcosds x'2y'2dt3asintcostdt,4所 (xL

y3)ds=

(x

y373a32(1cos22t)sin2tdt0

L

xa(tsint),ya(1 解：xa(1costyasintds t 所以y2ds2a3(1cost) dt

xyds,Lx2y2z2a2xyz0的交線LLx,yz的對稱性,LL所以 1(xyyz 31(x2y2z2)ds6解(xyyzzx)ds1(xyz2x2y 2aa2dsaL

xyzds,Lxt,y 1

2t3,z1t2(0t12解：x1y16所以ds16

2t2,z'1x'2y'2z'2dt11Lxyzds=01

2t2

1t2(1t)dt2

(10)

2y2z2ds,Lx2y2z2a2xy相交的圓周zz

x

， 2ay 2a2a22yds

x'2y'2z'2dy

ya22a22

所以

2y2z2ds2a 0

=2a

20

（y

2asin）a22ya22y2

(x2y2dsSxx2y

z1的邊界曲面(x2y2ds(x2y2ds(x2y2 其中S1z

x2y2x2y21，而Sz1x2y212 所以SSxoyDx2y22 11z 2對

(x2y2)ds 1z (x21z 所以(x

y

)ds(xD

y2

2dxdy

20d0

3dr21(x2y2)ds(x2y2)dxdy2d1r3dr21

所以(x2y2)ds(1 2) ,Sx2y2R2z0zH所截取的部分;x2yS解：前半柱面S1的方程為x Ry,RyR,0z R2y1x2xyR2y1x2xyzR2yyRS2x

R2y2

Ry

0zR2y1x2xR2y1x2xyzR2yyR x2y

x2y

2+

x2y

1

2H2R2yR2yR2yRR S

x3y2zds,Szx2y2z1割下部分1z2z 11z2z 14(x2y2 所以S

x3y2zdsx3y2(x2y2)D

14(x2y22cos3sin2d1r

14r20 1r15

4r2

z2ds,S是螺旋面的一部分xucosS

yusinv,zv(0ua,0v2 zu cos 解：

zv

usin

ucos 所以Ex2y2z21,Gx2y2z2u2 Fxuxvyuyvzuzv因此z2dsv

EGF2dudv2v2

1u2 43

a2 aa2x2y2ds,Sx2y2z2RSSxRcossinyRsin ，02，0 zR z

Rsinsin

Rcossin

Rcos Rsin

Ex2y2z2Rnsin2，GR2，F(xiàn) 所 (x2y2)dsR2sin2R4sin R42dsin3d8R Lxetcost，yetsint，zet

(0tt0)解(xyzk(x2y2z2，由(1,0,11k2所以(xyz1(x2y2z22

1(

2 21t0(e2te2t)3e2t2 3t0e3tdt3(e3t0 設(shè)有一質(zhì)量分布不均勻的半圓弧xrcos，yrsin

(0a（為常數(shù)，求它對原點(diǎn)(0,0)處質(zhì)量為m的質(zhì)點(diǎn)的引力FxFxyFy.任取弧長微元ds

dFr

ds其中kr0是向經(jīng)的單位矢量cos,sinads代入，得dFx,ydFkard(cos)kacosd r dFkard(sin)kasin r ka F cosd 0 F kasindka F2F2F

2所 F的大小

2，方向?yàn)?

cos4,sin4cos(4sin4)x 求螺線的一支Lxacost

yasint，z

t(0t2)x軸的轉(zhuǎn)動(dòng)慣量I y2z2ds.設(shè)此螺線的線密度是均勻的L2IL(2

z

)ds

(a

h2a2h4t a2h442( 22

2h24a4a2hz1(x2y20z1z2解：Mdszds1(x2y2

x2y2dxdy12d2r4dr42 D

2

y2z2a2，x0,

y0z0解：由對稱性，設(shè)重心坐標(biāo)為(x,yz)(ttatxds L：x2y2a2，z0；L：y2z2a2，x0；L：xz2

，y xds2acosada xds20ad xds2acosada 所以：y2z2ds2a2L2

t2a

4a，故重心坐標(biāo)為(4

,a,

,a,

a)x2y2z2a2(z0z軸的轉(zhuǎn)動(dòng)慣量1x2y2z2a2(z0zIz(x2y2)ds

d2a2sin2a2sin S4a3a2x2y求均勻球面z (x0,y0,a2x2y解：由對稱性可設(shè)重心坐標(biāo)為(kkla2x2ya2x2y由于xdsa2x2ya2x2y 4

2(1 2a2x2yzds a2x2ya2a2x2ya2x2ya2x2y2ds dxdyaaa2x2ya2x2y2 所 kS

24D4l ds21 所以：重心坐標(biāo)為

24

22若曲線以極坐標(biāo)給出：()(12)，試給出計(jì)算Lf(x,y)ds的 L

x2y2dsLa(04xdsLaekv(k0)ra內(nèi)的部分(xcos()所 ysin()sin

1

()cos()(sin)(cos()sin()sin()所以

f(x,y)ds

f(()cos,()sin

2()2(（1）L：a，故ds 22dex2y2ds4eaadae （2）Laek(k0)raaekra的交點(diǎn)坐標(biāo)為(0所以dsaek1k20

1k2a1k2a2kk2所 xdsaek

1

4k0xrcosyrsin，zr(02,0bra對位于曲面頂點(diǎn)(0,0,0的單位質(zhì)點(diǎn)的引力〉當(dāng)b0時(shí)，結(jié)果如何？解：對應(yīng)于半徑r處取斜交為dsds2rds2它與頂點(diǎn)(0,0,0的單位質(zhì)點(diǎn)的引力在ox軸和oy軸上合力的射影顯見為0而在ozdZ

2rdr0 r2z2

rr2zr2zX

Y

Zbk0drkln 當(dāng)b0時(shí)，由于

a，故在Z坐標(biāo)軸上引力的射影趨于b13.F(t)f(xyz)dsSxyztS1x2y2z2 x2y2z2f(x,y,z)xyz

當(dāng)x2y2z2333x2y2z213331x2y2z2

若t

f(x,y,z)

若txyz

x2y2z2 1 1xyxyt(xy)2

，記其圍成區(qū)域?yàn)镕(t)1x2y2t(xy)2

31t22(x2y2)2xy2t(x

xxt,yy32

2

1t2

x

2 3

t2

2

2

f3f

y 2xyF(t)

tt322xxyyxy則(1)222 2 t

y1tt13 1 3 3 記相應(yīng)的區(qū)域?yàn)?/p>

t2131

2

y2)于

t2

2

2F

3

131131t33最后，作廣義極坐標(biāo)變換，即x rcos, t2 1 t2F(t)1

3

31

22tt

d1rr3dr 3 t

3t

3F(t)考慮到函數(shù)u

t3F(t)3

(3t2)2 t3 t3§2第二型曲線積分與曲面積L(2ay)dxdy02aa(1cost)asinta(1cos

2a(a xdxydy,Lx2y2a2 x解L的參數(shù)方程xacosyasin,0xdx

2acos(asin)asinacos x2y a

Lxdxydyzdz,L為從(1,1,1到(2,3,4的直線段解L的參數(shù)方程為x1ty12tz13t,0t所以

xdxydyzdz1(1t)2(12t)3(13t)dt0(x22xy)dxy22xy)dy,Lyx2從(1,1到(1,1的一段L解：(x22xy)dxy22xy)dy1(x22xx2x42x32xdx

ydxxdyx2y2dz,Lxetyetzat從(1,1,0到(ee1aL

ydxxdy(x2y2)dz=1etetet(et)(e2te2t)0012a(e2te2t002a(e2e222asinh

(x2y2dxx2y2dy,LA(1,0B(2,0C(2,1D(1,1L針方向解LABBCCD其 AB:y0,1x2,起點(diǎn)對應(yīng)xBCx2,0y1,起點(diǎn)對應(yīng)yCDy1,1x2,起點(diǎn)對應(yīng)xDAx1,0y1,起點(diǎn)對應(yīng)y(x2y2)dx(x2y2L1021=2x2dx1(4y2)dy1(x21)dx0(1y2)dy1021計(jì)算曲線積分y2z2dxz2x2dyx2y2dzLLx2y2z21x0y0z0Lx2y2z2a2x2y2ax(a0的交線位于oxy平面上方的部分，x軸上(b,0,0)(ba點(diǎn)看去L是順時(shí)針方向.

LL1L2L1:x0,z

1y20y1,yL2:y0,zL3:z0,y

1x20x1x1x20x1x所 (y2z2)dx(z2x2)dy(x2y2L0

2 1

21y2121y21x2

)(y)

dy0(1

)x

10(1x2)(x2)1

2121xLL1L2,L1:y

axx2,z

a2ax,0xaxL2:y

axx2,z

a2ax,0xa起點(diǎn)x所以I (y2z2)dx(z2x2)dy(x2y2L22

=(y2z2)dx(z2x2)dy(x2y2

(y2z2)dx(z2x2)dy(x2y21L102axx2a2(a2axx2)a (2x2ax) axaxx2a2axa

a2axx2a2(a2axx2)2x (2x2ax) axaxx2a2ax0 axx2a(a2axx2)2xaxx20

axax

x0時(shí)

t

xa時(shí)

tx 1t2

dx

dt 03 0 24 24t

1t

33

dt202

dt303

44 2a3(I3II 其中In

1

n

2n

2n

In2(n1)(1t2I3a2

02(n1)In12(n1)

ydxLx2y2Lx2y2a2逆時(shí)針方向2L2a

y

L為以(0,0為中心,邊長為a,對邊平行于坐標(biāo)軸的正方形,順時(shí)針方向,解：(1)L的參數(shù)方程為xa

yasin

02,起點(diǎn)對應(yīng)ydx 所

Lx2y

a2

d(2)Lxa

ybsin

02起點(diǎn)對應(yīng)2.所ydxxdy0bsin(asin)acosbcosLx2y a2cos2b2sin22 0a2b2tan2 40

a2b2tan2

4arctanbtan2

a0 a0LL1L2L3L4,Lxa,aya,起點(diǎn)ya L:ya,axa,起點(diǎn)xa Lxa,aya,起點(diǎn)ya L:ya,axa,起點(diǎn)xa 1234所 ydxxdy 1234Lx

y

x

y a

a

aa

aa2

2

2 +

2 aa

a

a

a y4

2x24

y4

2x242a 2y2a4

d2y2

4a

a

21 a(4)LL1L2L3,L1:y

1x

xL2:y2xL3:y2x

1x0x

xx12所 ydxxdy 12Lx

y

3x2y31dx0(2x1)x2dx1(2x1)x1x2 x2(2x x2(2x1 1 15x24x2arctan2

05x24xF對運(yùn)動(dòng)的單位質(zhì)點(diǎn)所作的功,LAB點(diǎn)Fx2xy2y2x2yL為平面曲線yx2,A(0,0FxyxyL為平面曲線y11x,A(0,0

(xy,yz,zx),L的矢量形式為 2

r

t tk

Fy2z2x2LA(,0,0),B(,0,2)LLLL

xcostysintzt(,為正數(shù)

WFds(x2xy2)dx(y2x201(x2x3)(x22x4)01=6

WFLL(xy)dx011(2xx2)dx2[2x(2x)013

（3）WFds[(tt)(tt)2t(tt)

LL（4）WFdsydxzdyxLL2[2sin3ttcost2cos202PQ，RLL為光滑弧段，弧長為lLPdxQdy

MlMmax(x,y,y,z

P2Q2R2sLxy zl

0s所以LPdxQdyRdz0P(x(s),y(sz(s))dx(sQ(x(s),y(sR(x(s),y(s),l0P(x(s),y(s),z(s))dx(s)Q(x(s),y(s),z(s))dy(s)R(x(s),y(s),lP2Q2R dx2P2Q2R dx2dy2dzllLS上，Szf(x,yL在oxy平面上的投影曲線為lP(x,yzLLP(x,yz)dxlP(x,yz(x,證明：x作為參數(shù)，則Lxxyy(xzz(xlxxyy(xx1xx2，起點(diǎn)時(shí)應(yīng)x所

P(x,y,z)dxx2P(x,y(x),z(x, P(x,y,z(x,y))dxx2P(x,y(x),z(x, 因 lP(x,y,z(x,y))dxLP(x,y,

I

xyzdzL：x2y2z21yz卦限

x2y2Lx2y2z21yz相交的圓，故方程為

y2xcosy2

1sinz

1sin02，起點(diǎn)對應(yīng)0222xyzdz2cos1sin21cosd12sin2 222 22

y(xz)dydzx2dzdxy2xz)dxdySxyz0Sxyza，六個(gè)平面所圍的正立方體邊界的外側(cè)；SS上S下S左S右S前S后 y(xz)dydzx2dzdx(y2xz)dxdy(y2S y(xz)dydzx2dzdx(y2xz)dxdyyS y(xz)dydzx2dzdx(y2xz)dxdyxS y(xz)dydzx2dzdx(y2xz)dxdyxS y(xz)dydzx2dzdx(y2xz)dxdyy(aS y(xz)dydzx2dzdx(y2xz)dxdyS 所 y(xz)dydzx2dzdx(y2SS

S

S

S

S

S

dydzx2dzdx(y2 a(xy)dydzyz)dzdxzx)dxdyS是以質(zhì)點(diǎn)為中心，邊長為2S解:同（1）SS上S下S左S右S前S則(xy)dydzyz)dzdxzS S

S

S

S

S

S

(xy)dydz(yz)dzdx(z 243 x yzdxS為S

yb

zc

1解 SS左S右，其S左y

x1 x z1 x za c

zc

1S右y

x1 x z1 x za c

zc

1所以yzdzdxyzdzdx S S1 a c1 a c

11 a c 1 a1 a c

1r002abc22sind1r1r00zdxdyxdydzydzdxSx2y21z0z3S解 由于S在xoy平面上的投影為曲線x2y21，故zdxdyS對 xdydzxdydz S S

1y2dydz

1y2 D11

1y2dy3dz0 ydzd