大一下微積分課件-第九章習(xí)題答案

二重積分的概念與性習(xí)題 xy2dxy3dDxy xy1Dxy2xy3，所以xy2dxy3d DDxy2d與xy3dDDDx22y122所圍成Dxy2xy3，所以xy2dxy3d lnxydlnxy2dD為三角形閉區(qū)域，三頂點(diǎn)分別為10,1,1,20

DDxy3x50y1D

（1）IDxyxydDxy0x10yD0xyxy2，所以0I（2）IDsin2xsin2ydDxy0x0yD0sin2xsin2y1，所以0I2（3）IDxy1dDxy0x10y解：在D上，1xy14，所以2ID（4）Ix24y29dDxyx2y2DD上，13x24y2925，所以52I

x2y2d，其中DxDx2y2d1D

x1,yD3x2ydDxy22D3x2yd2Dx33x2yy3dDxy0x10yD

3x2yDxcosxyd，其中D為頂點(diǎn)分別為00,0和,的三角解：DxcosxyDyexydxdyDx2y2xy1D

yexydxdy

1222

2yexy

xy2dxdyDxpy22pxp02

xy2dxdy

p

2p2

exydxdyDx0x1y0y1D exydxdy1dx1D

xsinxydxdyDx0xy0y2

xsinxydxdy

dx2xsinxydy

xsinxcosxdx Dsinxycosx2y2dxdyDx2y2DDxysinxycosx2y2dxdyDDx3sinx2y2dxdyDx2y22DDyx的奇函數(shù)，所以x3sinx2y2dxdyDDD

1x2dxdyDx2y21212

dxdy1

dy121x21 1

dx3212DxydxdyDxy12xydxdy81dx1xxdy81xx2dx4 7D x2y2dxdyDx2y21x2y22x7D

解 x2y2dxdy

3

r3dr

203

r3dr DD

sinx2y2dxdyD為2x2y24sinx2y2dxdy2 2rsinr2dr

21cos21cos42

0 cos2cos42R2r2D R2x2y2dxdyDx2y2R2r2D

2

R2R2x2

dxdy

d

rdr

d D解：D

R2x2y2dxdyDx2y2RxR R2x2y2dxdy22d

R R2r2R2r222 1sin3d 30 Dx2xydxdyDxy1xy2yxy2xDxy解：做變換

x

u1

,x,y

111111

y

1

u,

1

1

11

xydxdy1 3du

41v3dv962313Dx3y3dxdyDx22yx23yxy2x2y22313Dx2

2

1

12解：做變換

xu3v3，則

x,y

u3

u3v3

1，所以

1

u,

2

21

yu3v 1u2v

3

7u

3v

3

u3v

ydxdy2

dv2

72du1442DxydDy

x,yx2 12 解：xyddx2xydy xxdx

03 Dxy2dDx2y24yD44

2

y4 Dxyd2

xydx22y2dy15 DexydDxy:xD

y1D10exyd0dxx1exydy1dx1xD100e2x1e1dx1ee2x1dxe1 Dx2y2xdDy2yxy2xD解 x2y2xd

2dyyx2y2xdx

219y33y2dy13

0 (5)D1xsinyd，其中D為頂點(diǎn)分別為00,10,12和0,1的梯形閉區(qū)域

1xsinyd1dxx11xsinydy11x1xcos1xdx 32sin2sin1cos2 2D（6）x2y2d，其中Dxy0ysinx0xD解 x2y2d

dxsinxx2y2dy

x2sinx1sin3xdx240

0 D(7)y23x6y9d，其中Dxyx2y2R2D2

x2

Rr D

3x6y9dD

9d

d0

92R49R2

20

d

9R 如果二重積分Dfx,ydxdy的被積函數(shù)fx,y是兩個(gè)函數(shù)f1x及f2y的乘積fxyf1xf2y，積分區(qū)Dx,yaxbcyd，證明此fx

ydxdybfxdx

dfydy.D

f1xf2ydxdybdxdf1xf2ydybdf2ydy b fxdxdfydy.b IDfx,ydyxy24xy x4解：I fx,ydyy x4

fx, 4xx2y2r2y0解I

r fx,ydy

r2 fx,y r2r2

r2yxx2y21x0x I1x

fx,ydy環(huán)形閉區(qū)域xy1x2y24；44 2 f444441 fx,41 1 fx, 4以O(shè)00,A20,B2,1,C0,1為頂點(diǎn)的矩形 I0dx0fxydy

f以O(shè)00,A10,B1,1為頂點(diǎn)的三角形 I0dx0fxydy0dyyf以O(shè)00A2a0B3aa,Caa為頂點(diǎn)的平行四邊形； I0

fx,ydy 0dy

fx,y I1

fx,yyx2xy2I

fx,y yx2,y4x2I

4 fx, xy2y2x2yx0 2 I dxxfx,ydy dxxfx,y 121220dyyfx,ydx1 fx,yy fxyDDyxyaxbba所圍成的閉區(qū)域。 adxafx,ydyadyyfx,y adxafxydyDfxyda

fx,yy（1）0

fx, 0dy0fxydx0dxxfx

2 fx,y 2解 fx,y 2

dx

fx,y

21（3）0 fx1解：0 fx,y22

fx,y（4）122x

fx,y解1dx2

fx,ydy0dy2

fx,y 1e1e

fx,ye解：1 fx,ydye

0 fx,

0dxsinxfx,y2 sin arcsin0dxsinxfxydy1dy2arcsinyfxydx0dyarcsin2

fx,y04（7）04

14dy4

fx,y

fx,

2 0dy 2

fx,ydx1 3 3

fx,y解：0 fx,ydx1

fx,y0dx

fx,y (10)0

fx,ydya 0dx0fxydy0

fx,y1(11)0 fx,y1 3y解：0 fx,ydy0 fx,y3yyay（12）0dyyfx,yy y1x解：0dyyfx,y1x

fx,11x

fx,y y 0dy0 fxdx0a證明： y y證明： y y D

emax

其中Dx0yayx圍成的

0dy0 fxd應(yīng)用二重積分證明：由射線,與曲線rrD2 r 證明：Sdd rdr 求心臟線ra1cos12

S20a1cos

d 2

x2y2

fydxdyxx解：x2y2

fydxdyxx

22

frsinrdrrcosrcos

cos2ftan22222

DydxdyD0xyxa2x2y2b2ba0Darctan b

arctanb3 Dydxdyarctandarsindrarctan sin b3a3 1 12 DarctanxdxdyDxyRyDarctanxdxdy

a2x2y2

3/

,D:0xa,0y

a

3/224d

3/2dr Da2x2y2

a2r2

x2y2x

xy

ux1

xu1

vy

yv x,y 0

xydxdy uv

，由對(duì)稱(chēng)性，u,v

x2y2x u2v22u2v22

uvdudv0，所

x2y2x

xydxdy

u2v22

dudv2

x4y4

x2y2 x2y2dxdy

2

4cos4sin4r3dr

x4y4

0cos4sin41 1tan d

1

dx

2 d(x ) 201tan4

01

0(x1)2 xxyuyvxDfxydxdy(Dx0y0xy1的公共部分）化為變量uv的累次積分。xy解：做變換

x

1

,x,y

111111

y

1

u,

1

1fx,ydxdy

1duf

,uv

1

1v 1y2pxy2qxx2ayx2by0pq0abx2

2

1

12

xu3v3

x,y，

u3

u3v3

1

1

u,

2

21 x

yu3v

u3v

u3v32313 baq2313所以Sddu1dv p 設(shè)平面薄片所占的閉區(qū)域D由直線xy2yxx軸所圍成，其面密度xyx2y2m

x2y2d

1dy2yx2y2dx

184y4y28y3dy4

x0,y0x1,y1z02x3yz解：V

62x3yd1

62x3ydy

192xdx711

0 x0,y0xy1z0x2y26z

1

4x3 VD6xyd0dx06xydy035x2x3dx6 zx22y2z62x2y2解：聯(lián)立z62x2

得投影區(qū)域Dxy0x2y2所以V

62x2y2x22y2d2d

263r2rdr 畫(huà)出積分區(qū)域，把積分Dfxydxdy Da（1）x,yx2y2a2aaDfxydxdy

d

frcos,rsin（2）x,yx2y2

fx,ydxdy

2 2

frcos,rsinb（3）xya2x2y2b2其中0abDfxydxdy

d

frcos,rsin（4）x,y0y1x,0x

fx,ydxdy2dcos

frcos,rsin（5）x,yx2y1,1x

fx,ydxdy4dcos2

frcos,rsin44

frcos,rsinrdr

34

cos2

frcos,rsin（6）x,yx2y2axa0,

fx,ydxdy

22

frcos,rsin（7）x,yx2y2byb0, Dfxydxdy0d

frcos,rsin(8)x2y24xx2y28xyxy2x Dfxydxdy4

frcos,rsin（9）x2y2axx2y2aya0所圍區(qū)域的公共部分。fx,ydxdy4d

a

frcos,rsinD24

frcos,rsin （1）0dx0fx,y dxfx,ydy4d

frcos,rsin24

frcos,rsin20（2）20

3xf

x2y2x解：2 3xfx

x2y2dy

3

2

fr

4

01

fx,y

fx,ydy2d frcos,rsinrdr.

cos（4）0

fx,y11

dx0fx,ydy

40

1cos2

frcos,rsin（1）0 xy

x2y2dy2d

2a

r3dr24a4cos4d

.2ax 2ax （2）0

x2y2 a

x2y2dy4dcosr2dr 32 2

3cos4a34

4

3ad a

a1dsin a1

dx

a2a

30cos

30(1

3

(1x 12

1

1x2 22

2 2

2

1201x 1x

1

1x 2

2 2

2

1201x 1x

1

1x a3ln32222

0x1 x1

x2

2

x2y22dy

40

cos2dr0

0cos2

d

2a（4）0aa

a2y2x2y2a2a2

解： x2y2dx2dr3dr d 0 fxy在閉區(qū)域Dxyx2y2yx0上連續(xù)fx,yfxy。Dfx,ydxdy

81x21x21x2D

fx,y

fx,ydxdy sin

1r2d1r

fx,ydxdyrdr

fx,y

所以

fx,ydxdy1.，fx,y

28

11x2Dex2y2dDx2y24Dx2

2

2e41 2 d0d0erdr0 de2 D(2)ln1x2y2dDx2y21D

1 ln 1ln1x2y2d2drln1r2dr2ln2

d

0

2

4（3）

arctanydDx2y24x2y21y0yxx

yd4d

rdr4

. 02 x2xDy2dDx2yxxy1 x y解： 2d1yx1x2

y2dy1

xdx

dDx1x2

1

1x2

11

2d2d

rdr 2D1x 01 2Dx2y2dDyxyxayay3aa0所圍成的閉D

3a

a3 Dxydadyyaxydxa2ayay3dy14a D x2y2d其中D為圓環(huán)形閉區(qū)域xya2x2y2b2D 解： x2y2d dr2dr b3a3 D2上一段弧0與直線 2 xyx2y2m

d2d

r3dr244d R2r2y0ykxk0zR2r2

R2x2R2x2

d d

rdr 3xOyx2y2axzx2y2

a

a4cos4

解：V

x2y2d

2

r3dr

d D（1）xy2sin2xydxdy,D為平行四邊形閉區(qū)域，它的四個(gè)頂點(diǎn)是D0,2,,2和0,1212xuv1212

xyxy

，則y

2u2

x,yu,

2 2

v2sin2

33sin2

xydxdy

dv

du Dx2y2dxdyDxy1xy2yxy4x所圍成Dxu,vxu,v

x,

1u2v122112213解：做變換y

，則y

u,

.，所以u(píng) uv2uv2 4 2

7lnDxydxdy1du12vdv1

ln2du 3

exydxdyDxyxy1xy

xu

x,y 解：做變換

y

，則y

u,v

，所以 u

exydxdy

eudv

y2

（4）Da2b2dxdy,其中Dx,y b2 xarcos解：做變換ybr

x,y，則r,

ab

arbr

，所以x2y2

b2dxdy0 Dxy4xy8xy35xy315

1 xy

xu2v2,x,

u2v 解：做變換

，則

xy3

y

12

u

1

3 2v2 dxdy8du151dv2ln 5Dyx3y4x3xy3x4y3

311

x

x,

u8v33解：做變換

u1

, u,

193 344

y

8v

u8v8所以

dxdy

1

434

dv18Dxy1x0y0cosxydxdy1

xy

xuv

xyxy

，則y

1211212u2

x,yu,

21.，所以 2xy

cosv dxdy

du dvusin1du

xy

11（1）

fxydxdy

fudu其中閉區(qū)域Dx,xuv

xy121211212

xyxy

，則y

2u2

x,yu,

12 .，所以1 2fxydxdy1du1fudv

fu1 1

faxbycdxdy

f

cdu, 1a2Dxyx2y21a21a2證明：做變換

a2b2 ，則a2b2a2b2 a2xaaaa2 a2

aubva2a2

x,yu,

y a2a2

a2faxbycdxdy

f

a2a2

cdv2

f

1a21a2習(xí)題Ifxyzdxdydz為三次積分，其中積分區(qū)域xyzxy10z0 Ifxyzdxdydz0

fx,y,zzx2y2z1I

fx,y,zdxdydz1 fx,y,z1 1

x2zx22y2z2x2I

fx,y,zdxdydz

fx,y,

x22 由曲面czxyc0a2

Ifxyzdxdydz

dy

fx,y,z

（1）z2dxdydz為兩個(gè)球：x2y2z2R2x2y2z22RzR0的x2y2z2R2解：由x2y2z22Rzz

R2

z2dxdydz

z2dxdy2

z2dxdyR

R R

2 2 2 2 2 2 2Rz dzR R dz02zlnx2y2z2

dxdydz，其中x2y2z21 xyz關(guān)于xOyzzlnx2y2z2

x2y2z2

dxdydzy2z2dxdydz，其中xOyy22xx軸旋轉(zhuǎn)而成的曲面x5所圍成的閉區(qū)域。

y2z2

。易得在yOzD0yzy2z210，所

y2

dxdydz

225y5y

2

y2z2

10

r2

2D5 yzdydz0d 52rdr321 11解：1

xyz

dxdydz，其中x0y0z0xyz111

dxdydz

0

1x0

xyz121 1

1dy

1ln21xln1xdx3ln

xy 2 (5)xydxdydz，其中zxyz0xy1

1x21

xydxdydz

xydz

x2y2dy

dx

xy2z3dxdydz，其中zxyyxx1z02

2

xx5

1 xyzdxdydz0dx0

xyzdz0

dy028dx364(7)xyzdxdydz，其中x2y2z21x0y0z0

xyzdxdydz2d2d

5sin3cossincosd1

xyzsinxyzdxdydz，其中x0,y0z0xyz2

x

xyzsinxyzdxdydz2dx

dy

2 2dx xyxysinxydy2

xxsinx

(9)lx2my2nz2dxdydz，其中x2y2z2a2所圍成的閉區(qū)域(lm為常數(shù)

lx2my2nz2dxdydz

lmnx2y2z2 lm

d0d0

sind

lmn.（10）

x2y2dxdydz,:x2y2z2,0z h

x2x2

dxdydz

d0drrrdz6（11）xydxdydz,:xy2z,z

2

ydxdydzddrr2rdz3 2

x2y2dxdydz,:x2y2z2

x2y2dxdydz

2x2y2z2 22dda4sind8a53

dxdydz,x2y2a2z0zhh0所圍成。x2x2y2

2 2

3

dxdydz

d

dz

ah2

a. x2y2

r2

3 （14）zdxdydz,:x2y2z22,x2y2x2y2z2解：由x2y2

z1 zdxdydz

dzDzdxdy0dzDzdxdy

1z2dz

22z2zdz7 （15）z2dxdydz,:x2y2z2a2,x2y2axaz2dxdydz

2

a

z2rdz

2

acos

ra2r22

acos

3

a2aa2

3334a5 822d

ra2r22dr22 1sin5d

0

15（16）

x2y2zdxdydz其中由2zx2y2z22

2

r4

xyzdxdydz ddrr2zrcossindz d28

cos

0 2128cos2sin2d32 R2x2x2y2z2R2x2

z

x2x2

x2y2z2dxdydz

d4d

4sind

2

2. z3dxdydz,:x2y2z2R2,x0,y0,z

z3dxdydz2d2d

5cos3sind

.

x2y2z2dxdydz其中x2y2z2z

x2y2z2dxdydz

d2d

3sind cos4sin d2 d xydxdydz,:axyzb,z

x2y2dxdydz

d2d

4sin3d

b5a5 x2y2 dxdydz其中xyz2x2y2

dxdydz

d2d

2a

sind1 x2y2 1 d22a2cos2sind

.

dxdydz,:a2x2y2z2b2bax2y2z21

b

1

2dxdydz

d0d

2d4 xyz

b（23）

xyzdxdydz,:x2y2z2

x2x2

xyzdxdydz

zdxdydz

d4

2a

3cossind d44a4cos5sind

. （24）

1x2y2z2dxdydz,:x2y2z2 1x21x2y21

dxdydz

d0d

sind （25）

dxdydz, xasincos解：做變換ybsinsin1z1

abc

1a2b2c2dxdydz

d0d

sin 4

xyzdxdydz,:xa2yb2zc2R2ux

uvwabc解：做變換vybwz

abcdudvdw

3

abc.設(shè)函數(shù)fx連續(xù)

fx2y2z2dvFt ,

fx2y2d fx2y2dGt ,tt

fx2dx其中txyzx2y2z2t2Dtxyx2y2t討論Ft在區(qū)間0內(nèi)的單調(diào)性證明：當(dāng)t0Ft2Gt（1）

fx2y2z2dv

2 tf22sindFt t 0 00

fx2y2d

0

tfr202tf22d0 0tfr20F't

2ft2

tfr2rdr2ft200tfr2rdr00

tf22d

2ft2ttfr2trrdr0tfr2rdr0 ，所以Ft在區(qū)間0內(nèi)單調(diào)上升 fx2y2

tfr2

tfr2（2）Gt

Dt t

tfx2

2fx2

fx2 .令Httfr2r2drtfr2drtfr2rdr2，則H0 .令

。當(dāng)t0時(shí)， H'tft2t2tfr2drft2tfr2r2dr2ft2ttfr

，所以0ft2ttr0

fr2drtfr2r2drtfr2drtfr2rdr2此等價(jià)于Ft2Gt 設(shè)有一物體，占有空間閉區(qū)域xyz0x10y10z1，在點(diǎn)xyzxyzxyzmxyz如果三重積分fxy

的被積函數(shù)fxyz

是三個(gè)函數(shù)f1x,f2y,f3z

fx,y,zf1xf2yf3z

x,y,zaxbcyd,l bfx,y,zdxdydzb bdxdf1xf2 daf1xdxcf2y

1xyz

，其中x0,y0z0xyz1

3 0dx

21x計(jì)算xzdxdydz，其中z0zyy1yx2所圍成的閉yOzx是奇函數(shù)，所以xzdxdydz

zdxdydz其中zx2x2

zhR0h0 0 解 z

x2y2z2dv其中x2y2z21

x2y2z2dv2dd14sind4 (2)zdv其中閉區(qū)域由不等式x2y2za2a2x2y2z2所確定

zdv

d4d

2a

3cossind

d44a4cos5sind

7. （1）

xydv其中x2y21z1z0x0y0

xydv2d

r3cossindz 2x2y2dv,其中為由曲面4z225x2y2z52x2y2dv

0

0

5r3dz8求下列區(qū)域V（1）V:x2y2a2,z0,zmxm

解：V

dxdydz

2

rdz

2

mr2cosdr 3

（2）V:a2b2c22,b2c2aax2y2z2 解：由

xay2z2

x2VVdxdydz dxDdydz0

bc2

2x427

a 6 （3）Vx2y2a2y2z2a2z2x2a2所圍成（9.33VVdxdydz

1sin34 4 cos（1）z6x2y2zx2x2

x2y2解：由

z26Vdxdydz0（2）x2y2z22aza0x2y2z2（z軸的部分x2y2z2解：由x2y2

zaVdxdydz0x2x2解：由

zx2y2x2y2z11zVdxdydz0dzDz55x2解：由

x2y24z5x2y2z1

xdydz

ay2a2az,x2y2 2

,z0Vdxdydz02a3a3sin20 y2a2az,x2y2ax,z0a r2sin2 V dxdydz

2

a

rdz

2

acosrarsin

a3cos2 a3sin2cos4

d

2 x2y2az4a2x2y2z24az解：由x2y2z2

za

zVdxdydzz

dzDdxdy

4az

azdz 232a39a337a3.所以上下之比為27 求球體ra位于錐面3

和23解：V

dxdydz

d3d

2sind

3.3 3x2y2z22zx2y2x2y2z212解：由zx212

z1V

dxdydz

dxdy0zdz

22z2dz8276zR的球體，在其上任意一點(diǎn)的密度的大小與這點(diǎn)到球心的距離成正6z 解mdvkx2y2z2dxdydzddkr3sindrk fxy

為由曲面zx2y2yx2y1z0 x2解 fx,y,zdxdydz fx,y,z9.4

習(xí)題x2y2z2a2x2y2ax內(nèi)部的那部分面積。11 22ax ax 1

a

r dxdy4a2d

a2x2

a2r4a2d

a

dr24a a2r x2x2y2z22x2x2x2y2z2x2

所割a

x2y2

得za.所以球面x2y2z22az被錐面z 所恰為半球面，面積為2旋轉(zhuǎn)拋物面2zx2y2x2y21解S

1x2y2dxdy2d

1r2rdr2223曲面azxyx2y2a231a1aay x

3S3

dxdy

d

rdr

22(4)x2y2R2x2z2R2y2z2R2所圍成的立體的表面；S

112R2x2

R2R2r2cos2

dxdy

4R R

224R24 d24222

2R2.0cos cos（5）zarctanyx2y21xS

1 1x2 1y 1y22 x2 x2

11rdr

2ln

22 2x2求錐面z 被柱面z22xx2解：由

x2y2

解得x

2y2

，所以在xOy面上的投影為1xyx 1xyx x S

dxdy

22

222cos2d 2x2y2R2x2z2R2S

dxdy

R2R

dr12R2x2R12R2x2R2r2cos2 S

dxdy 11ab c c a2b2a2c2Dy

2pxxx0y02222 23205x 0 23205

03x0

x00 0

x2x

x0

22x0020

1

32y

02322320x0D為半橢圓形閉區(qū)域x,y

21,y a解：x0;y a

a Dacosbcos0abd2d2 x

Dyx2yx所圍成，它在點(diǎn)xy處的面x,yx2y，求該薄片的質(zhì)心。 xD

x2

0

xx2 1x1設(shè)有一等腰直角三角形薄片，腰長(zhǎng)為a，各點(diǎn)處的面密度等于該點(diǎn)到直角頂點(diǎn)的距離的x

xx2y2

x2y2（1）z2x2y2,zx0y0z

a2（2）z A2x2y2,za2x0;y

d2d

3sincosdz

32A3a33A4a4 3A

AaAaa32A3a3 3

A2Aa（3）zx2y2,xya,x0,y0,z a x2 a xxdxdydz0a

xdz152a;y2aaa

0

a

x2 a x2

az a

0

a

x2 aax0;y

x2y2z2

d2d

2R

5cossind

z

x2y2z2dxdydz

4 2R4

32R54

d2d sind

R2 3 a，則0y

,可解得a 3 x2x2求質(zhì)量分布均勻的半個(gè)旋轉(zhuǎn)橢球體xyz

21z0

b 0 0 Dz

x0y0z

2a2b8 求高為h，底半徑為az ha2hz2z

0dzD

12 x0;y0;z

a2h4 設(shè)物體占據(jù)空間區(qū)域V0x10y10z1，在點(diǎn)Mxyz處密度為xyzmxyzdxdydzdxdyxyzdz3 xxxyzdxdydz0dx0dy0xxyzdz65.y5;z5 0 2 P0的距離的平方成正比（k，求此物質(zhì)球體質(zhì)心的位x0;y R2

2z

R2

2

R2 dd

k3cossin2sin2cos d

2

R2 ddk2sin2sin2cos

d 154R

2 設(shè)均勻薄片（面密度為常數(shù)1）D（1）Dx,yaa

I I2121bbI

x2dxdy

x2dyba4(2)Dy29xx2IIIx

2y2dxdy2

9 y2dy

.Iy

2x2dxdy2

92x2dy

9

9 （3）D為矩形閉區(qū)域x,y0xa0yb，求IxIy b

IxDydxdy0dx0ydy3.Iy3已知均勻矩形板（）的長(zhǎng)和寬分別為b和h，計(jì)算此