課件第十章重積分習(xí)題解答

習(xí)題10-2.

1212

x2y23d，且

{(x,y)|1x1,2y

xy|0x1,0y2I1I2之間的關(guān)系1112IDzx2y23的曲頂柱體I111222Dzx2y23的曲頂柱體221Dzx2y23yOzzOx面對(duì)稱，所以這兩個(gè)平面將1分成四個(gè)等體積的部分，第一卦限內(nèi)的就是2.由此可知：1I14I2（2）(xy)2d(xy)3d D是由圓周(x2)2y1)22所圍成區(qū)域yxy1DO1ln(xy)d與[ln(xyyxy1DO1 點(diǎn)是（1,0（1,1（2,0）.解（2）10-10圖，Dxy1內(nèi),從而(xy)2(xy)3所以(xD

(xy)3dD（3）D位于{(xy|1xy2內(nèi)，故積分區(qū)域上的點(diǎn)滿足0ln(xy)1，從而有：[ln(xy)]2ln(xy).因此

圖10-[ln(xy)]2dln(xy)d 本大題常見(jiàn)錯(cuò)誤是計(jì)算出二重積分的值進(jìn)行比較題意不清,二重積分的性質(zhì)不熟悉(x24y29)dDx2y24D解（4）因?yàn)?x2y24所 9x24y494(x2y2)925于 922(x24y29)d2522D

(x24y29)d100D本題常見(jiàn)錯(cuò)誤是計(jì)算出二重積分的值進(jìn)行估計(jì)題意不清,二重積分的性質(zhì)不熟悉習(xí)題10-(3x2y)d，其中D是由兩坐標(biāo)軸及直線xy2D(x33x2yy3)dDxy|0x0y1Dxcos(xy)dD的頂點(diǎn)是(0,0),(,0),(,)D角形閉區(qū)域 2解（2）(3x2y)d0dx0(3x2D202

2x2x2

=203（3）(x33x2yy3)dx3d3x2yd 1dy1x3dx1ydy13x2dx1y3dy1dx111

（4）xcos(xy)d0xdx0cos(xDx(sin2xsinx)dxxd(cosx1cos

xcosx cos2x2

0cosx2cos2xdx2 D

yd,Dy

xyx2xy2d,Dx2y24和yDexyd其中D（xD

(x2y2x)d其中D是由直線y2yxy2xD成的區(qū)域解（1）D(x

0x1,x2y

x于x 123x

dx2xydy y 2 D12

0

x (x40

x4)dx （2）D0x4y2，2y22xy2d2D

44y

xdx

12

2dy

64（3）exydexydexy 00exdxx1eydy1exdxx1eydye0常見(jiàn)錯(cuò)解

D1xy1x0y0exyd4exydxdy41exdx1x

e 11

(eex)dx42 2

(x2y2x)dD

0

y(x2y2219y3

2y

130 If(xy)dxdyD（1）Dy=xy24xy4yy2y4yy2O41y (x0x分析解此類題目,一般先畫(huà)出積分區(qū)解（1）10-11yx圖10-y24x交點(diǎn)是(0,0),(4,4), 2 或I或4

f(x,x

I

dyy2f(x,y)dx44①I0dx2

fxy)dy

I

y

f(x,y)dx2 ②Ixdx

f(xy)dy

I4

f(x,y)dx錯(cuò)誤原因①yx)的積分上下限顛倒了yx)的積分限的確定,位于上（或右）方的曲線位上限，下②（3）10-12D.若選擇先對(duì)y積分，積分區(qū)域 1y

yx

yD:

D1x1x

12,2 則f(xy)dxdy1dx1f(xy)dy 圖10-1xDDDD

Dyx2.1 1

y

21yf(x,y)dxdy=f(x,y)dxdy+f(x, =1dy1f(x,y)dx1dyyf(x,y)dx 2（1）0dy0f(x,y)dx（2）0dyy2f(x,y)dx2xx（3）0 f(x,y)dx（4）1dx2xf(x, ln

1

f(x,y)dy（6）0dxsinxf(x,y)dy2解（1）D={(xy|0xy,0y改寫(xiě)為Dxy|xy1,0x1 原式0dxxf(x,y)dy（2）D(xy)y2x2y,0y4DD(x,

xy24

x,0x4，于x原式0dxxf(xy)dy11（3）Dx

0y

x

1y2D(x

0y

1x21x1原式

f(x,y)dy2x2x 1x 0y1可改寫(xiě)為D:2yx112xx2xx1y1dx2

f(x,y)dy=0dy2

f(x,y)dxD(xy)1xe0ylnxD(x,y)0y1,eyx ln 1dx0f(x,y)dy0dyeyf(x,y)dxy1OysinD2xyy1OysinD2xysin2D:

0xDD1D22arcsinyxD1:

1y

arcsinyxarcsinD2:

sin

0y 0dxsinxf(x,2 arcsin=1dy2arcsinyf(x,y)dx+0dyarcsin

f(x,y)dx sin 常見(jiàn)錯(cuò)解0dxsinxf(xy)dy1dy2arcsinyf(x2+0dyarcsinyf(x,y)dx sin

arcsin②0dxsinxf(x,y)dy1dy2arcsinyf(x,y)dx2xysinx2y型區(qū)域的描述方式Dxy2,yxx軸所(x,y)x2y2,求該薄片的質(zhì)量.解如圖10-14所 M(x,y)d1dy2yx2y2 D

y

xy11x3xy22y 01

2O2 0

(2y)32y23

y3dy

圖10-x0y0x1y1z12x3yz6截得的立體的體積四個(gè)平面所圍成的立10-15，所求體積為V(62xD

z62x110dx0(62x11

116y2xy3y21

圖10-019

70

2xdx x0y0xy1z0x2y26z截得的立體的體積解xOyD{(xy|0y1x2,0x1z6x2y2V(6x2y2)dxdy (6x2y2)dy17 1 Dzx22y2及z62x2y2所圍成的立體體積解由 z62x

z

2

6

y2x2y22xOyx2y22因?yàn)榉exy軸均對(duì)稱xy都是偶函數(shù)所以VD

(62x2y2)(x22y222x

(63x23y222

(2

)dy

dx622x2把二重積分f(22x2D（1）D{(x,y)|x2y2a2}(a0)（2）D{(x,y)|x2y2解（1）D,|020a},f(x,y)df(cos,sinDaa0d

f(cos,sin)d （2）D,|2

,022 f(x,y)df(cos,sinD

0 22

f(cos,sin)d （1）dxf(x,y)dy；（2）

3xf

x（3）0dx1xf(x,y)dy（4）0dx0f(x,y)dyyy1yxD (,)0,0sec D(,),0csc 00原式4d00

f(cos,sin 24

f(cos,sin)d0常見(jiàn)錯(cuò) 原式2d0

f(cos,sin0或原式 0

d

f(cos,sin)d0搞不清在極坐標(biāo)系下極徑的變化區(qū)間怎樣確定y3y3O42D(,),02sec43432 2

2 f

x

3 f()d4D10-18所示

圖10-D(,11x

0

2cossin

y11y11sinO10dx1

f(x,圖10- 2d f(cos,sin)d. cos0yD:0x

10-yyO1xyyO1yx2sin2cos2secx1cos1secD:

0 4

圖10- x

000sec000sec

f(x,y)dy=4d

f(cos,sin)daxax

（2）

x2y20110

xdxx

x2y22dy；

a2aa2a

(x2y2D 解 (,)02,02acos2axx2axx所

(x2y2)dy2d

2acos2 31 2cos4d4a40

42

a44D 44

(,

0 ,0aseca于是a

0dx0

x2y2dy4d

xx3

04sec300

[6

222D 244

(,

0 0sectan secdx2(x2y2 dyd

1 4sectand202D

122

(,

0 ,0aa

a2(x2y2)dx2a2

a2d

a4

ex2y2dDx2y24Dln1x2y2dDx2y21Dy（3）arctanxyD

，其中D是由

4,

1解（1）D,|0202},

x2y

d

e

2d2

e e 21(e41)(e41)2（2）D(,

0,02

ln(1x2y2)d

02d0

ln(12)d

1(2ln21)

(2ln21)（3）D(,)0,124 4 yarctanxd

arctan(tan)ddD

4dd4dd

x2xy2d，其中D是由x2,yx及曲線xy1所圍DD

1x2 1x2y2

1D（4）ID(x,

x2y2dxdy,Da2x2y2D (xy)1x2,xyx，故

x

y2d1dx1y2dy1(xx)dx4 D,)001x2

1

1100100D

y2dD

1

2d2d

1

2 112

d 11112

1d1

112

arcsin210

121121 1x21x21x2

(2)常見(jiàn)錯(cuò)解x

1

0所以D

1x21x2y2 0錯(cuò)誤原因x2y21所圍成的平面區(qū)域，被積函數(shù)不僅在圓周上取值而且在圓內(nèi)取值.上述解法只考慮了在x2y21上取值，因此結(jié)果是錯(cuò)誤的.D(,)02ab Ix2y2dxdy=2dd d 21(b3a3)2(b3a3) 設(shè)平面薄片所占的閉區(qū)域D是由螺線

上一段弧(0

2)與直線2

所圍成,(x,y)x2y2D 210-202

(,

0 ,02

MD

(x,y)d 2 0

20

d420

4d 圖10- 圖10-求由y0ykx(k0),z0以及球心在原點(diǎn)，半徑為10-VD

R2x2y2dD

R22dRR22d1

R22d(R222 2 3

3

xOyx2y2ax所圍成的閉區(qū)域?yàn)榈?zx2y2為頂?shù)那斨w的體積解xOyDxy|x2y2D

(,

220acosV

(x2y2)dxdy 2

2x2 a3 a34 2cos4 2

a4

aV (x2y2)dxdyd 2 x2 40

a錯(cuò)誤原因的取值區(qū)間和區(qū)間的長(zhǎng)度了，的取值區(qū)間 長(zhǎng)度是，但在積分區(qū)域中是由 變

,0變到f(x,yz)dxdydz域xyzxy10z0所圍成的閉區(qū)域；（２）zx2y2z1zx22y2z2x2

czxy(c0),2a2

yb2

1z0解（1）的頂為xyzz0xOy面上的投影是由x軸,y軸及直線xy10圍成的平面區(qū)域.可以表示為x,y,z0zxy,0y1x,0x f(x,y,z)dxdydz0

f(x,y,z)dz（２）根據(jù)三重積分的“求圍定頂”定限法.由已知z1zx2y2,xOyx2y21.1x2y2z1

y

1x2,1x1x1x1f(x,y,

dx1x2dyx2y2f(x,y,z)dz根據(jù)三重積分的“求圍定頂”定限法.由已知z2x2zx22y2zx22y解方程組

z2x

z,x2y21.所以xOy::Dxy

y1x1x1x1x

.于是1x,y,zx22y2z2x212

y

1x2,1xf(x,y,z)dv

dxdyx22y2f(x,y,1=1dx1

2dy

f(x,y,z)dz根據(jù)三重積分的“求圍定頂”定限法.由已知czxyz0xOy面上的投影是由xOy面上的橢圓 a2

1及兩個(gè)坐標(biāo)軸圍成的第一象限的部分.(x,y,

0z

xy,0yc

,0xabaabaa2af(x,y,z)dxdydza

dx

a2x

dy

f(x,y,z)dz 注意利用直角坐標(biāo)計(jì)算三重積分時(shí),一般需要根據(jù)空間域的zx、yxOy面投影.設(shè)有一物體,占有空間閉區(qū)域0x1,0y1,0z1，在（xyz）xyzxyz計(jì)算該物體的質(zhì)量.解Mx,y,zdvxy

111 111

xy2z3dxdydz,是由曲面zxy與平面yxx1z0所圍成的閉區(qū)域xyz，z0,xOyxx1yx圍成的平面區(qū)域.可以表示為(x,y,z)|0zxy,0yx,0xzz

1x 1x

xy

2z41x 1x

4x y zdzx y zdz 11

6dy11x12dx14

28

,其中x0,y0z0

yxyz1解根據(jù)三重積分的“求圍定頂”定限法.由已知的頂為xyz1z0,xOyx軸,yxy1圍成的平面區(qū)域.x,y,z0z1xy,0y1x,0x(1x(1xy3

1x0

(1xy101

1

1x0

(1xy1x1

1x1

20 2

(1xyz)2

1 1x11

dy

11y1

8 8

2(1xy)2

0

21xy

1(1x)

1

1

1

(1x)21x

0

21x

8

11ln211ln2528 28計(jì)算xyzdxdydz,其中x2y2z21方法一x2y2z21z0x2y21，故

1x1x:0z

0x1x2y積分區(qū)域可表示為: 0y 1x

0x1x1xyzdxdydz01x1

zdz111x2y xyzdxdydz=2sincosd2sin3cosdr5dr 計(jì)算

,其中z0zyyyx2的頂zyz0,xOy面上的投影區(qū)域?yàn)閤2yDxy:1x10z積分區(qū)域可表示為:x2y1,1xxzdxdydz

yxzdz

y2dy0

x

2

x方法二因?yàn)閥Ozx xzdxdydz0x2x2y

,其中zhRh

zhR0h0解當(dāng)0zh時(shí)過(guò)(0,0zxOy面的平面Dz

x2y2

z.DRz2 2 R2z2

zdxdydz

dxdy

2

z3dz

．h h 常見(jiàn)錯(cuò) h Dz

=0zdzdxdyx2y2R2hD hD

2因

z

Rzdz Rh2Dzz有關(guān)的圓域,而不是x2y2R2.實(shí)際上Dzzzz截閉區(qū)域所2x2yzdv,2x2yzx2y2(x2y2dv,其中x2y22z22解（1）在柱面坐標(biāo)系下z

,z2,,

02,01,2z

2222zdv0d0d

211(224)d1(235)d70 （2）在柱面坐標(biāo)系下,的頂為z2,z

2,,

02,02

z2

(x2y2)dv

2dddz

223d 223212

216 （1）(x2y2z2dv,其中x2y2z2zdv,其中閉區(qū)域x2y2za)2a2x2y2z2所確定解（1）在球面坐標(biāo)系下,的邊界曲面為r1可以表示為r,,02,0,0r(x2y2z2)dv

r411

0d

sind

r4dr

45(x2y2z2)dv1dvdv4 因?yàn)槿胤e分的積分區(qū)域是空間立體,被積函數(shù)不僅在球面上取值而且在球體內(nèi)取值.上述解法只考慮了在球面x2y2z2（2在球面坐標(biāo)系下r2acos和4以表示為r,,

02,0,0r2acoszdv

44rcosr2

2a000 d4sincosd000

4sincos0

1(2acos)4478a 4sincos5d0

a46一般先把區(qū)域xOy面上投影,DxyDxy內(nèi)任取一z軸的直線自下而上穿過(guò)z1(,是z的積分下限,z2(,)z的積分上限，再根據(jù)Dxy與的變化范圍.若采用球面坐標(biāo)，化積分區(qū)域的邊界曲面為球面坐標(biāo)形式時(shí)，用從原點(diǎn)出發(fā)的射線穿過(guò),穿入的邊界曲面r1,r的積分下限,r2(,)r的積分上限.再根據(jù)的圖形特點(diǎn)確定與xydv，其中x2y21z1x0,y0

z0（2）

x2y2z2dv，其中x2y2z2z（3）(x2y2dv,其中是由曲面4z225(x2y2)z5所圍成的閉區(qū)域解（1）在直角坐標(biāo)系下,積分區(qū)域可表示為x,y,1

0x1,0y11x

1x2,0z11x1xxydv

dz

ydy 8 xydv2sincosd

3ddz118 18在球面的球面坐系下,rcos可r,,02,0,0rcos22

0000

x2y2z2dv0

d2d

rr2sin 00 d2sind00

r3sindr 0

sin

1cos44 2cos4dcos 在柱面坐標(biāo)系下,的頂為

z5,z52,,z02,02,5z5220(x2y2)dv0

0

2253d55222 355822 2

z6x2y2及z x25x2zx25x2解(1)在柱面坐標(biāo)系下,z62,z,,z02,02,z6 6Vdvdddz0d0d 22

23)d

323

5x2

(4)由

x2y2

消去得這兩曲面的交線關(guān) x2y24,于是xOy面上的投影域Dxy:x2y24在柱面坐標(biāo)系下,z

,z1255,,

02,0455

z 552Vdv

d0d1 4522 15 5 53

22

14.x2y2z22zx2y2所圍立體的體積.解x2y2z22z222曲面zx2y2的柱面坐標(biāo)方程z21，在柱面坐標(biāo)系下,積分區(qū)域,,

02,01,2z

2222Vdv22

d0d 1

習(xí)題10-x2y2z2a2x2y2ax內(nèi)部的那分析x2y2axzOx面對(duì)稱,xOyx軸對(duì)稱,而球面被柱面所截(含在柱面內(nèi))的部分對(duì)稱地1、4、7、8 設(shè)在第一卦限內(nèi),球面圓柱所截部分在xOy的投影為D1,得的面積為A1

z a2a2x21 zz a2x2面積元素dA

dxdy

a2x2Aa2x2

dxdy4a2d

acos

a2a204a2(aasin)d2a2(2)0x2求錐面z 被柱面z2x2x2x2 由z

x2y22xxOyD(x,y)x2y22x

z x21x21xyzz dA

dxdy dxdyx2

AD

2dxdy

2dxdyD

x2y2R2x2z2R2所解10-22x2z2R2上的那部分面積A，則由對(duì)稱性知,16A.11xyzz D

R2D

R2 R2DRR

R2x圖R2x圖10

dyR2故16R2（1）Dy

2pxxx0y0所圍 （2）D由半橢圓形閉區(qū)域x

a2b21,y02320解（1）設(shè)質(zhì)心為2320AD

dxdy

0dx

2 dy0

2pxdxxdxdyD

0xdx

225225202 2ydxdyD

ydy 02于是x= xdxdy3

，y ydxdy3yAD故所求質(zhì)心為

5

A 8D5x0,8y0 Dy軸，故質(zhì)心(xyyx0y

ydxdy

1

dx

a2a2x

4b A D故所求質(zhì)心為04b 31Dyx2yx所圍成.它在點(diǎn)(x,y(x,y)x2y,求該薄片的質(zhì)心.1解M

(x,y)dxdyD

0

xx2ydyx

11(x4x6)dx10 x

x(x,y)dxdy

1dxxx3ydy

11(x5x7)dx35MD

M

x

M0 y

y(x,y)dxdy 1dxxx2y2dy

11(x5x8)dxMD

M

M0 質(zhì)心坐標(biāo)為35,354854設(shè)有一等腰直角三角形薄片，腰長(zhǎng)為a,各點(diǎn)處的面密度等于解設(shè)直角頂點(diǎn)為坐標(biāo)原點(diǎn)，兩腰分別與兩坐標(biāo)軸重合，則斜邊xya.則面密度(x,y)x2y2xy.M(x2y2)dxdyadxax(x2y2)dy1a4 DMx(x2y2)dxdy (x2y2)dy1a5y yD因 xMy2a,yx2a 所求質(zhì)心 25a,5a （1（1）z2x2y2，z解（1）z2x2y2z1V113 xy0z1zdv1zdv12d1d1M V

V 12d112d1

3V

故所求質(zhì)心為0034 4 9．設(shè)均勻薄片（1）D如下，求指定的轉(zhuǎn) （1）D(xy)a2b21Iy解（1）Iy

x2dxdyD

a

ba2xa2xa2xba2ba

a2x2dx4ba

a2x2dx xasint上式a

2a2sin2tacosta0 44a3b2sin2tdt2sin4tdt)4

a3b 11.一均勻物體（為常量）占有的閉區(qū)域zx2y2z0xa,ya求物體的質(zhì)心(3)z解1)的曲頂為開(kāi)口向上的旋轉(zhuǎn)拋物面zx2y2xOy面的投影區(qū)域?yàn)镈xyaxa,ayVx2y2dxdyadx

x2y2dy8a4

DmV83

x2y

m40dx0

dz40dx0(xy4aax21a3dx41a41a48a4333 333 xy0

x2ymz zdvm

m0

a(x42x2y2y44a4 4a3aax42a3x2154a 0 31a62a61a6

328a6

7a24a4

4a

x2y Iz

y)dv40dx0

(x

00004adxax2y22dy4adxa(x42x2y2y40000428a6112a6 14.(x,y,z)x2y2R2,0zh求它對(duì)位于點(diǎn)M00,0aah解FFxFyFz},引力系數(shù)為GdFG ,

G , x2y2(za)2 x2y2(za)2dFz

(zxx2y2(za)2FxFy0F G z hh

(z

0(z

x2y2R

R x2y2(zx2y2(za)2h

d

2 2G

(za) R2(zR2(z

h1h

z

R2(R2(z h2Gh

R2(za)2R2R2(h

R2a2F{Fx,Fy,Fz}0,0,2Gh

R2a2R2(hR2(h小結(jié)1)在利用重積分求解應(yīng)用問(wèn)題時(shí),（211（2（31{(x,y,z)|x2y2z2R2,z0},12{(x,y,z)|x2y2z2R2,x0,y0,z0},2則有(A)xdv4xdv (B)ydv4ydv (C)zdv4zdv (D)xyzdv4xyzdv D(x,y)axa,xya則

D1(x,

0xa,xya(xycosxsiny)dxdy D(A)2cosxsin(C)4(xycosxsin

(B)(D) 設(shè)f(x)為連續(xù)函數(shù)，F(xiàn)(t)1dyyf(x)dx，則F(2) (A)2f

f(C)f (D)（1）zxy均是偶函數(shù)，1yOzxOz

故選(C)

xyxy(xy)dxdy0Dcosxsinyyx(cosxsiny)dxdy=0+2cosxsinydxdy.答案選

F(t1f(x)dx1dy1(x1f于是F(tt1f(t從而F(2f(2.故選（1）1xsinydD是頂點(diǎn)分別為(0,0),(1,0),(1,2)D(0,1)的梯形閉區(qū)域（2）x2y2d,D0ysinx,0xDD

R2x2y2d Dx2y2Rxy23x6y9d,D是閉區(qū)域x2y2R2D解（1）Dxy)0y1x,0x1,1(1x)sinyd1D

1dx

(1x)sin11x1xcos1022

ttcostdttt

tsintcost3sin1cos12sin2cos2由于x2dx2dxsin Dx2sinxdxx2cosx2xcosxdx240 sin

9y2d y2dy sin3xdx 2sin3xdx 9 0 3D9 x2y2d2409D利用極坐標(biāo)計(jì)算,R2x2y2dR2R22

R222(R(R22)2

cos12

3R

23 R34 R3

(1sin3

3 3利用對(duì)稱性可知3xd06yd0, 9d9R2,y2dd2sin2d R4 因 原式9R24常見(jiàn)錯(cuò)誤對(duì)于二重積分f(xy)dxdy，D：x2y2R2，D1Dx2y2R2,x0,y0，f(x,y)dxdy4f(x,y)dxdy 錯(cuò)誤原因應(yīng)用對(duì)稱性時(shí)，只注意到積分區(qū)域的對(duì)稱，沒(méi)考慮被３. 1(04（1）dy04

f(x,y)dx 2

3f2（2）0f2

fx,ydx

xx0

x,

4 x1(y4),0y442 2 D表示為(xy)2x0,2x4y4x2 1(

402402

f(x,y)dx

f(x,y)dy所給二次積分等于二重積分f(xy)dxdyDD D2D1xy|0y1,0x2yD2{(x,y)|0x3y,1y3}于是 D可以表示2

D(x,D3

y3x,0x2x2x2所 原式0dx2

f(x,y)dy2yy2yy原式= f(x,y)dx1

f(x,y)dx

adyemaxfxdxaxemaxfa 證emaxemaxfxdxdyaa

maxfD aaxemaxfxdx01x2f(xyD(xy)x2y2y1x2f(x,y)

8f(x,y)dxdyf(xy解

f(x,y)dxdyADf(x,y)dxdy 2 A12

1x2y2dxdy8AdxdyD1x2y2dxdyD在極坐標(biāo)系中,D(,

0sin,0221x2y2dxdy2，A1

1x2y2dxdy1

f(x,y)8.計(jì)算下列三重積分

821x21x22)

zln(x2y2z2x2y2z2x2y2z2

dv,其中

y2z2dv其中xOyy2xx5 2)因被積函數(shù)是關(guān)于z的奇函數(shù)，積分區(qū)域關(guān)于xOy面zln(x2y2z2對(duì)稱，所以

x2y2z2

dv3)y22xxy2z22x在柱面坐標(biāo)系下,,z02,0

2

z5y2z2dv2

3

2

0d

d2dx2020

5

d 2 xyz1被三坐標(biāo)面所割出的有限部分的面積 平面方程zccxcy，它被三坐標(biāo)軸所割出的有限部 xOyDx軸、yxy1A

1x1xyzz 1c ca