(x,y)的電荷，且(x,y)D上連續(xù)，試用二重積分表達(dá)該板上的全部電荷Q．解Dn個小閉區(qū)域i，其面積也記為i(i1,2,,n)

Q

(,)

(x,y)d 0 其中max{的直徑 1I1(x2y23dDxy)1x1,2y2}I2(x2y231 解由二重積分的幾何意義知，IDzx2y23的曲頂柱體 其中位于第一卦限的部分即為2．由此可知I1.)dD

kf(x,y)dkf(x,

f(xy)df(xy)df(xy)dDD1D2D1D2 證 由于被積函數(shù)f(x,y)1，故由二重積分定義dlimf(,)lim

lim 0i 0

kf(x,y)dlimkf(,)klimnf(,)

f(x,y)d 0i 0 )因?yàn)楹瘮?shù)f(x,y)D上可積，故不論把D變的，因此在分割D時，可以使D1和D2的公共邊界是一條分割線。這樣f(x,y)f(i,i)if(i,i)if(i,i)i 令所有i的直徑的最大值0f(x,y)df(x,y)df(x,y)dD1 (xy)2d與(xy)3dDxyxy1 (xy)2d與(xy)3dD是由圓周(x2)2y1)22 ln(xD

[ln(xy)]2D

D是三角形閉區(qū)域，三頂點(diǎn)分別為ln(xy)d與[ln(xy)]2dDxy3x5,0y1} 解D0xy1，故有(xy)3xy)2質(zhì)４，可得(xy)3d(xy)2d D位于半平面{(xy|xy1}D上有(xy)2xy)3．從而(xy)2d(xy)3d. 由于積分區(qū)域D位于條形區(qū)域{(xy|1xy2}D上的點(diǎn)滿足0ln(xy1，從而有[ln(xy)]2ln(xy．因此[ln(xy)]2dln(xy)d D位于半平面{(xy|xye}D上有l(wèi)n(xy1[ln(xy)]2ln(xy．因此[ln(xy)]2dln(xy)d Ixy(xy)dDxy)0x1,0y1}DIsin2xsin2ydDxy)0x0yDI(xy1)dDxy0x10y2}DI(x24y29)dDxy)x2y2D)D0x10y1，從而0xy(xy2D于1，因此0xy(xy)dDD0sinx10siny1，從而0sin2xsin2y1D的面積等于π2，因此0sin2xsin2ydπ2.DD0xy14D的面積等于2，因此2(xy1)dDD上，0x2y24，從而9x24y294(x2y2925,的面積等于4π，因此36π(x24y29)dD

(x2y2dDxy||x|1|y|1}DD(x33x2yy3dDxy|0x1,0y1}Dxcos(xy)dD是頂點(diǎn)分別為(0,0)(π0)和(ππD1 1 y3 1解(x2y2ddx(x2y2dyx2ydx(2x2)dx. D

3

D可用不等式表示為0y3x,0x2(3x2y)d2dx2x 2 2[3xyy2]2x D

2(42x2x2)dx20 (x33x2yy3)d0dy0(x33x2yy3D1

x3yy3

dy1(

yy3)dy0 0D可用不等式表示為0yπxcos(xy)d

y)dy

x[sin(xy)]xD 0xdx0 Dπx(sin2xsinx)dx3

xyd，其中D是由兩條拋物線y ，yx2所圍成的閉區(qū)域xDxxy2dDx2y24yDD(x2y2x)dDy2yxy2xD解)D可用不等式表示為x2y x,0x1，于xyd1xdx

xydy2 xdx

(x4-x)dx60D

3

xy 17 17

D可用不等式表示為0x 4y2 2y2，于4 2 12 4xyd2yD

xdx22y(4y)dy15DD1D2，其中D1{(x,y)|x1yx 1xD1xy|x1yx1,0x1}exydexydexy 0exdxx1eydy1exdxx1 x 0(e2x1e1)dx1(ee2x1)dxe Dyxy,0y22y 2y2yD2(xyx)d0dyyD22 2ydy219y33y2dy130

yxx

0 3 2y

6 2DIf(x,Dyxy24xxx2y2r2y0)yxx2y

1(x0)x環(huán)形閉區(qū)域{(xy|1x2y24}解yxy24x的交點(diǎn)為(0,0和(4,4)I4dx4xf(x,y)dy或I f(x, 4將D用不等式表示為0y r2x2,rxr，于是可將I化D用不等式表示為

Irr2rr2rr2r

f(x,y)dy,0yrIrrrI0rrr 1和(2,2)，于是I2dxxf(x,y)dy(2, I1dy1f(x,y)dx1dyyf(x, 4y44y44I24

f(x,y)dx1

f(x,4444 44或

f(x,y)dx

f(x,44444I2 f(x,y)dy1dy1x2f(x,4 444 f(x,y)dy1444 20dy0f(x,y)dx (2)0dyy2f(x,y)dx(3)0 f(x,y)dx

f(x,y)dy

ln1 f(x,y)dy

0dxsinxf(x,y)dy2二次積分等于二重積分f(xy)dDDxy|0xy,0y1}D可改寫為{(xy|xy1,0x1} 原式0dxxf(x 所給二次積分等于二重積分f(xy)dDDxy|y2x2y,0y2D可改寫為{(xy|xy2

0x4} 原式 dx2

f(x,所給二次積分等于二重積分f(xy)dD11Dxy| x ,011{(xy|0y1x2 11原式1 f(x,11所給二次積分等于二重積分f(xy)dDDxy|2xy2xx2,1x2}D1{(xy|2yx1 ,0y1 原式0dy2 f(x,所給二次積分等于二重積分f(xy)dDDxy|0ylnx,1xe}D可改寫為{(xy|eyxe,0y1} 原式0dyeyf(x DD1{(x,y)|arcsinyxπarcsiny,0y1}D2xy|2arcsinyxπ,1y0}原式

πarcsin f(x,y)dx

f(x,

2arcsinx0，y0，x1，y1z0及2x3yzxOy面上的閉區(qū)域Dxy|0y1,0x1}，頂是曲面z62x3y，因此所求立體的體積為V(62x3y)dxdy1dx1(62x3y)dy7 zx22y2z62x2y2解所求立體在xOyD{(x,y)|x2y2 V(62x2y2)d(x22y2 (63x23y2)d(632 00d00

2(632)dD){(x,y)|x2y2a2}(a0) (2){(x,y)|x2y22x}xy|a2x2y2b2}，其中0ab；(4){(xy|0y1x0x1}) f(x,y)df(cos,sin)dd

d0f(cos,sin) ππ}， f(x,y)d f(cos,sin)dd2

f(cos,sinbb2f(x,y)df(cos,sin)dd

daf(cos,sin1xy1sincosD{(,)|0

sin

,02 f(x,y)d

f(cos,sin)dd

sincosf(cos,sin

0dx0f(x,y)dy (2)0 f(x,y)dy (3)0dx1 f(x,y)dy (4)0dx0f(x,y)dy解yx將積分區(qū)域DD1D2}D{(,)|0sec,0π} D{(,)|0csc ππ 0原式40

f(cos,sin)d

2d

f(cos,sin 040 yx和y3x的方程分別是2sec,π4D,|02sec,

ππf} } 2

x2y2f( 原式304

f1sin

11 sin

1,0π}2} 原式2 f(cos,sin

sinsin2cos2tansec；兩者的交點(diǎn)與原點(diǎn)的連線的方程是π4}D{(,)|tansecsec,0}4 原式4d

f(cos,sin tan (x2y2)dy

x2y2dy a a )dx2(x2y2)2dy (4) (x2y2 )2 2a 原式2

2d

4)π原式π

4d

a

d

223a 223 在極坐標(biāo)中，yx2sin2cos2tansecyx的方程是πD,|0tansec,0π tansec 原式4

d }D{(,)|0a,0π}22原式πda2dπ2

ex2y2dDx2y24DarctanydDx2y24x2y21y0yx ) 原式2πd2e2dπ(e4 })}4 原式4

d

3.

D

y2d，其中D是由直線x2， x及曲線xy1所圍成的閉區(qū)域

11x21x2

(x2y2dDyxyxayay3a(a0)DD

x2y2dD是圓環(huán)形閉區(qū)域{(xy|a2x2y2)x x y2d1dx1y2dy4 )211x2y2 111 2 d11 11

d (π D

y)d0dyya(xy)dx0(2ayay

)dy14a3D,|01,0π}2x2y2ddd2πdb2d2π(b3a3 y0ykx(k0)z0R的上半球面所圍成的在第一卦限內(nèi)的立體的體積（．解R2VR2x2y2d R2 R2adR dRR2 If(xyz)dxdydz為三次積分，其中積分區(qū)域xyzxy10z0zx2y2z1zx22y2z2x2 由曲面czxy(c0) 0zxy,0y1x,0x1 I0dx0dy0f(x,y,1可用不等式表示為：x2y2z y1x2 1x1I1dx1x2dyx2y2f(x,y,1可用不等式表示為：x22y2z2x2 y1x2 1x1 2 I1dx1x2dyx22y2f(x,y,1 ,0yc

,0xaIadx

dycf(x,y, xy2z3dxdydz，其中zxyyxx1z0解0zxy,0yx,0x1 xy2z3dxdydz1xdxxy2dyxy 11xxx4dy11x10004(1x(1xy 3，其中為平面 0， 0， 0， z1所圍成解0z1xy,0y1x,0x1 1dx1xdy1x (1xy (1xy dx1x 1xydy

1x

1002(1xy1002(1xy00 2(1x

2

2 1

1 2

dx

1(ln20 2(1x 1解0z1x2y2,01

,0x11x

1x211

11y

0 y dy20x2(1x)4 011x(1-x2)2dx18 xzdxdydz，其中z0zyy1yx2解可用不等式表示為：0z x2y1,1x1，因 xzdxdydz1xdxx2dy0 1 1 1 1xdxx22dy61x(1x)dxx2 與平面zh(R0,hx2 解xOy

{(x,y)|x2y2R2}，{(x,y,z) z (x,y)hRx2 hRx2

zdxdydzdxdyhx2y2zdz2hR2(xy) D 1

2 R

2 2

dxdyR2

y)dxdy

2

2R20d0d4πRh 2x22x2

zx2y2(x2y2dv，其中x2y22zz22z22,01,02πzdvzdddz

2 2d0d2 1

6 d(224)d 2π2 2 6 )x2y22zz2zx2y24，從而知xOyDxyxy|x2y24}2, 2π.2z2,0 2, 2π.2 2 2 (xy)dvdddzdd2 6d2d

23(22)d2π4

16π 12 (x2y2z2dv，其中x2y2z21zdv，其中x2y2za)2a2x2y2z2(x2y2z2)dvr2 2

r5 dsind

r4dr2π[-cos]π05

在球面坐標(biāo)系中，不等式x2y2za)2a2x2y2z22azr22arcos，即r2acos；x2y2z2r2sin2r2cos2，即tan1，亦即4因此可表示為0r2acos,0π,02π4 rdr zdvrcos rdr 4sin24sin

2acos

xydv，其中x2y21z1z0x0y0

x2y2z2dv，其中x2y2z2z(x2y2dv，其中是由曲面4z225(x2y2z5x2y2(x2y2dx2y2

Az0)2 xydv2sincos 2sincosd1ddz10008x2y2z2zr2rcosrcos.示為0rcos,0π,02π2x2y2z2dvrr 22πdπ2

cosr3drπ

z5,02,02π2(x2y2)dv2 2πd23d 5dz2可表示為arA,0π,02π2 (x2y2)dvrsin2 22πdπsin3dr4dr4π(A5a52A x2y2z2a2x2y2axa2x2a2x2z

za2a2x2

a2a2x2a2x21xyzz 1xyzz2 A dxdy

a2x2a2x22πa2a2

dd

d2a2(πa2a2Dx2 x2解

x2y2x2解得x2y22x，故曲面在xOy面上的投影區(qū)域D{(x,y)|x2y22x}x21z1z2z2 y x21x2DAD

22dxdy22

2d

x2y2R2x2z2R2x2z2R2A，則由對稱性知全部表面的面積為16A．1R2A z 1R2DD

RRRR2R2

dxdyR

dyR2R2D故全部表面積為16R2Dy2pxxx0y0 D是半橢圓形閉區(qū)域(xy|a2b21,y0 D是介于兩個圓racos,rbcos(0ab)解設(shè)質(zhì)心為(xyA dxdy 0 230x 30xD

dy

xdxdy0 D

2pxdy22px5502502ydxdy0 ydy 0 2于是x1xdxdy3x y1ydxdy 3y,故所求質(zhì)心為 2 5 8 (x0,y0 y

ydxdy

a

ydy4bbabax因此所求質(zhì)心為(0,4b

A Dx軸，故質(zhì)心(xyxy0bAπ

a

π(b2a22 2

b xdxdy

cosdd2cosdacos2d8(b3a32 a2ab a2ab 故xAxdxdy 2(a .所求質(zhì)心為 2(a ,0 Dyx2yx所圍成，它在點(diǎn)(xy面密度(x,y)x2y解M x2ydxdy

1

ydy1x D

M y(x,y)dxdy x2y2dxdy1

xy2dy1

M x(x,y)dxdy x3ydxdy1

xydy1 于是xMy35 yMx35.所求質(zhì)心為(35, 48

解面密度(xyx2y2xyM(x2y2)dxdyadxax(x2y2)dy1MM x(xy)dxdy

a

6ax(x2y2)dy

于是xMy2 yxD2a.所求質(zhì)心為

(

z2x2y2,za2x2z A2x2y2,z (Aaa2x2zx2y2,xya,x0,y0,z0解z軸對稱，又由于它是勻質(zhì)的，因此它的質(zhì)心位于zxy0．立體的體積為V1π.z zdv zdzxx

V

xy

xy 12πd11(12)d3V 0

3)4zxy0．立體的體積為V2πA2a33z1zdv1rcosr2V V1 3(A4a4d2sincosdr3dr 3V 3(A4a4故所求質(zhì)心為(0,0,8(A3a3

8(Aa{(x,y,z)|0xa,0yax,0zx2 a x1Vdv0

dz6a 1 a x zVzdvV0 zdz30a 1 a xxVxdvV0 zdz5由于立體勻質(zhì)且關(guān)于平面yx對稱，故yx2a，所求質(zhì)心為 7a2)(a, xyz|x2y2z22Rz，它在內(nèi)部各點(diǎn)處的密度的大小0r2R 0π,02z軸對稱，故可知其質(zhì)心位于zxy0．2Mdv2πdπd2Rcosr2r2sindr32πR52z1zdv

22πdπd2Rcosr2rcosr2sindr52M M

5R)4設(shè)均勻薄片(面密度為常數(shù)D D(xy|a2b21Iy Dy29xx2II D為矩形閉區(qū)域{(xy|0xa0ybIxIya

2b 4bbaD解baD

x2dxdy

x2dxaaa

dy

a aa2x2dxa2x2dxa2x24b 上式 2a3sin2tcostacostdt4a3b2sin2tdt2sin4tdt a ))D

y3x,0xx22x2

2I y2dxdy 2

y2dy

2 3 3

x2 ;2 3 2I x2dxdy

2x2dx32dy

x2 .x2x27 27D 0３) Ixydxdy0D

ydy3 3Iyxdxdy0xdx0dy 3D1h1解Ix y2dxdy解Ix y2dxdy2dx2y2dy Iy x2dxdy2x2dx2dy

一均勻物體由曲面zx2y2和平面z0,|x|a|y|a所圍成,))) x2

a a3 8

2 0 y 40

a 4 x2 4 a 2 7)zMzdvV0dx0 zdzV0dx02(x2xyy)dy15a x2 ３) Iz

y)dv40dx0 (xy)dz

a度1 解(xyz|x2y2a2

hzh(,,z)|02π,0a,

hzh 2 2a 2a

I (x2y2)dv 2dddz

2dz2

設(shè)面密度為常量x2Dxy0)|R x2 解

G x (x2y2a2)2

G (x2y2z2)2FxG

d

33π33d

3333D(x2y2a2 (2a2 2G

2 R2R22R2a21R22

R2R21FzGa

d

2dR2 d π3D(x2y2a2 (2a2πGa

R2R22R21 .R2R2R2R22R2R2R2R22F

R2 2

R2

，占有閉區(qū)域xyz|x2y2R20zh，求它對于位于點(diǎn)M0(0,0,a)(ah)處的單位質(zhì)量的質(zhì)點(diǎn)的引力.FxFy0zFz z dvG(z 3R2(h [x2y2(za)2 x2y2R2[x2y2(zR2(h

h

2π

2πGh R2a2 [r2(za)2復(fù)習(xí)題設(shè)D是正方形區(qū)域{(x,y)|0x1,0y1}，則xydxdy .1 已知D是長方形區(qū)域{(xy|axb0y1}yf(x)dxdy1Dbaf(x)dx bDxy1和兩坐標(biāo)軸圍城的三角形區(qū)域，則二重積分f(x)dxdyD1為定積分f(x)dxdy0(x)dx，那么(x) D若

f(x,y)dy1dyx(y)f(x,y)dx，那么區(qū)間[x(y),x(y)] x 若

0a

x(y aa.f(x,y)dyd0rf(rcos,rsin)dr.則區(qū)間(

π,π Dykx(k0),y0x1所圍成的三角形區(qū)域，且xy2dxdy13;33;33k 3

點(diǎn)在(1,1)點(diǎn)，記

的外接圓區(qū)域，I1e2yx2y2則I1,I2,I3的大小順序?yàn)?)

I2e2yx2y2

I3e2yx2y2I1I2I3 B.I2I1I3 C.I3I1I2 I3I2 2I0d 化為直角坐標(biāo)系下的二次積分，則I I1 f(x,y)dx B.I0 2y2y2yC.I1 f(x,y)dx D.I1 f(x2yD是第二象限內(nèi)的一個有界閉區(qū)域，而且0y1I1yxd,I2y2xd

I3y2xd 則I1,I2,I3的大小順序?yàn)? A.I1I2I3 B.I2I1I3 C.I3I1I2 D.I3I2x2計算旋轉(zhuǎn)拋物面z1 在1z2那部分曲面的面積的是 2xyxy

1x2y2d 1x2y2d

xyxy

1x2y2d;1x2y2d計算重積分exdxdyDx0,yexy2所圍成的區(qū)域D2解exdxdy2dylnyexdx2y1)dy12 Dy計算重積分2dxdyDx2,yxxy1所圍成的區(qū)域yD 44

y解2dxdyyD

x

ydy2(xx)dx計算重積分(xy)dxdyDx2y22x2y22x所圍成的區(qū)域D解 (xy)dxdy2d (rcosrsin)rdr 2

(rcosrsin) 2 DD

4 2

rsin) 2將二重積分f(xy)dD給定如下D D是區(qū)域{(xy|a2b21,y0}a0,b0)D是區(qū)域{(xy|yx2y1x2Dyxyx3所圍成的區(qū)域Dy0,y1,yxyx2所圍成的區(qū)域 1解(1)f(x,y)d f(x,y)dydy2f(x, D

f(x,y)ddx

f(x,y)dy0

abyab

f(x,aba aba

f(x,y)d2dx

f(x,y)dy2 f(x,y)dx1 f(x,1 12 (4)f(x,y)ddx3f(x,y)dy

f(x,D

yf(x,y)d

f(x,y)dy1dx0f(x,y)dy2dxx2f(x,y)dy0 f(x, DD是區(qū)域{(xy|x2y22D是區(qū)域{(xy|x2y21,xy1}D是區(qū)域{(xy|1x2y24}2yDyx,y02y1解

f(x,y)dy

rf(rcos,rsin

0

f(x,

2y2y(2)

2

f(x,y)dy0dy1 f(x,y)dx0

sin 2

f(x,y)dy14 4

f(x,y)dy1 f(x,y)dy1

f(x,2

f(x,y)dy1 f(x,y)dy1 f(x,y)dy1 f(x,0d1rf(rcos,rsinxdxf(x,y)dy1dy 1

yf(x,y)dx

4

cosrf(rcos,rsin D是長方形區(qū)域{(xy|axbcyd，試證明 af

g(x)dxf(x)g(D

(f(x),g(x連續(xù) f(x)g(y)daD

f(x)g(y)dyaf(x)dxcg(y)dyaf(x)dxcgDf(x2y2D

D{(x,y)|0y R2x2 π解fD

y)dπ0rf(r)dr20fy10 f(x,y)dxy1

ln1 f(x,y)dy

20dxxf(x,y)dy 20dx0f(x,y)dy1 f(x,y)dy 20dx0f(x,y)dy1 f(x,y)dy (1)0dxx2f(x 0dyeyf(x, 0dyyf(x,y)dx2dyyf(x,

20dy f(x,1y0dy2f(x,yyy0 f(x,y2axa10 f(x,y)dy(a2axa10 f(x,y)dy

20dy f(x,y)dx 解

f y0rf

2a f(x,y)dx4dcosrf(rcos,rsin)dr2

rf(rcos,rsina a4f(x,y)dx rf(rcos,rsin210dy1xdxf(x,y)dy1x

2

f(x,y)dy4dcossinrf(rcos,rsin Dyex,ye2xx1所圍成 e2 解S

dD

0

dy (e1)2Dy2xxy2所圍成解S

d1

2ydx9 D

D由極坐標(biāo)下不等式ra(1cosra所確定解S1πa2 d1πa2 a(1cos)rdr5π

2πd

2a 1 1解

z1x2y2z0V(1x2y2)d2πd1(1r2)rdrπ zx2y2x2y2z22z1x21x2y2)d1x2

1r2)rdr7π解VD

1zx2y2xy1 解V(x2y2)d (x2y2)dy D

44

99解x rsinπy

所以質(zhì)心為0, I(x2y2z2dv,其中Gx2y2z22zG π 解：Id2d 22sindd2 cos5sind 0 將下列各題中三重積分f(xyz)dv（1）:1x2,2y1,0z12（2）:xyz1,x,y,z0（3）y x,y0,z0,xzπ2（4）zx2y2yx2y1,z01解：（1）f(xyz)dv2dx1dy2f(xyz)dz1 （2）f(x,y,z)dv

dy f(xyz)dz π（3）f(x,y,z)dv2π

πdy f(xyz)dz x2（4）f(x,y,z)dvdx2

f(x,y, 計算三重積分(x2y2dxdydz，其中z1(x2y2z2 2 2 (xy)dxdydzdrdrr2rdrπ x2y2 (a

0 dy0

R2R2R2xR2x2 2

xyzdz R2R2

66

zx2y2dz

2 2

zrdz （2）

22

πR2x2π

x2y2z2dz

d2d

Rrr2sindr

R2 R R2證明

復(fù)習(xí)題ay0ay

證明上式左端的二次積分等于二重積分em(axf(x)dxdyDD{(x,y)|0xy,0ya}{(x,y)|xya,0xaaay0ay

em(ax)f(x)dx0

em(ax)f(x)dya(ax)em(ax)f0把積分f(xyz)dxdydzzx2y2y