第六章多元數(shù)量值函數(shù)積分學(xué)3三重的計算

V 設(shè)平行于z軸穿過V的內(nèi)部將V投影到xoy面上,得閉區(qū)域Dxy,以Dxy的邊界曲線為準(zhǔn)線,作母線平行于z軸的柱面,該柱面與區(qū)域V的

Dxy S1:zz1(x,S2:zz2(x,z1x,y),z2x,y)在Dxy 且z1(x,y)z2(x,y)f(x,y,z)的某種物質(zhì),分別交曲面S1、S2于P1、P2兩點

Dxy (x,細(xì)桿,當(dāng)點P固定時,f(xyz)僅是z的函數(shù) zz1xy與zz2x因此,該細(xì)桿的質(zhì)量為 F(x,y) z2(x,yz1(x,y

f(x,y,的密度函數(shù),則空間立體

(x,

z2(x,y

f(x,y,

z1(x,y 這里對z積分時,先將x,y看成常數(shù) f(xyz)z的函數(shù) 表示為 ax

y(x)yy(m

y2(x)

z2(x,y

f(x,y,b y1(x z1(x,yb所求質(zhì)量就是所要求的三重積分的值f(x,y,z)dVf(x,y,

z2(x,yz1(x,y

f(x,y,bdxb

y2(x)dyz2(x,y

f(x,y, y1(x z1(x,y Vxyz10z1x解V0y110x1

ydV

1x

V11

1dx

y(1x 10[2(1

(1x)]dx 例2計算x2zdV其中Vyx2V與平面z0,y1,z1y1y 21 dxx dy21

x211211

x2

1

x2(1x61612例3

xx A(1,0,0),B(1,1,0),C(1,1,2),D(2,0,0),E(2,2,y梯形ACFD所在平面過x軸,其方程為:zyV:0z20y1xV:0z2 0y 1x dV 2

dy2yx2 V

x2 2xdx2x

2 y x2y 2[ln(2x2)lnx21ln例4If(x,yz)dxdydzV積分，其中積分區(qū)域Vzx22O及z2Ozx22解由

z2x2x2y21,

:x2y2x22y2z21111

y1x x22y2z2

y1111

1x22 I

1 dyx2

f(x,y,z)dz.zzxy

影區(qū)間[1,c2 z[c,c]用過z點且平行于 c計算二重積分fxyc

最后計算定積分c2

f(x,y,V坐標(biāo)面及平面xyz1所圍成的閉區(qū)域解

zdxdydz

0zdz 1 1dxdy1(1z)(111

0z

1(12

xz

yzy1 若V為R3中關(guān)于xoy面對稱的有界閉區(qū)域,fxyz)為V上的連續(xù)函數(shù),則當(dāng)fx,yz)關(guān)于z為奇函數(shù)時,fxyz)dVV當(dāng)fxyz)關(guān)于z為偶函數(shù)時f(x,y,z)dV2f(x,y, 其中V上為V在xoy面上方的部分 2c0z2z1z22c0z2z1z2xyb(12c2)ab(1c2

z2dxdydz

{(x,y)|

2z2a(1c2z2a(1c2)c

原式

ab(1c2

)zdz

abc 算ezdV,V:x2y2z2Vzy ezdV zy 021ezdz0xx21(1z2)ezdz02例8計算V

zln(x2y2z2x2y2z2

其中積分區(qū)域V{(x,y,z) x2y2z2zyxoy面對稱，z的zyV

zln(x2y2z2x2y2z2

dxdydzx,z(r,z)就叫點M的z規(guī)定：0r02z.0 x

M(x,y,(r,,y Pxrcos yrsin

(r,,r為常數(shù)z

平面 對r,z分別取得增量dr,d,dzdV f(x,y,Vf(rcos,rsin,V例1計算Izdxdydz,其中V是拋物面x2y2V與球面x2y2z24所圍的立體(取拋物面之上z0z0xx2y23z, r23z,x2y2z2r2z2yDxy:r 3,4r2 4r23V

0r303zdxdydzzV2

drr24r24r23

zzx2y2zx2y2z2z4 2z3 xyr3x.x2x2V

dV其中Vzx2y2與z1所圍成的閉區(qū)域 r2z0rV

0 yxx2y2dVrrdrdyxV rdz

例3計算I(x2y2 其中V是錐Vx2y2z2與平面za(a0)所圍的立體x2y2rz0r0

zr I2dardrar 2

r3(ar

a5 例4計算Ix2y2dxdydz,其中VVy22z, x0繞z軸旋轉(zhuǎn)一周而成的曲面與兩平面z2,z8所圍的立體.y2 由 xzx0yx2zx0yz8z82x0y 由x2y22z, z 2I(x2y2)dxdydz(x2y2

drdrr2

dz drdr2 2336

解2旋轉(zhuǎn)面方程為

2y22z, z zx0zx0y2zzVxzVx0y 2z 0r 02

3I1(3

y)dxdydz

drr

zxzx0yx 2z

0r 02 I2(

)dxdydz

drr2

rdz 6II

45 336 例4計算Ix2y2dxdydz,其中VVy22z, x0繞z軸旋轉(zhuǎn)一周而成的曲面與兩平面z2,z8所圍的立體.z8z82x0y 由z r2Dz:0r ,02I8dz(x2

y2222 2

r2 82

2168336例5計算(xyz)2 其中V是由拋V面zx2y2和球面x2y2z22所圍成的立體zx0y解(xyzx0yx2y2z22(xyyzV關(guān)于zox面對稱(xyyz)dVVI(xyz)2V[x2y2z22(xyyz VV關(guān)于yoz面對稱xzdV V由輪換對稱性x2dVy2dV I(2x2z2)dVV2rr22r投影區(qū)域

,zx0y0rzx0y02I(2x2z222r2 d

r(2r2cos2z222

設(shè)Mx,yz)為空間內(nèi)一點，則點M可用三個有次z從正z軸來看自x方向轉(zhuǎn)到有向線段OP的角這里P為點M在xoy面上的投影，這樣的三個數(shù)(,,)x

Pzzxsinz00

0

yy02 0. z 半平面及+d；

半徑為及d的球面 .xf(x,y,z)dVf(x,y,z)dxdydz V 是先對積分，再對積分，最后對積分. dVdxdydzrdrddz2sinddd 函數(shù)f(x,y,z)以及積分區(qū)域V所針對的特點.例1計算VR2x2R2x2

dV,其中Vx2y2x2y2z20yz x2y2所圍成的閉區(qū)域 0y解在球坐標(biāo)系中R2x2R2x22Rcos02Rcos0 02 V

x2y2z2

2R2sindV20

40

2R0

42

4

4R4cos4 0R4(8 5例2計算Ix2y2dxdydz,其中VVx2y2z2與平面za(a0)所圍的立體za

cosx2y2 4V:0

cos0 02 I(x2y2V2sin22sindd

4

cos4sin3d

524sin 50a5

cos

4 x例 計算 (x2y2z2)3V其中V R2x2y2z2R2 0RR R1R2,V02Icos32sinddd

z yR dsind2cos

2d 4(sinR3sinR3 x2例4求曲面x2y2z22x2解V由錐面與球面圍成，采用球面坐標(biāo).x2y2由x2y2z2 x2y2z

40 V0 02 xV 2d4d2a2sind 2

sin

d 4(

2 2xx2例5算(xz)d中V由x2V1x21x2V關(guān)于yozfxyz)x為x有xdVV(xz)dV

4d

cos2sind

V例6求