高數(shù)下934三重二重積分的被積函數(shù)是二元范

圍是平面有界閉域.三重積分可看作二重積分WWmxyz?C,W內(nèi)的物質(zhì)的M.WW“分割近似求和,n

DVkM

lfi0k定義fx,yzx,yz?W,Wn

作作lfi

Wfxy 則稱此極限為函數(shù)f(x,y,

fxyz)稱為被積函數(shù)WWdV=由定義可知，當(dāng)被積函數(shù)fxyzfxyz)dVV（積分區(qū)域W的體積W

Wf(x,y,z)d =f(x,h,z) f(x,y,z)dV=bdxddy f(x,y,z Wf(xyzf1(xf2yf3(z) f(x,y,z)dV=bf1(x)dxdf2(y)dy W

z=z2(x,方法1方法2

z=z1(x,方法1假設(shè)平行于z軸且邊界曲面S相交不

z=z2(x,z=z1(x,S：z=z1(x, S2:z=z2(x,z1xy)z2xy)都是D z1(x,y)￡z2(x,z1(x,y)￡z￡z2(x,

(x,y)?z2(x,y)f(x,y,z)dzxd

z=z2(x,z z1(x,

z=z1(x,Wf(x,y,z)d

z2(x,y)f(x,y,z)dzdxd

dxdD z1(x, DdxdyDdxdy2z(x,先對(duì)z,再對(duì)y,后對(duì) f(x,y,z)W

y=y2(xDy=y1(x b1 bdx

y2(x

(x,y

f(x,y,z) y1(x z1(x,y

z1(x,y)￡z￡z2(x,y) (x,y)?D:x1(y)￡x￡x2(

x=x2(D c￡y￡

x=x1( 先對(duì)z,再對(duì)x,后對(duì)y2d2d

x2(x1(

z2(x,z1(x,

f(x,y,z)例1設(shè)有一物體Ω=[0,1;0,1;0,1](方體)，它在點(diǎn)P(x,y,z)處的密度為點(diǎn)P到 M=(x2+y2+z2)dxdydz 1111yx1=1dx1dy1(x2

+z2 001dx00

1(x

3=1x2+2

03 03 例2計(jì)算三重積分Wxdxdydz, 三個(gè)坐標(biāo)面及平面x+2y+z=1所圍區(qū)域解:W為上下型z=1x-2zy112D DxD120￡z￡1-x-2 D122 0￡y￡1(1-20￡x￡ I=Wxdxd

11-x-21 xdx0=

例3Ifx,yz)dxdydzWW為zx22y2及z2x2解：zx22

,消去z由z=2-xDxy

x2+y2￡W：x2+2y2￡z￡2-x2-

￡y 1-1-\I=1dx-

2-x

f(x,y,z)dz.例4z8x2y2zx23解：z8x2 z=x2+3消去z，得Dxy:x2+2y2￡V=W00

yx(4-x2)/yx0

8-x2-y2 2例5化三重積分IfxyzdVWW:zxyxy1z0 x+ DI=

xyf(x,y,z) 0

1-x

xyf(x,y,z) 例6三重積分fxyzdVW其中積分區(qū)域Wzx2y2yx2y1z0 W:0￡z￡x2+y2x2￡y￡-1￡x￡

1y1D 1y1DI=1

12

dy

f(x,y,或I

y y21dy f(x,y,2y y

y2=

yy2(x,z)?y1(x,

y1=f(x,y,WDxz

y(x,zdxdz f(x,y,zy1(x,z例7fxyzdVWW及I=1dx1-x01

f(x,y,I=1dx1-x01

f(x,y,先對(duì)z,再對(duì)y,后對(duì)xI=1dx 0y-xf(x,y,02W是前后型區(qū)域,它

x1=

yx2(y,z)?x1(y, f(x,y,W

x2=Dyz

x(y,z f(x,y,zx1(y,z 計(jì)算ydxdydz,其中W是由球面x2+y2+z2=W與錐面y=x2z2所圍成的區(qū)域zy1-x2-zy1-x2-o1yxzx2+x2+x2+z2￡ y2x2+z1-x2-zI=x2+z1-x2-z

y=1-(x2+z2)

z=r 2 20 dq0

r2 22 2

rdr=42 42方法2.(zbWz以D為底,dz為高的柱形薄片質(zhì)量 z zaz f(x,y,z)dxdy)d zxaa

f(x,y,z)dxdbbaf(x,y,z)dxd3czczabx解:W

-c￡z￡ Dz:a2+b2￡1-zz2z

Dz面積為Szpab(1-2c cc

z2

0=20

DDdxdzz

pab(1z2cpab(1z2c2

z2=4p例10設(shè)Ωx2y2z2x2x所圍成,計(jì)算W(1+x4)dxdydz 2￡x￡ Dx:y2+z2￡ 2Dx面積為Sx=p 2I=4(1+x4)dx2px22=4(1+xpx22

y2+z2￡xdx=234020p

I=dxdyz2(x,y)f(x,y,z)d z1(x,

I=

bdz

f(x,y,z)dxd1aI=1a

y2(

z2(x, adxy(x)dyz(x, f(x,y,z)d y01y0111zx100 dxdx2+y2

ff aa設(shè)M(x,y,z)?R3 M在xy面上投影P(x,y,0，將x,y用極坐標(biāo)r,q代替,（r,q,

zM(x,y,zx=ry=r

0￡r<+￥ 0￡q￡2p

P( y,0z=

-￥<z<+￥