高數課件72二重積分計算

設有一立體它的底是xOy面上的閉區(qū)域D,D這里f(x,y)≥0且在D這種立體叫作曲頂柱體。fi,hiDsi是以Dsifi,hi為高的平頂柱體的體積Dsi的直徑很小時fi,hiDsi上變化也很小因此可將以Dsi為底以z=f(x,y)為頂的小曲頂柱體近似地看作平頂n n

iDsi

z=f,

zi=fi,ηi

fi

VVd=max{dDskfi0, n n,ydxdyiDsi dfi0

Di,hi

Dsiyyyy=f2?Dy=f1xoabdx=y1y

x=y2yD o

D=,

f1(x)￡y￡f2(x),a￡x￡b

y1(y)￡x￡y2(y),c￡y￡dx x y???y????xo

y=f2

,

f1(x)￡y￡f2(x),a￡x￡b}

Sx0f(x

x0b x0= f(x0,y)d f1(x0bbV=Df(x,y)ds=aS(x)d

y=f1(x)b=f2(b

f(x,y)d

dx

bdxf2(

f(x,

D=x,yy1y￡x￡y2y,c￡y￡d}x=y2(y x=y1

Dy

zxocSy=y2(y0)f(x,y)d

y y(y ddV= fx,ydxdy=

Syd

x=y1 yd d y1(y

fx,ydxd =

d d

y

x,yd則有f,ydxd

y=d

bDb=ad

f2(

fx,yd

y=f

x=y2y) c=dyc

1(

f(x,y)

byD=+yD=++ xY-型區(qū)域時,可以通過劃分, o f(x,y)ds=bf2(

f,ydx

bdxf2(

f(x,Daf1( D

,

f(x)￡y￡f(x),a￡x￡b f(x,y)ds

y2(

f(x,y)dxdy

ddy(

f,d dD

y1(y)￡x￡y2(y),c￡y￡d df(x,y)ds=bdxf2(d

f(x,

dy(

f,

例1.I22sin(x

y)dxdy p I=p

dy

x+yp p =-cosx+y d +y+cos0+y+cos0+yd20 2p=2[siny+cosy]d0p=[-cosy+siny]2=0例2計算Ixyds其中Dy＝1x＝2D

D

1￡y￡1￡x￡ I=dxxydy

1y2xd

y= 1 =2

x3-1xdx= 1

x

Dy￡x￡1￡y￡2I=d

2xydx=

x2 d

22y-

y3d y

1

=1

=D例3.計算Ixyds,其中D是拋物線y2=xy=x-D 2y2￡x￡y+ D:-1￡y￡

y2= - -y=x- y+Dxyds

1d

xyd=21y-1

dy=2

2yy+22-y5d 1

= +y3+2y2-y6 2

y4,y=y4,y=oD1y=x--x,

xDxDx x

￡y

x-2￡yxDxD 0￡x￡1 1￡x￡ yDxyds

xy

+

xy4 40-1=dx0-1

xyd

+dx

xyd11其中,既要考慮積分區(qū)域的形狀,又要考慮被積函數的特性。例4

y-

dxdy其中D為-1≤x≤1,0≤y1 在D上x2≤y 2D1：-1≤x≤1,0≤y≤x2D2：-1≤x≤1,x2≤y≤1;于是,根據二重積分的性質,有

y=y=x

y-

dxdy

x

-ydxd

+

y-x2dxdy2-12-=

1x0x1x0

-ydy

02dx02

y-x2d1=-1

x4dx

-x2+

x4

=-OxRDzy R2OxRDzy R2-xx2+y2=Dyx2+z2=xRoRx2+y2=R2,x2+z2=R2-R2-R2-0R2-xy?DR2-R2-x2RR2-R2-x2RR2-x2VR2-x2

dxdy=

dx d0=8R0

-x2d

3p2p例6.計算 d0 x

x

yd

x

yd

π πx,

x￡y￡2

yDp2 Dp,0￡x

4

x= 2D=x,y0￡x￡y2,0￡y￡ 2

y= 4p2p4dp

d

=2d

yx dyx

py dpy

yp=p

y-ycosydy=1

ysiny-cosy

= p=

p+1

8-8- I=

d 0

fx,ydy d

fx,yd0￡y￡1

y=1222￡y=122

D2:

+ =88-0￡x￡

0￡y

將D=D1+D2視為Y–,

28-D:28-0￡y￡8-8-(I=0d(

x,ydy 域D關于x軸對稱Ds1,Ds2, ,DsnDskk,hk和k,-hk

x則有fkhkDsk

x?xDskk=1,2,L,Dskfk,-

Dsk

xDnn

fk,hks

fk-hks

nkn

k

fx,ydxdy=limfk,hkskdfi0k

kDskdfi0k=111=11

fx,dxdy+

fx,-ydxd0

11=11

fx,y+fx,-dxd

=

fx,ydxd

yoDD關于xyoDfx-yfxy,((xy∈D)時f(x,y)dxd = f(x,y)dxd fx-yfxy ((xy∈D)時Df(x,y)dxdy=如果積分區(qū)域D關于y軸對稱, fx,y=fx,y

((xy∈D)時Df(x,y)dxd =

f(x,y)dxd

22fx,y=-fx,yDf(x,y)dxdy=

((xy∈D)時例10.

I=xlnyD

1+其中Dy=4-x2,y=-3x,1+解

fx,y=xlny 1+1+fx,y=-1+1+y+1+y2fx,-y=xy+1+y2

=-, 1+1+-

+1+y2 y1+y1+

=xln

=-f,y)y4y=y4y=4-y=-3o1x=

D=D1+D2則Ixy

f(-x,y)=-f(x,f(x,-y)=-f(x,1+y21+y2xdy+1+y21+y2

xdy=例11.I=y1+xfD其中Dy=x2y=1

+y2xd

1 y=x

令gx,y=xyfx2+y2)- D關于y軸對稱,gx,y=-xyfx2+y2=-gx,y yxfx2+y2dxdy=DI=y1+xf1D1

+y2xdy=ydxdy+yxfx2+y2dxd1 1=ydxdy=d

yd

=1

1-x4dx4

2例12.計算 x+2

dxdyD:xy￡1

x

+2xy+y2dxd

yx+y￡1

y==

x2dxdy+

2xydxdy+

y2dxd ①D既是關于x軸對稱又是關于y軸對稱

- DoD2-Dxydxdy=-Dx2dxdy=2x2dxdy=4x2dxdDD1¨ D同理y2dxdy4y2dxD②D1關于y=x對稱fx,ydxdyfyxdxdx2dxdyy2dxd

x+y2dxdy=

x2dxd

xy2

x2dxdy1x+y=1xoxy1x+y=1xox1

D1=0￡x￡1

1- 1 xdxdy1

80

xd

d

=80

1-xd 4 累次積分法

y=D例7證y=Da0da

ebx0

xdx

aebx-

xd 00D=x,y0￡y￡a,0￡x￡yD=x,y0￡x￡a,x￡y￡aaa

x a0da

ebxa0a

xd

=0d

ebx0x0

xda=a

ebx

x)xdyd

aebx-

xd例8

sinxdxdy,其中D是直線y=x,y=0 xx 因此取D為X–型域yy=Dx0￡yy=Dx0￡x￡sin

dxdy=

=psinxdx=-cosxp= P1342(1)(4),3(1)(3)(5),4(2)(4),6(1)(3)(5),9(2)(4),

定理:設f(x,y)在閉區(qū)域D上連續(xù) x=u, yT y

=u,

u,?D￠fi u滿足(1)xuv),yu一階導數連續(xù)在上Jcabi行列 Ju,Ju,?u,

DD Df(xydxdyD￠f(x(uvy(u

J(u,v)dud uov

v+v

uv)uDu

o￠

u+u+Du,v+Dv),u,v+ yP4PD2其對應頂點為ixiyii=P4PD22+2+u+Du,-u,uDv-u

Du+r u,vDv+ru,v?y?yu+Du,-u,uDv-u

P4PD2DP4PD2u,vDv+ru,v 14當△u△vP1P2P3P4近似于平行四邊形,14

1

uv-uuDv-uuv-uuDv?

DuDv=Ju,v)

ds=dxdy

Ju,v)dud Dfx,ydxdy=D￠fxu,v),yu,J(u,v)dud x=rcosq,y=rJ=x,=,q)

rsinq=rDfx,ydxdy=D￠frcosq,rsinqrdr Df(x,y)ds= rr=rr=r2θ)Dbor=r1θ)

在區(qū)間上連續(xù) Df(x,y)dxdDf(x,y)dxdybarq)2r1rDf(x,y)dxdybarq0 Df(x,y)dxdybarq0 o

r=rθ)DbDDf(x,y)dxdybarq)2r1rDDf(x,y)dxdybarq)0rr=rθDor=rθDo 0￡r￡rθD:0￡θ￡Df(x,y)dxdy2θfrcosq,rsinq)rdθ)f≡1則可求得D11

0

q

r=rθD

rr=rθDo (1)0￡q￡p

(2)-p￡q￡ y12o例y12o

x

dsxD=,q)1￡r￡2,0￡q￡2px

+y2dxd

=

rr1

rdr55 5

21例13.

x

y2dxdy,其中Dyx2+y2=4yxy

3y0,y-3x04解：x2+y2=2 r=2sinq域4x2+y2=4yr=y-3x=0q2= x-3y=

r,q6363\x+yd\x+ydxdy6

3r2rd3r2rdr=15 例14.

e-x2-y

dxdy

其中D

D0￡r￡a 00￡q￡原式

e-r2rdrdq=2p0

are-r2d0-1-

2 =

2

1- -由于e-

+￥e-x2 1D=(x,y)|x2+y2￡R2,x?0,y?012D=(x,y)|x2+y2￡2R2,x?0,y?02S=(x,y)|0￡x￡R,0￡y￡R

S x 顯然 D2

ex

e-x2-y2dxdy<e-x2-y2dxdy<e-x2-y2 因為e

-x2-y

dxdy

R

-x2

Re-y2

=

Re-x

e-x2-y2dxdy=p-e-R2

e-x2-y2dxdy=p-e-2R2

-x -e-R

￡1-e-2 令Rfi+￥上式兩端趨于同一極限p 從40e-2dx=2+￥e-x20事實上當D為R2時

dx p2p

e-x2-y

dxdy

+￥e-x2

dx+￥e-y2d2=4+￥e-x2dx 40e-22d 40e-22d =limafi1-2zoyzoy 2r,q 2 4a2-rDV=4 4a2-r2r4a2-rDp0=42dq0

0

rd

r==323

pp 1-0

3q D D =32a3p-2 例8

y-ey+xD

dxdy

其中Dxyyxy2解令uyxvyx,

x+y=Dx=v-u,y=v+

(Dfi 2J=?(x,

2- =

u=-

v=D￠u=2 2 2vy- 2v\

2ey2

dxdy=

ev

1

dudv=

d

d = e-e-1vdv=e-e- 例9.計算由y2px,y2qx,x2ay,x2by0pq0abDy2 Sx2=byDy2=qxy2=pxxx2=byDy2=qxy2=pxx2=ay p￡u￡D:a￡v￡b J=?(x,y)=

=- \S=Ddxd

o =

Jdudv

1 d d

=1(q-

- 3 例10.

x2+

+z2

￡1的體積V 取D: +

V=V=

112ya2-zdxdy=2dxdxarcosq,ybrsinq,則DD:r￡1,0￡q￡J=?(x,y)

ab

arbr

=abr1-r\V=21-r

abrdr =2 d

rdr=4p1-r1-r:

y=y2(x)Dy=y1(x) bxD=(x,y)a￡x￡b,y1(x)￡y￡y2

f(x,y)ds

bdxy2(x)f(x,y)d x=x2(x=x2(Dx D=(x,y)c￡y￡d,x1(y)￡x￡x2(y)} 則f(xydsdd

x2(y