數(shù)分相關(guān)十六

§16.21直角坐標(biāo)系下二重積分的計算z

=f

(x,y)為頂面的曲頂柱體21f

(

x

,

y

)

dyA

(

x

0

)

=j

(

x

)j

(

x

)2一、矩形區(qū)域上的二重積分的計算設(shè)

D

=

[a,

b]·[c,

d

],

f

:

D

fiR,如對"x

?

[a,b],函數(shù)

f

(

x, )

在[c,

d

]上可積,

則可得如下函數(shù)：dcf

(

x,

y)dy,

x

?

[a,

b].I

(

x)

=如果函數(shù)

I

(

x)

也在[a,

b]上可積,

則得積分

a

cb

b

da(

f

(

x,

y)dy)dx

.I

(

x)dx

=3ab

dcf

(

x,

y)dy.dx此積分稱為累次積分.

記為類似理解:(cd

badcbaf

(

x,

y)dx)dy

.f

(

x,

y)dx

=dy問題:

f

(

x,

y)dsDbadcf

(

x,

y)dy,dxdcbaf

(

x,

y)dx.dy4且對"x

?

[a,b],5積分定理1.設(shè)f

(x,y)在矩形區(qū)域D

=[a,b]·[c,d

]上可積,dcf

(x,y)dy

都存在,則累次積分ab

dcdxf

(

x,

y)dy

也存在,

且Df

(

x,

y)ds=ab

dcf

(

x,

y)dy.dx證明:f

(

x,

y)dy,dcx

?

[a,

b].I

(

x)

=對[a,b],[c,d

]的分割px

:

a

=

x0

<

x1

<

<

xn

=

b,p

y

:

c

=

y0

<

y1

<

<

ym

=

d

,6令

Ii

=

[

xi

-1

,

xi

],

i

=

1,,

n,J

j

=

[

y

j-1

,

y

j

],

j

=

1,,

m.因此子矩形Ii

·

J

j

形成了D的分割p

=

px

·py令A(yù)

=

fdsD由定義，"e

>0,$d

>0,當(dāng)分割p滿足p

<d時，有n

nA

-

e

<

f

(xi

,h

j

)Dxi

Dyj

<

A

+

e

(1)i

=1

j=127現(xiàn)取p

,p

<d

,則p

<dx

y在（1）中取n

ninf

f

(xi

,

J

j

)Dyj

sup

f

(xi

,

J

j

)Dyjj=1

j=1分別是f

(xi

,)在[c,d

]上的上和與下和n8lim

I

(xi

)Dxi

=

Apx

fi

0

j=1Df

(

x,

y)ds

=badcf

(

x,

y)dy.dxni

=1A

-

e

￡

I

(xi

)Dxi

￡

A

+

edcnji

jinf

f

(j=1f

(xi

,

J

j

)Dyjx

,

J

)Dy

￡nj=1￡

sup

f

(xi

,

J

j

)Dyj定理2.設(shè)f

(x,y)在矩形區(qū)域D

=[a,b]·[c,d

]上可積,9且對"y

?

[c,d

],積分baf

(x,y)dx

都存在,則累次Df

(

x,

y)ds積分cd

badyf

(

x,

y)dx

也存在,

且=dcbaf

(

x,

y)dx.dy證明:類似于定理1.

10=c

ady f

(

x,

y)dx.定理3.設(shè)f

(x,y)在矩形區(qū)域D

=[a,b]·[c,d

]上連續(xù),則有

f

(

x,

y)ds

=Dadb

dcbdxf

(

x,

y)dy累次積分交換順序的充分條件:f

(x,y)在D上可積,dcf

(x,y)dy

都存在,對"x

?

[a,b],對"y

?

[c,d

],積分積分baf

(x,y)dx

都存在.例1.設(shè)

f

(

x,

y)

=

1

-

x

-

yD計算

f

(x,y)ds

,其中D

=[0,1]·[0,1].解因為f

(x,y)滿足定理31010f

(

x,

y)dydxDf

(

x,

y)ds

=0x

x1y1x

+

y

=

111的條件,所以1010-

y)dy(1

-

xf

(

x,

y)dy

==1010ydy(1

-

x)dy

-= -

x1

12

2=

(1

-

x)

-而所以1212(10-

x)dx

=

0.Df

(

x,

y)ds

=二、一般區(qū)域上的二重積分的計算x型區(qū)域D

=

{(

x,

y)

|

y1

(

x)

￡

y

￡

y2

(

x),a

￡

x

￡

b}baxy

=

y1

(

x)y

=

y2

(

x)特點(diǎn)：穿過區(qū)域且平行于

yy軸的直線與區(qū)域邊界相交不多于兩個交點(diǎn).13y型區(qū)域D

=

{(

x,

y)

|

x1

(

y)

￡

x

￡

x2

(

y),c

￡

y

￡

d

}特點(diǎn)：穿過區(qū)域且平行于x

軸的直線與區(qū)域邊界相交不多于兩個交點(diǎn).xc14ydD3152DD1一般區(qū)域或分解成有限個無公共內(nèi)點(diǎn)的x型區(qū)域y型區(qū)域.因此一般區(qū)域上的二重積分計算問題歸結(jié)到x型區(qū)域或y型區(qū)域上的二重積分計算問題.定理

4.

設(shè)f

(

x,

y)在x型區(qū)域

D上連續(xù),

其中

y1

(

x)y2

(

x)在[a,

b]上連續(xù),

則Df

(

x,

y)ds=adxy

(

x

)b y2

(

x

)1f

(

x,

y)dy分析:baxy

=

y2

(

x)y

=

y1

(

x)cyd160, (

x,

y)

ˇ

D.

f

(

x,

y),(

x,

y)

?

D,F

(

x,

y)

=

170, (

x,

y)

ˇ

D.證

由于

y1

(

x),

y2

(

x)在[a,

b]上連續(xù),

故總存在矩形區(qū)域[a,

b]·[c,

d

]

D,

作定義在[a,

b]·[c,

d

]上的輔助函數(shù)F

(

x,

y)

=

f

(

x,

y),(

x,

y)

?

D,可以驗證F

(

x,

y)在[a,

b]·[c,

d

]上可積,

而且

f

(

x,

y)ds

=D[a

,b]·[c

,d

]F

(

x,

y)ds=ab

dcdxF

(

x,

y)dy=adxy

(

x

)b y2

(

x

)11F

(

x,

y)dy

=bay2

(

x

)y

(

x

)f

(

x,

y)dy.dx則18Df

(

x,

y)ds=dcdyx

(

y

)2x1

(

y

)f

(

x,

y)dx類似可證 若f

(

x,

y)在y型區(qū)域

D上連續(xù),

x1

(

y)x2

(y)在[c,d

]上連續(xù),積分限的問題注

意:務(wù)必保證:?--同一定積分?--累次積分下限￡

上限先定后積解兩曲線的交點(diǎn)2

(0,0) ,

(1,1),x

=

y

y

=

x2D(

x2

+

y)dxdy

=10x2x(

x2

+

y)dydx102=2

140[

x

(

x

-

x2

)

+

1

(

x

-

x4

)]dx

=

33

.x

=

y2y

=

x219例

2

求(

x2

+

y)dxdy，其中D

是由拋物線Dy

=x2和x

=y2所圍平面閉區(qū)域.x

=

y2y

=

x2

(

x2

+

y)dxdyD14033=

.=dyy2y(

x2

+

y)dx1013

31032332y

-

1

y6

)dy(

y

-

y

+=12074211825

01

15

4=yy

-

y

-Dx

e

dxdy2

-

y2=dxdy01

y0x2e-

y2e=-

y10y3321062y2edy

=-

y1

26

e(1

-

).dy2

=例

3.

計算

x

-2e

-

y2

ds

,

其中D是由x

=

0,

y

=

121D及y

=x

圍成的區(qū)域.2解

e-

y

dy

無法用初等函數(shù)表示\積分時必須考慮次序y

=

2

-

xy

=

2x

-

x2

dxdx2-

x0210例

4

改變積分10f

(

x,

y)dy

+2

x-

x2f

(x,y)dy的次序.22原式=102-

y1-

1-

y2dyf

(

x,

y)dx.解

積分區(qū)域如圖y

=

2ax解=ady02aa-

a2

-

y2y2f

(

x,

y)dx原式aa+

a

-

y+

dy02a2

22f

(

x,

y)dx

+

2aa2a2ayf

(

x,

y)dx.dyy

=

2ax

-

x2a2

-

y2

x

=

a

–2

aa2

aa223-f

(

x,

y)dy

(a>

0)dx2a02ax2ax

x例

5

改變積分的次序.例6.解11利用對稱關(guān)系V

=

8V

,

V

=R2

-

x

2

ds

,DR2

-

x

2

,0

￡

x

￡

R}.D

=

{(

x,

y)

0

￡

y

￡=R0R2

-

x

2

dyR2

-

x222(

)=V1

=

R

-

x

dsDRR

-

x

dx022233dx0R

.=所以3241631R

.V

=

8V

=z

=

6

-

2

x2

-

y2例725解:1、作出該立體的簡圖,并確定投影消去變量z得一垂直于xoy

面的柱面x2

+

y2

=

2立體鑲嵌在其中,立體在求由曲面z

=x

2

+2

y2所圍成的立體的體積D26V

=

6

ds

-

3

x

2

ds

-

3

y

2

dsD

D

ds

=

2pD曲面的投影區(qū)域就是該柱面在xoy面上所圍成的區(qū)域D

:x2

+

y2

￡

22、列出體積計算的表達(dá)式V

=

[(6

-

2x2

-

y2

)

-(

x2

+

2

y2

)]dsD=

(6

-

3

x

2

-

3

y

2

)dsD3、配置積分限,化二重積分為二次積分27D

D由x,y的對稱性

x2ds

=

y2ds2-

2D

-

22

-

x2

dx

dy

=

2

x2-

2-

x22 2-

x2

x2ds

=

x2dx2

p

2=

4

x2

2

-

x2

dx

=

4

4sin2