高數(shù)下冊(cè)10十章

1、若平面域D為[a,b]上的X型域,即：平面域D在x軸上的投影區(qū)間為[a,b],"x0

?

(a,b),x

=x0從下往上穿過D時(shí)與D的邊界至多有2個(gè)交點(diǎn)且固定地y

=j1

(x)為進(jìn)線y

=j2

(x)為出線,則：DX

={a

￡

x

￡b,j1(x)

￡

y

￡j2

(x)}復(fù)習(xí)：直標(biāo)系下平面域的解析表示2、若平面域D為[c,d

]上的Y型域,即：平面域D在y軸上的投影區(qū)間為[c,d

],"y0

?

(c,d

),y

=y0從左往右穿過D時(shí)與D的邊界至多有2個(gè)交點(diǎn)且固定地x

=w1

(y)為進(jìn)線x

=w

2

(y)為出線,則：DY

={c

￡

y

￡d,w1(y)￡

x

￡w2

(y)}注：任意平面域均能分塊成X或Y型域.如(1)D由y

=x與y

=x2圍成：X\

D=D

={0￡x

￡1,x2

￡

y

￡x}D=DY

={0￡

y

￡1,

y

￡x

￡

y}2y

=

x x

=

y,

y

=

x2

x

=–

y,y

=

xO(0,0),

A(1,1)y

=

xA(2)D由x

=y2與y

=x

-2圍成：

B(4,2)A(1,-1)y

=

x

-2x

=

y2Bx

=

y2

y

=–

x,

y

=

x

-2

x

=

y

+2,A\

D

=

D

={-1￡

y

￡

2,

y2

￡

x

￡

y

+2}YD

=DX

={0

￡

x

￡1,-

x

￡

y

￡

x}{1

￡

x

￡

4,

x

-

2

￡

y

￡

x}\

D=D

={0￡x

￡1,

x2

￡

y

￡1}XD=DY

={0￡y￡1，0￡x￡

y}(3)D由x

=0,y

=x2

,y

=1圍成：

y

=

x2

x

=

–

yD

A

y

=

x2

A(1,1),

B(-1,1)

y

=1BY\

D

=D

={-2

￡

y

￡1,

y2

￡

x

￡2-y}D=DX

={0￡x￡1,-

x

￡y￡

x}{1￡x￡2,-

x

￡y￡2-x}ABD(4)D由x

=y2

,x

=2

-y圍成x

=

y2

y

=–

x,

x

=

2

-

yy

=

2

-

x,

A(1,1),

B(4,-2)x

=

y2x

=

2

-

yD

=

DY

={0

￡

y

￡1,1-

y

￡

x

￡1}{1￡

y

￡

2,0

￡

x

￡1}(5)D由y

=0,y

=2；x

=1-y,x

=1圍成：

x

=1-y y

=1-x,

如圖，(1,2)(0,2)\

D=DX

={0￡x

￡1

,1-x

￡

y

￡2}(0,1)(1,0)(6)如圖：D

={(x,y)|

x

|

+|

y

|￡

1}(0,1)(-1,0)(1,0)(0,-1)x1、若D

=

DX

={a

￡

x

￡

b,j1(x)

￡

y

￡j2

(x)},則：10.2.1

二重積分的計(jì)算法(一)(2j

(

x)j1

(

x)f

(x,

y)dy

dxbD

af

(x,

y)ds

=

1f

(x,

y)dybadxj2

(

x)j

(

x)=2、若D

=

DY

={c

￡

y

￡

d,w1(y)

￡

x

￡w2

(y)},則：dD

f

(x,

y)dx

dy2w

(

y)c

w1

(

y)f

(x,

y)ds

==dcf

(x,

y)dxdy1w

2

(

y)w

(

y)xyds

,D例1(1)求2x

圍成的閉域.D：y

=x與y

=解：作D草圖XD=D

={0￡x

￡1,x2￡

y

￡x}=1012]dx2[(

xy

)2

xx=10521312-

x

)dx(

x1

1

18

12

24=

-=xydyx

x2\1D

0xyds

=

dx=10321

-

12

2y

)dy(

y=10212]dy[(

yx

)yy另解：作D草圖D=DY

={0￡

y

￡1,y

￡x

￡

y}

xydxyy

\1D

0xyds

=

dy1

1

18

12

24=

-=例1(2)求A

=D

xyds

,D：由x

=y2與y=x

-2圍成的閉域.解：作D草圖

B(4,2)A(1,-1)y

=

x

-2x

=

y2\

D

=

D

={-1￡

y

￡

2,

y2

￡

x

￡

y

+2}Y=2-112]dy

=[(

yx

)2

y+2y2ABxydxy

2

y+2\

A

=

dy-12例1(3)求A

=D

(x

+y)ds

,其中D：D1

={0

￡

y

￡1,

1

y

￡

x

￡

y}22D2

={0

￡

y

￡1,-

y

￡

x

￡-1

y}

\

A

=21DD(x

+

y)ds

+

(x

+

y)dsD2

D1y

=x2與y

=4x2及y

=1圍成閉域解：作D草圖D

=D1

D2103

2283

1103

221-

y

+

y

)dyy

+

y

)dy+

(83=

(13

252

1022=y

dy

=

y

0

=5

5(x

+

y)dxy-

yyy

=

dy2-1(

x

+

y)dx+

dy121010(x

+

y)ds(x

+

y)ds

+\

A

=21DD1)畫D的草圖；2)求邊界交點(diǎn)；選積分次序確定上下限： 后積先定限，總是點(diǎn)到點(diǎn)； 限內(nèi)畫射線，總是線到線；先交為下限，后交為上限；區(qū)域少分塊，積分計(jì)算簡(jiǎn)．計(jì)算兩次定積分.▲直標(biāo)系下求二重積分：f

(x,

y)dycb

dD

a

f

(x,

y)ds

=

dx3、若D={a

￡x

￡b,c

￡

y

￡d}=[a,b]·[c,d],則：f

(x,

y)dxdybadc=4、若D

={a

￡

x

￡b,c

￡

y

￡

d}=[a,b]·[c,d],且f

(x,y

)=f1

(x

)f

2

(y

),則：f2

(

y)dycb

dD

a

f

(x)dx1f

(x,

y)ds

=ydyx

dx

y

dy

-

dx

dy

-

xdx=210

121211

01

2020

0

1

13

2

3

2=[(1

x3

)

1

]-[(1

x2

)

1

]

[1

y3

)

2

]-[(1

y2

)

2

]1

7

3

73

6

2

3=-

- =

-2

2D(x

-

xy

-

y)ds

,例2(1)求A

=其中D

=[0,1]·[1,

2]

={0

￡

x

￡1,

1￡

y

￡

2}.解：D

={0

￡

x

￡1,

1￡

y

￡

2}

\

A

=D

Dx

ds

-

xy

ds

-Dyds222D

ay f

(x)ds例2(2)已知D

=[a,b]·[0,1]={a

￡

x

￡

b,0

￡

y

￡1}b且

=1,求

f

(x)dx.解：D

={a

￡

x

￡

b,0

￡

y

￡1}13f

(x)dx

=1ba=f

(x)dx

=

3\ba

\1

0

22y f

(x)dy

dybD

as

=

f

(x)dx\De

ds-(

x+

y

)=[(-e-x

)

2]

[(-e-y

)

1

]0

0=(1-e-2)(1-e-1)D：由x

=0,x

=2及y

=0,y

=1圍成.解：D

=[0,2]·[0,1]={0

￡

x

￡2，0

￡

y

￡1}且e-(x+y)

=e-x

e-ye

dye

dx-y-x=02

10例2(3)求D-(

x+

y

)e

ds

,其中2110已知：f

(x)=f

(x)dx.-

ye

dy,求A

=x(2110x所求A

=e dy

dx-

y2

e-y

關(guān)于y的原函數(shù)不能表成初等函數(shù),2

2而e-

y

關(guān)于x的原函數(shù)為xe-

y

,\可考慮改變積分順序求解.[分析]：畫D的草圖：四曲線圍成；改變D的類型：觀察草圖得.注1、D轉(zhuǎn)型后有時(shí)需分塊或合并.2、遇原函數(shù)不能表成初等函數(shù)時(shí)往往考慮將D轉(zhuǎn)型后求解.5、交換積分次序：

(1)A

=112f

(x,

y)dyxD

0f

(x,

y)ds

=

dx例3、改變積分次序：f

(x,

y)ds

=(2)A=yDf

(x,

y)dxdy2-y21-2

(3)A

=1101-yfdx+

dy1021Dfds

=fdxdyX解：D

=

D

={0

￡

x

￡1,

x2

￡

y

￡1}x

=0,x

=1；y

=x2

,y

=1圍成\

D

=

DY

={0

￡

y

￡1，0

￡

x

￡

y}y\

A=

dyf

(x,

y)dx010

(1)A

=112f

(x,

y)dyxD

0f

(x,

y)ds

=

dx例3、改變積分次序：畫D草圖：DY解：D

=D

={-2￡

y

￡1,

y2

￡

x

￡2-y}畫D草圖：

(2)A=yf

(x,

y)dx1D

-22-y2f

(x,

y)ds

=

dyf

(x,

y)dy+\

A=-

xdx1

x0\

D=DX

={0￡x￡1,-

x

￡y￡

x}{1￡x￡2,-

x

￡y￡2-x}-

xf

(x,

y)dydx2-x21y

=-2,y

=1；x

=y2

,x

=2

-y圍成A(1,1)B(4,-2)D

(3)A

=1101-yfdx+

dy102D

1fds

=

dyfdx解：D=DY

={0￡

y

￡1,1-y

￡x

￡1}{1￡

y

￡2,0￡x

￡1}畫D=D1+D2草圖：\

D=DX

={0￡x

￡1

,1-x

￡

y

￡2}\

A

=1

201-xfdydxy

=0,y

=1；x

=1-y,x

=1圍成D12y

=1,y

=2；x

=0,x

=1圍成D1DD2=D1

+D2,例4、求I

=1102y2xD

f

(x,

y)ds

=

dxxe

dy21012yye

dy=212014ye

d(y

)=1014=

1

ey24=

(e-1)

0

002[(e

1

x2

)

y

]dy10y21

y

y2\

I

=

dy

xe dx

=2解：D

=

DX

={0

￡

x

￡1,

x

￡

y

￡1},由例3(1)D

=

DY

={0

￡

y

￡1,0

￡

x

￡

y}10x

sin

y

dyyf

(x,

y)ds

=Dex1、求I

=xdxY解：D

=

D

={0

￡

y

￡1,

y2

￡

x

￡

y}\

I=dxdysinyy2

y1

y010=

[(1-

y)siny]dy==sin

ydy

-y

sin

ydy1010分部積分2110ex2、已知f

(x)=f

(x)dx.-

ye

dy,求A

=x(2110x解：所求A

=e dy

dx-

y=D-ye

ds2D

=

DX

={0

￡

x

￡1,

x

￡

y

￡1}=

DY

={0

￡

y

￡1,0

￡

x

￡

y}(2100y\

A

=e dx

dy-

y210ye

dy-

y=21-

y

2(1-e-1

)01212e d

(-y

)

==

-f

(x,

y)ds,

f

(x,-y)

=

f

(x,

y)f

(x,-y)

=-f

(x,

y)*20,DA

=(1)D關(guān)于x軸對(duì)稱,D*

={(x,

y)?

D,

y

?0}記A

=

D

f

(x,

y)ds6、對(duì)稱區(qū)域上二重積分的性質(zhì)：

\

A

=f

(x,

y)dybD

aj(x)-j(x)f

(x,

y)ds

=

dx證明：如圖y

=j(x)y

=

-j(x)設(shè)"x?[a,b],j(x)>0,則：abxD*D

=

DX

={a

￡

x

￡b,-j(x)

￡

y

￡j(x)}Df

(x,

y)dy

=0則A

=

bD

aj(x)-j(x)f

(x,

y)ds

=

dx(I)若f

(x,y)關(guān)于y為奇函數(shù)：f

(x,-y)=-f

(x,y),XD*

=

D*

={a

￡

x

￡b,0

￡

y

￡j(x)}(I)若f

(x,y)關(guān)于y為偶函數(shù)：f

(x,-y)=f

(x,y),

則A

=0f

(x,

y)dybD

af

(x,

y)dj(x)s

=

2

dx\

A

=

2D*

f

(x,

y)ds*20,f

(x,

y)ds,

f

(-x,

y)

=

f

(x,

y)DA

=(2)D關(guān)于y軸對(duì)稱,D*

={(x,

y)?

D,

x

?0}f

(-x,

y)

=-f

(x,

y)(3)若D關(guān)于直線y

=x對(duì)稱,則D

f

(x,y)ds

=D

f

(y,x)dsx2

yds

=

02

2x

+

y

￡2

x例如(1)xyds

=

0+

￡1y2(2)x2(3)x

+y

￡a(x

+

y3

sinx2

+2)ds2

2

2ds

=

2pa22

2

2=

0

+0

+2x

+y

￡a=(5)|x|+|y|￡1(x

+

y)2

ds(展開、分離、用對(duì)稱性)x?0,

y

?0=

8x+

y￡1x2

ds2001

x23=

8dxx dy

=2

2|x|+|

y|￡1(4)

x y

ds=

4x+y￡1x?0,

y?0x2

y2ds2x2

2x222ds

,其中D：|

x

|

+|

y

|￡1.x

+

yD

x

+

y例5、求A

=解：

D關(guān)于直線y

=x對(duì)稱,且S

(D)=2x2

y2\

A=D

x2

+y2

ds

=D

x2

+y2

dsx2

+y2x2

+y2ds

=S(D)

=2D2A=\

A

=1(0,1)(-1,0)

(1,0)(0,-1)x利用對(duì)稱性另解例1(3)：

D

=D1

D2關(guān)于y軸對(duì)稱2D1

={0

￡

y

￡1,

1

y

￡

x

￡

y}

\DD

Ds

=

xds

+

yds(x

+

y)d1=0+2Dydsydxy

1210=

2

ydy1120=2

y(

y-

y)dy

=2512D

D小結(jié)：1、直標(biāo)系下二重積分的計(jì)算：X型域、Y型域、特殊矩形域.2、交換積分順序的方法及其應(yīng)用.3、對(duì)稱區(qū)域上二重積分的性質(zhì)及其應(yīng)用.作業(yè)10--2.1：153--155頁1(2,3),2(1,3,4),6(1,4),8例6,7|

y

-

x2

|dsD例6、求I

=2 |

y

-x

|

=

y

-x2

,(x,

y)?

D2

x2

-y,(x,

y)?

D1D

={-1￡

x

￡1,0

￡

y

￡2}2D

={(x,y)?