課件9.2.2利用極坐標(biāo)系計(jì)算二重積分

1D9.2.2

利用極坐標(biāo)系計(jì)算二重積分

y

r

sinD

f

(r

cos

,

r

sin

)rdrd

.Dd

dr

rdx

r

cosxy

f

(

x,

y)d

f

(

x,

y)dxdyoDr

drrOxD

dd化為兩次定積分的定限口訣：2D

f

(

x,

y)d

f

(r

cos

,

r

sin

)rdrd

.D一般的,在極坐標(biāo)系下計(jì)算:先對(duì)r再對(duì)

積分后積先定（x

軸逆時(shí)針旋轉(zhuǎn)而定）上,下限均為常數(shù)先積后定（原點(diǎn)發(fā)出射線而定）進(jìn)是下限,出是上限.1r

(

)D

f

(r

cos

,

r

sin

)rdrdxOD1

(

)d2

(

)f

(r

cos

,

r

sin

)rdrOxD13r

(

)2r

(

)(1)積分區(qū)域D:(極點(diǎn)在區(qū)域外)r

2

(

)先對(duì)r再對(duì)

積分4DxOxODr

(

)(2)積分區(qū)域D:(極點(diǎn)在區(qū)域邊界上)r

(

)D

f

(r

cos

,

r

sin

)rdrdf

(r

cos

,

r

sin

)rdr

(

)

d

0先對(duì)r再對(duì)

積分Doxr

(

)(3)積分區(qū)域D:(極點(diǎn)在區(qū)域

)

f

(r

cos

,

r

sin

)rdrdD

(

)

d

f

(r

cos

,

r

sin

)rdr0

02先對(duì)r再對(duì)

積分5解Ddxdy

e

x2

y22a

(1

e

)例計(jì)算

e

x2

y2

dxdy,其中D是由中心在原點(diǎn),D半徑為a的圓周所圍成的閉區(qū)域.在極坐標(biāo)系下計(jì)算a

xOya0r

2e

dr122

2

02

r

a

[e

]e

rdrr

2a002d

y

r

sinx

r

cos

e

x

2

y

2

dxdy

x

2

y

2

a

2x0,

y06a

x7Oy

4

e

x

2

y

2

dxdyx

2

y

2

a

2x0,

y04

(1

ea2

)

e

x

2

y

2

dxdyx

2

y

2

a

2x0

e

x2

y2

dxdy

2Dx

2

y

2

a

2x0,

y0D4

e

x

2

y

2

dxdy

1

e

x

2

y

2

dxdy解36x

3

y

0

r

2sinDy

3x

0

D例

計(jì)算

(

x2

y2

)dxdy,

其中D為由圓3y

0,x2

y2

2

y,x2

y2

4

y及直線x

d

4sin

r

2

rdr2sin36y

3x

0所圍成的平面閉區(qū)域.yO

x2

y2

2

yx(

x2

y2

)dxdy

x2

y2

4

y

r

4sin

y

r

sinx

r

cos8(

x

y

)dxdyD2

22

3)d

4sin

r

2

rdr2sin633644sin2sinr

4d

441436

16sin

)d(256sin

34

606sin

d2(1

cos2

1536

)

d92

1536)d

15(1

cos4(1

2cos2

10x

r

cosDf

(

x,

y)dxdyf

(r

cos

,

r

sin

)rdrD其中積分區(qū)域

D

{(

x,

y)

1

x

y

1

x2

,0

x

1}解在極坐標(biāo)系下

y

r

sin圓方程為r

1直線方程為cos

sinr

1

11sin

cos2d0yO1Dy

1

x

2xy

1

x

r

cos1r

sin

1x2

y2

1例寫出積分

f

(x,y)dxdy

的極坐標(biāo)二次積分形式,11將直角坐標(biāo)系下累次積分:

21

0104

x24

x2

21

xf

(

x,

y)dydxf

(

x,

y)dy

dx化為極坐標(biāo)系下的累次積分.oxy解f

(r

cos

,

r

sin

)rdrd原式=y

4

x22

r

112012y

1

x2

r

2

12解(

x2

y2

)2

2a2

(

x2雙紐線(a

0)求曲線(x2

y2

)2

2a2

(x2

y2

)和x2

y2

a2

所圍成的圖形D的面積.例x2

y2

a2

1D

y2

)

r

2

2a2

cos

2x由r

ar

a2

cos

2得交點(diǎn)A

(a,

)6)3D

a2

( 3

rdr1DAa

2cos

2面積

dxdy

4dxdy

4

6d

0ayaO2a在極坐標(biāo)系下

x

r

cos

y

r

sinr

a對(duì)稱性質(zhì)13例求由不等式：x2

y2

z2

a2

與x2

y2

ax解所確定的空間幾何體的體積V1

f

(

x,

y)dxdyD1xyOa

a2aD1x2

y2

axaxyz

aO曲頂z

a2

x2

y2

D1a2

(

x2

y2

)dxdya2

r

2

rdrda

cos20014

/

20

01dacosa2

r

2

rdrV

00

/

2

23

acos1

/

2

0

[a

sin

a

]d3 3 33183

(a2

r

2

)2

d3

a

(I

)3

3

2)

a

(3

4)223

3

13

a

(

1

0

/

20d1212

3acosa2

r

2

d(a2

r

2

)15D11V1813

4)

a

(39V

4

f

(

x,

y)dxdy

2

a3

(3

4)對(duì)稱性質(zhì)aaxyz

aO16

(

x

y

)dxdyx2

y2

1解x2

y2

1例計(jì)算

4

xdxdy

4x

2

y2

1x0,

y01

xOy

(

x

y

)dxdy

4

(

x

y)dxdyx

2

y

2

1x0,

y0

xdxdyx

2

y

2

1x0,

y0

ydxdy

8x

2

y

2

1x0,

y0

xdxdy

ydxdyx

2

y

2

1

x

2

y

2

1x0,

y0

x0,

y017

(

x

y

)dxdyx2

y2

1

100r

cos

rdr

8

2

d

83

8

xdxdyx

2

y

2

1x0,

y01

xOy18計(jì)算x2

y2

a2解

3

ydxdy

2

dxdyx2

y2

a2

x2

y2

a2

x2dxdy

2xdxdy

x2

y2

a2

x2

y2

a2x

2

y2

a

2

(

x2

2x

3

y

2)dxdyx2

y2

a2

x2dxdy

2a2

1

(

x2

y2

)dxdy

2

a22

x

2

y2

a

22021a2r

rdr

2

a

0

d

224

2

a

a

4a

xO

(

x2

2x

3

y

2)dxdyy

x2dxdy

y2dxdyx2

y2

a2

x

2

y

2

a

202x

dx.例

求反常積分

e02e

dx

xa

xa

0e

dxlim220(a

x

2e

dx)

Syx

eSO

x

2

y

2分析aa00e

dye

dx

y

2

x

2adxdya若D為矩形域:a≤x≤b,c≤y≤d且f

(

x,

y)

f1

(

x)

f2

(

y)baf

(

x)dx1dcf

(

y)dy2D則

f

(x,y)d

19

e

x

2

y

2

dxdySD

:

x2

y2

a2

,

e

x2

y2

dxdy

(1

e

a

2

)DD122

e

x

y2(1

ea

)dxdy

1D

:

x2

y2

a2

,

x

0,

y

0對(duì)稱性質(zhì)4a

2a1DSD2y20xO2a1D

{(

x,

y)

|

x2

y2

a2

,

x

0,

y

0}2D

{(x,

y)

|x2

y2

2a2

,

x

0,

y

0}S

{(

x,

y)

|

0

x

a,0

y

a}

e

x2

y2S求反常積分02e

dx.

x例解顯然有D1

S

D2

0D1

D2

e

x2

y2

dxdy

e

x2

y2

dxdy

e

x2

y2

dxdyaD1SD2y21xO4

(1

ea2

)D1

e

x2

y2a2dxdy

(1

e

)41D

:

x2

y2

a2

,

x

0,

y

0D1I1

e

dxdy

x2

y22I

e

x2

y2

dxdy

(1

e2a2

)1D

{(

x,

y)

|

x2

y2

a2

,

x

0,

y

0}2D

{(x,

y)

|

x2

y2

2a2

,

x

0,

y

0}2014222a(1

e

)

Ie

dx)

D2(1

e

)

(

I

2a4

x

2a41224

42當(dāng)a

時(shí),

I

,

I

14I

24,

I

201222a(1

e

)

Ie

dx)

(1

e

)

(

I

2a

x

2a44概率積分定理當(dāng)a

時(shí),

4所求反常積分202e

dx

x0(

e

x

2

dx)2

20lim

(a

x2a

e

dx)23D

D24小結(jié)二重積分在極坐標(biāo)下的計(jì)算公式

f

(

x,

y)d

f

(r

cos

,

r

sin

)rdrd先對(duì)r再對(duì)

積分作業(yè)P389252(7)，

9，23(3),

25,22(2)(5)(8)

，26(3),

3026求曲面z

2

x2

y2與曲面z

x2

y2xyz

x2

y2o所圍成的

的體積2

z

z

2

x2

y2z

x2

y2zxyo分析zxy2oz

2

x2

y2z

x2

y2xyz

x2

y2o22z

x

yzyxyo2

z

z

2

x2

y2兩曲面的交線為:z

1

x2

y2

1