课件9.2.2利用极坐标系计算二重积分

1D9.2.2

利用极坐标系计算二重积分

y

r

sinD

f

(r

cos

,

r

sin

)rdrd

.Dd

dr

rdx

r

cosxy

f

(

x,

y)d

f

(

x,

y)dxdyoDr

drrOxD

dd化为两次定积分的定限口诀：2D

f

(

x,

y)d

f

(r

cos

,

r

sin

)rdrd

.D一般的,在极坐标系下计算:先对r再对

积分后积先定（x

轴逆时针旋转而定）上,下限均为常数先积后定（原点发出射线而定）进是下限,出是上限.1r

(

)D

f

(r

cos

,

r

sin

)rdrdxOD1

(

)d2

(

)f

(r

cos

,

r

sin

)rdrOxD13r

(

)2r

(

)(1)积分区域D:(极点在区域外)r

2

(

)先对r再对

积分4DxOxODr

(

)(2)积分区域D:(极点在区域边界上)r

(

)D

f

(r

cos

,

r

sin

)rdrdf

(r

cos

,

r

sin

)rdr

(

)

d

0先对r再对

积分Doxr

(

)(3)积分区域D:(极点在区域

)

f

(r

cos

,

r

sin

)rdrdD

(

)

d

f

(r

cos

,

r

sin

)rdr0

02先对r再对

积分5解Ddxdy

e

x2

y22a

(1

e

)例计算

e

x2

y2

dxdy,其中D是由中心在原点,D半径为a的圆周所围成的闭区域.在极坐标系下计算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

计算

(

x2

y2

)dxdy,

其中D为由圆3y

0,x2

y2

2

y,x2

y2

4

y及直线x

d

4sin

r

2

rdr2sin36y

3x

0所围成的平面闭区域.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其中积分区域

D

{(

x,

y)

1

x

y

1

x2

,0

x

1}解在极坐标系下

y

r

sin圆方程为r

1直线方程为cos

sinr

1

11sin

cos2d0yO1Dy

1

x

2xy

1

x

r

cos1r

sin

1x2

y2

1例写出积分

f

(x,y)dxdy

的极坐标二次积分形式,11将直角坐标系下累次积分:

21

0104

x24

x2

21

xf

(

x,

y)dydxf

(

x,

y)dy

dx化为极坐标系下的累次积分.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双纽线(a

0)求曲线(x2

y2

)2

2a2

(x2

y2

)和x2

y2

a2

所围成的图形D的面积.例x2

y2

a2

1D

y2

)

r

2

2a2

cos

2x由r

ar

a2

cos

2得交点A

(a,

)6)3D

a2

( 3

rdr1DAa

2cos

2面积

dxdy

4dxdy

4

6d

0ayaO2a在极坐标系下

x

r

cos

y

r

sinr

a对称性质13例求由不等式：x2

y2

z2

a2

与x2

y2

ax解所确定的空间几何体的体积V1

f

(

x,

y)dxdyD1xyOa

a2aD1x2

y2

axaxyz

aO曲顶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)对称性质aaxyz

aO16

(

x

y

)dxdyx2

y2

1解x2

y2

1例计算

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计算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

求反常积分

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为矩形域:a≤x≤b,c≤y≤d且f

(

x,

y)

f1

(

x)

f2

(

y)baf

(

x)dx1dcf

(

y)dy2D则

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对称性质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求反常积分02e

dx.

x例解显然有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当a

时,

I

,

I

14I

24,

I

201222a(1

e

)

Ie

dx)

(1

e

)

(

I

2a

x

2a44概率积分定理当a

时,

4所求反常积分202e

dx

x0(

e

x

2

dx)2

20lim

(a

x2a

e

dx)23D

D24小结二重积分在极坐标下的计算公式

f

(

x,

y)d

f

(r

cos

,

r

sin

)rdrd先对r再对

积分作业P389252(7)，

9，23(3),

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

，26(3),

3026求曲面z

2

x2

y2与曲面z

x2

y2xyz

x2

y2o所围成的

的体积2

z

z

2

x2

y2z

x2

y2zxyo分析zxy2oz

2

x2

y2z

x2

y2xyz

x2

y2o22z

x

yzyxyo2

z

z

2

x2

y2两曲面的交线为:z

1

x2

y2

1