微积分下习题集习题课答案

一.填空p.45p.45.一.填空题2

2ex

y

cos(

x

y)dxdy

x2

y2

r

211.

limr0

r

2.2

2ex

y

cos(

x

y)dxdyx2

y2

r

21r

0

r

2解lim(

2

2

r

2

)r0

r

2

lim

1

e

2

2

cos(

)

r

2

1.1x2100p.45.2.f

(x,y)dy的极坐标形式是dx二次积分形式是yx1.OI

00dr4sin

sec2

d2sec40sin

sec

f

(r

cos

,

r

sin

)rdrd.I

22011

xxdxx

y二次积分dy的极坐标2p.45.二.3.列出二重积分在极坐标系下的二次积分D

:x2

y2

1,则|

y

3

x

|

dxdy

8

,D.23532311r

2

(sin

3

cos

)dr

r

2

(sin

0033

cos

)drdd2210f

(

x,

y,

z)dzxdx0

2

x

xdy4.三次积分的柱面坐标形式是4f

(r

cos

,

r

sin

,

z)dzrdr0

2cos

r

cossec

0d

3I

xyzO4

x2

y2

z2

t

2f

(

x2

y2

z2

)dxdydz

31limt

0

t.2

2500sin

drttt

0解I

lim

12dd

f

(r

)r0

4

5

p.45.二.5.若f

(u)连续可f

(r

2

)r

2drt

50t

0

lim

4t4

f

(t

2

)t

24

f

(t

2

)42t

0t

0

lim

lim5t5tt

04

f

(t

2

)

2t

45

lim

.2

5t55.

4

.f

(

cos

,

sin

)

d

.402

secsin

sec

d2.

I

d

.d4ch7

.Ex.

一、1.

1.0sin

sec2

0I

3

cos

)

2

d

.3

cos

)

2

d3.

I

1(sin

031(sin

03

d523

d23

cos

d

02cossecf

(

cos

,

sin

,

z)dzd044.二.计算题p.45p.43.二.计算题1.计算

e|z|dv,其中

{(x,y,z)x2

y2

z2

1}.1|z|1解I

e

dzDz

:x2

y2

1

z2d10z

2

2

e

(1

z

)dz

2

.p.45.二.2.计算

z2dv,其中是由曲面z

1

x2

y2

与x2

y2

z

1

所围成的闭区域.1r

2r

1212解

I

0

d0rdrz

dz1

10

33[(1

r

2

)2

(r

1)3

]rdr

2

xz1y6

.O1p.46.二.3.y2

z2b2

c2

1.

(

a2x2

y2

z2

x2)dv,其中:b2

c2

a2计算x2

y2

z2

解I

a2

dv

b2

dv

c2

dvx2

y2

z2Dz

:a2

b2

1

c2cz2z2dv

dzdcc

c

2

2c

z2

z24

20

c2

ab(1

c2

)dz

15

abc.x2

y24同理

a2

dv

b2

dv

15

abc,

4I

abc.5p.46.二.3.解法二.x2

y2

z2a2b2

c2

1x2

y2

z2x2

y2

z2(

)dxdydza2

b2

c2

)dv

b2

c2

(

a2解22

2a2

b2

c2x2

y2

z2(

)abcda2

b2

c2x

y

z

1d

dx

y

za

b

cabc(

x2

y2

z2

)dxdydzx2

y2

z2

121r

2

sin

r

2dr

0004abc.5ddp.46.

二.4.设f

(u)连续,

F

(t

)

[z2

f

(

x2

y2

)]dv,t

y2

t

2

,0

z

h},t其中

{(

x,

y,

z)

x2t

2t

0求F

(t

)及lim

F

(t

).20200解F

(t

)th[z

f

(

2

)]dzdd

301

23[

h

f

(

2

)h]d

t3h

f

(t

2

)h],1F

(t

)

2

t[332t

2t

3

t

0t

0

t

0lim

F

(t

)

lim

F

(t

)

lim

1[

h

f

(t

2

)h]3

h3

hf

(0).p.46.二.5.一均匀物体(密度为常数)占有的闭区域是由曲面z

x2

y2和平面z

0,|

x

|

a,|

y

|

a(a

0)所围成,(1)求其体积; (2)求物体的重心;(3)求物体关于z轴的转动惯量.2解(1)V

a

aadxa(

x

y2

)dyOyxaza3

8

a4

;p.46.

二.5.(2)x

y

0,x2

y20a

a

adxdy

azdz

20zdv

20aa(

x

y2

)2

dydx45

56

a6

,Oyxaza56

a61538

a4

z

45

7

a2

;15重心:

(0,0,

7

a2

);p.46.

二.5.(3)Jz

(

x

y

)dv2

22

2(

x2

y2

)dz000

4a

adxdyx

y

200

4aa(

x

y2

)2

dydx

y4

)dy00

4ad2

y240

43

5a(ax

2

x2a3

1

a5

)dx45

112

a6

.习题课(课外习题)p.47一.填空p.47p.47.一.填空题1.设是由平面x

y

z

1

0,x

y

z

2

0,x

0,y

0,z

0所围I1

[ln(

x

y

z

3)]

dv与I2

(

x

y

z)

dv3

2

解(x,y,z)

,有

2

x

y

z

1,0

[ln(

x

y

z

3)]3

[ln

2]3

1

(

x

y

z)2

4,

I1

I2

.I1

I2

.的大小关系是

.p.47.2.设f

(x,y)连续,则更换积分Ox2I

x2100f

(

x,

y)dydx的积分次序后,.2

x2211f

(

x,

y)dydx0f

(

x,

y)dxyI

dy1 2

y2(1,1)y2a2ax2ax

x20f

(x,y)dy的极坐标dxp.47.二.3.二次积分.2a

sec2a

cos40f

(r

cos

,

r

sin

)rdrd2a

cos

csc2

24形式是2a

cosdf

(r

cos

,

r

sin

)rdryO2a(2a,2a)x2ax²+y²=2axp.47.二.4.23

214.ydxx1e

dy的值等于.221

y40112(e

1)ydye

dx

xzOy2322322y1101yydxx1e

dy

dye

dx解220ye

ydy120

1).2

2

1

e

y2

1

(e4p.47.一.5.三次积分I

a2

r

220

0f

(r

cos

,

r

sin

,

z)dzard

2

rdr

yO

ax22ax²+y²+z²=a²22a2

x2

y2f

(x,y,z)dz柱坐标形式是axI

dxa2

2

a

2

a22

x2dyx2

y2

.40002

add

f

(r

sin

cos

,r

sin

sin

,r

cos

)r

2

sin

dr

球坐标形式是I

x2

y2

z2

t

2x2f

(

y2

z2

)dxdydz

lim

1t

0

t

4.2400t

tt

0解

I

lim

12dd

f

(r

)r

sin

dr0

1p.47.二.6.设f

(u)连续可40tf

(r

)r

2drtt

0

lim

4f

(t

)t

23t

0t

0f

(t

)

f

(0)

lim

limttf

(0二.计算题p.47p.47.二.计算题1.将极坐标累次积分I

化成直角坐标形式.2

cos204f

(r

cos

,

r

sin

)rdrd21

20f

(

x,

y)dy解I

dx

x2f

(

x,

y)dy

dx0110f

(

x,

y)dx

f

(

x,

y)dx.dy1

1

y2

ydy1

1

y21

1

y21O2

x11

yp.48.二.2.计算

(x

y

z)dxdydz,其中是由曲面z2

x2

y2与平面z

1所围成的闭区域.解

xdxdydz

ydxdydz

0,

I

zdxdydzxz1y21

1

0

d0

rrdrzdzO11

10

2r(1

r

2

)dr

2

.42000p.48.二.3.计算

x2dxdydz,其中

:x2

y2

z2

2z.2

2cosd

解

I

d0325

2

cos5

sin3

d15

4

.xr

2

sin2

cos2

r

2

sin

drz1yOp.48.二.4.计算由曲面x2

y2

z2

2az与z2

x2

y2(含有z轴的部分)所围成的

的体积.2a

cos40020d解

I

dxdydz

d

a3

.yOx4

0

23

1

8a3

cos3

sin

dr

2

sin

drz2ax²+y²+z²=2azaap.48.二.5.由曲线x

0y

1

绕y轴旋转得一曲面,求z

y

1到y

3所解曲面方程为x2立体的重心坐标(设密度为常数).

z2

y

1,并且x

0,z

0,31m

dv

dydxdzDy

:x2

z2

y1

31

(

y

1)dy

2

,

Dy

:x2

z2

y131yM

ydv

ydy

31143

y(

y

1)dy

,

733y

7

,

重心为(0,

,0).xdxdzyz3O

1

y2

2zp.48.二.6.由曲线

x

0

绕z轴旋转得一曲面与两平面z

2,

z

8围成一

(密度为常数),求此

对z轴的转动惯量.2dz

k(

x2

y2

)dxdy8200dz

x2

y2

2

z22

zd

r

3dr

k

解曲面方程为Jz

k

(

x

y

)dxdydz,2

28

336

k.自测题p.49一.单项选择p.49CBCA1

x211

x2

y2

dy

(

C

).001.dx1

x211

x2

y2

dy

0061

4

8

3

.dx解(B)

4

; (C

)

;3

63(

A)

2

;8(

D)

.

y21102.e dy

(

B

).xdx22(

A)

1

e1;

(B)

1

(1

e1

); (C

)

1

e; (

D)

1

(1

e).22111000y

yxdxe dy

dye

dx

y解210e

ydy

y21012

y

e2

1

(1

e1

).3.设f

(x,y)连续,且f

(x,y)

xy

f

(x,y)dxdy,其中D由D8(

A)

xy;

(B)

2

xy; (C

)

xy

1

; (

D)

xy

1.C15012(12

3x

cx2

)dx

1

c

,y

0,y

x2

,x

1围成的闭区域,则f

(x,y)

()

.解设

f

(x,y)dxdy

c,则f

(x,y)

xy

c,D于是有c

1Oxyx2100(

xy

c)dyD(

xy

c)dxdy

dx

8

c

1

,8f

(

x,

y)

xy

1

.t4.设f

(t

)连续可导,f

(0)

0,

f

(0)

1,

:

x2

y2

t

2

,f

(

x2

y2

)dv

(

A

).(C

)

; (

D)

.4ttt

00

z

1,则lim

1f

(

x2

y2

)dv42124000t

0

t

lim

1f

(r

)dztttt

0解

lim

1drdr2401f

(r

)rdrttt

0

lim

22

f

(t

2

)t32t

0

lim4t2tt

0

[

f

(t

2

)

f

(0)]

2

lim

.(

A);

(B)

0;2二.填空p.491

11cos(

xy

)解I

dx1

11xe

sin(

xy)dy

0dx

0.

p.47.一.填空题1.

设D

:|

x

|

1,|

y

|

1,则

xecos(

xy

)

sin(

xy)dxdy

0

.D21

x0f

(

x,

y)dy

.xdx2.交换积分次序10dyyf

(

x,

y)dxyOx1y1y=xp.49.二.3.设

:x2

y2

z2

1,z

0,则

zdv

.

Ox1

2

1

1

.2

4

4120

z(1

z

)dz

.4120

0022d

r

cos

r

sin

dr

解

zdv

d10zdz

dzdv

Dz解二4zyx2

y2

被柱面x2

y2

2

y截下部p.49.二.4.锥面z

.分的面积为

2,x

yx

2y2zx2

2

2

z

x2

y2

y2z2dxdy

2

.D

Dyz²=x²+y²2xx²+y²=2y2DOD

:

x2

y2

2

yxy1

z

2

z

2

dxdy

A

2三.计算题p.49p.49.三.计算题1.计算I

|

y

x2

|

dxdy,

D

:

1

x

1,0

y

1.x1O1y1D解I

(x2

y)dxdy

(y

x2

)dxdyD1

D2211

122x21xd