高數(shù)定積分應(yīng)用d62幾何

(補(bǔ)充)第二節(jié)定積分在幾何學(xué)上的應(yīng)用第六章一、平面圖形的面積xbdA

=

f

(x)

dxA

=

a

f

(x)

dxox

+

dxy

=

f2

(xy

=

f1(x)f1

(x)

-

f2

(x)

dxA

=baa

x

x

+

d

x

b

x例1.xy2

=

xyy

=

x2o

x解:(1,1)1x

-

x2

dxd

A

==

131\

A

=

0xy

2

=

2x例2.y

2

=

2x解:(8,

4)y

=

x

-

4(2,-

2)y

)

dy(

y

+

4

-d

A

=212=18y

=

x

-

4yyy4-2\

A

=例3.解:d

A

=

y

dxo

x

x

+

d

xaA

=

40

y

d

x(0

￡

t

￡

2p

)

y

=

b

sin

t

x

=

a

cos

t當(dāng)a

=b

時(shí)得圓面積公式順時(shí)針?lè)较蚶?.dA

=2(令u

=t

)=

3p

a22p解:

A

=

0

a

(1

-

cos

t)

a

(1

-

cos

t)

d

tr

=

j

(q)qdq2dA

=

1

[j

(q)]2

dq2

aA

=

1

b

j

2

(q)

dqb例5.解:

A

=x2paoqdq22p

1

(aq

)2

dq0點(diǎn)擊圖片任意處播放開(kāi)始或暫停例6.解:2a

xodqq(利用對(duì)稱(chēng)性)2令t

=q2=

3pa2oxya心形線q點(diǎn)擊圖中任意點(diǎn)動(dòng)畫(huà)開(kāi)始或暫停???21

(1+

cos2q)yo

a例7.解:2A

=

1

pa2a例8.解:=

a2思考:答案:o二、平面曲線的弧長(zhǎng)定義:1AB

=定理:(證明略)nlfi

0

i=1s

=

lim

Mi-1Miaby

=

f

(x)1+

f

￠2

(x)

dxba=ds

=

(dx)2

+

(dy)2j￠2

(t)

+y

￠2

(t)

d

ts

=bads

=

(dx)2

+

(dy)2r

2

(q)

+

r￠2

(q)

dqbas

=(自己驗(yàn)證)(c

ch

x)￠c例9.(-b

￡

x

￡

b)cy

=

c

ch

x解:c=

2c

sh

b2ex

+

e-xch

x

=ex

-

e-x2sh

x

=(ch

x)

=

sh

x(sh

x)

=

ch

x-

bb例10.解:例11.解:=

8a2例12.解:r

=

aq

(a

>

0)2paor

=

aq三、已知平行截面面積函數(shù)的立體體積dV

=

A(x)

d

xbV

=

a

A(x)

d

xA(x)yo

a

b

xxyo

aby

=

f

(x)2p

[

f

(x)]dxV

=bay2p

[j

(

y)]dyV

=dcxyxdyc例13.解:方法1o

x(利用對(duì)稱(chēng)性)3=

4p

ab2a0V

=

22p

y

dx方法2a0V

=

2

p

y

2

dx3=

4p

ab2b

=

a34p

a3

.2例14.解:x02pa

2V

=

p

y

dx利用對(duì)稱(chēng)性=

5p

2a3aya

22a2px

=

x0注意上下限!注注分部積分關(guān)于

對(duì)稱(chēng)2p注(令u

=t

-p

)(利用“偶倍奇零”)說(shuō)明:x

x

+

d

xy偶函數(shù)奇函數(shù)2令u

=t2令v

=u

-p例15.證:tV

(t)

=

2p

f

(t)例16.R解:2A(x)

=

1

(R2

-

x2

)

tanaoxRoR思考:(xA(

y)

=提示:V

=abco例17.解:)d

xp

bc(1

-xa224=

3pabca0V

=

2xo12

xBy3AC例18.y

=

3

-

x2

-1(94

考研)解:202

2(x

-1) d

x-

2p15=

448p1yo

a

b

x四、旋轉(zhuǎn)體的側(cè)面積(補(bǔ)充)d

S

=

2p

y

d

s(x)

dxS

=

2pba2f

￠f

(x) 1

+xyoa

by

=

f

(x)abxabxd

S

=

2p

y

ds

?d

sdxx2py

(t)22+y

￠

(t)

d

tj

￠

(t)S

=ba注意:例19.S解:=

2p

R(x2

-

x1

)h＝2RS

=

4p

R2o例20.解:5=

12p

a2星形線點(diǎn)擊圖片任意處播放開(kāi)始或暫停內(nèi)容小結(jié)d

s

=

(d

x)2

+

(d

y)2d

s

=

r

2

(q)

+

r￠2

(q)

dq上下限按順時(shí)針?lè)较虼_定注意:求弧長(zhǎng)時(shí)積分上下限必須上大下小A(x)

=

p

y2d

S

=

2p

y

d

s(注意在不同坐標(biāo)系下ds

的表達(dá)式)A(x)

=

2p

x

y(柱殼法)思考與練習(xí)As提示:x

=

y2- -

=xo

13=

323直線段部分3(2

y

+

3)

-

y

2弧線段部分-1

1

+

4

y

2

dyb-

R

R提示:下-+=

2p

2

R2b方法1方法2b-

R

Ry說(shuō)明:下-+b-

R

R利用對(duì)稱(chēng)性R2

-

x2

)2p(b

+R2

-

x2

)2p(b

-下-+作業(yè)面積及弧長(zhǎng)部分：體積部分：備用題1.解：1xy

=

ey

=

exey

=

xxy

=

e2.解:A2A1