高等数学习题解答——定积分应用

高等数学(金路,童裕孙等编)(高等教育出版社)

p166

1.)µ{ 2= +2) : 1( 1, 1), 2(2,2), =

÷y¶© ∫È29. = 1[ ( 2 2)] =2

2.)µ Ô º: ( 2,5),= 2 4

=0? = 4 3, =3? = 10 +6

3ü : (2, 9)

¤¦¡È ∫∫32 =0[( 4 3) ( 4 3)] +3[( 10 +6) ( 2 4 3)] =9

43.)µ =∫1[2 ] +0∫22[2 ] =174.)µ

¤¦ 1 – Ü©¡È 4 "

31 – §¡È

= = 3 ( 3 )= 3 2 4 2

= , =0; =0, =

∫ ∫0 =40 =4 3 2 4 2 ∫42=12 0( 6 )

15 3 1 =12 2[3 ]2=3 ∫5|^0 4íúª.

5.)µ

12 = ∫ 312 =0 (1 6 )

2= 6.)µ2- 4 I § 2= ¤¦¡È 1 – Ü© 4 §

1

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2

2 221+ =401 =2 2 + 01+ 1∫∫∞2+1 = 2∞ ( )=2 2

0 =2 0( )2+2√2√ 1

∞= ∣0= 2

2½ö 2=4√∫124 =4023+ 4 = 2∫∫7.)µ

^R ux¶ ²¡ ^=N ¡¡È √√√22 ( )= [( + ( ]=4 ^=∫ N NÈ ∫ √ = ( ) =8 0 =8 1 2=2 2 2 10.)µ

.¡ √»3x¶þ§÷y¶ ¡È

( )=2

È© ∫ √∫ NÈ:23 =0 ( ) =20 =

11.)µ

²¡« þ0u[ , + ]Ü©§7y¶^=¤ ^=NNÈCq ü ÎNNÈ (p ( )):

Δ = ( ) [( +Δ )2 2]= ( )2 Δ + (Δ ), =2 ( ) ∫ È© =2 ( )

12.)µ

¤¦ ã9 l7x¶^=¤ ^=NNÈ Ú"∫ ∫ =0( )2 + ( 2 2)

3 3 2= [ 2 ∣+( ∣]0 2= 3(1 )

17.)µ

= , ′= 1, ”= 2,√1 =∣ ”∣

′, (1)=2, =2, ′(1)=1,(1+( ) (1,0)?{ 0=( 1)( 1), = +1.

d ”(1)<0 - 3P?þà§-Ç %3 me §3{ þ" ->1,K√Ç¥% ( , +1), √∣ ∣==2 =1±2,

高等数学(金路,童裕孙等编)(高等教育出版社)

vanishingtheorem3

u1 ) % (3, 2),l -Ç §

( 3)2+( +2)2= 2=8.

18.):11(1) =2 , ′= .√√1 =4 =2 =2 2 ∫ ∫ √3 =0 =4 0=8(( +1) 1).

(2)¤¦ 1 – Ü©^=¤ -¡¡È 2 "

33 = √ , = ,√ =( ′)2+( ′)2 =(3 2 )2+(3 2 )2 =3 ,

324 =2 =2 3 =6 ∫2. =206 2 4 =12 5(3) = (1+ ), = = (1+ ),√√ ∣ , ===2 ∣ =2 =2 (1+ ) 2 2

=16 2 42 2

5 ∫ 2 2 24 =016 2 2 = 32 5∣0=32 .5

19.):

±[ ¥% :§[ ¤3 x¶§[ R ²© ¶§ïá IX" : I (0, )"dé¡5 :¤ÉÚå3x þ©þ 0§3y þ©þ K "éAu[ [ , + ]Ü©§Úå =

È© ∫ = ( 2+ 2)∫ 2 = 0( 2+ 2)∫ = 1= 2 01 = 2 ( )2 = .20.)µ

\ 23\ 1mýa?"

d~K\màx? þ m :¤É 1 Úå =

\ 2éAu[ , + ]Ü©¤ÉÚå

2 1 =

22 = 1.112È© ∫ + ∫ + 211 = 2 = [ ] 121 1.1

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4

=+ 1)( + 2) ( . 1 2 ( + 1+ 2)

21.)µ

ᶤ3 y¶§R uY¡ e x¶§ý ¥%

2 2+=1,3Y x?§éA :§ïá IX"ý § u[ , + ] (Cq√ Ý/)¤ÉØå

= = 2 1 .È© ∫ √2 =2 0 1 =2 2 . 3

Ù¥ , ©O Y —Ý!-å\ Ý"2

22.)µ

¥% :§R e x¶§ïá IX.

Km© Y¡? = , ¥m þ§Y ? = .éAuY [ , √+ ]Ü°Y NÈ (Cq ÎN)

= = ,

òdÜ°Ä ©º Ñ-åI‰õ(£√Ä + )

= ( + )= ( + )l È© √∫ = ( + )23.)µ

\ ¥% :§R ²© y¶§ þ "3(0, )?>Ö 1.dé¡5\é:>Ö½åx©þ 0§y©þ § þ"

1 éAu[ , + ]Ü©§½å =

= 2 2( + )È© å ∫y

2 = 2 2 + )∫ 2( =2 022( + )∫ 2 = =2 0 =2 ( ) = 2.dO ü >Ö3(0, )?¤É½å ( )=2 ÷yÈ© 2+4 2

∫ ∫ 1

>|å‰ õ = ( ) =2 24+()√ 2 = 2 ( +4+ ∣

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vanishingtheorem5

=√ +2 [ ++ ].24.)µ

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