1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C Xϯî...

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1. ln 9 (= 2 ln 3) 2. 17 4 3. 6 27 - 5 p 5 4. 4 5 5. 64 9 6. < 4, 6, 8 > 7. 20 8. 10 9. 4 sin 2 + 21 10. -2

Transcript of 1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C Xϯî...

Page 1: 1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C XÏ¯î ¥ÄÌ– ⌘Ït –⇣– à‡ ⌅–⌧ ¸ L ⇠‹ƒ )•tp ⇠¿Ñ t ë@ –D

1. ln 9 (= 2 ln 3)

2.

17

4

3.

6

⇣27� 5

p5

4.

4

5

5.

64

9

6. < 4, 6, 8 >

7. 20

8. 10

9. 4 sin 2 + 21

10. �2

Page 2: 1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C XÏ¯î ¥ÄÌ– ⌘Ït –⇣– à‡ ⌅–⌧ ¸ L ⇠‹ƒ )•tp ⇠¿Ñ t ë@ –D

11. ¡På\| t©Xt Ä<î

V =

Z 1

0

Z 1+p1�x

2

1

Zy/(x2+y

2)

0

dzdydx

⇣î, V =

Z 2

1

Z p1�(y�1)2

0

Zy/(x2+y

2)

0

dzdxdy, ....

!<\ \⌅⌧‰.

¯¨‡, y = 1D ¸tå\\ \⌅Xt r = 1/ sin ✓ t‡,

x

2+ y

2 � 2y = 0@ r = 2 sin ✓ t¿\,

¸tå\| t©XÏ Ä<| ƒ∞Xî º⌘�Ñ@

V =

Z⇡/2

⇡/4

Z 2 sin ✓

1/ sin ✓

Z sin ✓/r

0

rdzdrd✓

<\ \⌅⌧‰.

¸tå\| \⌅⌧ ›D t©XÏ Ä<| lXt,

V =

Z⇡/2

⇡/4

Z 2 sin ✓

1/ sin ✓

sin ✓ drd✓ =

Z⇡/2

⇡/4

(2 sin

2✓ � 1) d✓

=

Z⇡/2

⇡/4

� cos 2✓ d✓ =

�1

2

sin 2✓

�⇡/2

⇡/4

=

1

2

t‰.

Page 3: 1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C XÏ¯î ¥ÄÌ– ⌘Ït –⇣– à‡ ⌅–⌧ ¸ L ⇠‹ƒ )•tp ⇠¿Ñ t ë@ –D

12. å⌅ ·tD ‰⌧¿⇠\ \⌅Xt,

r (u, v) =< sin u cos v, u, sin u sin v >

⇣0 u ⇡

2

, 0 v 2⇡

t‡,

r

u

= < cos u cos v, 1, cos u sin v >

r

v

= < � sin u sin v, 0, sin u cos v >

r

u

⇥ r

v

= < sin u cos v, � cos u sin u, sin u sin v >

| ru

⇥ r

v

| =

psin

2u cos

2v + cos

2u sin

2u+ sin

2u sin

2v = sin u

p1 + cos

2u

t‰. 0|⌧ ·t�ÑD ƒ∞Xt,

ZZ

E

p1� x

2 � z

2dS

=

Z 2⇡

0

Z⇡/2

0

p1� sin

2u cos

2v � sin

2u sin

2v sin u

p1 + cos

2u dudv

=

Z 2⇡

0

Z⇡/2

0

cos u sin u

p1 + cos

2u dudv

=

Z 2⇡

0

�1

3

(1 + cos

2u)

3/2

�u=⇡/2

u=0

dv

= �2⇡

3

⇣1� 2

p2

=

2⇡

3

⇣2

p2� 1

t‰.

Page 4: 1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C XÏ¯î ¥ÄÌ– ⌘Ït –⇣– à‡ ⌅–⌧ ¸ L ⇠‹ƒ )•tp ⇠¿Ñ t ë@ –D

13. C� XÏ¯î ¥Ä�Ì– ⌘Ït –⇣– à‡ ⌅–⌧ ¸ L ⇠‹ƒ )•tp ⇠¿Ñ

t ë@ –D ˜

C|‡ Xê. t –X ⇠¿ÑD a| Xt,

˜

CX ‰⌧ ¿⇠›@

˜

C(t) = (a cos t, a sin t) (0 t 2⇡)

<\ \⌅` ⇠ à‰.

P =

x

3+ xy

2 � 3y

x

2+ y

2, Q =

y

3+ x

2y + 3x

x

2+ y

2|‡ X‡, · C @ ˜

C\ Xϸ �ÌD D

|‡ Xt, Green �¨– XXÏ

I

C[�C̃

Pdx+Qdy =

ZZ

D

@Q

@x

� @P

@y

dxdy

t‰. 0|⌧

I

C

Pdx+Qdy =

I

Pdx+Qdy +

ZZ

D

✓@Q

@x

� @P

@y

◆dxdy

t‰. Ï0⌧,

@Q

@x

=

@

@x

✓y

3+ x

2y + 3x

x

2+ y

2

◆=

@

@x

✓y +

3x

x

2+ y

2

◆=

�3x

2+ 3y

2

(x

2+ y

2)

2

@P

@y

=

@

@y

✓x

3+ xy

2 � 3y

x

2+ y

2

◆=

@

@y

✓x� 3y

x

2+ y

2

◆=

�3x

2+ 3y

2

(x

2+ y

2)

2

t¿\,

ZZ

D

✓@Q

@x

� @P

@y

◆dxdy = 0 t‡,

I

Pdx+Qdy =

Z 2⇡

0

✓a cos t� 3a sin t

a

2

◆(�a sin t) +

✓a sin t+

3a cos t

a

2

◆(a cos t)dt

=

Z 2⇡

0

3(sin

2t+ cos

2t)dt = 6⇡

t‰. 0|⌧,

I

C

Pdx+Qdy = 6⇡

t‰.

Page 5: 1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C XÏ¯î ¥ÄÌ– ⌘Ït –⇣– à‡ ⌅–⌧ ¸ L ⇠‹ƒ )•tp ⇠¿Ñ t ë@ –D

14. yz-…t–⌧ y

2+ z

2 4 (z � 0)x �ÌD D|‡ Xê.

r · F = 3xt¿\ ⌧∞�¨| t©XÏ ·t�ÑD ƒ∞Xt,

ZZ

S

F · ndS =

ZZZ

E

r · F dV

=

ZZ

D

Z 4

y

2+z

2

3x dxdydz

=

ZZ

D

3

2

⇥16� (y

2+ z

2)

2⇤dydz

=

Z⇡

0

Z 2

0

3

2

�16� r

4�r drd✓

=

3⇡

2

8r

2 � 1

6

r

6

�2

0

= 32⇡

t‰.

Page 6: 1. ln9 (= 2ln3) 17 2.math.inha.ac.kr/~lecture/calculus/jokbo/15-2_ilsu_fin_sol.pdf · 13. C XÏ¯î ¥ÄÌ– ⌘Ït –⇣– à‡ ⌅–⌧ ¸ L ⇠‹ƒ )•tp ⇠¿Ñ t ë@ –D

15. 8 ⇣ (1, 0, 0), (0, 1, 0), (0, 0, 1)D ¿òî …tX )�›@ x+ y + z = 1t‰.

…tD ‰⌧ ¿⇠\ \⌅Xt,

r (x, y) =< x, y, 1� x� y > (0 x 1, 0 y 1� x)

t‰. t L, 0 x 1, 0 y 1� xx �ÌD D|‡ Xê. r

x

⇥ r

y

=< 1, 1, 1 >

t¿\ x+ y + z = 1– ⇠¡t‡ ⌅ΩD •Xî Ë⌅ ï °0| nt|‡ Xt,

n =

1p3

< 1, 1, 1 > t‡,

| rx

⇥ r

y

| =p3t‰.

⌧∞�¨– XXÏ,

Z

C

F · Tds =

ZZ

S

r⇥ F · ndS

=

Z 1

0

Z 1�x

0

(�x) dydx

=

Z 1

0

x

2 � x dx

=

1

3

x

3 � 1

2

x

2

�1

0

= �1

6

t‰.