8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + =...

26
8.4 – EXERCÍCIOS – pg. 344 Nos exercícios de 1 a 14, encontrar o comprimento de arco da curva dada. 1. 2 5 - = x y , 2 2 - x () ( ) . . 26 4 2 2 26 26 26 5 1 1 2 2 2 2 2 2 2 2 c u x dx dx dx x f s b a = + = = = + = + = - - - 2. 1 3 2 - = x y , 2 1 x 3 1 3 2 - = x y dx x dx x dx x x dx x s 3 1 2 1 2 1 3 2 2 1 3 2 3 2 2 1 3 2 . 3 1 . 4 9 9 4 9 9 4 1 - + = + = + = - + = - + = + = + = - 3 13 4 2 . 9 27 1 13 4 9 3 2 . 18 1 2 3 4 9 . 18 1 . 6 . 4 9 6 1 . 3 1 2 3 3 2 2 3 2 3 3 2 2 1 2 3 3 2 3 1 2 1 2 1 3 2 x x dx x x 3. ( ) 2 3 2 2 3 1 x y + = , 3 0 x ( ) x x y 2 . 2 2 3 . 3 1 2 1 2 + =

Transcript of 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + =...

Page 1: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

8.4 – EXERCÍCIOS – pg. 344

Nos exercícios de 1 a 14, encontrar o comprimento de arco da curva dada.

1. 25 −= xy , 22 ≤≤− x

( )

( ) ..2642226

262651

1

2

2

2

2

2

2

2

2

cu

xdxdx

dxxfs

b

a

=+=

==+=

′+=

−−−

∫∫

2. 132

−= xy , 21 ≤≤ x

31

3

2 −

=′ xy

dxxdxx

dx

x

xdx

x

s

31

2

1

21

32

2

13

2

322

13

2

.3

1.49

9

49

9

41

∫∫

+=

+=+=

+=

+=

+

=

+=

31342.927

11349

3

2.

18

1

23

49

.18

1

.6.496

1.

3

1

23

32

232

3

32

2

1

23

32

31

2

1

21

32

x

x

dxxx

3. ( ) 23

223

1xy += , 30 ≤≤ x

( ) xxy 2.22

3.

3

12

12+=′

Page 2: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

( )

( ) 1233

3

31

1

1

21

33

0

33

0

2

3

0

2

3

0

42

3

0

22

=+=+=+=

+=

++=

++=

xx

dxx

dxx

dxxx

dxxxs

4. 32

32

32

2=+ yx

=

=

tseny

tx

3

3

2

cos2

( )

..12

2.24cos.24

coscos.64

cos36cos364

2

0

22

0

2

0

2222

2

0

2424

cu

tsendtttsen

dttsentttsen

dtttsentsents

=

==

+=

+=

ππ

π

π

5. 2

4

8

1

4

1

xxy += , 21 ≤≤ x

( ) 33 28

1

4

1 −−+=′ xxy

Page 3: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( ) dxxx

dxx

xx

dxx

xx

dxxx

xx

dxx

xs

+=

++=

−+=

+−+=

−+=

2

1

26

3

2

1

6

126

2

1

6

36

2

1

3

36

2

1

2

3

3

144

1

16

1168

16

1.21

616

1

4

1.21

4

11

( )

( )

32

123

2

11

2.2

12

4

1

24.4

4

1

44

1

144

1

2

4

2

1

24

2

1

33

2

1

26

3

=

+−−=

+=

+=

+=

xx

dxxx

dxxx

6. y

yx4

1

3

1 3 += , 31 ≤≤ y

( )

2

4

2

2

22

4

14

4

1

.14

13.

3

1

y

y

yy

yyx

−=

−=

−+=′ −

Page 4: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )4

24

4

48

4

484

4

482

2

4

16

14

16

1816

16

181616

16

18161

4

141

y

y

y

yy

y

yyy

y

yy

y

y

−=

++=

+−+=

+−+=

−+

( )

6

53

4

1

3

1

3.4

1

3

3

1.

4

1

3

4

1

4

14

16

14

3

3

1

13

3

1

22

3

1

2

4

3

1

4

24

=

+−−=

−+=

+=

+=

+=

∫∫

yy

dyyydyy

y

dyy

ys

7. ( )xxeey

−+=2

1 de ( )1,0 a

+ −

2,1

1ee

( )xxeey

−−=′2

1

( )

( )

( )

dxee

dxee

dxee

dxeeee

dxees

xx

xx

xx

xxxx

xx

−−

+−+=

+−+=

+−+=

+−+=

−+=

1

0

22

1

0

22

1

0

22

1

0

22

1

0

2

4

24

4

1

2

1

4

11

24

11

..24

11

4

11

Page 5: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

( )

( )

( )

( )

1

112

1

2

1

2

1

11

2

1

11

2

1

12

1

1

0

1

0

1

0

2

1

0

22

1

0

2

2

hsen

ee

ee

dxee

dxee

dxee

dxe

e

xx

xx

x

x

x

x

x

x

=

+−−=

+=

+=

+=

+=

++=

8. xy ln= , 83 ≤≤ x

xy

1=′

dxx

xdx

xs ∫∫

+=+=

8

3

28

3

2

111

( )

( ) dtttdx

tx

tx

tx

2.12

1

1

1

1

21

2

21

2

22

22

−=

−=

−=

=+

Page 6: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( ) ( )

( ) ( )∫ ∫

∫∫

+−+=

−+=

−=

−−

=+

=

11

1

11

1

1

.

1

1

2

2

2

21

221

2

2

tt

dtdt

dtt

t

dtt

t

dtt

t

tdx

x

xI

Cx

xx

Ct

tt

Cttt

dtt

dtt

t

+++

−+++=

++

−+=

++−−+=

+−

−+= ∫∫

11

11ln

2

11

1

1ln

2

1

1ln2

11ln

2

1

1

21

1

21

2

22

2

3ln

2

11

3

1ln

2

12

4

2ln

2

13

11

11ln

2

11

8

3

2

22

+=

−−+=

++

−+++=

x

xxs

9. ( )xseny ln1−= , 46

ππ≤≤ x

xsen

xy

cos−=′

Page 7: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

∫∫

=

=+

=

+=

4

6

4

6

4

6

2

22

4

6

2

2

cos

cos

cos1

π

π

π

π

π

π

π

π

dxxec

sen

dxdx

xsen

xxsen

dxxsen

xs

x

..332

12ln

262

632ln

132

3.

2

22ln

3

32ln1

2

2ln

cotcosln 4

6

cu

xgec

−=

−=

−=

−−−=

−=π

π

10. 3xy = , de ( )0,00P ate ( )8,41P

21

2

3xy =′

( ) ..1101027

8

14.4

91

3

2.

9

4

23

4

91

9

4

4

91

23

4

0

23

4

0

cu

x

dxxs

−=

+=

+

=

+= ∫

Page 8: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

11. 24 3 += xy de ( )2,00P ate ( )6,11P

21

21

62

3.4 xxy ==′

( )

( )

( ) ..1373754

1

137373

2.

36

1

23

361

36

1

361

1

0

23

1

0

cu

x

dxxs

−=

−=

+=

+= ∫

12. ( )16 3 2 −= xy de ( )0,10P ate ( )6,221P

31

31

4

3

2.6

=

=′

x

xy

dxxx

dxx

x

dx

x

dxxs

31

22

1

21

32

22

13

2

32

22

13

2

22

1

32

16

16

161

161

−−

+=

+=

+=

+=

Page 9: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

..1717254

17171818

172216

2

3

16

2

3

232

3

32

22

1

23

32

cu

x

−=

−=

+=

+

=

13. ( ) ( )3241 +=− xy de ( )2,30 −P ate ( )9,01P

( )

( )

( ) 21

23

23

42

3

41

41

+=′

++=

+=−

xy

xy

xy

( )

( )( )

...27

13131080

4

1310

3

2.

9

4

23

34

91

9

4

44

91

23

23

0

3

23

0

3

cu

x

dxxs

−=

−=

++=

++=

14. 32 yx = , de ( )0,00P ate ( )4,81P

31

32

3

2 −

=′

=

xy

xy

Page 10: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

dxxx

dx

x

x

dxxs

+=

+=

+=

8

0

312

1

32

8

0 32

32

8

0

32

.3.49

9

49

9

41

31

32

3

2.9

49

=

+=

xdu

xu

( )

( ) ..8108027

1

4440403

2.

18

1

23

49

.18

1

8

0

23

32

cu

x

−=

−=

+

=

Nos exercícios de 15 a 21, estabelecer a integral que da o comprimento de arco da curva

dada.

15. 2xy = , 20 ≤≤ x

xy 2=′

∫ +=2

0

241 dxxs

16. x

y1

= de

4,

4

10P ate

4

1,41P

2

1

xy

−=′

Page 11: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

+=

+=

+=

4

41

2

4

4

41

4

4

4

41

4

1

1

11

dxx

x

dxx

x

dxx

s

17. 122 =− yx de ( )22,30 −P ate ( )22,30P

( ) ( ) yyxyx

yx

2.12

11

1

21

221

2

22

+=′⇒+=

+=

dyy

y

dyy

yy

dyy

ys

+

+=

+

++=

++=

22

22

2

2

22

22

2

22

22

22

2

2

1

21

1

1

11

18. xey = , de ( )1,00P ate ( )2

1 ,2 eP

xey =′

dxesx

∫ +=

2

0

21

19. 122 −+= xxy , 10 ≤≤ x

22 +=′ xy

Page 12: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

dxxx

dxxx

dxxs

++=

+++=

++=

1

0

2

1

0

2

1

0

2

584

4841

221

20. xy = , 42 ≤≤ x

21

2

1 −=′ xy

∫ +=

4

24

11 dx

xs

21. xseny 3= , π20 ≤≤ x

xy 3cos3=′

dxxs ∫ +=

π2

0

2 3cos91

Nos exercícios de 22 a 29, calcular o comprimento de arco da curva dada na forma

paramétrica.

22.

=

=

2

3

ty

tx, 31 ≤≤ t

Page 13: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( ) ( )

( ) ..1313858527

1

131385853

2.

18

1

23

49

18

1

49

49

3

1

23

2

3

1

2

3

1

24

cu

t

dttt

dttts

−=

−=+

=

+=

+=

23. ( )

( )

−=

−=

ty

tsentx

cos12

2, [ ]π,0∈t

( )

tseny

tx

2

cos12

=′

−=′

( )

( )

( ) dtt

dtt

dttsentt

dttsents

−=

−=

++−=

+−=

π

π

π

π

0

0

0

22

0

22

cos12

cos222

coscos214

4cos14

Page 14: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

..8

108

0cos2

cos82

cos.2.4

2222

2222

cos122

0

0

0

2

0

cu

t

dtt

sen

dtt

sen

dtt

=

−−=

−−=−=

=

=

−=

ππ

π

π

π

24.

=

−=

ty

tsenx

cos, [ ]π2,0∈t

ππ

π

2

cos

2

0

2

0

22

==

+= ∫

t

dttsents

25.

=

=

tty

tsentx

cos, [ ]π,0∈t

22

0

22

0

2

0

222222

1ln2

11

2

1ln2

11

2

1

cos2coscos2cos

ππππ

π

π

π

++++=

++++=

+=

+−+++=

tttt

dtt

dttsentttsentttsentsentttts

26.

−=

+=

1

23

ty

tx, [ ]2,0∈t

Page 15: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

10210

91

2

0

2

0

==

+= ∫

t

dts

27.

=

=

2

3

2

1

3

1

ty

tx

, 20 ≤≤ t

=′

=′

ty

tx

2.2

1

3.3

1 2

dtttdttts ∫∫ +=+=

2

0

2

2

0

24 1

dttdu

tu

2

12

=

+=

( ) ( )

( ) ..1553

1

153

2.

2

1

23

1

2

11

23

2

0

23

22

0

21

2

cu

tdttts

−=

−=

+=+= ∫

28.

=

=

tseney

tex

t

t cos, 21 ≤≤ t

+=′

+−=′

tt

tt

etsentey

ettsenex

cos

cos

Page 16: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( ) ( )

( ) ..22

2

coscos

22

1

2

1

2

1

2222

cueee

dte

tsentetsentes

t

t

tt

−==

=

++−=

29.

−=

+=

tttseny

tsenttx

cos22

2cos2,

20

π≤≤ t

tsenty

ttx

2

cos2

=′

=′

===

+=

2

0

22

0

2

2

0

2222

..42

22

4cos4

π π

π

πcu

tdtt

dttsenttts

30. Achar o comprimento da hipociclóide

=

=

ty

tsenx

3cos4

4 3

, [ ]π2,0∈t

−=′

=′

tsenty

ttsenx

.cos3.4

cos.3.4

2

2

Page 17: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

..24

012

48

2.12.4

cos124

cos12cos124

2

0

2

2

0

2

0

24242

cu

tsen

dtttsen

dttsentttsens

=

−=

=

=

+=

π

π

π

31. Achar o comprimento da circunferência.

=

=

tsenay

tax cos, [ ]π2,0∈t

tay

senax

cos=′

−=′

..244

cos4

2

0

2

0

2222

cuaatdta

dttatsenas

π

π

π

===

+=

32. Calcular o comprimento da parte da circunferência que está no 1° quadrante

=

=

47

4cos7

tseny

tx

..4

7

4

7

4

7

4cos

4

7

44

7

2

0

2

0

2

0

2

2

2

2

cutdt

dttt

sens

π

ππ

π

===

+

=

Page 18: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

Nos exercícios de 33 a 35, calcular a área da região limitada pelas seguintes curvas, dadas

na forma paramétrica.

33.

=

=

tseny

tx cos e

=

=

tseny

tx

2

1

cos

-1 1

-1.5

-1

-0.5

0.5

1

1.5

x

y

( ) ( )

..2

1

2

1

2

144

2

0

2

0

cu

dttsentsendttsentsenA

π

ππ

ππ

=

−=

−+−−= ∫∫

34.

=

=

tseny

tx

3

3

2

cos2 e

=

=

tseny

tx

2

cos2

Page 19: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

-2 -1 1 2

-2.5

-2

-1.5

-1

-0.5

0.5

1

1.5

2

2.5

x

y

( ) ( )

( )

−−=

−=

−−−−−=

∫ ∫

2

0

24

2

0

242

0

2

0

2

0

23

1122

.24

cos1222

1

2

1.44

cos3.2.2224

π

ππ

π π

πdttsentsen

dtttsentsent

dttsenttsendttsentsenA

( )

..2

5

2

38

2

34

2

164

4

3cos

4

184

482

548cos

6

484

486

5cos

6

1484

484

2

0

2

0

23

2

0

2

0

442

0

5

2

0

445

2

0

64

cu

t

dttsenttsen

dttsendttsenttsen

dttsendttsenttsen

dttsentsen

πππππ

π

π

π

π

π

π

π

π ππ

π

π

=−

=−=

−=

+−−=

−+−=

+−+=

−−=

∫ ∫

∫∫

35.

=

=

2ty

tx e

+=

+=

ty

tx

31

1

Page 20: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

1 2

-2

-1

1

2

3

4

x

y

23

2

−=

=

xy

xy

2=x e 1=x

( )( ) 24,2

11,1

=→

=→

t

t

( )( ) 14,2

01,1

=→

=→

t

t

3

7

3

.1.

2

1

3

2

1

2

1

=

=

= ∫

t

dttA

( )

2

5

23

..31

1

0

2

1

0

1

=

+=

+= ∫

tt

dttA

..6

1

3

7

2

5cuA =−=

36. Calcular a área da arte da circunferência

tseny

tx

2

cos2

=

=

que está acima da reta 1=y

Page 21: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

( )62

3cos3cos21,3

22,0

π

π

=∴=∴=→

=→

ttt

t

( )

6

334

6

6332

2

3

3

22

3

2

1

62

2

1

232

1

62

22

12

2

2cos14

22

6

2

6

2

6

2

1

+=

++−=

++−=

−−−=

+−−−=

−−=

−−=

−=

πππ

ππ

ππ

ππππ

π

π

π

π

π

π

sensen

tsentdtt

dttsentsenA

..6

334

6

36334

36

334

cu

A

−=

−+=

−+

=

π

π

π

37. Calcular a área da região delimitada pela elipse

Page 22: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

-3 -2 -1 1 2 3

-1

1

x

y

=

=

tseny

tx cos3

( )

..4

3

3

3

0

2

2

0

2

1

cu

dttsen

dttsentsenA

π

π

π

=

=

−=

auA .34

3.4 π

π==

38. Calcular a área da região limitada à direita pela elipse

=

=

tseny

tx

2

cos3 e a esquerda pela

reta 2

33=x

Page 23: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

6cos

2

3cos3

2

33 π=∴∴= ttt

( )

..34

3

22

3.

2

3

2

34

1

6.

2

16

24

1

2

16

632

6

0

0

6

2

0

6

1

au

sen

tsent

dttsendttsentsenA

−=−=

−=

−=

=−= ∫∫

ππ

ππ

π

ππ

39. Calcular a área da região entre as curvas

=

=

tseny

tx

2

cos4 e

=

=

tseny

tx cos

Page 24: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

-4 -3 -2 -1 1 2 3 4

-2

-1

1

2

x

y

( ) ππ

8244

0

2

1 =−= ∫ dttsentsenA

π

ππ

=

+−=−= ∫

0

2

0

2

22

1cos

2

14 tttsendttsentsenA

..78 auA πππ =−=

40. Calcular a área entre o arco da hipociclóide

=

=

tseny

tx

3

3

3

cos3, [ ]

2,0 π∈t

e a reta 3=+ yx

Page 25: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

( )

( )

ππ

π

π

π

π

π

32

27

2.

6

1.27

16

1cos

16

1cos

24

1cos

6

127

27

127

cos27

.cos3.33

2

0

35

0

2

64

2

0

2

4

2

0

2

4

3

0

2

3

1

==

+−−=

−=

−=

=

−=

tttsenttsenttsen

dttsentsen

dttsentsen

dtttsen

dttsenttsenA

..32

27144

32

27

2

9auA

ππ

−=−=

41. Calcular a área delimitada pela hipociclóide

=

=

ty

tsenx

3

3

cos4

4

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

x

y

Page 26: 8.4 – EXERCÍCIOS – pg. 344 · ( ) 3 12 3 3 3 1 1 1 1 2 3 3 0 3 3 2 3 0 2 3 0 2 4 3 0 2 2 = + = + = + = = + = + + = + + ∫ ∫ ∫ ∫ x x x dx x dx x x dx s x x dx 4. 3 2 3

( )

( )

( )

..2

3

coscos48

cos1cos48

cos48

cos..3.4.cos4

2

0

64

2

0

24

2

0

24

22

0

3

au

dttt

dttt

dttsent

dtttsentA

π

π

π

π

π

=

−=

−=

=

=

auA .62

3.4 π

π==

42. Calcular a área da região S , hachurada na figura 8.12

( )( )tky

tsentkx

cos1−=

−=

( ) ( )

( )

( )

..3

3.

2.2

12

2

1cos

2

12

coscos21

cos1

cos1cos1

2

2

2

2

0

2

2

0

22

2

0

22

2

0

auk

k

k

ttsenttsentk

dtttk

dttk

dttktkA

π

π

ππ

π

π

π

π

=

=

+=

++−=

+−=

−=

−−=