4.- Hallar el área de la región encerrada por la (s) curva (s): a) r2 = a2 Sen2  Solución: A = 4       2 1  2 0  (a2 Sen 2 ) ...

4.- Hallar el área de la región encerrada por la (s) curva (s):
a) r2 = a2 Sen2 
Solución:
A = 4 





2
1

2
0

(a2 Sen 2 ) d
A = 2 a2 





2
1

0
2
2
2


 





−
Sen

A = 2
2
2
u
a 

b) r 2 = a 2 Sen 2 
Solución:
A =) 4 





4
1

4
0

(a2 Sen 2 2) d (2 )
A = 





2
2a

0
4
2
4
2


 





−
Sen

A = 2
2
4
u
a 

c) r = Tg 2  ;  = 0 = 





8


Solución:
A = 





2
1

8
0


(Tg2 2 ) d
A = 





2
1

8
0


(Tg2 2 ) d ( 2 )
A =
4
1
( )
0
8
22





−Tg

A = 2
16
4
u




 − 

d) r = 4 Cos  - Sec  ;  = 0 ,  = 





3



Solución:
A = 





2
1

3
0


(4 Cos  - Sec )2 d
A = 





2
1

3
0


(16 Cos2  - 8 + Sec2 ) d
A = 8 
3
0


(Cos 2 ) d - 4 
3
0


d + 





2
1

3
0


(Sec2 ) d
A = 4
0
3
2
2





 





+
Sen
- ( )
0
3
4



 + 





2
1
( )
0
3



Tg

A = 2
2
33
u








h) r = YejeelySec 





22
1 2 

Solución:
A = 2 





2
1

2
0




2
2
22
1












Sec d

A = 2 





4
1

2
0














2
4 
Sec d 





2


A = 





2
1

2
0















2
2 
Sec d 





2

+ 





2
1

2
0





























222
22 
TgSec

A = 





2
1

0
2
9


