Integrating, we get
the whole area enclosed by the given circle
= 
2
2
2
–1
0
4
sin
2
2
a
x
a
x
a
x
a


−
+




= 
2
1
4
0
sin 1
0
2
2
a
a
−




× +
−








 =  
2
2
4
2
2
a
a


π
 =π

 


 

Fig 8 5
Fig  8 4
362
MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8 6, the whole area of the
region enclosed by circle
= 
40
∫a xdy
 = 
2
2
40
a
a
y
dy
−
∫
(Why