7, the area of the region ABA′B′A bounded by the ellipse
= 
in
4
,
0,
area of theregion AOBA
the first quadrant bounded
bythecurve x
axis and theordinates x
x
a




−
=
=


(as the ellipse is symmetrical about both x-axis and y-axis)
= 
40
(takingverticalstrips)
a ydx
∫
Now 
2
2
2
2
x
y
a
+b
 = 1 gives 
2
2
b
y
a
x
=±a
−
, but as the region AOBA lies in the first
quadrant, y is taken as positive So, the required area is
= 
2
2
40
a b
a
x dx
a
−
∫
= 
2
2
2
–1
0
4
sin
2
2
a
b x
a
x
a
x
a
a


−
+




 (Why )
= 
2
1
4
0
sin
1
0
2
2
b
a
a
a
−




× +
−










= 
2
4
2 2
b a
ab
a
π =π
Fig 8 6
Fig 8