22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD Thus, we have the required area
= 
3π
π
2π
2
2
3π
π
0
2
2
cos
cos
cos
xdx
xdx
x dx
+
+
∫
∫
∫
= [
]
[
]
[
]
3
2
2
2
3
0
2
2
sin
sin
sin
x
x
x
π
π
π
π
π
+
+
= 1 + 2 + 1 = 4
Example 13 Prove that the curves y2 = 4x and x2 = 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts Solution Note that the point of intersection of the
parabolas y2 = 4x and x2 = 4y are (0, 0) and (4, 4) as
Fig 8 22
374
MATHEMATICS
Y'
R
O
X
X'
x =�2
T
S
P�(0,1) x =�1
Y
Q(1,2)
shown in the Fig 8