Example.1
Find the area enclosed between the circle x2
+ y2 – 2x + 4y –11 = 0 and the parabola, y= -x2 + 2x + 1
-2
.
Solution
The given curves are
(x -1)2 + (y + 2)2 = 16 ….(1).
y =-(x –1)2 -2
+ 2 ….(2)
Solving (1) and (2) we get x = -1, 3
Hence required area,
A = 
=
=
=
sq. units
Example.2
Let f(x) = max 
then
determine the area of the region bounded by the curves y = f(x), x-axis, y axis
and x = 2p
Solution
Given f(x) = max 
Hence required area = 
=
= 
= 
sq. units
Example.3
Find the area enclosed by |x| +|y| = 1.
Solution
Due to the modulus sign we consider four different
cases :
Case-I:
x ≥
0 , y ≥ 0, then |x| = x and
|y| = y,
and the given equation becomes x + y =1 . . .
. . . (1).
Case-II:
x ≥ 0 , y < 0 then
|x| = x, |y| = – y and the given equation becomes
x - y = 1 .
. . . . (2).
Case-III:
x < 0 , y ≥
0 then |x| = -x , |y| = y and the given equation becomes
– x + y = 1 . .
. . . (3).
Case-IV:
x < 0 , y < 0 then |x| = – x , |y| = – y
and the given equation becomes – x – y = 1 . . . .
. (4).
Taking into the considerations of these (1)-(4)
equations, the graph of |x|+|y| =1is as shown :
Obviously the quadrilateral ABCD is a square, whose side
is of length 
. Hence required area is 2 sq. units.
Example.4
Find the area enclosed between the curves y = ln(x + e), x =
ln(1/y) and the x-axis . .
Solution
The
area enclosed between the curves is as shown in the fig. .
Required area
= 
= 
=
= 
= (0 + 1 – e + e) – (0 + 0 –1 + 0) = 2 .
Example.5
ABCD
is a square of side 2a, circles of radius 2a are drawn with centers at A, B, C,
D then find the common area of these circles.
Solution
Given
AB = BC = CD = DA = 2a
Co-ordinates
of A, B, C, D are (a, a): (-a, a); (-a, -a); (a, -a) respectively
Equations
of circles if centres are at A, B, C, D are
(x
-a)2 + (y -a)2 = 4a2 …..(1).
(x
+a)2 + (y -a)2 = 4a2 …..(2).
(x
+a)2 + (y +a)2 = 4a2 …..(3).
(x
-a)2 + (y +a)2 = 4a2 …..(4).
Solving
these we get
P
(0, a(-1)); Q (-a(-1),
0); R(0, -a(-1)); S(a(-1),0)
Required
area PQRSP = 4 area of OSPO
= 
= 
=
= 
=
sq. units
