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An area problem revisited
Yue Kwok Choy
Question
Given a square ABCD with 1 as length of each side.




Find the area enclosed by arcs EF , FG , GH and HE .
There are many ways to solve this problem.
Here is a way in which definite integration is employed.
Solution
Place the square in the grid.
A(0, 0), B(1, 0), C(1, 1), D(0,1) .
There are 4 circles as in the diagram.
The equations of the 4 arcs of circles
lying inside the given square where
0  x  1, are shown below:
Centre A,
CA :
y  1 x 2
Centre B,
CB :
y  1  1  x 
2
Centre C,
CC :
y  1  1  1  x 
Centre D,
2
CD :
Solving CB
y  1 1 x 2
and CC , we can get the point
Solving CA and CB,
we can get the point

3 1
, .
H = 1 
2 2 

1
3
.
G =  , 1 

2
2


1
Required area = S
= 2  [area under the curve HG in circle CB – area under the curve HE of circle CC]
1
 1  1  x 2  1  1  1  x 2 dx



 2 2
3
1
2
1
2 1  1  x 2  1dx


 2 2
3
1
2
Now, for
Put
I
1  1  x  dx
1  x  sin ,
1  1  x  
2
2
dx   cos d
1  sin 2   cos 
 I   cos  cos d  


1
1  cos 2d   1   sin 2   c   1 sin 1 1  x   sin  cos   c

2
2
2 
2
1
2
  sin 1 1  x   1  x  1  1  x    c

2 
S  2 
1
1
2
1
2 1  1  x 

2
3
2
2
2
 1dx   2sin 1 1  x   1  x  1  1  x    2x 


 
1
3
2
2

2
 
  
 3   
3 
  1 1 1
 1    1    1  3   3 




  2 sin

1     2    2 sin 
   2  1   2    21  2 

2
2
2
2
2







 


  

 
  




3 
3 
3 
 
 
 

  2 

  1   2 
  21 
2 

 6 4  
 
 3 4  

 1 3 

3
2
A better way to use calculus is to use parametric integration.
Place the axes and origin O as in the figure in the left.
We like to find the shaded area OFG in the first quadrant
and multiply by 4.
The arc FG in fact is part of the circle centre A, radius AB.
The Cartesian equation for this circle is
O
1 2
1 2
(x + ) + (y + ) = 1
2
2
The parametric equation is therefore:
1
x = cos t − 2
{
1
y = sin t − 2
Point
B is where t = 0, Point F is at t =
π
6
. Point G is at t =
π
3
.
Therefore required area is
π/3
π/3
dx
π/3
1
1
S = 4 ∫π/6 ydx = −4 ∫π/6 y dt dt = 4 ∫π/6 (sin t − 2) d (cos t − 2)
π/3
1
π/3
1
π
= −4 ∫π/6 (sin t − 2) (−sin t)dt = 4 ∫π/6 (sin2 t − 2 sin t) dt = 1 − √3 + 3
3