Quick Question Area Under Curves

[leflyi]

Just how you go about it,

I realize that each arc can be defined as separate ellipses/circular functions, such as:

  or or or

Just where can i go from there, or am I looking at it completely wrong ?

[brightsky]

Okay I'm going to try find the area of the region obtained when you add one of the pink regions (P)and one of the blue regions (B).
Let the base of square be the base of a triangle with vertex being the top of the yellow region. Draw this triangle.
Now you will see that the sides of this triangle are all 1 units in length (since they are all radii of the same circle). Now it's hard to explain this next bit without a diagram but hopefully you can follow:
P+B = (quarter of circle with radius 1) -(squilateral triangle) - 2(sector of triangle with radius 1 which spans an angle of 60) - (equilateral triangle))
= pi/4 - 1/2*1*1*sin(60) - 2 (60/360*pi - 1/2*1*1*sin(60))
=pi/4 - 1/2*(sqrt(3)/2) - 2(pi/6 - 1/2*(sqrt(3)/2))
=sqrt(3)/4 -pi/12
Now let the area of the req yellow region be Y.
Y = 1 - 4P - 4B
= 1-4(P+B)
= 1 - 4(sqrt(3)/4 - pi/12)
= 1 - sqrt(3) + pi/3
= 0.315...units^2