Consider square with vertices at (1,1), (-1,1), (-1, -1), and (1, -1).

S is the region which consists of all points inside the square which are nearer to the origin than any edge of this square.

Determine the area of S.

At first glance, if you measured all distances radially, the half way point from the origin to the square would seem to be just a smaller square with 1/2 the side length and 1/4 the area.  So the area of S would be exactly 1.  But this is incorrect because distance measurements made from the region border to the square are made orthogonally.  Clearly for the point (1/2, 1/2) the distance to the square is 1/2 but the distance to the origin is √2/2, so that point is not in the region of interest.  We expect the final answer to be somewhat less than 1.

Consider the square divided into 8 octants whose borders are the x and y axes and the lines y=x and y= -x.
The distance from the origin to any point (x,y) is just √(x^2 + y^2)

Distance from a point, (x,y), to the square:  need to know which limb of the square is closest; i.e. which octant is it in?  
Consider just the first 2 octants, then multiply by 4.
In the 1st octant, this distance is 1-x,
in the 2nd octant, this distance is 1-y.
The equations intersect at (√2 - 1, √2 - 1) because at that point,
1 - x = √2x  or 1 = (1+√2)x or x = 1/(1+√2 or x = √2 - 1)

The equation for the 1st octant is
 (1-x) = √(x^2 + y^2)  or y = √(1-2x)
 integral is:  (-1/3)(1-2x)^(3/2)
 the limits:  from x = √2 - 1 to x =1/2
 evaluates to 0.0236892706218
 
The equation for the 2nd octant is
 (1-y) = √(x^2 + y^2) or y = (1-x^2)/2
 integral is: (1/2)(x - (x^3)/3 )
 the limits:  from x = 0 to x= √2 - 1.
 evaluates to 0.195262145876

Multiplying the sum of the 2 integrals by 4 yields S:
0.8758056659898403

Posted by Larry on 2023-08-19 10:11:16