Solution to Problem #5

Tye Rattenbury of the University of Colorado found the area of the region. John Shonder of Oak Ridge TN, Ross Millikan of San Mateo CA, James D. Beltz of St. Charles MO, and Brian MacBride of Palm Springs CA found the area and the perimeter. Here is John Shonder's solution:

 
The shaded region is the bounded by the four parabolas:

	x = 1/4 - y^2
	y = 1/4 - x^2
	x = y^2 - 1/4
	y = x^2 - 1/4

and the "corners" of the shaded region are:

	x = (sqrt(2)-1)/2, y = (sqrt(2)-1)/2
	x = (sqrt(2)-1)/2, y = -(sqrt(2)-1)/2
	x = -(sqrt(2)-1)/2, y = -(sqrt(2)-1)/2
	x = -(sqrt(2)-1)/2, y = (sqrt(2)-1)/2

The problem is one of simple integration. The area of the figure is:

	(4*sqrt(2)-5)/3

or 0.21895 approximately.

Arc length is found in the usual way, by integrating sqrt(1+(dy/dx)^2). The
perimeter of the figure is:

	2*LN(sqrt(4-2*sqrt(2))+sqrt(2)-1)-sqrt(16-8*sqrt(2))+sqrt(32-16*sqrt2)

or 1.70308 approximately.

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