Question 542455
<pre>
The only way

&#8970;x&#8971;² + &#8970;y&#8971;² = 1

can be true is for values of (x,y) such that 
either 

&#8970;x&#8971;² = 0 and &#8970;y&#8971;² = 1 or 
&#8970;y&#8971;² = 0 and &#8970;x&#8971;² = 1 or

&#8970;x&#8971;² = 0 when and only when  0 &#8806; x < 1
&#8970;x&#8971;² = 1 when and only when -1 &#8806; x &#8806; 1
&#8970;y&#8971;² = 0 when and only when  0 &#8806; y < 1
&#8970;y&#8971;² = 1 when and only when -1 &#8806; y &#8806; 1


&#8970;x&#8971;² = 0 and &#8970;y&#8971;² = 1 when and only when 0 &#8806; x < 1 and -1 &#8806; y &#8806; 1
&#8970;y&#8971;² = 0 and &#8970;x&#8971;² = 1 when and only when 0 &#8806; y < 1 and -1 &#8806; x &#8806; 1
 
Thus it is true at all points interior and on the square excpt at the
four corners

{{{drawing(400,400,-1.5,1.5,-1.5,1.5, graph(400,400,-1.5,1.5,-1.5,1.5),
line(-1,-.98,-1,.98),line(-.98,1,.98,1),
line(1,.98,1,-.98), circle(1,1,.01),circle(1,-1,.01),circle(-1,1,.01),circle(-1,-1,.01), 
line(1,-.98,-1,-.98) ) }}}

Its area is 2x2 or 4 square units.

Edwin</pre>