Question 587691
<pre>{{{drawing(400,240,-2.5,2.5,-.5,2.5,

graph(400,240,-2.5,2.5,-.5,2.5),arc(0,0,4,-4,0,180),

line(-2cos(40*pi/180), 2sin(40*pi/180),2cos(40*pi/180), 2sin(40*pi/180)),
line(-2cos(40*pi/180), 0,-2cos(40*pi/180), 2sin(40*pi/180)),
line(2cos(40*pi/180), 0,2cos(40*pi/180), 2sin(40*pi/180)),
locate(2cos(40*pi/180), 2sin(40*pi/180)+.2,"(x,y)"),

locate(2cos(40*pi/180)/2, 2sin(40*pi/180)+.2,x),

locate(-2cos(40*pi/180)/2, 2sin(40*pi/180)+.2,x),

locate(2cos(40*pi/180)/2, .2,x),

locate(-2cos(40*pi/180)/2, .2,x),

locate(2cos(40*pi/180)+.05, sin(40*pi/180)+.1,y),

locate(-2cos(40*pi/180)-.15, sin(40*pi/180)+.1,y)





     )}}} 

As you can see by the picture the rectangle has length 2x
(the horizontal measurement) and width y (the vertical
measurement).

              Area = length·width

              Area = (2x)·y

But since y = {{{sqrt(4-x^2)}}},

              Area = (2x)·{{{sqrt(4-x^2)}}}

              Area =  2x{{{sqrt(4-x^2)}}} 

And now since we have area expressed as a function of x, we can write

              A(x) =  2x{{{sqrt(4-x^2)}}}

The domain of 2x{{{sqrt(4-x^2)}}} is the set of all values of x for
which 2x{{{sqrt(4-x^2)}}} is a real number, which will be whenever

              4 - x² <u>></u> 0

      (2 - x)(2 + x) <u>></u> 0

Critical numbers are 2 and -2, so we have to get test points in
the intervals ({{{-infinity}}}, -2), (-2, 2), (2, {{{infinity}}}). We find
that the only interval in which 4 - x² is non-negative is (-2, 2).

We must exclude the endpoints -2, and +2 because the function would be 0
there, and no rectangle can have 0 area (unless we allow that a horizontal
line segment could be called "a rectangle with width 0" or that a
vertical line segment could be called "a rectangle with length 0").  So the
domain is

                 (-2, 2)

Edwin</pre>