Originally Posted by

**billh** I think it's more elegant to do the indefinite integral in polar coordinates, recognizing this is the upper half of the circle with radius=2 centered at the origin as in

integral sqrt(4-x^2) = integral integral 2r dr dT

where r = radius and T = theta = angle formed with the positive x axis. The conversion to polar coordinates uses r^2 = (x/2)^2 + (y/2)^2, and we know from the original eqn that r=2. The extra r in the double integral is from the Jacobian necessary in converting to polar coord. Then finding this integral is easy:

= integral r^2 dT = Tr^2 = 4T + C

which I guess would work for finding areas of sectors of the circle of radius 2.

What I'm curious is if I convert back to x and y, if I get the solution above, using x = rcosT, and r^2 = (x/2)^2 + (y/2)^2

4T = 4arccos(x/r) = 4arccos[x/(sqrt((x/2)^2 + (y/2)^2))] =

= 4arccos[x/(sqrt((x/2)^2 + ((4-x^2)/2)^2))]

does this simplify to above???