Using polar coordinates:
#x= rho cos theta#
#y= rho sin theta#
we have to integrate for #0<=rho <=1# and #0 <= theta <= 2pi#, so:
#S = int_0^1int_0^(2pi) rho d rho d theta#
#S = int_0^1 rho d rho int_0^(2pi) d theta#
#S = 2pi int_0^1 rho d rho#
#S = 2pi [rho^2/2]_0^1 #
#S = pi#
Alternatively we have that for #x,y >=0#:
#x^2+y^2 = 1#
#y=sqrt(1-x^2)#
For symmetry reasons:
#S = 4int_0^1 sqrt(1-x^2)dx#
substitute:
#x = sint# with #t in [0,pi/2]#
#dx = cost dt#
As in the interval the cosine is positive:
#sqrt(1-x^2) = sqrt(1-sin^2t) = sqrt(cos^2t) = cost#
So:
#S = 4int_0^(pi/2) cos^2tdt#
#S = 4int_0^(pi/2) (1+cos(2t))/2dt#
#S = 2int_0^(pi/2) dt + 2int_0^(pi/2)cos(2t)dt#
#S = 2[t]_0^(pi/2) + [sin2t]_0^(pi/2)#
#S = 2(pi/2-0) + sinpi-sin0 = pi#
