The points of intersection are given by
#r = a sin theta = a ( 1 + cos theta )#, giving #sin theta - cos theta = 1#
Reorganized, #sin (theta-pi/4)=1/sqrt2= sin (pi/4)#.
#theta = 0 and pi/2#, for one period #[0, 2pi].#
The area comprises one part bounded by
#theta = 0, r = a sintheta and theta = pi/2#
and the second part bounded by
#theta =pi/2, r = a (1+costheta) and theta = pi#.
Now, the area is
#1/2int r^2 d theta#, for the first area,
# + 1/2 int r^2 d theta#, for the second area.
#=a^2/2( int sin^2 theta d theta#, with #theta# from 0 to #pi/2#
#+int (1+costheta)^2 d theta#), with #theta# from #pi/2# to #pi#.
#=a^2/2(int 1/2(1-cos 2theta) d theta#, with #theta# from #0# to #pi/2#
#+ int(1+2 cos theta+1/2(1+cos 2theta ) d theta)#,
with #theta# from #pi/2 to pi#
#=a^2/2(1/2[theta-1/2 sin 2theta]#, between #theta=0 and pi/2#,
# +[theta+2sin theta+1/2(theta+1/2sin2theta)]#,
between #theta =pi/2 and pi#)
#=a^2/2([ pi/4]+[(pi-pi/2)-2+1/2(pi-pi/2)])#
#=a^2/2(pi-2)#
#=(pi/2-1)a^2#
graph{(x^2+y^2-y)(x^2+y^2-sqrt(x^2+y^2)-x)=0 [-4.034, 4.044, -2.02, 2.018]}
