The area of the enclosed region is #5/24 pi-sqrt{3}/4#.
It looks like this:
#A=int_0^{pi/3}int_0^{sin theta}rdrd theta
+int_{pi/3}^{pi/2}int_0^{sqrt{3}cos theta}rdrd theta#
#=int_0^{pi/3} [r^2/2]_0^{sin theta} d theta
+ int_{pi/3}^{pi/2}[r^2/2]_0^{sqrt{3}cos theta} d theta#
#=int_0^{pi/3} {sin^2theta}/2 d theta
+ int_{pi/3}^{pi/2}{3cos^2theta}/2 d theta#
#=1/4int_0^{pi/3} (1-sin2 theta) d theta
+ 3/4int_{pi/3}^{pi/2}(1+cos2 theta) d theta#
#=1/4[theta-{sin2theta}/2]_0^{pi/3}
+3/4[theta+{sin2theta}/2]_{pi/3}^{pi/2}#
#=1/4(pi/3-sqrt{3}/4)+3/4(pi/6-sqrt{3}/4)#
#=5/24 pi-sqrt{3}/4#