If the pole r = 0 is not outside the region, the area is given by
#(1/2) int r^2 d theta#, with appropriate limits.
The given curve is a closed curve called cardioid.
It passes through the pole r = 0 and is symmetrical about the initial
line #theta = 0#.
As #r = f(cos theta)#, r is periodic with period #2pi#.
And so the area enclosed by the cardioid is
#(1/2) int r^2 d theta#, over #theta in [0, 2pi]#.
#(1/2)(2) int (1+cos theta)^2 d theta, theta in [0, pi]#, using symmetry about #theta=0#
#=int (1+cos theta)^2 d theta, theta in [0, pi]#
#=int (1+2 cos theta + cos^2theta) d theta, theta in [0, pi]#
#=int (1+2 cos theta + (1+cos 2theta)/2) d theta, theta in [0, pi]#
#=[3/2theta+2sin theta]+(1/2)(1/2)sin 2theta]#,
between limits #0 and pi#
#=(3pi)/2+0+0#
