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#