What is the area inside the polar curve #r=1#, but outside the polar curve #r=2costheta#?

Here is the graph of the two curves. The shaded area, #A#, is the area of interest:

This is a symmetrical problems so we only need find the shaded area, #B# and subtract twice this from that of a unit circle (#r=1#).

We can find the polar coordinate of the point of intersection in Q1 by simultaneously solving the polar equations:

# r=2cos theta #
# r=1 #

From which we get:

# 2cos theta = 1 =>cos theta = 1/2#
# :. theta = pi/3 #

So we can easily calculate the area, #B#, which is that of the a circle sector #C# and that bounded by the curve #r=2costheta# where #theta in (pi/3,pi/2)#

The area, #C# is simply #1/2r^2theta#:

# A_C = 1/2 * 1 * pi/3 #
# \ \ \ \ \ = pi/6 #

And we calculate the area of the segment #B#, via Calculus using:

# A = int \ 1/2r^2 \ d theta#

Thus:

# A_D = int_(pi/3)^(pi/2) \ 1/2(2costheta)^2 \ d theta#
# \ \ \ \ \ = int_(pi/3)^(pi/2) \ 1/2 \ 4cos^2 theta \ d theta#
# \ \ \ \ \ = 2int_(pi/3)^(pi/2) \ cos^2 theta \ d theta#
# \ \ \ \ \ = 2int_(pi/3)^(pi/2) \ 1/2(cos2theta+1) \ d theta#
# \ \ \ \ \ = int_(pi/3)^(pi/2) \ cos2theta+1 \ d theta#
# \ \ \ \ \ = [1/2sin2theta + theta]_(pi/3)^(pi/2)#
# \ \ \ \ \ = (1/2sin(pi)+pi/2) - (1/2sin((2pi)/3)+pi/3)#
# \ \ \ \ \ = (0+pi/2) - (1/2 sqrt(3)/2+pi/3)#
# \ \ \ \ \ = pi/2 - sqrt(3)/4-pi/3#
# \ \ \ \ \ = pi/6 - sqrt(3)/4#

So then the total area we require, is given by:

# A_A = pi(1)^2 - 2(A_C+A_D) #
# \ \ \ \ \ = pi - 2(pi/6 + pi/6 - sqrt(3)/4) #
# \ \ \ \ \ = pi - 2(pi/3 - sqrt(3)/4) #
# \ \ \ \ \ = pi - (2pi)/3 + (2sqrt(3))/4 #

# \ \ \ \ \ = pi/3 + sqrt(3)/2 #