To find the area of the region that is inside the circle r = 2 cos(q) and outside the cardioid r = 2(1 – cos(q)), we need to determine the bounds of integration and set up the integral accordingly.

1. Determine the limits of q:
The cardioid r = 2(1 – cos(q)) is the inner curve, and the circle r = 2 cos(q) is the outer curve. To find the limits of q, we need to find the points where these two curves intersect.

Setting the two equations equal to each other:
2(1 – cos(q)) = 2 cos(q)

Simplifying, we get:
1 – cos(q) = cos(q)
1 = 2cos(q)
cos(q) = 1/2
q = π/3, 5π/3

Therefore, the limits of integration for q are π/3 to 5π/3.

2. Set up the integral:
The area of a region in polar coordinates is given by the integral ∫(1/2)r^2 dq, where r is the radius of the curve.

The cardioid r = 2(1 – cos(q)) is the inner curve, so we subtract the area enclosed by it from the area enclosed by the circle r = 2 cos(q).

The integral for the area is therefore:
A = ∫[π/3, 5π/3] [(1/2)(2cos(q))^2 - (1/2)(2(1 – cos(q)))^2] dq

Simplifying the expression inside the integral:
A = ∫[π/3, 5π/3] (2cos^2(q) - (1 – cos(q))^2) dq
A = ∫[π/3, 5π/3] (2cos^2(q) - (1 – 2cos(q) + cos^2(q))) dq
A = ∫[π/3, 5π/3] (3cos^2(q) - 2cos(q) + 1) dq

3. Evaluate the integral:
Integrating with respect to q, we get:
A = [3/2 sin(q) - 2sin(q) + q] [π/3, 5π/3]
A = [3/2 sin(5π/3) - 2sin(5π/3) + 5π/3] - [3/2 sin(π/3) - 2sin(π/3) + π/3]
A = [3/2 (-√3/2) - 2(-√3/2) + 5π/3] - [3/2 (√3/2) - 2(√3/2) + π/3]
A = [-9√3/4 + 15π/3] - [9√3/4 - 6√3/2 + π/3]
A = [-9√3/4 + 5π] - [9√3/4 - 6√3/2 + π/3]
A = 9√3/4 - 6√3/2 + π/3

Therefore, the correct answer is option D)