# Area of a region given in polar coordinates

Determine the area of the following regions given in polar coordinates:

$$A)$$ $$D=\{(r, \theta):\, 1+\cos \theta \leq r \leq 3\cos \theta \}$$

$$B)$$ $$D=\{(r, \theta):\, r\leq 3\sin \theta,\, r\leq -5.2\cos \theta\}$$

Well, I know the formula to calculate the area given in polar coordinates, but if I want to use it I have to know the interval of $$\theta$$ and the radius, and here I don't have them, I just have an "interval of the radius". Can you tell me what to do, please?

I believe that the allowed $$\theta$$ are supposed to be all possible values of $$\theta \in[0,2\pi]$$ for which there exists $$r\ge 0$$ that satisfies given inequalities.
That is in A): $$D = \left\{(r,\theta): 1+\cos\theta \le 3\cos\theta, r \in [1+\cos\theta, 3\cos\theta] \right\} = \\ = \left\{(r,\theta): \cos\theta \ge \frac12, r \in [1+\cos\theta, 3\cos\theta] \right\} = \\ = \left\{(r,\theta): \theta \in [0,\frac\pi 3]\cup[\frac{5\pi}{3},2\pi], r \in [1+\cos\theta, 3\cos\theta] \right\}$$ and in B): $$D= \left\{(r,\theta): 3\sin\theta > 0, -5.2\cos\theta > 0, r \in [0,\min(3\sin\theta, -5.2\cos\theta)] \right\} = \\ = \left\{(r,\theta): \theta \in [\frac\pi 2,\pi], r \in [0,\min(3\sin\theta, -5.2\cos\theta)] \right\}$$