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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?

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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): $$ D= \left\{(r,\theta): 3\sin\theta > 0, -5.2\cos\theta > 0, r \in [0,\min(3\sin\theta, -5.2\cos\theta)] \right\} = \\ = \left\{(r,\theta): \theta \in [\frac\pi 2,\pi], r \in [0,\min(3\sin\theta, -5.2\cos\theta)] \right\}$$

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