<s>[INST] finding an $r$-simple polar coordinate paramtrization of the area bounded by $y \geq 12$ and inside the circle $x^2+(y-9)^2=9^2$ I want to find an $r$-simple polar coordinate paramtrization of the area bounded by $y \geq 12$ and inside the circle $x^2+(y-9)^2=9^2$, meaning find two constants $a$ and $b$ and two functions $f$ and $g$ such that this area is given by $a \leq \theta \leq b$ and $f(\theta) \leq r \leq g(\theta)$
Solving $x^2+(y-9)^2=9^2$ where $y=12$ gives me $x=6\sqrt{2}$ and $x=-2\sqrt{6}$. But I'm not sure if I've interpreted the question right, as it seems like $\theta$ should be between $0$ and $\pi$ if we're talking about angles. But in that case I don't understand how I'm supposed to find the functions $f$ and $g$...
any ideas?
 [/INST] First things first, the region above the line $y = 12$ in polar coordinates is
$$ r\sin\theta \ge 12 \implies r \ge 12\csc\theta $$
Second, for the region inside the circle
$$ x^2 + (y-9)^2 \le 9^2 $$
$$ x^2 + y^2 - 18y \le 0 $$
$$ r^2 - 18r\sin\theta \le 0 $$
$$ r \le 18\sin\theta $$
Combining them together gives you the bounds for $r$
$$ 12\csc\theta \le r \le 18\sin\theta $$
As for the angles, note that the two points of intersection are $(\pm 6\sqrt{2}, 12)$, corresponding with $r = 6\sqrt{6}$ and $\tan\theta = \pm\sqrt{2}$, or $\theta = \tan^{-1}\sqrt{2}$ (first quadrant) and $\theta = \pi - \tan^{-1}\sqrt{2}$ (fourth quadrant). So the bounds for $\theta$ are
$$ \tan^{-1}\sqrt{2} \le \theta \le \pi - \tan^{-1}\sqrt{2} $$
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