finding an $r$-simple polar coordinate paramtrization of the area bounded by $y \geq 12$ and inside the circle $x^2+(y-9)^2=9^2$

Question

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?

Answer 1

$$x^2+(y-9)^2=9^2 \implies x^2 + y^2 -18y=0 \implies r^2= 18rsin(\theta)$$

Thus the equation of the circle is $$ r=18 sin(\theta )$$

The polar equation of the line $y=12$, is $$ rsin(\theta ) =12 $$ which is $$ r=12 csc(\theta ) $$

Points of intersection of the line and the circle are found by solving $$ 12csc(\theta )=18 sin(\theta )$$

$$sin ^2(\theta)=2/3 $$ or $$ \theta _1 = sin^{-1} (\sqrt{2/3}) $$

$$ \theta _2 = \pi - sin^{-1} (\sqrt{2/3}) $$

Thus the region is described by $$ \theta_1\le \theta \le \theta _2$$

and $$ 12csc(\theta)\le r \le 18 sin(\theta)$$

Answer 2

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 $$ \tan^{-1}\sqrt{2} \le \theta \le \pi - \tan^{-1}\sqrt{2} $$