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I consider the region inside $x^2+(y-1)^2=1$ but outside $x^2+y^2=1$ and I want to write this region in terms of polar coordinates.

I have started with equaling the two equations. After doing the calculations I get $y=\frac{1}{2}$. I know that in pola coordinates $x=r\cos\theta$ and $y=r\sin\theta$, so $r\sin\theta =\frac{1}{2}$. After this I am stuck. Can someone give me a hand?

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Inside $$x^2+(y-1)^2=1 \implies x^2+y^2-2y+1<1 \implies\rho^2-2\rho \sin \theta<0\implies \rho<2 \sin \theta$$

Outside $$x^2+y^2=1 \implies x^2+y^2>1 \implies\rho^2>1\implies\rho>1$$

Thus:

$$\begin{cases} \sin \theta > \frac12 \implies\theta \in (\frac16\pi,\frac56\pi)\\ 1< \rho <2 \sin \theta \end{cases}$$

enter image description here

http://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%3E1+and+x%5E2%2B(y-1)%5E2%3C1

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When $y=1/2$, $x=\pm \sqrt{3}/2$.

So there are two points: $(\sqrt{3}/2,1/2)$ and $(-\sqrt{3}/{2},1/2)$.

The corresponding angles are $\theta=\pi/6$ and $5\pi/6$.

You have $1 \le r \le 2\sin\theta$ and $\pi/6 \le \theta \le5\pi/6$

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    $\begingroup$ $r=1$ :) (that's the radius!) Geometrically we just take the arctangent of y over x. I didnt even consider r. $\endgroup$ – David Peterson Dec 14 '17 at 15:58

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