Region in Polar Coords

Question

Hi I am intersted in the following question regarding polar corodinates:

Can anyone see how the region inside the circle $$(x-1)^{2} + (y-1)^{2} = 1$$ is described in polar coordinates?

Thanks for any assistance.

Answer 1

I guess that it is $$0\leq \theta \leq \frac{\pi}{2}\quad \text{and}\quad \cos \theta+\sin \theta -\sqrt{\sin 2 \theta}\leq r \leq \cos \theta+\sin \theta +\sqrt{\sin 2 \theta}. $$

Answer 2

The region inside your circle can be parametrized by setting $x = 1 + r\cos \theta$, $y = 1 + r\sin\theta$, for $0 < \theta < 2\pi$ and $0 < r < 1$. The $(r,\theta)$ parameters are polar coordinates centered at $(1,1)$.

If you want the polar coordinates $(r,\theta)$ to be centered at $(0,0)$, then by sketching the graph of the circle, you find that $0 < \theta < \pi/2$. The $r$-variable is subject to the relation $$r^2 + 2\sqrt{2}r \cos\left(\theta - \frac{\pi}{4}\right) + 1 < 0.$$

This can be found by substituting $r\cos\theta$ for $x$ and $r\sin \theta$ for $y$ in the inequality $(x - 1)^2 + (y - 1)^2 < 1$. This results in

$$\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) - \sqrt{\sin 2\theta} < r < \sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right) + \sqrt{\sin 2\theta}.$$