How do I solve this exercise of double integral $\iint_{D}\frac{1}{\sqrt{x^{2}+y^{2}}}dA$?

Question

I have the double integral $\iint_{D}\frac{1}{\sqrt{x^{2}+y^{2}}}dA$ where $D$ is the domain given by:

$D = \left \{ (x, y ) \in \mathbb R^{2} : x^{2}+y^{2} \le 2y\right \} $

I tried this in polar coordinates: $$\left ( r cos \Theta \right )^{2} + \left ( r sin \Theta - 1 \right )^{2} = 1 $$

$$r = 2sin\Theta $$ $$\int_{-\Pi /4}^{\Pi /4}\int_{0}^{2sin\Theta } \frac{1}{r} rdrd\Theta$$

Am I the right way? How to resolve then?

Answer 1

Draw a diagram: the correct range for $\theta$ is from $\pi/4$ to $3\pi/4$, giving$$\int_{\pi/4}^{3\pi/4}\int_0^{2\sin\theta}drd\theta=\int_{\pi/4}^{3\pi/4}2\sin\theta d\theta=[-2\cos\theta]_{\pi/4}^{3\pi/4}=2\sqrt{2}.$$

Answer 2

$$x^2+y^2\le 2y$$ $$\Rightarrow x^2+y^2-2y\le 0$$ $$x^2+(y-1)^2-1\le 0$$ $$x^2+(y-1)^2\le 1$$ so this is the circular region you are integrating over