Evaluating double integral. Find the double integeral  $$I=\iint_D y dy dx$$ where D is area bounded by 
$$D= \{(x,y): x^2+y^2\leq 1, x^2+y^2\leq2x, y\leq0 \}$$
First off, function i am evaluating is $$z=y$$ and it's just one plane in $\mathbb{R}^3$.
Now, this is the $D$ i am looking for:

Now, i should integrate over the intersection of these two circles where $y<0$. Obviously, i should use polar coordinates so $$x=rcos\theta \\ y=rsin\theta$$, Jacobian is $r$ so it remains now to find the bounds of integration, however i am not quite sure how to do that, since the intersection point of two circles is $$(x,y)=(\frac{1}{2}, -\frac{\sqrt3}{2})$$ i suppose that angle should go from $-\frac{\pi}{6}$ to $0$ but then, i could write out the equations of circles to get bounds for $r$, where $r$ should go from the blue circle to red circle, which means $$r \in [2cos\theta, 1]$$ , but i am not quite sure is that correct way to do it because when i calculate it this way i get the positive result $$I=\frac{5}{12}$$, which is not what i expected since i am integrating over an area where $x$ is positive and $y$ is negative which means that i am actually finding a volume of body that's either in fourth or eighth octant, but, when i take a look at the function i am integrating i see that $z=y$ meaning that sign of $y$ will determine the sign of $z$ so i am finding a volume of a body that's located in eighth octant, so i expected a negative value here. Obviously, something is wrong, it might be $D$ or bounds of integration or calculations or ,the worst case probably, my reasoning. Any help is appreciated.
 A: No need to start with polar coordinates.
Your integral can be written (using the geometry of the domain), as 
$$I=\int_{\frac{-\sqrt{3}}{2}}^{0} \int_{1-\sqrt{1-y^2}}^{  \sqrt{1-y^2}} y dx dy = \int_{\frac{-\sqrt{3}}{2}}^{0} y \left(2 \sqrt{1-y^2}-1 \right) dy$$
Edit following OP's comment pointing out that the domain contains only region below x-axis:
Making the substitution $y=\sin\theta$, 
$$I=\int_{\frac{-\pi}{3}}^0 \sin\theta (2\cos \theta-1) \cos\theta d\theta = -\frac{5}{24}$$
A: The length of the section of the area at ordinate $y$ is $\sqrt{1-y^2}-(1-\sqrt{1-y^2})$, and it cancels at $y=-\sqrt3/2$.
Then 
$$I=\int_{y=-\sqrt3/2}^0 y\,\left(2\sqrt{1-y^2}-1\right)\,dy=\left.-\frac23(1-y^2)^{3/2}-\frac{y^2}2\right|_{y=-\sqrt3/2}^0.$$
A: The solution I propose is based on the Green's theorem. Let R the interior of the region delimited by D, and let (P(x,y),Q(x,y)) a differentiable vector field. Then
\begin{equation}
\int_C P dx + Q dy = \int \int_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}) dx dy
\end{equation}
Now if we assume $P = -y^2/2$ and $Q = 0$ we have
\begin{equation}
-\int_C \frac{y^2}{2} dx = \int \int_R y dx dy
\end{equation}
We can then evaluate the integral over C. To do so we use the polar coordinated. We have two circles, but the only difference is that one is translated along the x-axis with respect to the other. If we consider the circle centered into the origin we have:
\begin{equation}
x(\theta)= \cos(\theta)
\end{equation}
and
\begin{equation}
y(\theta)= \sin(\theta)
\end{equation}
So
\begin{equation}
dx= -\sin(\theta) d\theta
\end{equation}
Then
\begin{equation}
-\int_{C_1} \frac{y^2}{2} dx = \frac{1}{2}\int \sin^{3}(\theta)
\end{equation}
where $C_1$ is the circle centered into the origin.
For the second circle $dx$ is the same so you can find exactly the same expression. Once integrating with with the correct integration limits you get the value -5/24.
