Need a bit of help trying to figure out the boundary for $\theta \:\:$and to see if I did the one for r correct. No need to evaluate the integral, just need to learn how to do the boundary.
$$\int \int _D\frac{1}{\sqrt{1-x^2-y^2}}\:dxdy$$ D: $x^2+y^2\ge \frac{1}{4}$ , $x^2+y^2-x\le 0$
I did the following so far:
$x^2+y^2=\frac{1}{4}$ circle with the center $O(0,0)$ and $R= \frac{1}{2}$
$x^2+y^2=x$ circle with the center $O(\frac{1}{2},0)$ and $R= \frac{1}{2}$
$\rightarrow$ $x=\frac{1}{4}$ , $ y=+-\frac{\sqrt{3}}{4}$ $\rightarrow$ $A(\frac{1}{4},\frac{\sqrt{3}}{4})$ and $B\left(\frac{1}{4},-\frac{\sqrt{3}}{4}\right)$
$$T:D'\rightarrow D,T:x=rcos\theta ,y=r sin\theta$$
I tried finding the $\:\theta \:$ from $\frac{1}{4}=cos\theta \:,\:\frac{\sqrt{3}}{4}=sin\:\theta \:\:and\:\frac{1}{4}=cos\theta \:\:,\:-\frac{\sqrt{3}}{4}=sin\:\theta \:\:$ as shown in an example from one of my textbooks but that one was so easy and this one I legit have no clue how to proceed.
I have found the boundary of r from the domain equations thus getting $\frac{1}{2}\le r\le cos\:\theta \:\:$
desmos: 
Indeed, the constraint of $ r $ is $ \frac12 \leq r \leq \cos \theta $. As for the range of $ \theta $, notice that we only need the region outside the blue circle and inside the blue circle. Therefore it suffice to find the angle at the intersection points. For $A$, we may compute $$ \tan \theta_A = \frac{\sqrt{3}/4}{1/4} = \sqrt{3} \implies \theta = \frac\pi3. $$ A similar computation can be done for point $ B $ to obtain $ \theta_B$. The range of $ \theta $ is then $ \theta_B \leq \theta \leq \theta_A $.