Orientation preserving homeomorphism between half disk and waxing moon-shaped surface I need to prove that the connected sum of two orientable (top.) manifolds is orientable. To do so, I need to find an atlas for the connected sum, and to find it, I need to provide an explicit orientation preserving homeomorphism between half closed disk and waxing moon-shaped surface (see figure)

Obviously they are homeo, but I cannot find an explicit homeo between the two, I find some problems even in finding a kind of parametrisation of the waxing-moon surface.
I tried with cartesian and polar coordinate, but once I found two separate parametrisations for the arcs, I do not know how to combine them in an unique parametrisation
Can somebody provide any hints?
addendum the upper arc of the waxing moon is a circular arc

 A: Let's let $C$ be the crescent moon shape:
$$
C = \{ (x, y) \mid x^2 + (y-1)^2 \leq 2 \textrm{ and } x^2 + y^2 \geq 1 \textrm{ and } y \geq 0\}
$$
And let $D$ be the upper half-disc of radius $\sqrt{2}$, centered at $(0,0)$.
$$
D = \{ (x, y) \mid x^2 + y^2 \leq 2 \textrm{ and } y \geq 0\}
$$
Note that the boundary of $D$ has a top half and a bottom half, and for each half of the boundary, $y$ is a function of $x$.  (Top half is $y =  \sqrt{2 - x^2}$; bottom half is $y = 0$.)  It turns out that the boundary of $C$ is also made of a top half and a bottom half (though the bottom half is slightly more complicated-looking; along the bottom half of the boundary, $y$ would be defined as a piecewise function of $x$).
Now, one way to get a homeomorphism $h: D\to C$ is to calculate, for each point $(x,y) \in D$, how far above the bottom boundary $(x,y)$ is, as a proportion (percentage) of the distance between the top and bottom edge of $D$ at that $x$-value.  Then we'll basically match that same proportion between the top and bottom boundaries of $C$, and use that to define $h$.
Given any point $(x, y)\in D$, this proportion is just:
$$
\frac{y-0}{\sqrt{(2-x^2)}-0} = \frac{y}{\sqrt{2-x^2}}
$$
Of course, $h$ is going to end up being a piecewise function. If $x < -1$ or $x > 1$ then the top and bottom boundaries of $C$ are $y=1+\sqrt{2-x^2}$ and $y=1-\sqrt{2-x^2}$, so for these $x$-values we get the following for the $y$-value (start at bottom boundary of $C$, and add the proper proportion of the distance to the top  boundary of $C$):
$$
(1-\sqrt{2-x^2}) + \frac{y}{\sqrt{2-x^2}}(1+\sqrt{2-x^2} - (1-\sqrt{2-x^2}))
$$
Simplifying, we see that for $x < -1$ or $x > 1$, we get:
$$
h(x,y) = \Big( x, (1-\sqrt{2-x^2}) + 2y \Big),
$$
which is also valid at $(x,y)=(\pm \sqrt{2},0)$.
On the other hand, if $-1 \leq x \leq 1$, then you'll get a different formula, due to the different lower boundary of $C$.
Putting both pieces together, you should be able to show that $h: D\to C$ is a bijection (use the fact that $h$ doesn't change $x$, the first coordinate). To show $h$ is continuous, you could calculate the limit in both pieces as $x \to \pm 1$ and show they agree.  Finally, to show $h^{-1}$ is continuous: https://www.proofwiki.org/wiki/Continuous_Bijection_from_Compact_to_Hausdorff_is_Homeomorphism
