Projection of a Sphere onto a Cylinder 

Source: Tristan Needham VDG

I've been trying to determine the metric for the projection of a sphere onto a cylinder, where $0 \leq x \leq 2\pi R$ and $-R \leq y \leq R$
The book says to reason geometrically that $d\hat{s}^2 = \frac{\,dy^2}{R^2-y^2} + (R^2-y^2)\,dx^2$ where $d\hat{s}$ is the infinitesimal displacement on the sphere.
I keep getting stuck at trying to determine what the individual changes $d\hat{s_1}$ due to $dx$ and $d\hat{s_2}$ due to $dy$ are, but I keep ending up with an $R^2$ in the denominator of the $dx$ term and an $R^2$ in the numerator of the $dy$ term.
Any help would be greatly appreciated, I've beens staring at this one for quite a while!
EDIT: I have discovered in the Errata for the book that my original solution was actually correct. Thank you everyone for taking the time to address my question!
 A: 


In the above picture, suppose I fix $y$ and rotate $r$ by $d \theta$, what is the length gained?
$ \delta s= r' \delta \theta$
For that, we go back to highschool trignometry and draw a diagram:


We can write:

$$ r' = \sqrt{R^2 - y^2} \tag{1}$$
Putting that back in the distance equation:
$$ \delta s = \sqrt{R^2 - y^2} \delta \theta$$
Or,
$$ \frac{ \delta s}{\delta \theta} = \sqrt{R^2-y^2} \tag{2}$$
Next we need to check how much distance is gained when $y$ is increased, in (1), we can take the differential:
$$ \delta r'= \frac{-2y \delta y}{\sqrt{R^2 -y^2}} \tag{3}$$
What does $\delta r'$ represent? The shrinking of the radius of circle as we increase $y$:

The arclength (light blue) should be:
$$\delta s = \sqrt{\delta y^2-\delta r^2}
 \tag{4}$$
Using (3) in (4) , we have:
$$ \delta s = \frac{ R^2\delta y}{\sqrt{R^2-y^2}}$$
Or,
$$ \frac{\delta s}{\delta y} = \frac{R^2}{\sqrt{R^2 -y^2}}$$
Since geometrically we can see that the metric is orthogonal (page-37), we can write:
$$ ds^2 =  \frac{R^2dy^2}{R^2-y^2} +  (R^2-y^2)d \theta^2$$
A: Vectorially,
$ \vec{r_C }= ( R \cos(x), R \sin(x) , y ) $
Its projection on the sphere is
$ \vec{r_S} = ( \sqrt{R^2 - y^2} \cos(x), \sqrt{R^2 - y^2} \sin(x) , y ) $
Then, taking the differentials,
$ d \vec{r_S} = \left( \dfrac{d \vec{r_S}}{dx} \right) dx + \left( \dfrac{d \vec{r_S}}{dy} \right) dy $
and this evaluates to
$ d \vec{r_S} = \left( \sqrt{R^2 - y^2} (-sin(x) ), \sqrt{R^2 - y^2} \cos(x), 0  \right) dx + \left( -\dfrac{y\cos(x)}{\sqrt{R^2 - y^2}} , - \dfrac{y\sin(x)}{\sqrt{R^2 - y^2}} , 1 \right) dy $
Since the vector multiplying $dx$ is orthogonal to the vector multiplying $dy$ , then
$ \|d \vec{r_S} \|^2 = (dx)^2 (R^2 - y^2) + (dy)^2 ( \dfrac{y^2}{R^2 - y^2} + 1 ) $
And this simplifies to
$ \| d \vec{r_S} \|^2 = (dx)^2 (R^2 - y^2) + (dy)^2 \dfrac{R^2}{R^2 - y^2} $
