Coordinates $ (\alpha, \beta, R) $ with $ -1 \ge \alpha,\space \beta \le1,\space R \lt 1 $ are related to Cartesian co-ordinates $ (x, y, z) $ via $ x= R \alpha, \space y= r \beta $ and $ \space z=(1- \alpha^2-\beta^2)^{\frac{1}{2}}R, \space R=|r|. $
Show that the volume element of the coordinate system is
$ dV= \frac{R^2\space d\alpha\space d\beta \space d R}{(1-\alpha^2-\beta^2)^{1/2}}$ .
The answer states to find the volume element I must find the following cross product:
$dV=d\alpha\space d\beta\space dR \times det(\bf {e}_1, \bf {e}_2, e_3) $
where $\bf e_1, e_2, e_3 $ are the tangent vectors $\bf{e_i} $ ($= \frac{d\bf r}{d\alpha}, ....) $ .
I have found the three tangent vectors, and I am able to solve for the determinant and find the volume, however I do not understand how to get to the above calculation of $dV=d\alpha\space d\beta\space dR \times det(\bf {e}_1, \bf {e}_2, e_3) $ or why to find the volume we must carry out this calculation.
If you have found the determinant, then what is the problem? Did it not match the solution given? Sorry, I don't understand.
But to answer your second question, ergo, why is this calculation needed, rather than attempting to give a rigorous answer (this is a theorem that is quite unpleasant to prove in my experience), I'll give you a mathematically incorrect, but rather intuitive answer.
The volume element is basically an infinitesimal paralelepiped with edges that point along the coordinate lines. Let us introduce vectors $\mathbf{d\alpha}=d\alpha\ \mathbf{e}_1$, $\mathbf{d\beta}=d\beta\ \mathbf{e}_2$ and $\mathbf{dR}=dR\ \mathbf{e}_3$. These are "infinitesimal vectors" pointing along the coordinate lines, so we can take the volume element $dV$ to be the scalar triple product $$dV=\mathbf{d\alpha}\cdot(\mathbf{d\beta}\times\mathbf{dR})=d\alpha d\beta dR\ \mathbf{e}_1\cdot(\mathbf{e}_2\times\mathbf{e}_3), $$ which is exactly the determinant you need. Obviously this was pure heuristic, but should nontheless convince you that this is the correct formula.