Curved space problem

from here: $$dl^2 = dr^2 + \frac{r^2 dr^2}{\kappa^{-1}R^2 -r^2}$$ i got $$dl^2 = dr^2 -\frac{(dr)^2}{1-\frac {R^2}{\kappa r^2}}$$

but here is different: $$dl^2 = \frac{dr^2}{1 - \kappa\frac{r^2}{R^2}} + r^2d\theta^2 + r^2\sin^2\theta d\phi^2,$$ which one is correct $$(1- \frac {R^2}{\kappa r^2}),$$

$${1 - \kappa\frac{r^2}{R^2}}$$

and why?

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where do your formulae come from? In particular, where do you get the one on the sixth line? – wisefool Dec 17 '12 at 11:41
@wisefool en.wikipedia.org/wiki/Curved_space – Neo Dec 17 '12 at 11:43

Ok, I think I understood your question. You started from $$dl^2=dx^2+dy^2+dz^w+\frac{(xdx+ydy+zdz)^2}{\kappa^{-1}R^2-x^2-y^2-z^2}$$ and tried to pass to spherical coordinates, as it is done on the wiki page.
Set $x=r\sin\theta\cos\phi$, $y=r\sin\theta\sin\phi$, $z=r\cos\theta$, then the Jacobian matrix is $$J=\begin{pmatrix}\sin\theta\cos\phi&r\cos\theta\cos\phi&-r\sin\theta\sin\phi\\ \sin\theta\sin\phi&r\cos\theta\sin\phi&r\sin\theta\cos\phi\\\cos\theta&-r\sin\theta&0\end{pmatrix}$$ so the metric $dx^2+dy^2+dz^2$ in these new coordinates is given by $$g=J^tJ=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2\theta\end{pmatrix}$$ i.e. $dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2$.
So $$dl^2=dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2+\frac{r^2dr^2}{\kappa^{-1}R^2-r^2}$$ from which $$dl^2=dr^2\left(1+\frac{1}{\frac{R^2}{\kappa r^2}-1}\right)+r^2d\theta^2+r^2\sin^2\theta d\phi^2$$ and $$1+\frac{1}{\frac{R^2}{\kappa r^2}-1}=\frac{\frac{R^2}{\kappa r^2}}{\frac{R^2}{\kappa r^2}-1}=\frac{1}{1-\frac{\kappa r^2}{R^2}}$$ so, plugging this into the previous, you obtain $$dl^2=\frac{dr^2}{1-\frac{\kappa r^2}{R^2}}+r^2d\theta^2+r^2\sin^2\theta d\phi^2\;.$$ Does this answer to your doubt?